3.1.20 \(\int (a+b \log (c x^n))^3 \log (1+e x) \, dx\) [20]

3.1.20.1 Optimal result
3.1.20.2 Mathematica [A] (verified)
3.1.20.3 Rubi [A] (verified)
3.1.20.4 Maple [F]
3.1.20.5 Fricas [F]
3.1.20.6 Sympy [F(-1)]
3.1.20.7 Maxima [F]
3.1.20.8 Giac [F]
3.1.20.9 Mupad [F(-1)]

3.1.20.1 Optimal result

Integrand size = 19, antiderivative size = 327 \[ \int \left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x) \, dx=-12 a b^2 n^2 x+24 b^3 n^3 x-12 b^3 n^2 x \log \left (c x^n\right )-6 b^2 n^2 x \left (a+b \log \left (c x^n\right )\right )+6 b n x \left (a+b \log \left (c x^n\right )\right )^2-x \left (a+b \log \left (c x^n\right )\right )^3-\frac {6 b^3 n^3 (1+e x) \log (1+e x)}{e}+\frac {6 b^2 n^2 (1+e x) \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{e}-\frac {3 b n (1+e x) \left (a+b \log \left (c x^n\right )\right )^2 \log (1+e x)}{e}+\frac {(1+e x) \left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x)}{e}+\frac {6 b^3 n^3 \operatorname {PolyLog}(2,-e x)}{e}-\frac {6 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,-e x)}{e}+\frac {3 b n \left (a+b \log \left (c x^n\right )\right )^2 \operatorname {PolyLog}(2,-e x)}{e}+\frac {6 b^3 n^3 \operatorname {PolyLog}(3,-e x)}{e}-\frac {6 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(3,-e x)}{e}+\frac {6 b^3 n^3 \operatorname {PolyLog}(4,-e x)}{e} \]

output
-12*a*b^2*n^2*x+24*b^3*n^3*x-12*b^3*n^2*x*ln(c*x^n)-6*b^2*n^2*x*(a+b*ln(c* 
x^n))+6*b*n*x*(a+b*ln(c*x^n))^2-x*(a+b*ln(c*x^n))^3-6*b^3*n^3*(e*x+1)*ln(e 
*x+1)/e+6*b^2*n^2*(e*x+1)*(a+b*ln(c*x^n))*ln(e*x+1)/e-3*b*n*(e*x+1)*(a+b*l 
n(c*x^n))^2*ln(e*x+1)/e+(e*x+1)*(a+b*ln(c*x^n))^3*ln(e*x+1)/e+6*b^3*n^3*po 
lylog(2,-e*x)/e-6*b^2*n^2*(a+b*ln(c*x^n))*polylog(2,-e*x)/e+3*b*n*(a+b*ln( 
c*x^n))^2*polylog(2,-e*x)/e+6*b^3*n^3*polylog(3,-e*x)/e-6*b^2*n^2*(a+b*ln( 
c*x^n))*polylog(3,-e*x)/e+6*b^3*n^3*polylog(4,-e*x)/e
 
3.1.20.2 Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 584, normalized size of antiderivative = 1.79 \[ \int \left (a+b \log \left (c x^n\right )\right )^3 \log (1+e x) \, dx=\frac {-a^3 e x+6 a^2 b e n x-18 a b^2 e n^2 x+24 b^3 e n^3 x-3 a^2 b e x \log \left (c x^n\right )+12 a b^2 e n x \log \left (c x^n\right )-18 b^3 e n^2 x \log \left (c x^n\right )-3 a b^2 e x \log ^2\left (c x^n\right )+6 b^3 e n x \log ^2\left (c x^n\right )-b^3 e x \log ^3\left (c x^n\right )+a^3 \log (1+e x)-3 a^2 b n \log (1+e x)+6 a b^2 n^2 \log (1+e x)-6 b^3 n^3 \log (1+e x)+a^3 e x \log (1+e x)-3 a^2 b e n x \log (1+e x)+6 a b^2 e n^2 x \log (1+e x)-6 b^3 e n^3 x \log (1+e x)+3 a^2 b \log \left (c x^n\right ) \log (1+e x)-6 a b^2 n \log \left (c x^n\right ) \log (1+e x)+6 b^3 n^2 \log \left (c x^n\right ) \log (1+e x)+3 a^2 b e x \log \left (c x^n\right ) \log (1+e x)-6 a b^2 e n x \log \left (c x^n\right ) \log (1+e x)+6 b^3 e n^2 x \log \left (c x^n\right ) \log (1+e x)+3 a b^2 \log ^2\left (c x^n\right ) \log (1+e x)-3 b^3 n \log ^2\left (c x^n\right ) \log (1+e x)+3 a b^2 e x \log ^2\left (c x^n\right ) \log (1+e x)-3 b^3 e n x \log ^2\left (c x^n\right ) \log (1+e x)+b^3 \log ^3\left (c x^n\right ) \log (1+e x)+b^3 e x \log ^3\left (c x^n\right ) \log (1+e x)+3 b n \left (a^2-2 a b n+2 b^2 n^2+2 b (a-b n) \log \left (c x^n\right )+b^2 \log ^2\left (c x^n\right )\right ) \operatorname {PolyLog}(2,-e x)-6 b^2 n^2 \left (a-b n+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(3,-e x)+6 b^3 n^3 \operatorname {PolyLog}(4,-e x)}{e} \]

input
Integrate[(a + b*Log[c*x^n])^3*Log[1 + e*x],x]
 
output
(-(a^3*e*x) + 6*a^2*b*e*n*x - 18*a*b^2*e*n^2*x + 24*b^3*e*n^3*x - 3*a^2*b* 
e*x*Log[c*x^n] + 12*a*b^2*e*n*x*Log[c*x^n] - 18*b^3*e*n^2*x*Log[c*x^n] - 3 
*a*b^2*e*x*Log[c*x^n]^2 + 6*b^3*e*n*x*Log[c*x^n]^2 - b^3*e*x*Log[c*x^n]^3 
+ a^3*Log[1 + e*x] - 3*a^2*b*n*Log[1 + e*x] + 6*a*b^2*n^2*Log[1 + e*x] - 6 
*b^3*n^3*Log[1 + e*x] + a^3*e*x*Log[1 + e*x] - 3*a^2*b*e*n*x*Log[1 + e*x] 
+ 6*a*b^2*e*n^2*x*Log[1 + e*x] - 6*b^3*e*n^3*x*Log[1 + e*x] + 3*a^2*b*Log[ 
c*x^n]*Log[1 + e*x] - 6*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] + 3*a^2*b*e*x*Log[c*x^n]*Log[1 + e*x] - 6*a*b^2*e*n*x*Lo 
g[c*x^n]*Log[1 + e*x] + 6*b^3*e*n^2*x*Log[c*x^n]*Log[1 + e*x] + 3*a*b^2*Lo 
g[c*x^n]^2*Log[1 + e*x] - 3*b^3*n*Log[c*x^n]^2*Log[1 + e*x] + 3*a*b^2*e*x* 
Log[c*x^n]^2*Log[1 + e*x] - 3*b^3*e*n*x*Log[c*x^n]^2*Log[1 + e*x] + b^3*Lo 
g[c*x^n]^3*Log[1 + e*x] + b^3*e*x*Log[c*x^n]^3*Log[1 + e*x] + 3*b*n*(a^2 - 
 2*a*b*n + 2*b^2*n^2 + 2*b*(a - b*n)*Log[c*x^n] + b^2*Log[c*x^n]^2)*PolyLo 
g[2, -(e*x)] - 6*b^2*n^2*(a - b*n + b*Log[c*x^n])*PolyLog[3, -(e*x)] + 6*b 
^3*n^3*PolyLog[4, -(e*x)])/e
 
3.1.20.3 Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 303, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2817, 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.

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

\(\Big \downarrow \) 2817

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

\(\Big \downarrow \) 2009

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

input
Int[(a + b*Log[c*x^n])^3*Log[1 + e*x],x]
 
output
-(x*(a + b*Log[c*x^n])^3) + ((1 + e*x)*(a + b*Log[c*x^n])^3*Log[1 + e*x])/ 
e - 3*b*n*(4*a*b*n*x - 8*b^2*n^2*x + 4*b^2*n*x*Log[c*x^n] + 2*b*n*x*(a + b 
*Log[c*x^n]) - 2*x*(a + b*Log[c*x^n])^2 + (2*b^2*n^2*(1 + e*x)*Log[1 + e*x 
])/e - (2*b*n*(1 + e*x)*(a + b*Log[c*x^n])*Log[1 + e*x])/e + ((1 + e*x)*(a 
 + b*Log[c*x^n])^2*Log[1 + e*x])/e - (2*b^2*n^2*PolyLog[2, -(e*x)])/e + (2 
*b*n*(a + b*Log[c*x^n])*PolyLog[2, -(e*x)])/e - ((a + b*Log[c*x^n])^2*Poly 
Log[2, -(e*x)])/e - (2*b^2*n^2*PolyLog[3, -(e*x)])/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)
 

3.1.20.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2817
Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. 
)]*(b_.))^(p_.), x_Symbol] :> With[{u = IntHide[Log[d*(e + f*x^m)^r], 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, r, m, n}, x] && IGtQ[p, 0] 
&& RationalQ[m] && (EqQ[p, 1] || (FractionQ[m] && IntegerQ[1/m]) || (EqQ[r, 
 1] && EqQ[m, 1] && EqQ[d*e, 1]))
 
3.1.20.4 Maple [F]

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

input
int((a+b*ln(c*x^n))^3*ln(e*x+1),x)
 
output
int((a+b*ln(c*x^n))^3*ln(e*x+1),x)
 
3.1.20.5 Fricas [F]

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

input
integrate((a+b*log(c*x^n))^3*log(e*x+1),x, algorithm="fricas")
 
output
integral(b^3*log(c*x^n)^3*log(e*x + 1) + 3*a*b^2*log(c*x^n)^2*log(e*x + 1) 
 + 3*a^2*b*log(c*x^n)*log(e*x + 1) + a^3*log(e*x + 1), x)
 
3.1.20.6 Sympy [F(-1)]

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

input
integrate((a+b*ln(c*x**n))**3*ln(e*x+1),x)
 
output
Timed out
 
3.1.20.7 Maxima [F]

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

input
integrate((a+b*log(c*x^n))^3*log(e*x+1),x, algorithm="maxima")
 
output
-(b^3*e*x - (b^3*e*x + b^3)*log(e*x + 1))*log(x^n)^3/e + integrate((3*(b^3 
*e*log(c)^2 + 2*a*b^2*e*log(c) + a^2*b*e)*x*log(e*x + 1)*log(x^n) + (b^3*e 
*log(c)^3 + 3*a*b^2*e*log(c)^2 + 3*a^2*b*e*log(c) + a^3*e)*x*log(e*x + 1) 
+ 3*(b^3*e*n*x - (b^3*n + ((e*n - e*log(c))*b^3 - a*b^2*e)*x)*log(e*x + 1) 
)*log(x^n)^2)/x, x)/e
 
3.1.20.8 Giac [F]

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

input
integrate((a+b*log(c*x^n))^3*log(e*x+1),x, algorithm="giac")
 
output
integrate((b*log(c*x^n) + a)^3*log(e*x + 1), x)
 
3.1.20.9 Mupad [F(-1)]

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

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