\(\int x^3 (a+b \log (c x^n))^3 \log (1+e x) \, dx\) [23]

Optimal result

Integrand size = 22, antiderivative size = 710 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x) \, dx=\frac {15 a b^2 n^2 x}{8 e^3}-\frac {255 b^3 n^3 x}{128 e^3}+\frac {45 b^3 n^3 x^2}{256 e^2}-\frac {175 b^3 n^3 x^3}{3456 e}+\frac {3}{128} b^3 n^3 x^4+\frac {15 b^3 n^2 x \log \left (c x^n\right )}{8 e^3}+\frac {3 b^2 n^2 x \left (a+b \log \left (c x^n\right )\right )}{32 e^3}-\frac {21 b^2 n^2 x^2 \left (a+b \log \left (c x^n\right )\right )}{64 e^2}+\frac {37 b^2 n^2 x^3 \left (a+b \log \left (c x^n\right )\right )}{288 e}-\frac {9}{128} b^2 n^2 x^4 \left (a+b \log \left (c x^n\right )\right )-\frac {15 b n x \left (a+b \log \left (c x^n\right )\right )^2}{16 e^3}+\frac {9 b n x^2 \left (a+b \log \left (c x^n\right )\right )^2}{32 e^2}-\frac {7 b n x^3 \left (a+b \log \left (c x^n\right )\right )^2}{48 e}+\frac {3}{32} b n x^4 \left (a+b \log \left (c x^n\right )\right )^2+\frac {x \left (a+b \log \left (c x^n\right )\right )^3}{4 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^3}{8 e^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )^3}{12 e}-\frac {1}{16} x^4 \left (a+b \log \left (c x^n\right )\right )^3+\frac {3 b^3 n^3 \log (1+e x)}{128 e^4}-\frac {3}{128} b^3 n^3 x^4 \log (1+e x)-\frac {3 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{32 e^4}+\frac {3}{32} b^2 n^2 x^4 \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)}{16 e^4}-\frac {3}{16} b n x^4 \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)}{4 e^4}+\frac {1}{4} x^4 \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)}{32 e^4}+\frac {3 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,-e x)}{8 e^4}-\frac {3 b n \left (a+b \log \left (c x^n\right )\right )^2 \operatorname {PolyLog}(2,-e x)}{4 e^4}-\frac {3 b^3 n^3 \operatorname {PolyLog}(3,-e x)}{8 e^4}+\frac {3 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(3,-e x)}{2 e^4}-\frac {3 b^3 n^3 \operatorname {PolyLog}(4,-e x)}{2 e^4} \] Output:

-1/16*x^4*(a+b*ln(c*x^n))^3+15/8*a*b^2*n^2*x/e^3+15/8*b^3*n^2*x*ln(c*x^n)/ 
e^3+3/32*b^2*n^2*x*(a+b*ln(c*x^n))/e^3-21/64*b^2*n^2*x^2*(a+b*ln(c*x^n))/e 
^2+37/288*b^2*n^2*x^3*(a+b*ln(c*x^n))/e-15/16*b*n*x*(a+b*ln(c*x^n))^2/e^3+ 
9/32*b*n*x^2*(a+b*ln(c*x^n))^2/e^2-7/48*b*n*x^3*(a+b*ln(c*x^n))^2/e+3/2*b^ 
2*n^2*(a+b*ln(c*x^n))*polylog(3,-e*x)/e^4+3/8*b^2*n^2*(a+b*ln(c*x^n))*poly 
log(2,-e*x)/e^4-3/4*b*n*(a+b*ln(c*x^n))^2*polylog(2,-e*x)/e^4-3/32*b^2*n^2 
*(a+b*ln(c*x^n))*ln(e*x+1)/e^4+3/32*b^2*n^2*x^4*(a+b*ln(c*x^n))*ln(e*x+1)+ 
3/16*b*n*(a+b*ln(c*x^n))^2*ln(e*x+1)/e^4-3/16*b*n*x^4*(a+b*ln(c*x^n))^2*ln 
(e*x+1)-255/128*b^3*n^3*x/e^3+45/256*b^3*n^3*x^2/e^2-175/3456*b^3*n^3*x^3/ 
e-3/2*b^3*n^3*polylog(4,-e*x)/e^4-3/8*b^3*n^3*polylog(3,-e*x)/e^4-3/32*b^3 
*n^3*polylog(2,-e*x)/e^4+3/128*b^3*n^3*ln(e*x+1)/e^4-3/128*b^3*n^3*x^4*ln( 
e*x+1)-9/128*b^2*n^2*x^4*(a+b*ln(c*x^n))+3/32*b*n*x^4*(a+b*ln(c*x^n))^2+3/ 
128*b^3*n^3*x^4+1/4*x^4*(a+b*ln(c*x^n))^3*ln(e*x+1)-1/8*x^2*(a+b*ln(c*x^n) 
)^3/e^2+1/4*x*(a+b*ln(c*x^n))^3/e^3+1/12*x^3*(a+b*ln(c*x^n))^3/e-1/4*(a+b* 
ln(c*x^n))^3*ln(e*x+1)/e^4
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 1144, normalized size of antiderivative = 1.61 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x) \, dx =\text {Too large to display} \] Input:

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

Output:

(1728*a^3*e*x - 6480*a^2*b*e*n*x + 13608*a*b^2*e*n^2*x - 13770*b^3*e*n^3*x 
 - 864*a^3*e^2*x^2 + 1944*a^2*b*e^2*n*x^2 - 2268*a*b^2*e^2*n^2*x^2 + 1215* 
b^3*e^2*n^3*x^2 + 576*a^3*e^3*x^3 - 1008*a^2*b*e^3*n*x^3 + 888*a*b^2*e^3*n 
^2*x^3 - 350*b^3*e^3*n^3*x^3 - 432*a^3*e^4*x^4 + 648*a^2*b*e^4*n*x^4 - 486 
*a*b^2*e^4*n^2*x^4 + 162*b^3*e^4*n^3*x^4 + 5184*a^2*b*e*x*Log[c*x^n] - 129 
60*a*b^2*e*n*x*Log[c*x^n] + 13608*b^3*e*n^2*x*Log[c*x^n] - 2592*a^2*b*e^2* 
x^2*Log[c*x^n] + 3888*a*b^2*e^2*n*x^2*Log[c*x^n] - 2268*b^3*e^2*n^2*x^2*Lo 
g[c*x^n] + 1728*a^2*b*e^3*x^3*Log[c*x^n] - 2016*a*b^2*e^3*n*x^3*Log[c*x^n] 
 + 888*b^3*e^3*n^2*x^3*Log[c*x^n] - 1296*a^2*b*e^4*x^4*Log[c*x^n] + 1296*a 
*b^2*e^4*n*x^4*Log[c*x^n] - 486*b^3*e^4*n^2*x^4*Log[c*x^n] + 5184*a*b^2*e* 
x*Log[c*x^n]^2 - 6480*b^3*e*n*x*Log[c*x^n]^2 - 2592*a*b^2*e^2*x^2*Log[c*x^ 
n]^2 + 1944*b^3*e^2*n*x^2*Log[c*x^n]^2 + 1728*a*b^2*e^3*x^3*Log[c*x^n]^2 - 
 1008*b^3*e^3*n*x^3*Log[c*x^n]^2 - 1296*a*b^2*e^4*x^4*Log[c*x^n]^2 + 648*b 
^3*e^4*n*x^4*Log[c*x^n]^2 + 1728*b^3*e*x*Log[c*x^n]^3 - 864*b^3*e^2*x^2*Lo 
g[c*x^n]^3 + 576*b^3*e^3*x^3*Log[c*x^n]^3 - 432*b^3*e^4*x^4*Log[c*x^n]^3 - 
 1728*a^3*Log[1 + e*x] + 1296*a^2*b*n*Log[1 + e*x] - 648*a*b^2*n^2*Log[1 + 
 e*x] + 162*b^3*n^3*Log[1 + e*x] + 1728*a^3*e^4*x^4*Log[1 + e*x] - 1296*a^ 
2*b*e^4*n*x^4*Log[1 + e*x] + 648*a*b^2*e^4*n^2*x^4*Log[1 + e*x] - 162*b^3* 
e^4*n^3*x^4*Log[1 + e*x] - 5184*a^2*b*Log[c*x^n]*Log[1 + e*x] + 2592*a*b^2 
*n*Log[c*x^n]*Log[1 + e*x] - 648*b^3*n^2*Log[c*x^n]*Log[1 + e*x] + 5184...
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 663, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.

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

\(\Big \downarrow \) 2824

\(\displaystyle -3 b n \int \left (-\frac {1}{16} \left (a+b \log \left (c x^n\right )\right )^2 x^3+\frac {1}{4} \left (a+b \log \left (c x^n\right )\right )^2 \log (e x+1) x^3+\frac {\left (a+b \log \left (c x^n\right )\right )^2 x^2}{12 e}-\frac {\left (a+b \log \left (c x^n\right )\right )^2 x}{8 e^2}+\frac {\left (a+b \log \left (c x^n\right )\right )^2}{4 e^3}-\frac {\left (a+b \log \left (c x^n\right )\right )^2 \log (e x+1)}{4 e^4 x}\right )dx-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )^3}{4 e^4}+\frac {x \left (a+b \log \left (c x^n\right )\right )^3}{4 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^3}{8 e^2}+\frac {1}{4} x^4 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )^3+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )^3}{12 e}-\frac {1}{16} x^4 \left (a+b \log \left (c x^n\right )\right )^3\)

\(\Big \downarrow \) 2009

\(\displaystyle -3 b n \left (\frac {\operatorname {PolyLog}(2,-e x) \left (a+b \log \left (c x^n\right )\right )^2}{4 e^4}-\frac {b n \operatorname {PolyLog}(2,-e x) \left (a+b \log \left (c x^n\right )\right )}{8 e^4}-\frac {b n \operatorname {PolyLog}(3,-e x) \left (a+b \log \left (c x^n\right )\right )}{2 e^4}-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )^2}{16 e^4}+\frac {b n \log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{32 e^4}+\frac {5 x \left (a+b \log \left (c x^n\right )\right )^2}{16 e^3}-\frac {b n x \left (a+b \log \left (c x^n\right )\right )}{32 e^3}-\frac {3 x^2 \left (a+b \log \left (c x^n\right )\right )^2}{32 e^2}+\frac {7 b n x^2 \left (a+b \log \left (c x^n\right )\right )}{64 e^2}+\frac {1}{16} x^4 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )^2-\frac {1}{32} b n x^4 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac {7 x^3 \left (a+b \log \left (c x^n\right )\right )^2}{144 e}-\frac {37 b n x^3 \left (a+b \log \left (c x^n\right )\right )}{864 e}-\frac {1}{32} x^4 \left (a+b \log \left (c x^n\right )\right )^2+\frac {3}{128} b n x^4 \left (a+b \log \left (c x^n\right )\right )-\frac {5 a b n x}{8 e^3}-\frac {5 b^2 n x \log \left (c x^n\right )}{8 e^3}+\frac {b^2 n^2 \operatorname {PolyLog}(2,-e x)}{32 e^4}+\frac {b^2 n^2 \operatorname {PolyLog}(3,-e x)}{8 e^4}+\frac {b^2 n^2 \operatorname {PolyLog}(4,-e x)}{2 e^4}-\frac {b^2 n^2 \log (e x+1)}{128 e^4}+\frac {85 b^2 n^2 x}{128 e^3}-\frac {15 b^2 n^2 x^2}{256 e^2}+\frac {1}{128} b^2 n^2 x^4 \log (e x+1)+\frac {175 b^2 n^2 x^3}{10368 e}-\frac {1}{128} b^2 n^2 x^4\right )-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )^3}{4 e^4}+\frac {x \left (a+b \log \left (c x^n\right )\right )^3}{4 e^3}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )^3}{8 e^2}+\frac {1}{4} x^4 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )^3+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )^3}{12 e}-\frac {1}{16} x^4 \left (a+b \log \left (c x^n\right )\right )^3\)

Input:

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

Output:

(x*(a + b*Log[c*x^n])^3)/(4*e^3) - (x^2*(a + b*Log[c*x^n])^3)/(8*e^2) + (x 
^3*(a + b*Log[c*x^n])^3)/(12*e) - (x^4*(a + b*Log[c*x^n])^3)/16 - ((a + b* 
Log[c*x^n])^3*Log[1 + e*x])/(4*e^4) + (x^4*(a + b*Log[c*x^n])^3*Log[1 + e* 
x])/4 - 3*b*n*((-5*a*b*n*x)/(8*e^3) + (85*b^2*n^2*x)/(128*e^3) - (15*b^2*n 
^2*x^2)/(256*e^2) + (175*b^2*n^2*x^3)/(10368*e) - (b^2*n^2*x^4)/128 - (5*b 
^2*n*x*Log[c*x^n])/(8*e^3) - (b*n*x*(a + b*Log[c*x^n]))/(32*e^3) + (7*b*n* 
x^2*(a + b*Log[c*x^n]))/(64*e^2) - (37*b*n*x^3*(a + b*Log[c*x^n]))/(864*e) 
 + (3*b*n*x^4*(a + b*Log[c*x^n]))/128 + (5*x*(a + b*Log[c*x^n])^2)/(16*e^3 
) - (3*x^2*(a + b*Log[c*x^n])^2)/(32*e^2) + (7*x^3*(a + b*Log[c*x^n])^2)/( 
144*e) - (x^4*(a + b*Log[c*x^n])^2)/32 - (b^2*n^2*Log[1 + e*x])/(128*e^4) 
+ (b^2*n^2*x^4*Log[1 + e*x])/128 + (b*n*(a + b*Log[c*x^n])*Log[1 + e*x])/( 
32*e^4) - (b*n*x^4*(a + b*Log[c*x^n])*Log[1 + e*x])/32 - ((a + b*Log[c*x^n 
])^2*Log[1 + e*x])/(16*e^4) + (x^4*(a + b*Log[c*x^n])^2*Log[1 + e*x])/16 + 
 (b^2*n^2*PolyLog[2, -(e*x)])/(32*e^4) - (b*n*(a + b*Log[c*x^n])*PolyLog[2 
, -(e*x)])/(8*e^4) + ((a + b*Log[c*x^n])^2*PolyLog[2, -(e*x)])/(4*e^4) + ( 
b^2*n^2*PolyLog[3, -(e*x)])/(8*e^4) - (b*n*(a + b*Log[c*x^n])*PolyLog[3, - 
(e*x)])/(2*e^4) + (b^2*n^2*PolyLog[4, -(e*x)])/(2*e^4))
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

-1/48*(3*b^3*e^4*x^4 - 4*b^3*e^3*x^3 + 6*b^3*e^2*x^2 - 12*b^3*e*x - 12*(b^ 
3*e^4*x^4 - b^3)*log(e*x + 1))*log(x^n)^3/e^4 + 1/16*integrate((48*(b^3*e^ 
4*log(c)^2 + 2*a*b^2*e^4*log(c) + a^2*b*e^4)*x^4*log(e*x + 1)*log(x^n) + 1 
6*(b^3*e^4*log(c)^3 + 3*a*b^2*e^4*log(c)^2 + 3*a^2*b*e^4*log(c) + a^3*e^4) 
*x^4*log(e*x + 1) + (3*b^3*e^4*n*x^4 - 4*b^3*e^3*n*x^3 + 6*b^3*e^2*n*x^2 - 
 12*b^3*e*n*x + 12*((4*a*b^2*e^4 - (e^4*n - 4*e^4*log(c))*b^3)*x^4 + b^3*n 
)*log(e*x + 1))*log(x^n)^2)/x, x)/e^4
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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