\(\int (d+e (F^{c (a+b x)})^n)^2 (f+g x)^m \, dx\) [53]

Optimal result

Integrand size = 25, antiderivative size = 228 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^m \, dx=\frac {d^2 (f+g x)^{1+m}}{g (1+m)}+\frac {2^{-1-m} e^2 F^{2 c \left (a-\frac {b f}{g}\right ) n-2 c n (a+b x)} \left (F^{a c+b c x}\right )^{2 n} (f+g x)^m \Gamma \left (1+m,-\frac {2 b c n (f+g x) \log (F)}{g}\right ) \left (-\frac {b c n (f+g x) \log (F)}{g}\right )^{-m}}{b c n \log (F)}+\frac {2 d e F^{c \left (a-\frac {b f}{g}\right ) n-c n (a+b x)} \left (F^{a c+b c x}\right )^n (f+g x)^m \Gamma \left (1+m,-\frac {b c n (f+g x) \log (F)}{g}\right ) \left (-\frac {b c n (f+g x) \log (F)}{g}\right )^{-m}}{b c n \log (F)} \] Output:

d^2*(g*x+f)^(1+m)/g/(1+m)+2^(-1-m)*e^2*F^(2*c*(a-b*f/g)*n-2*c*n*(b*x+a))*( 
F^(b*c*x+a*c))^(2*n)*(g*x+f)^m*GAMMA(1+m,-2*b*c*n*(g*x+f)*ln(F)/g)/b/c/n/l 
n(F)/((-b*c*n*(g*x+f)*ln(F)/g)^m)+2*d*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)*ln(F)/g)^m)
 

Mathematica [F]

\[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^m \, dx=\int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^m \, dx \] Input:

Integrate[(d + e*(F^(c*(a + b*x)))^n)^2*(f + g*x)^m,x]
 

Output:

Integrate[(d + e*(F^(c*(a + b*x)))^n)^2*(f + g*x)^m, x]
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 228, 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.

\(\displaystyle \int (f+g x)^m \left (e \left (F^{c (a+b x)}\right )^n+d\right )^2 \, dx\)

\(\Big \downarrow \) 2614

\(\displaystyle \int \left (2 d e (f+g x)^m \left (F^{a c+b c x}\right )^n+e^2 (f+g x)^m \left (F^{a c+b c x}\right )^{2 n}+d^2 (f+g x)^m\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 d e (f+g x)^m \left (F^{a c+b c x}\right )^n F^{c n \left (a-\frac {b f}{g}\right )-c n (a+b x)} \left (-\frac {b c n \log (F) (f+g x)}{g}\right )^{-m} \Gamma \left (m+1,-\frac {b c n (f+g x) \log (F)}{g}\right )}{b c n \log (F)}+\frac {e^2 2^{-m-1} (f+g x)^m \left (F^{a c+b c x}\right )^{2 n} F^{2 c n \left (a-\frac {b f}{g}\right )-2 c n (a+b x)} \left (-\frac {b c n \log (F) (f+g x)}{g}\right )^{-m} \Gamma \left (m+1,-\frac {2 b c n (f+g x) \log (F)}{g}\right )}{b c n \log (F)}+\frac {d^2 (f+g x)^{m+1}}{g (m+1)}\)

Input:

Int[(d + e*(F^(c*(a + b*x)))^n)^2*(f + g*x)^m,x]
 

Output:

(d^2*(f + g*x)^(1 + m))/(g*(1 + m)) + (2^(-1 - m)*e^2*F^(2*c*(a - (b*f)/g) 
*n - 2*c*n*(a + b*x))*(F^(a*c + b*c*x))^(2*n)*(f + g*x)^m*Gamma[1 + m, (-2 
*b*c*n*(f + g*x)*Log[F])/g])/(b*c*n*Log[F]*(-((b*c*n*(f + g*x)*Log[F])/g)) 
^m) + (2*d*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)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2614

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]
 

Maple [F]

\[\int {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{2} \left (g x +f \right )^{m}d x\]

Input:

int((d+e*(F^(c*(b*x+a)))^n)^2*(g*x+f)^m,x)
 

Output:

int((d+e*(F^(c*(b*x+a)))^n)^2*(g*x+f)^m,x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.85 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^m \, dx=\frac {4 \, {\left (d e g m + d e g\right )} e^{\left (-\frac {g m \log \left (-\frac {b c n \log \left (F\right )}{g}\right ) + {\left (b c f - a c g\right )} n \log \left (F\right )}{g}\right )} \Gamma \left (m + 1, -\frac {{\left (b c g n x + b c f n\right )} \log \left (F\right )}{g}\right ) + {\left (e^{2} g m + e^{2} g\right )} e^{\left (-\frac {g m \log \left (-\frac {2 \, b c n \log \left (F\right )}{g}\right ) + 2 \, {\left (b c f - a c g\right )} n \log \left (F\right )}{g}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (b c g n x + b c f n\right )} \log \left (F\right )}{g}\right ) + 2 \, {\left (b c d^{2} g n x + b c d^{2} f n\right )} {\left (g x + f\right )}^{m} \log \left (F\right )}{2 \, {\left (b c g m + b c g\right )} n \log \left (F\right )} \] Input:

integrate((d+e*(F^((b*x+a)*c))^n)^2*(g*x+f)^m,x, algorithm="fricas")
 

Output:

1/2*(4*(d*e*g*m + d*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) + (e^2*g*m + e^2 
*g)*e^(-(g*m*log(-2*b*c*n*log(F)/g) + 2*(b*c*f - a*c*g)*n*log(F))/g)*gamma 
(m + 1, -2*(b*c*g*n*x + b*c*f*n)*log(F)/g) + 2*(b*c*d^2*g*n*x + b*c*d^2*f* 
n)*(g*x + f)^m*log(F))/((b*c*g*m + b*c*g)*n*log(F))
 

Sympy [F]

\[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^m \, dx=\int \left (d + e \left (F^{a c + b c x}\right )^{n}\right )^{2} \left (f + g x\right )^{m}\, dx \] Input:

integrate((d+e*(F**((b*x+a)*c))**n)**2*(g*x+f)**m,x)
 

Output:

Integral((d + e*(F**(a*c + b*c*x))**n)**2*(f + g*x)**m, x)
 

Maxima [F]

\[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^m \, dx=\int { {\left ({\left (F^{{\left (b x + a\right )} c}\right )}^{n} e + d\right )}^{2} {\left (g x + f\right )}^{m} \,d x } \] Input:

integrate((d+e*(F^((b*x+a)*c))^n)^2*(g*x+f)^m,x, algorithm="maxima")
 

Output:

(g*x + f)^(m + 1)*d^2/(g*(m + 1)) + integrate((2*F^(b*c*n*x)*F^(a*c*n)*d*e 
 + F^(2*b*c*n*x)*F^(2*a*c*n)*e^2)*(g*x + f)^m, x)
 

Giac [F]

\[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^m \, dx=\int { {\left ({\left (F^{{\left (b x + a\right )} c}\right )}^{n} e + d\right )}^{2} {\left (g x + f\right )}^{m} \,d x } \] Input:

integrate((d+e*(F^((b*x+a)*c))^n)^2*(g*x+f)^m,x, algorithm="giac")
 

Output:

integrate(((F^((b*x + a)*c))^n*e + d)^2*(g*x + f)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^m \, dx=\int {\left (f+g\,x\right )}^m\,{\left (d+e\,{\left (F^{c\,\left (a+b\,x\right )}\right )}^n\right )}^2 \,d x \] Input:

int((f + g*x)^m*(d + e*(F^(c*(a + b*x)))^n)^2,x)
 

Output:

int((f + g*x)^m*(d + e*(F^(c*(a + b*x)))^n)^2, x)
 

Reduce [F]

\[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^2 (f+g x)^m \, dx=\frac {f^{2 b c n x +2 a c n} \left (g x +f \right )^{m} e^{2} g m +f^{2 b c n x +2 a c n} \left (g x +f \right )^{m} e^{2} g +4 f^{b c n x +a c n} \left (g x +f \right )^{m} d e g m +4 f^{b c n x +a c n} \left (g x +f \right )^{m} d e g +2 \left (g x +f \right )^{m} \mathrm {log}\left (f \right ) b c \,d^{2} f n +2 \left (g x +f \right )^{m} \mathrm {log}\left (f \right ) b c \,d^{2} g n x -f^{2 a c n} \left (\int \frac {f^{2 b c n x} \left (g x +f \right )^{m}}{g x +f}d x \right ) e^{2} g^{2} m^{2}-f^{2 a c n} \left (\int \frac {f^{2 b c n x} \left (g x +f \right )^{m}}{g x +f}d x \right ) e^{2} g^{2} m -4 f^{a c n} \left (\int \frac {f^{b c n x} \left (g x +f \right )^{m}}{g x +f}d x \right ) d e \,g^{2} m^{2}-4 f^{a c n} \left (\int \frac {f^{b c n x} \left (g x +f \right )^{m}}{g x +f}d x \right ) d e \,g^{2} m}{2 \,\mathrm {log}\left (f \right ) b c g n \left (m +1\right )} \] Input:

int((d+e*(F^((b*x+a)*c))^n)^2*(g*x+f)^m,x)
 

Output:

(f**(2*a*c*n + 2*b*c*n*x)*(f + g*x)**m*e**2*g*m + f**(2*a*c*n + 2*b*c*n*x) 
*(f + g*x)**m*e**2*g + 4*f**(a*c*n + b*c*n*x)*(f + g*x)**m*d*e*g*m + 4*f** 
(a*c*n + b*c*n*x)*(f + g*x)**m*d*e*g + 2*(f + g*x)**m*log(f)*b*c*d**2*f*n 
+ 2*(f + g*x)**m*log(f)*b*c*d**2*g*n*x - f**(2*a*c*n)*int((f**(2*b*c*n*x)* 
(f + g*x)**m)/(f + g*x),x)*e**2*g**2*m**2 - f**(2*a*c*n)*int((f**(2*b*c*n* 
x)*(f + g*x)**m)/(f + g*x),x)*e**2*g**2*m - 4*f**(a*c*n)*int((f**(b*c*n*x) 
*(f + g*x)**m)/(f + g*x),x)*d*e*g**2*m**2 - 4*f**(a*c*n)*int((f**(b*c*n*x) 
*(f + g*x)**m)/(f + g*x),x)*d*e*g**2*m)/(2*log(f)*b*c*g*n*(m + 1))