\(\int x (a+b \log (c x^n))^3 \log (1+e x) \, dx\) [25]

Optimal result

Integrand size = 20, antiderivative size = 530 \[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x) \, dx=\frac {9 a b^2 n^2 x}{2 e}-\frac {45 b^3 n^3 x}{8 e}+\frac {3}{4} b^3 n^3 x^2+\frac {9 b^3 n^2 x \log \left (c x^n\right )}{2 e}+\frac {3 b^2 n^2 x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac {9}{8} b^2 n^2 x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {9 b n x \left (a+b \log \left (c x^n\right )\right )^2}{4 e}+\frac {3}{4} b n x^2 \left (a+b \log \left (c x^n\right )\right )^2+\frac {x \left (a+b \log \left (c x^n\right )\right )^3}{2 e}-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )^3+\frac {3 b^3 n^3 \log (1+e x)}{8 e^2}-\frac {3}{8} b^3 n^3 x^2 \log (1+e x)-\frac {3 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{4 e^2}+\frac {3}{4} b^2 n^2 x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)+\frac {3 b n \left (a+b \log \left (c x^n\right )\right )^2 \log (1+e x)}{4 e^2}-\frac {3}{4} b n x^2 \left (a+b \log \left (c x^n\right )\right )^2 \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x)}{2 e^2}+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x)-\frac {3 b^3 n^3 \operatorname {PolyLog}(2,-e x)}{4 e^2}+\frac {3 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,-e x)}{2 e^2}-\frac {3 b n \left (a+b \log \left (c x^n\right )\right )^2 \operatorname {PolyLog}(2,-e x)}{2 e^2}-\frac {3 b^3 n^3 \operatorname {PolyLog}(3,-e x)}{2 e^2}+\frac {3 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(3,-e x)}{e^2}-\frac {3 b^3 n^3 \operatorname {PolyLog}(4,-e x)}{e^2} \] Output:

9/2*a*b^2*n^2*x/e-45/8*b^3*n^3*x/e+3/4*b^3*n^3*x^2+9/2*b^3*n^2*x*ln(c*x^n) 
/e+3/4*b^2*n^2*x*(a+b*ln(c*x^n))/e-9/8*b^2*n^2*x^2*(a+b*ln(c*x^n))-9/4*b*n 
*x*(a+b*ln(c*x^n))^2/e+3/4*b*n*x^2*(a+b*ln(c*x^n))^2+1/2*x*(a+b*ln(c*x^n)) 
^3/e-1/4*x^2*(a+b*ln(c*x^n))^3+3/8*b^3*n^3*ln(e*x+1)/e^2-3/8*b^3*n^3*x^2*l 
n(e*x+1)-3/4*b^2*n^2*(a+b*ln(c*x^n))*ln(e*x+1)/e^2+3/4*b^2*n^2*x^2*(a+b*ln 
(c*x^n))*ln(e*x+1)+3/4*b*n*(a+b*ln(c*x^n))^2*ln(e*x+1)/e^2-3/4*b*n*x^2*(a+ 
b*ln(c*x^n))^2*ln(e*x+1)-1/2*(a+b*ln(c*x^n))^3*ln(e*x+1)/e^2+1/2*x^2*(a+b* 
ln(c*x^n))^3*ln(e*x+1)-3/4*b^3*n^3*polylog(2,-e*x)/e^2+3/2*b^2*n^2*(a+b*ln 
(c*x^n))*polylog(2,-e*x)/e^2-3/2*b*n*(a+b*ln(c*x^n))^2*polylog(2,-e*x)/e^2 
-3/2*b^3*n^3*polylog(3,-e*x)/e^2+3*b^2*n^2*(a+b*ln(c*x^n))*polylog(3,-e*x) 
/e^2-3*b^3*n^3*polylog(4,-e*x)/e^2
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 806, normalized size of antiderivative = 1.52 \[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x) \, dx=\frac {4 a^3 e x-18 a^2 b e n x+42 a b^2 e n^2 x-45 b^3 e n^3 x-2 a^3 e^2 x^2+6 a^2 b e^2 n x^2-9 a b^2 e^2 n^2 x^2+6 b^3 e^2 n^3 x^2+12 a^2 b e x \log \left (c x^n\right )-36 a b^2 e n x \log \left (c x^n\right )+42 b^3 e n^2 x \log \left (c x^n\right )-6 a^2 b e^2 x^2 \log \left (c x^n\right )+12 a b^2 e^2 n x^2 \log \left (c x^n\right )-9 b^3 e^2 n^2 x^2 \log \left (c x^n\right )+12 a b^2 e x \log ^2\left (c x^n\right )-18 b^3 e n x \log ^2\left (c x^n\right )-6 a b^2 e^2 x^2 \log ^2\left (c x^n\right )+6 b^3 e^2 n x^2 \log ^2\left (c x^n\right )+4 b^3 e x \log ^3\left (c x^n\right )-2 b^3 e^2 x^2 \log ^3\left (c x^n\right )-4 a^3 \log (1+e x)+6 a^2 b n \log (1+e x)-6 a b^2 n^2 \log (1+e x)+3 b^3 n^3 \log (1+e x)+4 a^3 e^2 x^2 \log (1+e x)-6 a^2 b e^2 n x^2 \log (1+e x)+6 a b^2 e^2 n^2 x^2 \log (1+e x)-3 b^3 e^2 n^3 x^2 \log (1+e x)-12 a^2 b \log \left (c x^n\right ) \log (1+e x)+12 a b^2 n \log \left (c x^n\right ) \log (1+e x)-6 b^3 n^2 \log \left (c x^n\right ) \log (1+e x)+12 a^2 b e^2 x^2 \log \left (c x^n\right ) \log (1+e x)-12 a b^2 e^2 n x^2 \log \left (c x^n\right ) \log (1+e x)+6 b^3 e^2 n^2 x^2 \log \left (c x^n\right ) \log (1+e x)-12 a b^2 \log ^2\left (c x^n\right ) \log (1+e x)+6 b^3 n \log ^2\left (c x^n\right ) \log (1+e x)+12 a b^2 e^2 x^2 \log ^2\left (c x^n\right ) \log (1+e x)-6 b^3 e^2 n x^2 \log ^2\left (c x^n\right ) \log (1+e x)-4 b^3 \log ^3\left (c x^n\right ) \log (1+e x)+4 b^3 e^2 x^2 \log ^3\left (c x^n\right ) \log (1+e x)-6 b n \left (2 a^2-2 a b n+b^2 n^2-2 b (-2 a+b n) \log \left (c x^n\right )+2 b^2 \log ^2\left (c x^n\right )\right ) \operatorname {PolyLog}(2,-e x)+12 b^2 n^2 \left (2 a-b n+2 b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(3,-e x)-24 b^3 n^3 \operatorname {PolyLog}(4,-e x)}{8 e^2} \] Input:

Integrate[x*(a + b*Log[c*x^n])^3*Log[1 + e*x],x]
 

Output:

(4*a^3*e*x - 18*a^2*b*e*n*x + 42*a*b^2*e*n^2*x - 45*b^3*e*n^3*x - 2*a^3*e^ 
2*x^2 + 6*a^2*b*e^2*n*x^2 - 9*a*b^2*e^2*n^2*x^2 + 6*b^3*e^2*n^3*x^2 + 12*a 
^2*b*e*x*Log[c*x^n] - 36*a*b^2*e*n*x*Log[c*x^n] + 42*b^3*e*n^2*x*Log[c*x^n 
] - 6*a^2*b*e^2*x^2*Log[c*x^n] + 12*a*b^2*e^2*n*x^2*Log[c*x^n] - 9*b^3*e^2 
*n^2*x^2*Log[c*x^n] + 12*a*b^2*e*x*Log[c*x^n]^2 - 18*b^3*e*n*x*Log[c*x^n]^ 
2 - 6*a*b^2*e^2*x^2*Log[c*x^n]^2 + 6*b^3*e^2*n*x^2*Log[c*x^n]^2 + 4*b^3*e* 
x*Log[c*x^n]^3 - 2*b^3*e^2*x^2*Log[c*x^n]^3 - 4*a^3*Log[1 + e*x] + 6*a^2*b 
*n*Log[1 + e*x] - 6*a*b^2*n^2*Log[1 + e*x] + 3*b^3*n^3*Log[1 + e*x] + 4*a^ 
3*e^2*x^2*Log[1 + e*x] - 6*a^2*b*e^2*n*x^2*Log[1 + e*x] + 6*a*b^2*e^2*n^2* 
x^2*Log[1 + e*x] - 3*b^3*e^2*n^3*x^2*Log[1 + e*x] - 12*a^2*b*Log[c*x^n]*Lo 
g[1 + e*x] + 12*a*b^2*n*Log[c*x^n]*Log[1 + e*x] - 6*b^3*n^2*Log[c*x^n]*Log 
[1 + e*x] + 12*a^2*b*e^2*x^2*Log[c*x^n]*Log[1 + e*x] - 12*a*b^2*e^2*n*x^2* 
Log[c*x^n]*Log[1 + e*x] + 6*b^3*e^2*n^2*x^2*Log[c*x^n]*Log[1 + e*x] - 12*a 
*b^2*Log[c*x^n]^2*Log[1 + e*x] + 6*b^3*n*Log[c*x^n]^2*Log[1 + e*x] + 12*a* 
b^2*e^2*x^2*Log[c*x^n]^2*Log[1 + e*x] - 6*b^3*e^2*n*x^2*Log[c*x^n]^2*Log[1 
 + e*x] - 4*b^3*Log[c*x^n]^3*Log[1 + e*x] + 4*b^3*e^2*x^2*Log[c*x^n]^3*Log 
[1 + e*x] - 6*b*n*(2*a^2 - 2*a*b*n + b^2*n^2 - 2*b*(-2*a + b*n)*Log[c*x^n] 
 + 2*b^2*Log[c*x^n]^2)*PolyLog[2, -(e*x)] + 12*b^2*n^2*(2*a - b*n + 2*b*Lo 
g[c*x^n])*PolyLog[3, -(e*x)] - 24*b^3*n^3*PolyLog[4, -(e*x)])/(8*e^2)
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 494, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2824, 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.

\(\displaystyle \int x \log (e x+1) \left (a+b \log \left (c x^n\right )\right )^3 \, dx\)

\(\Big \downarrow \) 2824

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

\(\Big \downarrow \) 2009

\(\displaystyle -3 b n \left (-\frac {b n \operatorname {PolyLog}(2,-e x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {b n \operatorname {PolyLog}(3,-e x) \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {\operatorname {PolyLog}(2,-e x) \left (a+b \log \left (c x^n\right )\right )^2}{2 e^2}+\frac {b n \log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{4 e^2}-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )^2}{4 e^2}-\frac {b n x \left (a+b \log \left (c x^n\right )\right )}{4 e}+\frac {3 x \left (a+b \log \left (c x^n\right )\right )^2}{4 e}-\frac {1}{4} b n x^2 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} x^2 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )^2+\frac {3}{8} b n x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )^2-\frac {3 a b n x}{2 e}-\frac {3 b^2 n x \log \left (c x^n\right )}{2 e}+\frac {b^2 n^2 \operatorname {PolyLog}(2,-e x)}{4 e^2}+\frac {b^2 n^2 \operatorname {PolyLog}(3,-e x)}{2 e^2}+\frac {b^2 n^2 \operatorname {PolyLog}(4,-e x)}{e^2}-\frac {b^2 n^2 \log (e x+1)}{8 e^2}+\frac {1}{8} b^2 n^2 x^2 \log (e x+1)+\frac {15 b^2 n^2 x}{8 e}-\frac {1}{4} b^2 n^2 x^2\right )-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )^3}{2 e^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )^3}{2 e}+\frac {1}{2} x^2 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )^3-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )^3\)

Input:

Int[x*(a + b*Log[c*x^n])^3*Log[1 + e*x],x]
 

Output:

(x*(a + b*Log[c*x^n])^3)/(2*e) - (x^2*(a + b*Log[c*x^n])^3)/4 - ((a + b*Lo 
g[c*x^n])^3*Log[1 + e*x])/(2*e^2) + (x^2*(a + b*Log[c*x^n])^3*Log[1 + e*x] 
)/2 - 3*b*n*((-3*a*b*n*x)/(2*e) + (15*b^2*n^2*x)/(8*e) - (b^2*n^2*x^2)/4 - 
 (3*b^2*n*x*Log[c*x^n])/(2*e) - (b*n*x*(a + b*Log[c*x^n]))/(4*e) + (3*b*n* 
x^2*(a + b*Log[c*x^n]))/8 + (3*x*(a + b*Log[c*x^n])^2)/(4*e) - (x^2*(a + b 
*Log[c*x^n])^2)/4 - (b^2*n^2*Log[1 + e*x])/(8*e^2) + (b^2*n^2*x^2*Log[1 + 
e*x])/8 + (b*n*(a + b*Log[c*x^n])*Log[1 + e*x])/(4*e^2) - (b*n*x^2*(a + b* 
Log[c*x^n])*Log[1 + e*x])/4 - ((a + b*Log[c*x^n])^2*Log[1 + e*x])/(4*e^2) 
+ (x^2*(a + b*Log[c*x^n])^2*Log[1 + e*x])/4 + (b^2*n^2*PolyLog[2, -(e*x)]) 
/(4*e^2) - (b*n*(a + b*Log[c*x^n])*PolyLog[2, -(e*x)])/(2*e^2) + ((a + b*L 
og[c*x^n])^2*PolyLog[2, -(e*x)])/(2*e^2) + (b^2*n^2*PolyLog[3, -(e*x)])/(2 
*e^2) - (b*n*(a + b*Log[c*x^n])*PolyLog[3, -(e*x)])/e^2 + (b^2*n^2*PolyLog 
[4, -(e*x)])/e^2)
 

Defintions of rubi rules used

rule 2009

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

rule 2824

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_ 
.))^(p_.)*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q*Log[d* 
(e + f*x^m)], x]}, Simp[(a + b*Log[c*x^n])^p   u, x] - Simp[b*n*p   Int[(a 
+ b*Log[c*x^n])^(p - 1)/x   u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, 
 q}, x] && IGtQ[p, 0] && RationalQ[m] && RationalQ[q] && NeQ[q, -1] && (EqQ 
[p, 1] || (FractionQ[m] && IntegerQ[(q + 1)/m]) || (IGtQ[q, 0] && IntegerQ[ 
(q + 1)/m] && EqQ[d*e, 1]))
 

Maple [F]

\[\int x {\left (a +b \ln \left (c \,x^{n}\right )\right )}^{3} \ln \left (e x +1\right )d x\]

Input:

int(x*(a+b*ln(c*x^n))^3*ln(e*x+1),x)
 

Output:

int(x*(a+b*ln(c*x^n))^3*ln(e*x+1),x)
 

Fricas [F]

\[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{3} x \log \left (e x + 1\right ) \,d x } \] Input:

integrate(x*(a+b*log(c*x^n))^3*log(e*x+1),x, algorithm="fricas")
 

Output:

integral(b^3*x*log(c*x^n)^3*log(e*x + 1) + 3*a*b^2*x*log(c*x^n)^2*log(e*x 
+ 1) + 3*a^2*b*x*log(c*x^n)*log(e*x + 1) + a^3*x*log(e*x + 1), x)
 

Sympy [F(-1)]

Timed out. \[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x) \, dx=\text {Timed out} \] Input:

integrate(x*(a+b*ln(c*x**n))**3*ln(e*x+1),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{3} x \log \left (e x + 1\right ) \,d x } \] Input:

integrate(x*(a+b*log(c*x^n))^3*log(e*x+1),x, algorithm="maxima")
 

Output:

-1/4*(b^3*e^2*x^2 - 2*b^3*e*x - 2*(b^3*e^2*x^2 - b^3)*log(e*x + 1))*log(x^ 
n)^3/e^2 + 1/4*integrate((12*(b^3*e^2*log(c)^2 + 2*a*b^2*e^2*log(c) + a^2* 
b*e^2)*x^2*log(e*x + 1)*log(x^n) + 4*(b^3*e^2*log(c)^3 + 3*a*b^2*e^2*log(c 
)^2 + 3*a^2*b*e^2*log(c) + a^3*e^2)*x^2*log(e*x + 1) + 3*(b^3*e^2*n*x^2 - 
2*b^3*e*n*x + 2*(b^3*n + (2*a*b^2*e^2 - (e^2*n - 2*e^2*log(c))*b^3)*x^2)*l 
og(e*x + 1))*log(x^n)^2)/x, x)/e^2
 

Giac [F]

\[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )}^{3} x \log \left (e x + 1\right ) \,d x } \] Input:

integrate(x*(a+b*log(c*x^n))^3*log(e*x+1),x, algorithm="giac")
 

Output:

integrate((b*log(c*x^n) + a)^3*x*log(e*x + 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x) \, dx=\int x\,\ln \left (e\,x+1\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^3 \,d x \] Input:

int(x*log(e*x + 1)*(a + b*log(c*x^n))^3,x)
 

Output:

int(x*log(e*x + 1)*(a + b*log(c*x^n))^3, x)
 

Reduce [F]

\[ \int x \left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x) \, dx =\text {Too large to display} \] Input:

int(x*(a+b*log(c*x^n))^3*log(e*x+1),x)
 

Output:

(4*int(log(x**n*c)**3/(e*x**2 + x),x)*b**3*n + 12*int(log(x**n*c)**2/(e*x* 
*2 + x),x)*a*b**2*n - 6*int(log(x**n*c)**2/(e*x**2 + x),x)*b**3*n**2 + 12* 
int(log(x**n*c)/(e*x**2 + x),x)*a**2*b*n - 12*int(log(x**n*c)/(e*x**2 + x) 
,x)*a*b**2*n**2 + 6*int(log(x**n*c)/(e*x**2 + x),x)*b**3*n**3 + 4*log(e*x 
+ 1)*log(x**n*c)**3*b**3*e**2*n*x**2 + 12*log(e*x + 1)*log(x**n*c)**2*a*b* 
*2*e**2*n*x**2 - 6*log(e*x + 1)*log(x**n*c)**2*b**3*e**2*n**2*x**2 + 12*lo 
g(e*x + 1)*log(x**n*c)*a**2*b*e**2*n*x**2 - 12*log(e*x + 1)*log(x**n*c)*a* 
b**2*e**2*n**2*x**2 + 6*log(e*x + 1)*log(x**n*c)*b**3*e**2*n**3*x**2 + 4*l 
og(e*x + 1)*a**3*e**2*n*x**2 - 4*log(e*x + 1)*a**3*n - 6*log(e*x + 1)*a**2 
*b*e**2*n**2*x**2 + 6*log(e*x + 1)*a**2*b*n**2 + 6*log(e*x + 1)*a*b**2*e** 
2*n**3*x**2 - 6*log(e*x + 1)*a*b**2*n**3 - 3*log(e*x + 1)*b**3*e**2*n**4*x 
**2 + 3*log(e*x + 1)*b**3*n**4 - log(x**n*c)**4*b**3 - 4*log(x**n*c)**3*a* 
b**2 - 2*log(x**n*c)**3*b**3*e**2*n*x**2 + 4*log(x**n*c)**3*b**3*e*n*x + 2 
*log(x**n*c)**3*b**3*n - 6*log(x**n*c)**2*a**2*b - 6*log(x**n*c)**2*a*b**2 
*e**2*n*x**2 + 12*log(x**n*c)**2*a*b**2*e*n*x + 6*log(x**n*c)**2*a*b**2*n 
+ 6*log(x**n*c)**2*b**3*e**2*n**2*x**2 - 18*log(x**n*c)**2*b**3*e*n**2*x - 
 3*log(x**n*c)**2*b**3*n**2 - 6*log(x**n*c)*a**2*b*e**2*n*x**2 + 12*log(x* 
*n*c)*a**2*b*e*n*x + 12*log(x**n*c)*a*b**2*e**2*n**2*x**2 - 36*log(x**n*c) 
*a*b**2*e*n**2*x - 9*log(x**n*c)*b**3*e**2*n**3*x**2 + 42*log(x**n*c)*b**3 
*e*n**3*x - 2*a**3*e**2*n*x**2 + 4*a**3*e*n*x + 6*a**2*b*e**2*n**2*x**2...