\(\int \frac {(d+e (F^{c (a+b x)})^n)^2}{(f+g x)^3} \, dx\) [32]

Optimal result

Integrand size = 25, antiderivative size = 286 \[ \int \frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2}{(f+g x)^3} \, dx=-\frac {d^2}{2 g (f+g x)^2}-\frac {d e \left (F^{a c+b c x}\right )^n}{g (f+g x)^2}-\frac {e^2 \left (F^{a c+b c x}\right )^{2 n}}{2 g (f+g x)^2}-\frac {b c d e \left (F^{a c+b c x}\right )^n n \log (F)}{g^2 (f+g x)}-\frac {b c e^2 \left (F^{a c+b c x}\right )^{2 n} n \log (F)}{g^2 (f+g x)}+\frac {b^2 c^2 d e F^{c \left (a-\frac {b f}{g}\right ) n-c n (a+b x)} \left (F^{a c+b c x}\right )^n n^2 \operatorname {ExpIntegralEi}\left (\frac {b c n (f+g x) \log (F)}{g}\right ) \log ^2(F)}{g^3}+\frac {2 b^2 c^2 e^2 F^{2 c \left (a-\frac {b f}{g}\right ) n-2 c n (a+b x)} \left (F^{a c+b c x}\right )^{2 n} n^2 \operatorname {ExpIntegralEi}\left (\frac {2 b c n (f+g x) \log (F)}{g}\right ) \log ^2(F)}{g^3} \] Output:

-1/2*d^2/g/(g*x+f)^2-d*e*(F^(b*c*x+a*c))^n/g/(g*x+f)^2-1/2*e^2*(F^(b*c*x+a 
*c))^(2*n)/g/(g*x+f)^2-b*c*d*e*(F^(b*c*x+a*c))^n*n*ln(F)/g^2/(g*x+f)-b*c*e 
^2*(F^(b*c*x+a*c))^(2*n)*n*ln(F)/g^2/(g*x+f)+b^2*c^2*d*e*F^(c*(a-b*f/g)*n- 
c*n*(b*x+a))*(F^(b*c*x+a*c))^n*n^2*Ei(b*c*n*(g*x+f)*ln(F)/g)*ln(F)^2/g^3+2 
*b^2*c^2*e^2*F^(2*c*(a-b*f/g)*n-2*c*n*(b*x+a))*(F^(b*c*x+a*c))^(2*n)*n^2*E 
i(2*b*c*n*(g*x+f)*ln(F)/g)*ln(F)^2/g^3
 

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.76 \[ \int \frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2}{(f+g x)^3} \, dx=-\frac {d^2 g^2-2 b^2 c^2 d e F^{-\frac {b c n (f+g x)}{g}} \left (F^{c (a+b x)}\right )^n n^2 (f+g x)^2 \operatorname {ExpIntegralEi}\left (\frac {b c n (f+g x) \log (F)}{g}\right ) \log ^2(F)-4 b^2 c^2 e^2 F^{-\frac {2 b c n (f+g x)}{g}} \left (F^{c (a+b x)}\right )^{2 n} n^2 (f+g x)^2 \operatorname {ExpIntegralEi}\left (\frac {2 b c n (f+g x) \log (F)}{g}\right ) \log ^2(F)+2 d e \left (F^{c (a+b x)}\right )^n g (g+b c n (f+g x) \log (F))+e^2 \left (F^{c (a+b x)}\right )^{2 n} g (g+2 b c n (f+g x) \log (F))}{2 g^3 (f+g x)^2} \] Input:

Integrate[(d + e*(F^(c*(a + b*x)))^n)^2/(f + g*x)^3,x]
 

Output:

-1/2*(d^2*g^2 - (2*b^2*c^2*d*e*(F^(c*(a + b*x)))^n*n^2*(f + g*x)^2*ExpInte 
gralEi[(b*c*n*(f + g*x)*Log[F])/g]*Log[F]^2)/F^((b*c*n*(f + g*x))/g) - (4* 
b^2*c^2*e^2*(F^(c*(a + b*x)))^(2*n)*n^2*(f + g*x)^2*ExpIntegralEi[(2*b*c*n 
*(f + g*x)*Log[F])/g]*Log[F]^2)/F^((2*b*c*n*(f + g*x))/g) + 2*d*e*(F^(c*(a 
 + b*x)))^n*g*(g + b*c*n*(f + g*x)*Log[F]) + e^2*(F^(c*(a + b*x)))^(2*n)*g 
*(g + 2*b*c*n*(f + g*x)*Log[F]))/(g^3*(f + g*x)^2)
 

Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 286, 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.

\(\displaystyle \int \frac {\left (e \left (F^{c (a+b x)}\right )^n+d\right )^2}{(f+g x)^3} \, dx\)

\(\Big \downarrow \) 2614

\(\displaystyle \int \left (\frac {2 d e \left (F^{a c+b c x}\right )^n}{(f+g x)^3}+\frac {e^2 \left (F^{a c+b c x}\right )^{2 n}}{(f+g x)^3}+\frac {d^2}{(f+g x)^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^2 c^2 d e n^2 \log ^2(F) \left (F^{a c+b c x}\right )^n F^{c n \left (a-\frac {b f}{g}\right )-c n (a+b x)} \operatorname {ExpIntegralEi}\left (\frac {b c n (f+g x) \log (F)}{g}\right )}{g^3}+\frac {2 b^2 c^2 e^2 n^2 \log ^2(F) \left (F^{a c+b c x}\right )^{2 n} F^{2 c n \left (a-\frac {b f}{g}\right )-2 c n (a+b x)} \operatorname {ExpIntegralEi}\left (\frac {2 b c n (f+g x) \log (F)}{g}\right )}{g^3}-\frac {b c d e n \log (F) \left (F^{a c+b c x}\right )^n}{g^2 (f+g x)}-\frac {d e \left (F^{a c+b c x}\right )^n}{g (f+g x)^2}-\frac {b c e^2 n \log (F) \left (F^{a c+b c x}\right )^{2 n}}{g^2 (f+g x)}-\frac {e^2 \left (F^{a c+b c x}\right )^{2 n}}{2 g (f+g x)^2}-\frac {d^2}{2 g (f+g x)^2}\)

Input:

Int[(d + e*(F^(c*(a + b*x)))^n)^2/(f + g*x)^3,x]
 

Output:

-1/2*d^2/(g*(f + g*x)^2) - (d*e*(F^(a*c + b*c*x))^n)/(g*(f + g*x)^2) - (e^ 
2*(F^(a*c + b*c*x))^(2*n))/(2*g*(f + g*x)^2) - (b*c*d*e*(F^(a*c + b*c*x))^ 
n*n*Log[F])/(g^2*(f + g*x)) - (b*c*e^2*(F^(a*c + b*c*x))^(2*n)*n*Log[F])/( 
g^2*(f + g*x)) + (b^2*c^2*d*e*F^(c*(a - (b*f)/g)*n - c*n*(a + b*x))*(F^(a* 
c + b*c*x))^n*n^2*ExpIntegralEi[(b*c*n*(f + g*x)*Log[F])/g]*Log[F]^2)/g^3 
+ (2*b^2*c^2*e^2*F^(2*c*(a - (b*f)/g)*n - 2*c*n*(a + b*x))*(F^(a*c + b*c*x 
))^(2*n)*n^2*ExpIntegralEi[(2*b*c*n*(f + g*x)*Log[F])/g]*Log[F]^2)/g^3
 

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]

\[\int \frac {{\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{2}}{\left (g x +f \right )^{3}}d x\]

Input:

int((d+e*(F^(c*(b*x+a)))^n)^2/(g*x+f)^3,x)
 

Output:

int((d+e*(F^(c*(b*x+a)))^n)^2/(g*x+f)^3,x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.11 \[ \int \frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2}{(f+g x)^3} \, dx=-\frac {d^{2} g^{2} - \frac {4 \, {\left (b^{2} c^{2} e^{2} g^{2} n^{2} x^{2} + 2 \, b^{2} c^{2} e^{2} f g n^{2} x + b^{2} c^{2} e^{2} f^{2} n^{2}\right )} {\rm Ei}\left (\frac {2 \, {\left (b c g n x + b c f n\right )} \log \left (F\right )}{g}\right ) \log \left (F\right )^{2}}{F^{\frac {2 \, {\left (b c f - a c g\right )} n}{g}}} - \frac {2 \, {\left (b^{2} c^{2} d e g^{2} n^{2} x^{2} + 2 \, b^{2} c^{2} d e f g n^{2} x + b^{2} c^{2} d e f^{2} n^{2}\right )} {\rm Ei}\left (\frac {{\left (b c g n x + b c f n\right )} \log \left (F\right )}{g}\right ) \log \left (F\right )^{2}}{F^{\frac {{\left (b c f - a c g\right )} n}{g}}} + {\left (e^{2} g^{2} + 2 \, {\left (b c e^{2} g^{2} n x + b c e^{2} f g n\right )} \log \left (F\right )\right )} F^{2 \, b c n x + 2 \, a c n} + 2 \, {\left (d e g^{2} + {\left (b c d e g^{2} n x + b c d e f g n\right )} \log \left (F\right )\right )} F^{b c n x + a c n}}{2 \, {\left (g^{5} x^{2} + 2 \, f g^{4} x + f^{2} g^{3}\right )}} \] Input:

integrate((d+e*(F^((b*x+a)*c))^n)^2/(g*x+f)^3,x, algorithm="fricas")
 

Output:

-1/2*(d^2*g^2 - 4*(b^2*c^2*e^2*g^2*n^2*x^2 + 2*b^2*c^2*e^2*f*g*n^2*x + b^2 
*c^2*e^2*f^2*n^2)*Ei(2*(b*c*g*n*x + b*c*f*n)*log(F)/g)*log(F)^2/F^(2*(b*c* 
f - a*c*g)*n/g) - 2*(b^2*c^2*d*e*g^2*n^2*x^2 + 2*b^2*c^2*d*e*f*g*n^2*x + b 
^2*c^2*d*e*f^2*n^2)*Ei((b*c*g*n*x + b*c*f*n)*log(F)/g)*log(F)^2/F^((b*c*f 
- a*c*g)*n/g) + (e^2*g^2 + 2*(b*c*e^2*g^2*n*x + b*c*e^2*f*g*n)*log(F))*F^( 
2*b*c*n*x + 2*a*c*n) + 2*(d*e*g^2 + (b*c*d*e*g^2*n*x + b*c*d*e*f*g*n)*log( 
F))*F^(b*c*n*x + a*c*n))/(g^5*x^2 + 2*f*g^4*x + f^2*g^3)
 

Sympy [F]

\[ \int \frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2}{(f+g x)^3} \, dx=\int \frac {\left (d + e \left (F^{a c + b c x}\right )^{n}\right )^{2}}{\left (f + g x\right )^{3}}\, dx \] Input:

integrate((d+e*(F**((b*x+a)*c))**n)**2/(g*x+f)**3,x)
 

Output:

Integral((d + e*(F**(a*c + b*c*x))**n)**2/(f + g*x)**3, x)
 

Maxima [F]

\[ \int \frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2}{(f+g x)^3} \, dx=\int { \frac {{\left ({\left (F^{{\left (b x + a\right )} c}\right )}^{n} e + d\right )}^{2}}{{\left (g x + f\right )}^{3}} \,d x } \] Input:

integrate((d+e*(F^((b*x+a)*c))^n)^2/(g*x+f)^3,x, algorithm="maxima")
 

Output:

F^(2*a*c*n)*e^2*integrate(F^(2*b*c*n*x)/(g^3*x^3 + 3*f*g^2*x^2 + 3*f^2*g*x 
 + f^3), x) + 2*F^(a*c*n)*d*e*integrate(F^(b*c*n*x)/(g^3*x^3 + 3*f*g^2*x^2 
 + 3*f^2*g*x + f^3), x) - 1/2*d^2/(g^3*x^2 + 2*f*g^2*x + f^2*g)
 

Giac [F]

\[ \int \frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2}{(f+g x)^3} \, dx=\int { \frac {{\left ({\left (F^{{\left (b x + a\right )} c}\right )}^{n} e + d\right )}^{2}}{{\left (g x + f\right )}^{3}} \,d x } \] Input:

integrate((d+e*(F^((b*x+a)*c))^n)^2/(g*x+f)^3,x, algorithm="giac")
 

Output:

integrate(((F^((b*x + a)*c))^n*e + d)^2/(g*x + f)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2}{(f+g x)^3} \, dx=\int \frac {{\left (d+e\,{\left (F^{c\,\left (a+b\,x\right )}\right )}^n\right )}^2}{{\left (f+g\,x\right )}^3} \,d x \] Input:

int((d + e*(F^(c*(a + b*x)))^n)^2/(f + g*x)^3,x)
 

Output:

int((d + e*(F^(c*(a + b*x)))^n)^2/(f + g*x)^3, x)
 

Reduce [F]

\[ \int \frac {\left (d+e \left (F^{c (a+b x)}\right )^n\right )^2}{(f+g x)^3} \, dx =\text {Too large to display} \] Input:

int((d+e*(F^((b*x+a)*c))^n)^2/(g*x+f)^3,x)
 

Output:

(2*f**(2*a*c*n)*int(f**(2*b*c*n*x)/(log(f)*b*c*f**4*n + 3*log(f)*b*c*f**3* 
g*n*x + 3*log(f)*b*c*f**2*g**2*n*x**2 + log(f)*b*c*f*g**3*n*x**3 - 2*f**3* 
g - 6*f**2*g**2*x - 6*f*g**3*x**2 - 2*g**4*x**3),x)*log(f)*b*c*e**2*f**3*g 
*n + 4*f**(2*a*c*n)*int(f**(2*b*c*n*x)/(log(f)*b*c*f**4*n + 3*log(f)*b*c*f 
**3*g*n*x + 3*log(f)*b*c*f**2*g**2*n*x**2 + log(f)*b*c*f*g**3*n*x**3 - 2*f 
**3*g - 6*f**2*g**2*x - 6*f*g**3*x**2 - 2*g**4*x**3),x)*log(f)*b*c*e**2*f* 
*2*g**2*n*x + 2*f**(2*a*c*n)*int(f**(2*b*c*n*x)/(log(f)*b*c*f**4*n + 3*log 
(f)*b*c*f**3*g*n*x + 3*log(f)*b*c*f**2*g**2*n*x**2 + log(f)*b*c*f*g**3*n*x 
**3 - 2*f**3*g - 6*f**2*g**2*x - 6*f*g**3*x**2 - 2*g**4*x**3),x)*log(f)*b* 
c*e**2*f*g**3*n*x**2 - 4*f**(2*a*c*n)*int(f**(2*b*c*n*x)/(log(f)*b*c*f**4* 
n + 3*log(f)*b*c*f**3*g*n*x + 3*log(f)*b*c*f**2*g**2*n*x**2 + log(f)*b*c*f 
*g**3*n*x**3 - 2*f**3*g - 6*f**2*g**2*x - 6*f*g**3*x**2 - 2*g**4*x**3),x)* 
e**2*f**2*g**2 - 8*f**(2*a*c*n)*int(f**(2*b*c*n*x)/(log(f)*b*c*f**4*n + 3* 
log(f)*b*c*f**3*g*n*x + 3*log(f)*b*c*f**2*g**2*n*x**2 + log(f)*b*c*f*g**3* 
n*x**3 - 2*f**3*g - 6*f**2*g**2*x - 6*f*g**3*x**2 - 2*g**4*x**3),x)*e**2*f 
*g**3*x - 4*f**(2*a*c*n)*int(f**(2*b*c*n*x)/(log(f)*b*c*f**4*n + 3*log(f)* 
b*c*f**3*g*n*x + 3*log(f)*b*c*f**2*g**2*n*x**2 + log(f)*b*c*f*g**3*n*x**3 
- 2*f**3*g - 6*f**2*g**2*x - 6*f*g**3*x**2 - 2*g**4*x**3),x)*e**2*g**4*x** 
2 + 4*f**(a*c*n)*int(f**(b*c*n*x)/(log(f)*b*c*f**4*n + 3*log(f)*b*c*f**3*g 
*n*x + 3*log(f)*b*c*f**2*g**2*n*x**2 + log(f)*b*c*f*g**3*n*x**3 - f**3*...