Integrand size = 19, antiderivative size = 72 \[ \int F^{c (a+b x)} \left ((d+e x)^n\right )^m \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left ((d+e x)^n\right )^m \Gamma \left (1+m n,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-m n}}{b c \log (F)} \] Output:
F^(c*(a-b*d/e))*((e*x+d)^n)^m*GAMMA(m*n+1,-b*c*(e*x+d)*ln(F)/e)/b/c/ln(F)/ ((-b*c*(e*x+d)*ln(F)/e)^(m*n))
Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} \left ((d+e x)^n\right )^m \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left ((d+e x)^n\right )^m \Gamma \left (1+m n,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-m n}}{b c \log (F)} \] Input:
Integrate[F^(c*(a + b*x))*((d + e*x)^n)^m,x]
Output:
(F^(c*(a - (b*d)/e))*((d + e*x)^n)^m*Gamma[1 + m*n, -((b*c*(d + e*x)*Log[F ])/e)])/(b*c*Log[F]*(-((b*c*(d + e*x)*Log[F])/e))^(m*n))
Time = 0.37 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2045, 2612}
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.
\(\displaystyle \int F^{c (a+b x)} \left ((d+e x)^n\right )^m \, dx\) |
\(\Big \downarrow \) 2045 |
\(\displaystyle \left ((d+e x)^n\right )^m \left (\frac {e x}{d}+1\right )^{-m n} \int F^{c (a+b x)} \left (\frac {e x}{d}+1\right )^{m n}dx\) |
\(\Big \downarrow \) 2612 |
\(\displaystyle \frac {\left ((d+e x)^n\right )^m F^{c \left (a-\frac {b d}{e}\right )} \left (-\frac {b c \log (F) (d+e x)}{e}\right )^{-m n} \Gamma \left (m n+1,-\frac {b c (d+e x) \log (F)}{e}\right )}{b c \log (F)}\) |
Input:
Int[F^(c*(a + b*x))*((d + e*x)^n)^m,x]
Output:
(F^(c*(a - (b*d)/e))*((d + e*x)^n)^m*Gamma[1 + m*n, -((b*c*(d + e*x)*Log[F ])/e)])/(b*c*Log[F]*(-((b*c*(d + e*x)*Log[F])/e))^(m*n))
Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Simp[Si mp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/a))^(p*q)] Int[u*(1 + b*(x^n/a))^(p*q) , x], x] /; FreeQ[{a, b, c, n, p, q}, x] && !GeQ[a, 0]
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c + d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d) )^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m + 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] && !IntegerQ[m]
\[\int F^{c \left (b x +a \right )} \left (\left (e x +d \right )^{n}\right )^{m}d x\]
Input:
int(F^(c*(b*x+a))*((e*x+d)^n)^m,x)
Output:
int(F^(c*(b*x+a))*((e*x+d)^n)^m,x)
Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int F^{c (a+b x)} \left ((d+e x)^n\right )^m \, dx=\frac {e^{\left (-\frac {e m n \log \left (-\frac {b c \log \left (F\right )}{e}\right ) + {\left (b c d - a c e\right )} \log \left (F\right )}{e}\right )} \Gamma \left (m n + 1, -\frac {{\left (b c e x + b c d\right )} \log \left (F\right )}{e}\right )}{b c \log \left (F\right )} \] Input:
integrate(F^((b*x+a)*c)*((e*x+d)^n)^m,x, algorithm="fricas")
Output:
e^(-(e*m*n*log(-b*c*log(F)/e) + (b*c*d - a*c*e)*log(F))/e)*gamma(m*n + 1, -(b*c*e*x + b*c*d)*log(F)/e)/(b*c*log(F))
\[ \int F^{c (a+b x)} \left ((d+e x)^n\right )^m \, dx=\int F^{c \left (a + b x\right )} \left (\left (d + e x\right )^{n}\right )^{m}\, dx \] Input:
integrate(F**((b*x+a)*c)*((e*x+d)**n)**m,x)
Output:
Integral(F**(c*(a + b*x))*((d + e*x)**n)**m, x)
\[ \int F^{c (a+b x)} \left ((d+e x)^n\right )^m \, dx=\int { {\left ({\left (e x + d\right )}^{n}\right )}^{m} F^{{\left (b x + a\right )} c} \,d x } \] Input:
integrate(F^((b*x+a)*c)*((e*x+d)^n)^m,x, algorithm="maxima")
Output:
integrate(((e*x + d)^n)^m*F^((b*x + a)*c), x)
\[ \int F^{c (a+b x)} \left ((d+e x)^n\right )^m \, dx=\int { {\left ({\left (e x + d\right )}^{n}\right )}^{m} F^{{\left (b x + a\right )} c} \,d x } \] Input:
integrate(F^((b*x+a)*c)*((e*x+d)^n)^m,x, algorithm="giac")
Output:
integrate(((e*x + d)^n)^m*F^((b*x + a)*c), x)
Timed out. \[ \int F^{c (a+b x)} \left ((d+e x)^n\right )^m \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\left ({\left (d+e\,x\right )}^n\right )}^m \,d x \] Input:
int(F^(c*(a + b*x))*((d + e*x)^n)^m,x)
Output:
int(F^(c*(a + b*x))*((d + e*x)^n)^m, x)
\[ \int F^{c (a+b x)} \left ((d+e x)^n\right )^m \, dx=\frac {f^{a c} \left (f^{b c x} \left (e x +d \right )^{m n}-\left (\int \frac {f^{b c x} \left (e x +d \right )^{m n}}{e x +d}d x \right ) e m n \right )}{\mathrm {log}\left (f \right ) b c} \] Input:
int(F^((b*x+a)*c)*((e*x+d)^n)^m,x)
Output:
(f**(a*c)*(f**(b*c*x)*(d + e*x)**(m*n) - int((f**(b*c*x)*(d + e*x)**(m*n)) /(d + e*x),x)*e*m*n))/(log(f)*b*c)