`inte^(ax)cos(b x+c)dx`

Explanation

Integrating by parts, taking `e^(ax)` as the second function, we get
`inte^(ax)cos(bx+c)dx=cos(bx+c)*(e^(ax))/(a)-int{-bsin(bx+c)*(e^(ax))/(a)}dx`
`=(e^(ax))/(a)cos(bx+c)+(b)/(a)inte^(ax)sin(bx+c)dx`
`=(e^(ax))/(a)*cos(bx+c)+(b)/(a){:[sin(bx+c)*(e^(ax))/(a)-int{bcos(bx+c)*(e^(ax))/(a)}]:}dx+C` [integrating `e^(ax)sin(bx+c)` parts]
`=(e^(ax))/(a)*cos(bx+c)+(b)/(a^(2))e^(ax)sin(bx+c)-(b^(2))/(a^(2))inte^(ax)cos(bx+c)dx+C`
`:.(1+(b^(2))/(a^(2)))inte^(ax)cos(bx+c)dx=(e^(ax))/(a)cos(bx+c)+(b)/(a^(2))e^(ax)sin(bx+c)+C`
`orinte^(ax)cos(bx+c)dx=e^(ax){:[(acos(bx+c)+bsin(bx+c))/((a^(2)+b^(2)))]:}+C'`.