\(\int \frac {(c+d x)^3}{a+b (F^{g (e+f x)})^n} \, dx\) [46]

Optimal result

Integrand size = 25, antiderivative size = 192 \[ \int \frac {(c+d x)^3}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {(c+d x)^4}{4 a d}-\frac {(c+d x)^3 \log \left (1+\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f g n \log (F)}-\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^2 g^2 n^2 \log ^2(F)}+\frac {6 d^2 (c+d x) \operatorname {PolyLog}\left (3,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^3 g^3 n^3 \log ^3(F)}-\frac {6 d^3 \operatorname {PolyLog}\left (4,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^4 g^4 n^4 \log ^4(F)} \] Output:

1/4*(d*x+c)^4/a/d-(d*x+c)^3*ln(1+b*(F^(g*(f*x+e)))^n/a)/a/f/g/n/ln(F)-3*d* 
(d*x+c)^2*polylog(2,-b*(F^(g*(f*x+e)))^n/a)/a/f^2/g^2/n^2/ln(F)^2+6*d^2*(d 
*x+c)*polylog(3,-b*(F^(g*(f*x+e)))^n/a)/a/f^3/g^3/n^3/ln(F)^3-6*d^3*polylo 
g(4,-b*(F^(g*(f*x+e)))^n/a)/a/f^4/g^4/n^4/ln(F)^4
 

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.86 \[ \int \frac {(c+d x)^3}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {-(c+d x)^3 \log \left (1+\frac {a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )+\frac {3 d \left (f^2 g^2 n^2 (c+d x)^2 \log ^2(F) \operatorname {PolyLog}\left (2,-\frac {a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )+2 d \left (f g n (c+d x) \log (F) \operatorname {PolyLog}\left (3,-\frac {a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )+d \operatorname {PolyLog}\left (4,-\frac {a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )\right )\right )}{f^3 g^3 n^3 \log ^3(F)}}{a f g n \log (F)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2615, 2620, 3011, 7163, 2720, 7143}

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.

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

\(\Big \downarrow \) 2615

\(\displaystyle \frac {(c+d x)^4}{4 a d}-\frac {b \int \frac {\left (F^{g (e+f x)}\right )^n (c+d x)^3}{b \left (F^{g (e+f x)}\right )^n+a}dx}{a}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {(c+d x)^4}{4 a d}-\frac {b \left (\frac {(c+d x)^3 \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{b f g n \log (F)}-\frac {3 d \int (c+d x)^2 \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )dx}{b f g n \log (F)}\right )}{a}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {(c+d x)^4}{4 a d}-\frac {b \left (\frac {(c+d x)^3 \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{b f g n \log (F)}-\frac {3 d \left (\frac {2 d \int (c+d x) \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )dx}{f g n \log (F)}-\frac {(c+d x)^2 \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{f g n \log (F)}\right )}{b f g n \log (F)}\right )}{a}\)

\(\Big \downarrow \) 7163

\(\displaystyle \frac {(c+d x)^4}{4 a d}-\frac {b \left (\frac {(c+d x)^3 \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{b f g n \log (F)}-\frac {3 d \left (\frac {2 d \left (\frac {(c+d x) \operatorname {PolyLog}\left (3,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{f g n \log (F)}-\frac {d \int \operatorname {PolyLog}\left (3,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )dx}{f g n \log (F)}\right )}{f g n \log (F)}-\frac {(c+d x)^2 \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{f g n \log (F)}\right )}{b f g n \log (F)}\right )}{a}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {(c+d x)^4}{4 a d}-\frac {b \left (\frac {(c+d x)^3 \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{b f g n \log (F)}-\frac {3 d \left (\frac {2 d \left (\frac {(c+d x) \operatorname {PolyLog}\left (3,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{f g n \log (F)}-\frac {d \int F^{-g (e+f x)} \operatorname {PolyLog}\left (3,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )dF^{g (e+f x)}}{f^2 g^2 n \log ^2(F)}\right )}{f g n \log (F)}-\frac {(c+d x)^2 \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{f g n \log (F)}\right )}{b f g n \log (F)}\right )}{a}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {(c+d x)^4}{4 a d}-\frac {b \left (\frac {(c+d x)^3 \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{b f g n \log (F)}-\frac {3 d \left (\frac {2 d \left (\frac {(c+d x) \operatorname {PolyLog}\left (3,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{f g n \log (F)}-\frac {d \operatorname {PolyLog}\left (4,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{f^2 g^2 n^2 \log ^2(F)}\right )}{f g n \log (F)}-\frac {(c+d x)^2 \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{f g n \log (F)}\right )}{b f g n \log (F)}\right )}{a}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2615

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x 
_))))^(n_.)), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(a*d*(m + 1)), x] - Simp[ 
b/a   Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n)), x] 
, x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[m, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3074\) vs. \(2(190)=380\).

Time = 0.32 (sec) , antiderivative size = 3075, normalized size of antiderivative = 16.02

method	result	size
risch	\(\text {Expression too large to display}\)	\(3075\)

Input:

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

Output:

-1/n/g^4/f^4/ln(F)^4*d^3*(ln(F^(g*(f*x+e)))-g*(f*x+e)*ln(F))^3/a*ln(F^(n*g 
*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n)+1/n/g^4/f^4/ln(F)^4*d^3*(ln(F^(g*(f* 
x+e)))-g*(f*x+e)*ln(F))^3/a*ln((F^(g*(f*x+e)))^n*F^(-n*g*f*x)*F^(n*g*f*x)* 
b+a)-1/n/g/f/ln(F)*d^3/a*ln(1+b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n 
/a)*x^3-3/n^2/g^2/f^2/ln(F)^2*d^3/a*polylog(2,-b*F^(n*g*f*x)*F^(-n*g*f*x)* 
(F^(g*(f*x+e)))^n/a)*x^2-1/n/g^4/f^4/ln(F)^4*d^3/a*ln(1+b*F^(n*g*f*x)*F^(- 
n*g*f*x)*(F^(g*(f*x+e)))^n/a)*(ln(F^(g*(f*x+e)))-g*(f*x+e)*ln(F))^3-3/n^2/ 
g^2/f^2/ln(F)^2*c^2*d/a*polylog(2,-b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e 
)))^n/a)-1/n/g/f^4/ln(F)*d^3/a*ln(1+b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+ 
e)))^n/a)*e^3+3/g/f/ln(F)*c^2*d/a*x*(ln(F^(g*(f*x+e)))-g*(f*x+e)*ln(F))-3/ 
g^2/f^2/ln(F)^2*c*d^2/a*x*(ln(F^(g*(f*x+e)))-g*(f*x+e)*ln(F))^2+3/g^2/f^3/ 
ln(F)^2*d^3/a*x*e*(ln(F^(g*(f*x+e)))-g*(f*x+e)*ln(F))^2-3/n/g/f^2/ln(F)*c^ 
2*d/a*ln(1+b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n/a)*e-3/n/g^2/f^2/l 
n(F)^2*c^2*d/a*ln(1+b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n/a)*(ln(F^ 
(g*(f*x+e)))-g*(f*x+e)*ln(F))-6/g/f^2/ln(F)*c*d^2/a*x*e*(ln(F^(g*(f*x+e))) 
-g*(f*x+e)*ln(F))-3/n/g/f/ln(F)*c*d^2/a*ln(1+b*F^(n*g*f*x)*F^(-n*g*f*x)*(F 
^(g*(f*x+e)))^n/a)*x^2+3/n/g/f^3/ln(F)*c*d^2/a*ln(1+b*F^(n*g*f*x)*F^(-n*g* 
f*x)*(F^(g*(f*x+e)))^n/a)*e^2+3/n/g^3/f^3/ln(F)^3*c*d^2/a*ln(1+b*F^(n*g*f* 
x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n/a)*(ln(F^(g*(f*x+e)))-g*(f*x+e)*ln(F))^2 
-6/n^2/g^2/f^2/ln(F)^2*c*d^2/a*polylog(2,-b*F^(n*g*f*x)*F^(-n*g*f*x)*(F...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 413 vs. \(2 (189) = 378\).

Time = 0.07 (sec) , antiderivative size = 413, normalized size of antiderivative = 2.15 \[ \int \frac {(c+d x)^3}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {4 \, {\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} g^{3} n^{3} \log \left (F^{f g n x + e g n} b + a\right ) \log \left (F\right )^{3} + {\left (d^{3} f^{4} g^{4} n^{4} x^{4} + 4 \, c d^{2} f^{4} g^{4} n^{4} x^{3} + 6 \, c^{2} d f^{4} g^{4} n^{4} x^{2} + 4 \, c^{3} f^{4} g^{4} n^{4} x\right )} \log \left (F\right )^{4} - 4 \, {\left (d^{3} f^{3} g^{3} n^{3} x^{3} + 3 \, c d^{2} f^{3} g^{3} n^{3} x^{2} + 3 \, c^{2} d f^{3} g^{3} n^{3} x + {\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2}\right )} g^{3} n^{3}\right )} \log \left (F\right )^{3} \log \left (\frac {F^{f g n x + e g n} b + a}{a}\right ) - 12 \, {\left (d^{3} f^{2} g^{2} n^{2} x^{2} + 2 \, c d^{2} f^{2} g^{2} n^{2} x + c^{2} d f^{2} g^{2} n^{2}\right )} {\rm Li}_2\left (-\frac {F^{f g n x + e g n} b + a}{a} + 1\right ) \log \left (F\right )^{2} - 24 \, d^{3} {\rm polylog}\left (4, -\frac {F^{f g n x + e g n} b}{a}\right ) + 24 \, {\left (d^{3} f g n x + c d^{2} f g n\right )} \log \left (F\right ) {\rm polylog}\left (3, -\frac {F^{f g n x + e g n} b}{a}\right )}{4 \, a f^{4} g^{4} n^{4} \log \left (F\right )^{4}} \] Input:

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

Output:

1/4*(4*(d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*g^3*n^3*log(F^( 
f*g*n*x + e*g*n)*b + a)*log(F)^3 + (d^3*f^4*g^4*n^4*x^4 + 4*c*d^2*f^4*g^4* 
n^4*x^3 + 6*c^2*d*f^4*g^4*n^4*x^2 + 4*c^3*f^4*g^4*n^4*x)*log(F)^4 - 4*(d^3 
*f^3*g^3*n^3*x^3 + 3*c*d^2*f^3*g^3*n^3*x^2 + 3*c^2*d*f^3*g^3*n^3*x + (d^3* 
e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2)*g^3*n^3)*log(F)^3*log((F^(f*g*n*x + e 
*g*n)*b + a)/a) - 12*(d^3*f^2*g^2*n^2*x^2 + 2*c*d^2*f^2*g^2*n^2*x + c^2*d* 
f^2*g^2*n^2)*dilog(-(F^(f*g*n*x + e*g*n)*b + a)/a + 1)*log(F)^2 - 24*d^3*p 
olylog(4, -F^(f*g*n*x + e*g*n)*b/a) + 24*(d^3*f*g*n*x + c*d^2*f*g*n)*log(F 
)*polylog(3, -F^(f*g*n*x + e*g*n)*b/a))/(a*f^4*g^4*n^4*log(F)^4)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 480 vs. \(2 (189) = 378\).

Time = 0.11 (sec) , antiderivative size = 480, normalized size of antiderivative = 2.50 \[ \int \frac {(c+d x)^3}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=c^{3} {\left (\frac {f g n x + e g n}{a f g n} - \frac {\log \left (F^{f g n x + e g n} b + a\right )}{a f g n \log \left (F\right )}\right )} - \frac {3 \, {\left (f g n x \log \left (\frac {F^{f g n x} F^{e g n} b}{a} + 1\right ) \log \left (F\right ) + {\rm Li}_2\left (-\frac {F^{f g n x} F^{e g n} b}{a}\right )\right )} c^{2} d}{a f^{2} g^{2} n^{2} \log \left (F\right )^{2}} - \frac {3 \, {\left (f^{2} g^{2} n^{2} x^{2} \log \left (\frac {F^{f g n x} F^{e g n} b}{a} + 1\right ) \log \left (F\right )^{2} + 2 \, f g n x {\rm Li}_2\left (-\frac {F^{f g n x} F^{e g n} b}{a}\right ) \log \left (F\right ) - 2 \, {\rm Li}_{3}(-\frac {F^{f g n x} F^{e g n} b}{a})\right )} c d^{2}}{a f^{3} g^{3} n^{3} \log \left (F\right )^{3}} - \frac {{\left (f^{3} g^{3} n^{3} x^{3} \log \left (\frac {F^{f g n x} F^{e g n} b}{a} + 1\right ) \log \left (F\right )^{3} + 3 \, f^{2} g^{2} n^{2} x^{2} {\rm Li}_2\left (-\frac {F^{f g n x} F^{e g n} b}{a}\right ) \log \left (F\right )^{2} - 6 \, f g n x \log \left (F\right ) {\rm Li}_{3}(-\frac {F^{f g n x} F^{e g n} b}{a}) + 6 \, {\rm Li}_{4}(-\frac {F^{f g n x} F^{e g n} b}{a})\right )} d^{3}}{a f^{4} g^{4} n^{4} \log \left (F\right )^{4}} + \frac {d^{3} f^{4} g^{4} n^{4} x^{4} \log \left (F\right )^{4} + 4 \, c d^{2} f^{4} g^{4} n^{4} x^{3} \log \left (F\right )^{4} + 6 \, c^{2} d f^{4} g^{4} n^{4} x^{2} \log \left (F\right )^{4}}{4 \, a f^{4} g^{4} n^{4} \log \left (F\right )^{4}} \] Input:

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

Output:

c^3*((f*g*n*x + e*g*n)/(a*f*g*n) - log(F^(f*g*n*x + e*g*n)*b + a)/(a*f*g*n 
*log(F))) - 3*(f*g*n*x*log(F^(f*g*n*x)*F^(e*g*n)*b/a + 1)*log(F) + dilog(- 
F^(f*g*n*x)*F^(e*g*n)*b/a))*c^2*d/(a*f^2*g^2*n^2*log(F)^2) - 3*(f^2*g^2*n^ 
2*x^2*log(F^(f*g*n*x)*F^(e*g*n)*b/a + 1)*log(F)^2 + 2*f*g*n*x*dilog(-F^(f* 
g*n*x)*F^(e*g*n)*b/a)*log(F) - 2*polylog(3, -F^(f*g*n*x)*F^(e*g*n)*b/a))*c 
*d^2/(a*f^3*g^3*n^3*log(F)^3) - (f^3*g^3*n^3*x^3*log(F^(f*g*n*x)*F^(e*g*n) 
*b/a + 1)*log(F)^3 + 3*f^2*g^2*n^2*x^2*dilog(-F^(f*g*n*x)*F^(e*g*n)*b/a)*l 
og(F)^2 - 6*f*g*n*x*log(F)*polylog(3, -F^(f*g*n*x)*F^(e*g*n)*b/a) + 6*poly 
log(4, -F^(f*g*n*x)*F^(e*g*n)*b/a))*d^3/(a*f^4*g^4*n^4*log(F)^4) + 1/4*(d^ 
3*f^4*g^4*n^4*x^4*log(F)^4 + 4*c*d^2*f^4*g^4*n^4*x^3*log(F)^4 + 6*c^2*d*f^ 
4*g^4*n^4*x^2*log(F)^4)/(a*f^4*g^4*n^4*log(F)^4)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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