\(\int \frac {(a+b (F^{g (e+f x)})^n)^3}{c+d x} \, dx\) [43]

Optimal result

Integrand size = 25, antiderivative size = 200 \[ \int \frac {\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3}{c+d x} \, dx=\frac {3 a^2 b F^{\left (e-\frac {c f}{d}\right ) g n-g n (e+f x)} \left (F^{e g+f g x}\right )^n \operatorname {ExpIntegralEi}\left (\frac {f g n (c+d x) \log (F)}{d}\right )}{d}+\frac {3 a b^2 F^{2 \left (e-\frac {c f}{d}\right ) g n-2 g n (e+f x)} \left (F^{e g+f g x}\right )^{2 n} \operatorname {ExpIntegralEi}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right )}{d}+\frac {b^3 F^{3 \left (e-\frac {c f}{d}\right ) g n-3 g n (e+f x)} \left (F^{e g+f g x}\right )^{3 n} \operatorname {ExpIntegralEi}\left (\frac {3 f g n (c+d x) \log (F)}{d}\right )}{d}+\frac {a^3 \log (c+d x)}{d} \] Output:

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

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3}{c+d x} \, dx=\frac {3 a^2 b F^{-\frac {f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^n \operatorname {ExpIntegralEi}\left (\frac {f g n (c+d x) \log (F)}{d}\right )+3 a b^2 F^{-\frac {2 f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^{2 n} \operatorname {ExpIntegralEi}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right )+b^3 F^{-\frac {3 f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^{3 n} \operatorname {ExpIntegralEi}\left (\frac {3 f g n (c+d x) \log (F)}{d}\right )+a^3 \log (c+d x)}{d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 200, 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.

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

\(\Big \downarrow \) 2614

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.70 \[ \int \frac {\left (a+b \left (F^{g (e+f x)}\right )^n\right )^3}{c+d x} \, dx=\frac {F^{\frac {3 \, {\left (d e - c f\right )} g n}{d}} b^{3} {\rm Ei}\left (\frac {3 \, {\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) + 3 \, F^{\frac {2 \, {\left (d e - c f\right )} g n}{d}} a b^{2} {\rm Ei}\left (\frac {2 \, {\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) + 3 \, F^{\frac {{\left (d e - c f\right )} g n}{d}} a^{2} b {\rm Ei}\left (\frac {{\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) + a^{3} \log \left (d x + c\right )}{d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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