Integrand size = 25, antiderivative size = 202 \[ \int \frac {\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2}{(c+d x)^2} \, dx=-\frac {a^2}{d (c+d x)}-\frac {2 a b \left (F^{e g+f g x}\right )^n}{d (c+d x)}-\frac {b^2 \left (F^{e g+f g x}\right )^{2 n}}{d (c+d x)}+\frac {2 a b f F^{\left (e-\frac {c f}{d}\right ) g n-g n (e+f x)} \left (F^{e g+f g x}\right )^n g n \operatorname {ExpIntegralEi}\left (\frac {f g n (c+d x) \log (F)}{d}\right ) \log (F)}{d^2}+\frac {2 b^2 f F^{2 \left (e-\frac {c f}{d}\right ) g n-2 g n (e+f x)} \left (F^{e g+f g x}\right )^{2 n} g n \operatorname {ExpIntegralEi}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right ) \log (F)}{d^2} \]
-a^2/d/(d*x+c)-2*a*b*(F^(f*g*x+e*g))^n/d/(d*x+c)-b^2*(F^(f*g*x+e*g))^(2*n) /d/(d*x+c)+2*a*b*f*F^((e-c*f/d)*g*n-g*n*(f*x+e))*(F^(f*g*x+e*g))^n*g*n*Ei( f*g*n*(d*x+c)*ln(F)/d)*ln(F)/d^2+2*b^2*f*F^(2*(e-c*f/d)*g*n-2*g*n*(f*x+e)) *(F^(f*g*x+e*g))^(2*n)*g*n*Ei(2*f*g*n*(d*x+c)*ln(F)/d)*ln(F)/d^2
Time = 0.67 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.67 \[ \int \frac {\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2}{(c+d x)^2} \, dx=\frac {-\frac {d \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2}{c+d x}+2 a b f F^{-\frac {f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^n g n \operatorname {ExpIntegralEi}\left (\frac {f g n (c+d x) \log (F)}{d}\right ) \log (F)+2 b^2 f F^{-\frac {2 f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^{2 n} g n \operatorname {ExpIntegralEi}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right ) \log (F)}{d^2} \]
(-((d*(a + b*(F^(g*(e + f*x)))^n)^2)/(c + d*x)) + (2*a*b*f*(F^(g*(e + f*x) ))^n*g*n*ExpIntegralEi[(f*g*n*(c + d*x)*Log[F])/d]*Log[F])/F^((f*g*n*(c + d*x))/d) + (2*b^2*f*(F^(g*(e + f*x)))^(2*n)*g*n*ExpIntegralEi[(2*f*g*n*(c + d*x)*Log[F])/d]*Log[F])/F^((2*f*g*n*(c + d*x))/d))/d^2
Time = 0.57 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2614, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2}{(c+d x)^2} \, dx\) |
\(\Big \downarrow \) 2614 |
\(\displaystyle \int \left (\frac {a^2}{(c+d x)^2}+\frac {2 a b \left (F^{e g+f g x}\right )^n}{(c+d x)^2}+\frac {b^2 \left (F^{e g+f g x}\right )^{2 n}}{(c+d x)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2}{d (c+d x)}+\frac {2 a b f g n \log (F) \left (F^{e g+f g x}\right )^n F^{g n \left (e-\frac {c f}{d}\right )-g n (e+f x)} \operatorname {ExpIntegralEi}\left (\frac {f g n (c+d x) \log (F)}{d}\right )}{d^2}-\frac {2 a b \left (F^{e g+f g x}\right )^n}{d (c+d x)}+\frac {2 b^2 f g n \log (F) \left (F^{e g+f g x}\right )^{2 n} F^{2 g n \left (e-\frac {c f}{d}\right )-2 g n (e+f x)} \operatorname {ExpIntegralEi}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right )}{d^2}-\frac {b^2 \left (F^{e g+f g x}\right )^{2 n}}{d (c+d x)}\) |
-(a^2/(d*(c + d*x))) - (2*a*b*(F^(e*g + f*g*x))^n)/(d*(c + d*x)) - (b^2*(F ^(e*g + f*g*x))^(2*n))/(d*(c + d*x)) + (2*a*b*f*F^((e - (c*f)/d)*g*n - g*n *(e + f*x))*(F^(e*g + f*g*x))^n*g*n*ExpIntegralEi[(f*g*n*(c + d*x)*Log[F]) /d]*Log[F])/d^2 + (2*b^2*f*F^(2*(e - (c*f)/d)*g*n - 2*g*n*(e + f*x))*(F^(e *g + f*g*x))^(2*n)*g*n*ExpIntegralEi[(2*f*g*n*(c + d*x)*Log[F])/d]*Log[F]) /d^2
3.1.37.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*(F ^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n}, x] && IGtQ[p, 0]
\[\int \frac {{\left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )}^{2}}{\left (d x +c \right )^{2}}d x\]
Time = 0.28 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2}{(c+d x)^2} \, dx=-\frac {2 \, F^{f g n x + e g n} a b d + F^{2 \, f g n x + 2 \, e g n} b^{2} d - 2 \, {\left (b^{2} d f g n x + b^{2} c f g n\right )} F^{\frac {2 \, {\left (d e - c f\right )} g n}{d}} {\rm Ei}\left (\frac {2 \, {\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) \log \left (F\right ) - 2 \, {\left (a b d f g n x + a b c f g n\right )} F^{\frac {{\left (d e - c f\right )} g n}{d}} {\rm Ei}\left (\frac {{\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) \log \left (F\right ) + a^{2} d}{d^{3} x + c d^{2}} \]
-(2*F^(f*g*n*x + e*g*n)*a*b*d + F^(2*f*g*n*x + 2*e*g*n)*b^2*d - 2*(b^2*d*f *g*n*x + b^2*c*f*g*n)*F^(2*(d*e - c*f)*g*n/d)*Ei(2*(d*f*g*n*x + c*f*g*n)*l og(F)/d)*log(F) - 2*(a*b*d*f*g*n*x + a*b*c*f*g*n)*F^((d*e - c*f)*g*n/d)*Ei ((d*f*g*n*x + c*f*g*n)*log(F)/d)*log(F) + a^2*d)/(d^3*x + c*d^2)
\[ \int \frac {\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2}{(c+d x)^2} \, dx=\int \frac {\left (a + b \left (F^{e g + f g x}\right )^{n}\right )^{2}}{\left (c + d x\right )^{2}}\, dx \]
\[ \int \frac {\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2}{(c+d x)^2} \, dx=\int { \frac {{\left ({\left (F^{{\left (f x + e\right )} g}\right )}^{n} b + a\right )}^{2}}{{\left (d x + c\right )}^{2}} \,d x } \]
F^(2*e*g*n)*b^2*integrate(F^(2*f*g*n*x)/(d^2*x^2 + 2*c*d*x + c^2), x) + 2* F^(e*g*n)*a*b*integrate(F^(f*g*n*x)/(d^2*x^2 + 2*c*d*x + c^2), x) - a^2/(d ^2*x + c*d)
\[ \int \frac {\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2}{(c+d x)^2} \, dx=\int { \frac {{\left ({\left (F^{{\left (f x + e\right )} g}\right )}^{n} b + a\right )}^{2}}{{\left (d x + c\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2}{(c+d x)^2} \, dx=\int \frac {{\left (a+b\,{\left (F^{g\,\left (e+f\,x\right )}\right )}^n\right )}^2}{{\left (c+d\,x\right )}^2} \,d x \]