#int sqrt(2+(tanx)^2)*dx#
=#int (sqrt(2+(tanx)^2)*((tanx)^2+1)*dx)/((tanx)^2+1)#
=#int (sqrt(2+(tanx)^2)*(secx)^2*dx)/((tanx)^2+1)#
After using #tanx=sqrt2*tany# and #(secx)^2*dx=sqrt2*(secy)^2*dy# transforms, this integral became
#int (2(secy)^3*dy)/(2(tany)^2+1)#
=#int (2secy*dy)/(2(siny)^2+(cosy)^2)#
=#int (2*dy)/[cosy*(2(siny)^2+1-(siny)^2)]#
=#int (2cosy*dy)/[(cosy)^2*((siny)^2+1)]#
=#int (2cosy*dy)/([1-(siny)^2]*[(siny)^2+1])#
After using #z=siny# and #dz=cosy*dy# transforms, it became
#int (2*dz)/[(1-z^2)*(z^2+1)]#
=#int dz/(1-z^2)+int dz/(z^2+1)#
=#1/2int dz/(1+z)+1/2int dz/(1-z)+arctanz+C#
=#1/2ln(1+z)-1/2ln(1-z)+arctanz+C#
=#1/2ln((1+z)/(1-z))+arctanz+C#
=#1/2ln((1+siny)/(1-siny))+arctan(siny)+C#
After using #tanx=sqrt2*tany#, #tany=tanx/sqrt2#, #secy=sqrt((tanx)^2+2)/sqrt2# and #siny=tany/secy=tanx/sqrt((tanx)^2+2)# inverse transforms, I found
#int sqrt(2+(tanx)^2)*dx#
=#1/2ln((1+tanx/sqrt((tanx)^2+2))/(1-tanx/sqrt((tanx)^2+2)))+arctan(tanx/sqrt((tanx)^2+2))+C#
=#1/2ln((sqrt((tanx)^2+2)+tanx)/(sqrt((tanx)^2+2)-tanx))#+#arctan(tanx/sqrt((tanx)^2+2))#+#C#
