\(\int \frac {(a+b (F^{g (e+f x)})^n)^3}{(c+d x)^3} \, dx\) [45]

Optimal result

Integrand size = 25, antiderivative size = 447 \[ \int \frac {\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3}{(c+d x)^3} \, dx=-\frac {a^3}{2 d (c+d x)^2}-\frac {3 a^2 b \left (F^{e g+f g x}\right )^n}{2 d (c+d x)^2}-\frac {3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{2 d (c+d x)^2}-\frac {b^3 \left (F^{e g+f g x}\right )^{3 n}}{2 d (c+d x)^2}-\frac {3 a^2 b f \left (F^{e g+f g x}\right )^n g n \log (F)}{2 d^2 (c+d x)}-\frac {3 a b^2 f \left (F^{e g+f g x}\right )^{2 n} g n \log (F)}{d^2 (c+d x)}-\frac {3 b^3 f \left (F^{e g+f g x}\right )^{3 n} g n \log (F)}{2 d^2 (c+d x)}+\frac {3 a^2 b f^2 F^{\left (e-\frac {c f}{d}\right ) g n-g n (e+f x)} \left (F^{e g+f g x}\right )^n g^2 n^2 \operatorname {ExpIntegralEi}\left (\frac {f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)}{2 d^3}+\frac {6 a b^2 f^2 F^{2 \left (e-\frac {c f}{d}\right ) g n-2 g n (e+f x)} \left (F^{e g+f g x}\right )^{2 n} g^2 n^2 \operatorname {ExpIntegralEi}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)}{d^3}+\frac {9 b^3 f^2 F^{3 \left (e-\frac {c f}{d}\right ) g n-3 g n (e+f x)} \left (F^{e g+f g x}\right )^{3 n} g^2 n^2 \operatorname {ExpIntegralEi}\left (\frac {3 f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)}{2 d^3} \] Output:

-1/2*a^3/d/(d*x+c)^2-3/2*a^2*b*(F^(f*g*x+e*g))^n/d/(d*x+c)^2-3/2*a*b^2*(F^ 
(f*g*x+e*g))^(2*n)/d/(d*x+c)^2-1/2*b^3*(F^(f*g*x+e*g))^(3*n)/d/(d*x+c)^2-3 
/2*a^2*b*f*(F^(f*g*x+e*g))^n*g*n*ln(F)/d^2/(d*x+c)-3*a*b^2*f*(F^(f*g*x+e*g 
))^(2*n)*g*n*ln(F)/d^2/(d*x+c)-3/2*b^3*f*(F^(f*g*x+e*g))^(3*n)*g*n*ln(F)/d 
^2/(d*x+c)+3/2*a^2*b*f^2*F^((e-c*f/d)*g*n-g*n*(f*x+e))*(F^(f*g*x+e*g))^n*g 
^2*n^2*Ei(f*g*n*(d*x+c)*ln(F)/d)*ln(F)^2/d^3+6*a*b^2*f^2*F^(2*(e-c*f/d)*g* 
n-2*g*n*(f*x+e))*(F^(f*g*x+e*g))^(2*n)*g^2*n^2*Ei(2*f*g*n*(d*x+c)*ln(F)/d) 
*ln(F)^2/d^3+9/2*b^3*f^2*F^(3*(e-c*f/d)*g*n-3*g*n*(f*x+e))*(F^(f*g*x+e*g)) 
^(3*n)*g^2*n^2*Ei(3*f*g*n*(d*x+c)*ln(F)/d)*ln(F)^2/d^3
 

Mathematica [A] (verified)

Time = 1.43 (sec) , antiderivative size = 325, normalized size of antiderivative = 0.73 \[ \int \frac {\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3}{(c+d x)^3} \, dx=-\frac {a^3 d^2-3 a^2 b f^2 F^{-\frac {f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^n g^2 n^2 (c+d x)^2 \operatorname {ExpIntegralEi}\left (\frac {f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)-12 a b^2 f^2 F^{-\frac {2 f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^{2 n} g^2 n^2 (c+d x)^2 \operatorname {ExpIntegralEi}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)-9 b^3 f^2 F^{-\frac {3 f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^{3 n} g^2 n^2 (c+d x)^2 \operatorname {ExpIntegralEi}\left (\frac {3 f g n (c+d x) \log (F)}{d}\right ) \log ^2(F)+3 a^2 b d \left (F^{g (e+f x)}\right )^n (d+f g n (c+d x) \log (F))+3 a b^2 d \left (F^{g (e+f x)}\right )^{2 n} (d+2 f g n (c+d x) \log (F))+b^3 d \left (F^{g (e+f x)}\right )^{3 n} (d+3 f g n (c+d x) \log (F))}{2 d^3 (c+d x)^2} \] Input:

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

Output:

-1/2*(a^3*d^2 - (3*a^2*b*f^2*(F^(g*(e + f*x)))^n*g^2*n^2*(c + d*x)^2*ExpIn 
tegralEi[(f*g*n*(c + d*x)*Log[F])/d]*Log[F]^2)/F^((f*g*n*(c + d*x))/d) - ( 
12*a*b^2*f^2*(F^(g*(e + f*x)))^(2*n)*g^2*n^2*(c + d*x)^2*ExpIntegralEi[(2* 
f*g*n*(c + d*x)*Log[F])/d]*Log[F]^2)/F^((2*f*g*n*(c + d*x))/d) - (9*b^3*f^ 
2*(F^(g*(e + f*x)))^(3*n)*g^2*n^2*(c + d*x)^2*ExpIntegralEi[(3*f*g*n*(c + 
d*x)*Log[F])/d]*Log[F]^2)/F^((3*f*g*n*(c + d*x))/d) + 3*a^2*b*d*(F^(g*(e + 
 f*x)))^n*(d + f*g*n*(c + d*x)*Log[F]) + 3*a*b^2*d*(F^(g*(e + f*x)))^(2*n) 
*(d + 2*f*g*n*(c + d*x)*Log[F]) + b^3*d*(F^(g*(e + f*x)))^(3*n)*(d + 3*f*g 
*n*(c + d*x)*Log[F]))/(d^3*(c + d*x)^2)
 

Rubi [A] (verified)

Time = 1.61 (sec) , antiderivative size = 447, 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.

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

\(\Big \downarrow \) 2614

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^3}{2 d (c+d x)^2}+\frac {3 a^2 b f^2 g^2 n^2 \log ^2(F) \left (F^{e g+f g x}\right )^n F^{g n \left (e-\frac {c f}{d}\right )-g n (e+f x)} \operatorname {ExpIntegralEi}\left (\frac {f g n (c+d x) \log (F)}{d}\right )}{2 d^3}-\frac {3 a^2 b f g n \log (F) \left (F^{e g+f g x}\right )^n}{2 d^2 (c+d x)}-\frac {3 a^2 b \left (F^{e g+f g x}\right )^n}{2 d (c+d x)^2}+\frac {6 a b^2 f^2 g^2 n^2 \log ^2(F) \left (F^{e g+f g x}\right )^{2 n} F^{2 g n \left (e-\frac {c f}{d}\right )-2 g n (e+f x)} \operatorname {ExpIntegralEi}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right )}{d^3}-\frac {3 a b^2 f g n \log (F) \left (F^{e g+f g x}\right )^{2 n}}{d^2 (c+d x)}-\frac {3 a b^2 \left (F^{e g+f g x}\right )^{2 n}}{2 d (c+d x)^2}+\frac {9 b^3 f^2 g^2 n^2 \log ^2(F) \left (F^{e g+f g x}\right )^{3 n} F^{3 g n \left (e-\frac {c f}{d}\right )-3 g n (e+f x)} \operatorname {ExpIntegralEi}\left (\frac {3 f g n (c+d x) \log (F)}{d}\right )}{2 d^3}-\frac {3 b^3 f g n \log (F) \left (F^{e g+f g x}\right )^{3 n}}{2 d^2 (c+d x)}-\frac {b^3 \left (F^{e g+f g x}\right )^{3 n}}{2 d (c+d x)^2}\)

Input:

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

Output:

-1/2*a^3/(d*(c + d*x)^2) - (3*a^2*b*(F^(e*g + f*g*x))^n)/(2*d*(c + d*x)^2) 
 - (3*a*b^2*(F^(e*g + f*g*x))^(2*n))/(2*d*(c + d*x)^2) - (b^3*(F^(e*g + f* 
g*x))^(3*n))/(2*d*(c + d*x)^2) - (3*a^2*b*f*(F^(e*g + f*g*x))^n*g*n*Log[F] 
)/(2*d^2*(c + d*x)) - (3*a*b^2*f*(F^(e*g + f*g*x))^(2*n)*g*n*Log[F])/(d^2* 
(c + d*x)) - (3*b^3*f*(F^(e*g + f*g*x))^(3*n)*g*n*Log[F])/(2*d^2*(c + d*x) 
) + (3*a^2*b*f^2*F^((e - (c*f)/d)*g*n - g*n*(e + f*x))*(F^(e*g + f*g*x))^n 
*g^2*n^2*ExpIntegralEi[(f*g*n*(c + d*x)*Log[F])/d]*Log[F]^2)/(2*d^3) + (6* 
a*b^2*f^2*F^(2*(e - (c*f)/d)*g*n - 2*g*n*(e + f*x))*(F^(e*g + f*g*x))^(2*n 
)*g^2*n^2*ExpIntegralEi[(2*f*g*n*(c + d*x)*Log[F])/d]*Log[F]^2)/d^3 + (9*b 
^3*f^2*F^(3*(e - (c*f)/d)*g*n - 3*g*n*(e + f*x))*(F^(e*g + f*g*x))^(3*n)*g 
^2*n^2*ExpIntegralEi[(3*f*g*n*(c + d*x)*Log[F])/d]*Log[F]^2)/(2*d^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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 475, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3}{(c+d x)^3} \, dx=-\frac {a^{3} d^{2} - 9 \, {\left (b^{3} d^{2} f^{2} g^{2} n^{2} x^{2} + 2 \, b^{3} c d f^{2} g^{2} n^{2} x + b^{3} c^{2} f^{2} g^{2} n^{2}\right )} F^{\frac {3 \, {\left (d e - c f\right )} g n}{d}} {\rm Ei}\left (\frac {3 \, {\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) \log \left (F\right )^{2} - 12 \, {\left (a b^{2} d^{2} f^{2} g^{2} n^{2} x^{2} + 2 \, a b^{2} c d f^{2} g^{2} n^{2} x + a b^{2} c^{2} f^{2} g^{2} n^{2}\right )} F^{\frac {2 \, {\left (d e - c f\right )} g n}{d}} {\rm Ei}\left (\frac {2 \, {\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) \log \left (F\right )^{2} - 3 \, {\left (a^{2} b d^{2} f^{2} g^{2} n^{2} x^{2} + 2 \, a^{2} b c d f^{2} g^{2} n^{2} x + a^{2} b c^{2} f^{2} g^{2} n^{2}\right )} F^{\frac {{\left (d e - c f\right )} g n}{d}} {\rm Ei}\left (\frac {{\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) \log \left (F\right )^{2} + {\left (b^{3} d^{2} + 3 \, {\left (b^{3} d^{2} f g n x + b^{3} c d f g n\right )} \log \left (F\right )\right )} F^{3 \, f g n x + 3 \, e g n} + 3 \, {\left (a b^{2} d^{2} + 2 \, {\left (a b^{2} d^{2} f g n x + a b^{2} c d f g n\right )} \log \left (F\right )\right )} F^{2 \, f g n x + 2 \, e g n} + 3 \, {\left (a^{2} b d^{2} + {\left (a^{2} b d^{2} f g n x + a^{2} b c d f g n\right )} \log \left (F\right )\right )} F^{f g n x + e g n}}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \] Input:

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

Output:

-1/2*(a^3*d^2 - 9*(b^3*d^2*f^2*g^2*n^2*x^2 + 2*b^3*c*d*f^2*g^2*n^2*x + b^3 
*c^2*f^2*g^2*n^2)*F^(3*(d*e - c*f)*g*n/d)*Ei(3*(d*f*g*n*x + c*f*g*n)*log(F 
)/d)*log(F)^2 - 12*(a*b^2*d^2*f^2*g^2*n^2*x^2 + 2*a*b^2*c*d*f^2*g^2*n^2*x 
+ a*b^2*c^2*f^2*g^2*n^2)*F^(2*(d*e - c*f)*g*n/d)*Ei(2*(d*f*g*n*x + c*f*g*n 
)*log(F)/d)*log(F)^2 - 3*(a^2*b*d^2*f^2*g^2*n^2*x^2 + 2*a^2*b*c*d*f^2*g^2* 
n^2*x + a^2*b*c^2*f^2*g^2*n^2)*F^((d*e - c*f)*g*n/d)*Ei((d*f*g*n*x + c*f*g 
*n)*log(F)/d)*log(F)^2 + (b^3*d^2 + 3*(b^3*d^2*f*g*n*x + b^3*c*d*f*g*n)*lo 
g(F))*F^(3*f*g*n*x + 3*e*g*n) + 3*(a*b^2*d^2 + 2*(a*b^2*d^2*f*g*n*x + a*b^ 
2*c*d*f*g*n)*log(F))*F^(2*f*g*n*x + 2*e*g*n) + 3*(a^2*b*d^2 + (a^2*b*d^2*f 
*g*n*x + a^2*b*c*d*f*g*n)*log(F))*F^(f*g*n*x + e*g*n))/(d^5*x^2 + 2*c*d^4* 
x + c^2*d^3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*f**(3*e*g*n)*int(f**(3*f*g*n*x)/(log(f)**2*c**5*f**2*g**2*n**2 + 3*log( 
f)**2*c**4*d*f**2*g**2*n**2*x + 3*log(f)**2*c**3*d**2*f**2*g**2*n**2*x**2 
+ log(f)**2*c**2*d**3*f**2*g**2*n**2*x**3 - 3*log(f)*c**4*d*f*g*n - 9*log( 
f)*c**3*d**2*f*g*n*x - 9*log(f)*c**2*d**3*f*g*n*x**2 - 3*log(f)*c*d**4*f*g 
*n*x**3 + 2*c**3*d**2 + 6*c**2*d**3*x + 6*c*d**4*x**2 + 2*d**5*x**3),x)*lo 
g(f)**2*b**3*c**4*d*f**2*g**2*n**2 + 4*f**(3*e*g*n)*int(f**(3*f*g*n*x)/(lo 
g(f)**2*c**5*f**2*g**2*n**2 + 3*log(f)**2*c**4*d*f**2*g**2*n**2*x + 3*log( 
f)**2*c**3*d**2*f**2*g**2*n**2*x**2 + log(f)**2*c**2*d**3*f**2*g**2*n**2*x 
**3 - 3*log(f)*c**4*d*f*g*n - 9*log(f)*c**3*d**2*f*g*n*x - 9*log(f)*c**2*d 
**3*f*g*n*x**2 - 3*log(f)*c*d**4*f*g*n*x**3 + 2*c**3*d**2 + 6*c**2*d**3*x 
+ 6*c*d**4*x**2 + 2*d**5*x**3),x)*log(f)**2*b**3*c**3*d**2*f**2*g**2*n**2* 
x + 2*f**(3*e*g*n)*int(f**(3*f*g*n*x)/(log(f)**2*c**5*f**2*g**2*n**2 + 3*l 
og(f)**2*c**4*d*f**2*g**2*n**2*x + 3*log(f)**2*c**3*d**2*f**2*g**2*n**2*x* 
*2 + log(f)**2*c**2*d**3*f**2*g**2*n**2*x**3 - 3*log(f)*c**4*d*f*g*n - 9*l 
og(f)*c**3*d**2*f*g*n*x - 9*log(f)*c**2*d**3*f*g*n*x**2 - 3*log(f)*c*d**4* 
f*g*n*x**3 + 2*c**3*d**2 + 6*c**2*d**3*x + 6*c*d**4*x**2 + 2*d**5*x**3),x) 
*log(f)**2*b**3*c**2*d**3*f**2*g**2*n**2*x**2 - 6*f**(3*e*g*n)*int(f**(3*f 
*g*n*x)/(log(f)**2*c**5*f**2*g**2*n**2 + 3*log(f)**2*c**4*d*f**2*g**2*n**2 
*x + 3*log(f)**2*c**3*d**2*f**2*g**2*n**2*x**2 + log(f)**2*c**2*d**3*f**2* 
g**2*n**2*x**3 - 3*log(f)*c**4*d*f*g*n - 9*log(f)*c**3*d**2*f*g*n*x - 9...