Integrand size = 23, antiderivative size = 116 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x)^m \, dx=\frac {a (c+d x)^{1+m}}{d (1+m)}+\frac {b F^{\left (e-\frac {c f}{d}\right ) g n-g n (e+f x)} \left (F^{e g+f g x}\right )^n (c+d x)^m \Gamma \left (1+m,-\frac {f g n (c+d x) \log (F)}{d}\right ) \left (-\frac {f g n (c+d x) \log (F)}{d}\right )^{-m}}{f g n \log (F)} \] Output:
a*(d*x+c)^(1+m)/d/(1+m)+b*F^((e-c*f/d)*g*n-g*n*(f*x+e))*(F^(f*g*x+e*g))^n* (d*x+c)^m*GAMMA(1+m,-f*g*n*(d*x+c)*ln(F)/d)/f/g/n/ln(F)/((-f*g*n*(d*x+c)*l n(F)/d)^m)
Time = 1.97 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x)^m \, dx=\frac {(c+d x)^m \left (\left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x)-b F^{-\frac {f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^n (c+d x) \Gamma \left (2+m,-\frac {f g n (c+d x) \log (F)}{d}\right ) \left (-\frac {f g n (c+d x) \log (F)}{d}\right )^{-1-m}\right )}{d (1+m)} \] Input:
Integrate[(a + b*(F^(g*(e + f*x)))^n)*(c + d*x)^m,x]
Output:
((c + d*x)^m*((a + b*(F^(g*(e + f*x)))^n)*(c + d*x) - (b*(F^(g*(e + f*x))) ^n*(c + d*x)*Gamma[2 + m, -((f*g*n*(c + d*x)*Log[F])/d)]*(-((f*g*n*(c + d* x)*Log[F])/d))^(-1 - m))/F^((f*g*n*(c + d*x))/d)))/(d*(1 + m))
Time = 0.53 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, 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.
| \(\displaystyle \int (c+d x)^m \left (a+b \left (F^{g (e+f x)}\right )^n\right ) \, dx\) |
| \(\Big \downarrow \) 2614 |
| \(\displaystyle \int \left (a (c+d x)^m+b (c+d x)^m \left (F^{e g+f g x}\right )^n\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a (c+d x)^{m+1}}{d (m+1)}+\frac {b (c+d x)^m \left (F^{e g+f g x}\right )^n F^{g n \left (e-\frac {c f}{d}\right )-g n (e+f x)} \left (-\frac {f g n \log (F) (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {f g n (c+d x) \log (F)}{d}\right )}{f g n \log (F)}\) |
Input:
Int[(a + b*(F^(g*(e + f*x)))^n)*(c + d*x)^m,x]
Output:
(a*(c + d*x)^(1 + m))/(d*(1 + m)) + (b*F^((e - (c*f)/d)*g*n - g*n*(e + f*x ))*(F^(e*g + f*g*x))^n*(c + d*x)^m*Gamma[1 + m, -((f*g*n*(c + d*x)*Log[F]) /d)])/(f*g*n*Log[F]*(-((f*g*n*(c + d*x)*Log[F])/d))^m)
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]
\[\int \left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right ) \left (d x +c \right )^{m}d x\]
Input:
int((a+b*(F^(g*(f*x+e)))^n)*(d*x+c)^m,x)
Output:
int((a+b*(F^(g*(f*x+e)))^n)*(d*x+c)^m,x)
Time = 0.08 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.96 \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x)^m \, dx=\frac {{\left (b d m + b d\right )} e^{\left (\frac {{\left (d e - c f\right )} g n \log \left (F\right ) - d m \log \left (-\frac {f g n \log \left (F\right )}{d}\right )}{d}\right )} \Gamma \left (m + 1, -\frac {{\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) + {\left (a d f g n x + a c f g n\right )} {\left (d x + c\right )}^{m} \log \left (F\right )}{{\left (d f g m + d f g\right )} n \log \left (F\right )} \] Input:
integrate((a+b*(F^(g*(f*x+e)))^n)*(d*x+c)^m,x, algorithm="fricas")
Output:
((b*d*m + b*d)*e^(((d*e - c*f)*g*n*log(F) - d*m*log(-f*g*n*log(F)/d))/d)*g amma(m + 1, -(d*f*g*n*x + c*f*g*n)*log(F)/d) + (a*d*f*g*n*x + a*c*f*g*n)*( d*x + c)^m*log(F))/((d*f*g*m + d*f*g)*n*log(F))
\[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x)^m \, dx=\int \left (a + b \left (F^{e g + f g x}\right )^{n}\right ) \left (c + d x\right )^{m}\, dx \] Input:
integrate((a+b*(F**(g*(f*x+e)))**n)*(d*x+c)**m,x)
Output:
Integral((a + b*(F**(e*g + f*g*x))**n)*(c + d*x)**m, x)
\[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x)^m \, dx=\int { {\left ({\left (F^{{\left (f x + e\right )} g}\right )}^{n} b + a\right )} {\left (d x + c\right )}^{m} \,d x } \] Input:
integrate((a+b*(F^(g*(f*x+e)))^n)*(d*x+c)^m,x, algorithm="maxima")
Output:
F^(e*g*n)*b*integrate(e^(f*g*n*x*log(F) + m*log(d*x + c)), x) + (d*x + c)^ (m + 1)*a/(d*(m + 1))
\[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x)^m \, dx=\int { {\left ({\left (F^{{\left (f x + e\right )} g}\right )}^{n} b + a\right )} {\left (d x + c\right )}^{m} \,d x } \] Input:
integrate((a+b*(F^(g*(f*x+e)))^n)*(d*x+c)^m,x, algorithm="giac")
Output:
integrate(((F^((f*x + e)*g))^n*b + a)*(d*x + c)^m, x)
Timed out. \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x)^m \, dx=\int \left (a+b\,{\left (F^{g\,\left (e+f\,x\right )}\right )}^n\right )\,{\left (c+d\,x\right )}^m \,d x \] Input:
int((a + b*(F^(g*(e + f*x)))^n)*(c + d*x)^m,x)
Output:
int((a + b*(F^(g*(e + f*x)))^n)*(c + d*x)^m, x)
\[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x)^m \, dx=\frac {f^{f g n x +e g n} \left (d x +c \right )^{m} b d m +f^{f g n x +e g n} \left (d x +c \right )^{m} b d +\left (d x +c \right )^{m} \mathrm {log}\left (f \right ) a c f g n +\left (d x +c \right )^{m} \mathrm {log}\left (f \right ) a d f g n x -f^{e g n} \left (\int \frac {f^{f g n x} \left (d x +c \right )^{m}}{d x +c}d x \right ) b \,d^{2} m^{2}-f^{e g n} \left (\int \frac {f^{f g n x} \left (d x +c \right )^{m}}{d x +c}d x \right ) b \,d^{2} m}{\mathrm {log}\left (f \right ) d f g n \left (m +1\right )} \] Input:
int((a+b*(F^(g*(f*x+e)))^n)*(d*x+c)^m,x)
Output:
(f**(e*g*n + f*g*n*x)*(c + d*x)**m*b*d*m + f**(e*g*n + f*g*n*x)*(c + d*x)* *m*b*d + (c + d*x)**m*log(f)*a*c*f*g*n + (c + d*x)**m*log(f)*a*d*f*g*n*x - f**(e*g*n)*int((f**(f*g*n*x)*(c + d*x)**m)/(c + d*x),x)*b*d**2*m**2 - f** (e*g*n)*int((f**(f*g*n*x)*(c + d*x)**m)/(c + d*x),x)*b*d**2*m)/(log(f)*d*f *g*n*(m + 1))