'Fix' the denominator
First 'complete the square' in the denominator:
sqrt(x^2-2x+10)=sqrt((x-1)^2+9)
This is needed in order to put it into a form that a trig substitution can simplify by use of an identity - we need to simplify the denominator here in order to tackle the integral.
Then set x-1=3tan theta, making the integral
int (1+3 tan theta)^3/sqrt(9 tan^2 theta + 9) dx/(d theta) d theta
From our substitution formula, dx/(d theta)=3sec^2theta, the standard derivative of tan. Thus the integral is
int (1+3 tan theta)^3/sqrt(9 tan^2 theta + 9) * 3sec^2theta d theta
=int (sec^2theta(1+3 tan theta)^3)/sqrt(tan^2 theta + 1) d theta
(Recall the identity 1+tan^2 x=sec^2)
=int (sec^2theta(1+3 tan theta)^3)/sqrt(sec^2 theta) d theta
=int sec theta(1+3 tan theta)^3d theta
(Multiply out terms, recalling the binomial theorem for coefficients)
=int sec theta+9 sec theta tan theta + 27 sec theta tan^2 theta + 27 sec theta tan^3 theta d theta
So this is how it is done in general - you simplify the denominator until it is gone, at which point you find out exactly how messy you have made your numerator. In this case, quite messy, but it is solvable term by term.
Solve what's left
For the first term, it will suffice here to state that int sec theta d theta=log|sec theta + tan theta|+C. This answer is growing long, and there are proofs of this online, for example here: https://www.math.ubc.ca/~feldman/m121/secx.pdf
Note that d/dx secx=secxtanx, by the quotient rule and identity. So the second term is easy: int 9 sec theta tan theta dx = 9 sec theta+C.
Recall "integration by parts", a reformulation of the product rule of differentiation:
int u (dv)/(d theta) d theta = uv - int v (du)/(d theta) d theta
For the third term, use the sec derivative that was so useful for the second term; set u=tan theta and (dv)/(d theta)=27 sec theta tan theta. Then v=27 sec theta, (du)/(d theta)=sec^2 theta and the integral of the third term is 27 sec theta tan theta - 27int sec^3 theta d theta.
The above link also explains how to integrate sec^3 theta. Taking this result as given, we see that int27 sec theta tan^2 theta d theta=27/2 sec theta tan theta-27/2log|sec theta +tan theta|+C.
Finally, for the fourth term, we take the same approach with the sec derivative. Set (dv)/(d theta) = 27 sec theta tan theta; set u=tan^2 theta. Then v=27sec theta and (du)/(d theta)=2 tan theta sec^2 theta, so the fourth term becomes
27 sec theta tan^2 theta - int 54tan theta sec^3 theta d theta.
Repeat integration by parts, this time with (du)/(d theta)=54 sec theta tan theta and v=sec^2 theta. Then u=54 sec theta and (dv)/(d theta)=2 tan theta sec theta. The fourth term is now
27 sec theta tan^2 theta - 54 sec^3 theta + int 108 tan theta sec^2 theta d theta.
A third integration by parts now (have patience!)... Again, (du)/(d theta)=108 sec theta tan theta. v=sec theta. So u=108 sec theta and (dv)/(d theta)=sec theta tan theta. The fourth term is now
27 sec theta tan^2 theta - 54 sec^3 theta + 108 sec^2 theta - int 108 tan theta sec^2 theta d theta.
But this is where we can break the chain of integrations (hurrah, I hear you cry - it's been pretty tedious) - we now have the same integral as we did at the previous step, so we can employ a trick.
This step showed us that
int tan theta sec^2 theta d theta=sec^2 theta - int tan theta sec^2 theta d theta
So
int tan theta sec^2 theta d theta=1/2 sec^2 theta
Therefore the fourth term is
27 sec theta tan^2 theta - 54 sec^3 theta + 54 sec^2 theta
Using the identity tan^2 theta = sec^2 theta-1, we can simplify this:
27 sec^3 theta -27 sec theta - 54 sec^3 theta + 54 sec^2 theta
-27 sec theta + 54 sec^2 theta - 27 sec^3 theta
Combining the four terms, we obtain the answer:
log|sec theta + tan theta|+9 sec x+27/2 sec theta tan theta-27/2log|sec theta +tan theta|-27 sec theta + 54 sec^2 theta - 27 sec^3 theta+C
which simplifies a little to
-25/2log|sec theta + tan theta|+27/2 sec theta tan theta-18 sec theta+ 54 sec^2 theta - 27 sec^3 theta+C
But we aren't quite done yet - we have to get back to our original variable, x. Recall our substitution: x-1=3tan theta. This was an elegant substitution that would have made for a nice integral if the numerator hadn't been so awkward. So tan theta = (x-1)/3 and sec theta=sqrt(1+(x-1)^2/9).
This leaves us with
-25/2log|(x-1)/3+sqrt(1+(x-1)^2/9)|+9/2 (x-5) sqrt(1+(x-1)^2/9) + 54 (1+(x-1)^2/9) - 27 (1+(x-1)^2/9)^(3/2)+C
Not a beautiful expression!