\(\int \frac {(f+g x)^3}{a+b \log (c (d+e x)^n)} \, dx\) [88]

Optimal result

Integrand size = 24, antiderivative size = 299 \[ \int \frac {(f+g x)^3}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {a}{b n}} (e f-d g)^3 (d+e x) \left (c (d+e x)^n\right )^{-1/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )}{b e^4 n}+\frac {3 e^{-\frac {2 a}{b n}} g (e f-d g)^2 (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^4 n}+\frac {3 e^{-\frac {3 a}{b n}} g^2 (e f-d g) (d+e x)^3 \left (c (d+e x)^n\right )^{-3/n} \operatorname {ExpIntegralEi}\left (\frac {3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^4 n}+\frac {e^{-\frac {4 a}{b n}} g^3 (d+e x)^4 \left (c (d+e x)^n\right )^{-4/n} \operatorname {ExpIntegralEi}\left (\frac {4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )}{b e^4 n} \] Output:

(-d*g+e*f)^3*(e*x+d)*Ei((a+b*ln(c*(e*x+d)^n))/b/n)/b/e^4/exp(a/b/n)/n/((c* 
(e*x+d)^n)^(1/n))+3*g*(-d*g+e*f)^2*(e*x+d)^2*Ei(2*(a+b*ln(c*(e*x+d)^n))/b/ 
n)/b/e^4/exp(2*a/b/n)/n/((c*(e*x+d)^n)^(2/n))+3*g^2*(-d*g+e*f)*(e*x+d)^3*E 
i(3*(a+b*ln(c*(e*x+d)^n))/b/n)/b/e^4/exp(3*a/b/n)/n/((c*(e*x+d)^n)^(3/n))+ 
g^3*(e*x+d)^4*Ei(4*(a+b*ln(c*(e*x+d)^n))/b/n)/b/e^4/exp(4*a/b/n)/n/((c*(e* 
x+d)^n)^(4/n))
 

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.89 \[ \int \frac {(f+g x)^3}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {e^{-\frac {4 a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-4/n} \left (e^{\frac {3 a}{b n}} (e f-d g)^3 \left (c (d+e x)^n\right )^{3/n} \operatorname {ExpIntegralEi}\left (\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )+g (d+e x) \left (3 e^{\frac {2 a}{b n}} (e f-d g)^2 \left (c (d+e x)^n\right )^{2/n} \operatorname {ExpIntegralEi}\left (\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )-g (d+e x) \left (-3 e^{\frac {a}{b n}} (e f-d g) \left (c (d+e x)^n\right )^{\frac {1}{n}} \operatorname {ExpIntegralEi}\left (\frac {3 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )-g (d+e x) \operatorname {ExpIntegralEi}\left (\frac {4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{b n}\right )\right )\right )\right )}{b e^4 n} \] Input:

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

Output:

((d + e*x)*(E^((3*a)/(b*n))*(e*f - d*g)^3*(c*(d + e*x)^n)^(3/n)*ExpIntegra 
lEi[(a + b*Log[c*(d + e*x)^n])/(b*n)] + g*(d + e*x)*(3*E^((2*a)/(b*n))*(e* 
f - d*g)^2*(c*(d + e*x)^n)^(2/n)*ExpIntegralEi[(2*(a + b*Log[c*(d + e*x)^n 
]))/(b*n)] - g*(d + e*x)*(-3*E^(a/(b*n))*(e*f - d*g)*(c*(d + e*x)^n)^n^(-1 
)*ExpIntegralEi[(3*(a + b*Log[c*(d + e*x)^n]))/(b*n)] - g*(d + e*x)*ExpInt 
egralEi[(4*(a + b*Log[c*(d + e*x)^n]))/(b*n)]))))/(b*e^4*E^((4*a)/(b*n))*n 
*(c*(d + e*x)^n)^(4/n))
 

Rubi [A] (verified)

Time = 1.35 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2846, 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.

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

\(\Big \downarrow \) 2846

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2846

Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.) 
]*(b_.)), x_Symbol] :> Int[ExpandIntegrand[(f + g*x)^q/(a + b*Log[c*(d + e* 
x)^n]), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0] & 
& IGtQ[q, 0]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 3.34 (sec) , antiderivative size = 9346, normalized size of antiderivative = 31.26

method	result	size
risch	\(\text {Expression too large to display}\)	\(9346\)

Input:

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

Output:

result too large to display
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.02 \[ \int \frac {(f+g x)^3}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {{\left (g^{3} \operatorname {log\_integral}\left ({\left (e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}\right )} e^{\left (\frac {4 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}\right ) + 3 \, {\left (e f g^{2} - d g^{3}\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )} \operatorname {log\_integral}\left ({\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} e^{\left (\frac {3 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}\right ) + 3 \, {\left (e^{2} f^{2} g - 2 \, d e f g^{2} + d^{2} g^{3}\right )} e^{\left (\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )} \operatorname {log\_integral}\left ({\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} e^{\left (\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}\right ) + {\left (e^{3} f^{3} - 3 \, d e^{2} f^{2} g + 3 \, d^{2} e f g^{2} - d^{3} g^{3}\right )} e^{\left (\frac {3 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )} \operatorname {log\_integral}\left ({\left (e x + d\right )} e^{\left (\frac {b \log \left (c\right ) + a}{b n}\right )}\right )\right )} e^{\left (-\frac {4 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}}{b e^{4} n} \] Input:

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

Output:

(g^3*log_integral((e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4 
)*e^(4*(b*log(c) + a)/(b*n))) + 3*(e*f*g^2 - d*g^3)*e^((b*log(c) + a)/(b*n 
))*log_integral((e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)*e^(3*(b*log(c) + 
 a)/(b*n))) + 3*(e^2*f^2*g - 2*d*e*f*g^2 + d^2*g^3)*e^(2*(b*log(c) + a)/(b 
*n))*log_integral((e^2*x^2 + 2*d*e*x + d^2)*e^(2*(b*log(c) + a)/(b*n))) + 
(e^3*f^3 - 3*d*e^2*f^2*g + 3*d^2*e*f*g^2 - d^3*g^3)*e^(3*(b*log(c) + a)/(b 
*n))*log_integral((e*x + d)*e^((b*log(c) + a)/(b*n))))*e^(-4*(b*log(c) + a 
)/(b*n))/(b*e^4*n)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 582, normalized size of antiderivative = 1.95 \[ \int \frac {(f+g x)^3}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {f^{3} {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e n} - \frac {3 \, d f^{2} g {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e^{2} n} + \frac {3 \, d^{2} f g^{2} {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e^{3} n} - \frac {d^{3} g^{3} {\rm Ei}\left (\frac {\log \left (c\right )}{n} + \frac {a}{b n} + \log \left (e x + d\right )\right ) e^{\left (-\frac {a}{b n}\right )}}{b c^{\left (\frac {1}{n}\right )} e^{4} n} + \frac {3 \, f^{2} g {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{n} + \frac {2 \, a}{b n} + 2 \, \log \left (e x + d\right )\right ) e^{\left (-\frac {2 \, a}{b n}\right )}}{b c^{\frac {2}{n}} e^{2} n} - \frac {6 \, d f g^{2} {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{n} + \frac {2 \, a}{b n} + 2 \, \log \left (e x + d\right )\right ) e^{\left (-\frac {2 \, a}{b n}\right )}}{b c^{\frac {2}{n}} e^{3} n} + \frac {3 \, d^{2} g^{3} {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{n} + \frac {2 \, a}{b n} + 2 \, \log \left (e x + d\right )\right ) e^{\left (-\frac {2 \, a}{b n}\right )}}{b c^{\frac {2}{n}} e^{4} n} + \frac {3 \, f g^{2} {\rm Ei}\left (\frac {3 \, \log \left (c\right )}{n} + \frac {3 \, a}{b n} + 3 \, \log \left (e x + d\right )\right ) e^{\left (-\frac {3 \, a}{b n}\right )}}{b c^{\frac {3}{n}} e^{3} n} - \frac {3 \, d g^{3} {\rm Ei}\left (\frac {3 \, \log \left (c\right )}{n} + \frac {3 \, a}{b n} + 3 \, \log \left (e x + d\right )\right ) e^{\left (-\frac {3 \, a}{b n}\right )}}{b c^{\frac {3}{n}} e^{4} n} + \frac {g^{3} {\rm Ei}\left (\frac {4 \, \log \left (c\right )}{n} + \frac {4 \, a}{b n} + 4 \, \log \left (e x + d\right )\right ) e^{\left (-\frac {4 \, a}{b n}\right )}}{b c^{\frac {4}{n}} e^{4} n} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(f+g x)^3}{a+b \log \left (c (d+e x)^n\right )} \, dx=\frac {\left (\int \frac {x^{4}}{\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b d +\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b e x +a d +a e x}d x \right ) b \,e^{2} g^{3} n +\left (\int \frac {x^{3}}{\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b d +\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b e x +a d +a e x}d x \right ) b d e \,g^{3} n +3 \left (\int \frac {x^{3}}{\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b d +\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b e x +a d +a e x}d x \right ) b \,e^{2} f \,g^{2} n +3 \left (\int \frac {x^{2}}{\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b d +\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b e x +a d +a e x}d x \right ) b d e f \,g^{2} n +3 \left (\int \frac {x^{2}}{\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b d +\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b e x +a d +a e x}d x \right ) b \,e^{2} f^{2} g n +3 \left (\int \frac {x}{\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b d +\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b e x +a d +a e x}d x \right ) b d e \,f^{2} g n +\left (\int \frac {x}{\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b d +\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b e x +a d +a e x}d x \right ) b \,e^{2} f^{3} n +\mathrm {log}\left (\mathrm {log}\left (\left (e x +d \right )^{n} c \right ) b +a \right ) d \,f^{3}}{b e n} \] Input:

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

Output:

(int(x**4/(log((d + e*x)**n*c)*b*d + log((d + e*x)**n*c)*b*e*x + a*d + a*e 
*x),x)*b*e**2*g**3*n + int(x**3/(log((d + e*x)**n*c)*b*d + log((d + e*x)** 
n*c)*b*e*x + a*d + a*e*x),x)*b*d*e*g**3*n + 3*int(x**3/(log((d + e*x)**n*c 
)*b*d + log((d + e*x)**n*c)*b*e*x + a*d + a*e*x),x)*b*e**2*f*g**2*n + 3*in 
t(x**2/(log((d + e*x)**n*c)*b*d + log((d + e*x)**n*c)*b*e*x + a*d + a*e*x) 
,x)*b*d*e*f*g**2*n + 3*int(x**2/(log((d + e*x)**n*c)*b*d + log((d + e*x)** 
n*c)*b*e*x + a*d + a*e*x),x)*b*e**2*f**2*g*n + 3*int(x/(log((d + e*x)**n*c 
)*b*d + log((d + e*x)**n*c)*b*e*x + a*d + a*e*x),x)*b*d*e*f**2*g*n + int(x 
/(log((d + e*x)**n*c)*b*d + log((d + e*x)**n*c)*b*e*x + a*d + a*e*x),x)*b* 
e**2*f**3*n + log(log((d + e*x)**n*c)*b + a)*d*f**3)/(b*e*n)