Integrand size = 23, antiderivative size = 116 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^m \, dx=\frac {d (f+g x)^{1+m}}{g (1+m)}+\frac {e F^{c \left (a-\frac {b f}{g}\right ) n-c n (a+b x)} \left (F^{a c+b c x}\right )^n (f+g x)^m \Gamma \left (1+m,-\frac {b c n (f+g x) \log (F)}{g}\right ) \left (-\frac {b c n (f+g x) \log (F)}{g}\right )^{-m}}{b c n \log (F)} \] Output:
d*(g*x+f)^(1+m)/g/(1+m)+e*F^(c*(a-b*f/g)*n-c*n*(b*x+a))*(F^(b*c*x+a*c))^n* (g*x+f)^m*GAMMA(1+m,-b*c*n*(g*x+f)*ln(F)/g)/b/c/n/ln(F)/((-b*c*n*(g*x+f)*l n(F)/g)^m)
Time = 2.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^m \, dx=\frac {(f+g x)^m \left (\left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)-e F^{-\frac {b c n (f+g x)}{g}} \left (F^{c (a+b x)}\right )^n (f+g x) \Gamma \left (2+m,-\frac {b c n (f+g x) \log (F)}{g}\right ) \left (-\frac {b c n (f+g x) \log (F)}{g}\right )^{-1-m}\right )}{g (1+m)} \] Input:
Integrate[(d + e*(F^(c*(a + b*x)))^n)*(f + g*x)^m,x]
Output:
((f + g*x)^m*((d + e*(F^(c*(a + b*x)))^n)*(f + g*x) - (e*(F^(c*(a + b*x))) ^n*(f + g*x)*Gamma[2 + m, -((b*c*n*(f + g*x)*Log[F])/g)]*(-((b*c*n*(f + g* x)*Log[F])/g))^(-1 - m))/F^((b*c*n*(f + g*x))/g)))/(g*(1 + m))
Time = 0.54 (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.
| \(\displaystyle \int (f+g x)^m \left (e \left (F^{c (a+b x)}\right )^n+d\right ) \, dx\) |
| \(\Big \downarrow \) 2614 |
| \(\displaystyle \int \left (e (f+g x)^m \left (F^{a c+b c x}\right )^n+d (f+g x)^m\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e (f+g x)^m \left (F^{a c+b c x}\right )^n F^{c n \left (a-\frac {b f}{g}\right )-c n (a+b x)} \left (-\frac {b c n \log (F) (f+g x)}{g}\right )^{-m} \Gamma \left (m+1,-\frac {b c n (f+g x) \log (F)}{g}\right )}{b c n \log (F)}+\frac {d (f+g x)^{m+1}}{g (m+1)}\) |
Input:
Int[(d + e*(F^(c*(a + b*x)))^n)*(f + g*x)^m,x]
Output:
(d*(f + g*x)^(1 + m))/(g*(1 + m)) + (e*F^(c*(a - (b*f)/g)*n - c*n*(a + b*x ))*(F^(a*c + b*c*x))^n*(f + g*x)^m*Gamma[1 + m, -((b*c*n*(f + g*x)*Log[F]) /g)])/(b*c*n*Log[F]*(-((b*c*n*(f + g*x)*Log[F])/g))^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]
\[\int \left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right ) \left (g x +f \right )^{m}d x\]
Input:
int((d+e*(F^(c*(b*x+a)))^n)*(g*x+f)^m,x)
Output:
int((d+e*(F^(c*(b*x+a)))^n)*(g*x+f)^m,x)
Time = 0.07 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^m \, dx=\frac {{\left (e g m + e g\right )} e^{\left (-\frac {g m \log \left (-\frac {b c n \log \left (F\right )}{g}\right ) + {\left (b c f - a c g\right )} n \log \left (F\right )}{g}\right )} \Gamma \left (m + 1, -\frac {{\left (b c g n x + b c f n\right )} \log \left (F\right )}{g}\right ) + {\left (b c d g n x + b c d f n\right )} {\left (g x + f\right )}^{m} \log \left (F\right )}{{\left (b c g m + b c g\right )} n \log \left (F\right )} \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)*(g*x+f)^m,x, algorithm="fricas")
Output:
((e*g*m + e*g)*e^(-(g*m*log(-b*c*n*log(F)/g) + (b*c*f - a*c*g)*n*log(F))/g )*gamma(m + 1, -(b*c*g*n*x + b*c*f*n)*log(F)/g) + (b*c*d*g*n*x + b*c*d*f*n )*(g*x + f)^m*log(F))/((b*c*g*m + b*c*g)*n*log(F))
\[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^m \, dx=\int \left (d + e \left (F^{a c + b c x}\right )^{n}\right ) \left (f + g x\right )^{m}\, dx \] Input:
integrate((d+e*(F**((b*x+a)*c))**n)*(g*x+f)**m,x)
Output:
Integral((d + e*(F**(a*c + b*c*x))**n)*(f + g*x)**m, x)
\[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^m \, dx=\int { {\left ({\left (F^{{\left (b x + a\right )} c}\right )}^{n} e + d\right )} {\left (g x + f\right )}^{m} \,d x } \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)*(g*x+f)^m,x, algorithm="maxima")
Output:
F^(a*c*n)*e*integrate(e^(b*c*n*x*log(F) + m*log(g*x + f)), x) + (g*x + f)^ (m + 1)*d/(g*(m + 1))
\[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^m \, dx=\int { {\left ({\left (F^{{\left (b x + a\right )} c}\right )}^{n} e + d\right )} {\left (g x + f\right )}^{m} \,d x } \] Input:
integrate((d+e*(F^((b*x+a)*c))^n)*(g*x+f)^m,x, algorithm="giac")
Output:
integrate(((F^((b*x + a)*c))^n*e + d)*(g*x + f)^m, x)
Timed out. \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^m \, dx=\int {\left (f+g\,x\right )}^m\,\left (d+e\,{\left (F^{c\,\left (a+b\,x\right )}\right )}^n\right ) \,d x \] Input:
int((f + g*x)^m*(d + e*(F^(c*(a + b*x)))^n),x)
Output:
int((f + g*x)^m*(d + e*(F^(c*(a + b*x)))^n), x)
\[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right ) (f+g x)^m \, dx=\frac {f^{b c n x +a c n} \left (g x +f \right )^{m} e g m +f^{b c n x +a c n} \left (g x +f \right )^{m} e g +\left (g x +f \right )^{m} \mathrm {log}\left (f \right ) b c d f n +\left (g x +f \right )^{m} \mathrm {log}\left (f \right ) b c d g n x -f^{a c n} \left (\int \frac {f^{b c n x} \left (g x +f \right )^{m}}{g x +f}d x \right ) e \,g^{2} m^{2}-f^{a c n} \left (\int \frac {f^{b c n x} \left (g x +f \right )^{m}}{g x +f}d x \right ) e \,g^{2} m}{\mathrm {log}\left (f \right ) b c g n \left (m +1\right )} \] Input:
int((d+e*(F^((b*x+a)*c))^n)*(g*x+f)^m,x)
Output:
(f**(a*c*n + b*c*n*x)*(f + g*x)**m*e*g*m + f**(a*c*n + b*c*n*x)*(f + g*x)* *m*e*g + (f + g*x)**m*log(f)*b*c*d*f*n + (f + g*x)**m*log(f)*b*c*d*g*n*x - f**(a*c*n)*int((f**(b*c*n*x)*(f + g*x)**m)/(f + g*x),x)*e*g**2*m**2 - f** (a*c*n)*int((f**(b*c*n*x)*(f + g*x)**m)/(f + g*x),x)*e*g**2*m)/(log(f)*b*c *g*n*(m + 1))