3.3.10 \(\int (a+b \log (c x^n)) \operatorname {PolyLog}(2,e x) \, dx\) [210]

3.3.10.1 Optimal result

Integrand size = 16, antiderivative size = 106 \[ \int \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=3 b n x-x \left (a+b \log \left (c x^n\right )\right )+\frac {2 b n (1-e x) \log (1-e x)}{e}-\frac {(1-e x) \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{e}-\frac {b n \operatorname {PolyLog}(2,e x)}{e}-b n x \operatorname {PolyLog}(2,e x)+x \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \]

output

3*b*n*x-x*(a+b*ln(c*x^n))+2*b*n*(-e*x+1)*ln(-e*x+1)/e-(-e*x+1)*(a+b*ln(c*x 
^n))*ln(-e*x+1)/e-b*n*polylog(2,e*x)/e-b*n*x*polylog(2,e*x)+x*(a+b*ln(c*x^ 
n))*polylog(2,e*x)
 

3.3.10.2 Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.07 \[ \int \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right ) \left (-x+\left (-\frac {1}{e}+x\right ) \log (1-e x)+x \operatorname {PolyLog}(2,e x)\right )+\frac {b n (3 e x+2 \log (1-e x)-2 e x \log (1-e x)+\log (x) (-e x+(-1+e x) \log (1-e x))+(-1-e x+e x \log (x)) \operatorname {PolyLog}(2,e x))}{e} \]

input

Integrate[(a + b*Log[c*x^n])*PolyLog[2, e*x],x]
 

output

(a + b*(-(n*Log[x]) + Log[c*x^n]))*(-x + (-e^(-1) + x)*Log[1 - e*x] + x*Po 
lyLog[2, e*x]) + (b*n*(3*e*x + 2*Log[1 - e*x] - 2*e*x*Log[1 - e*x] + Log[x 
]*(-(e*x) + (-1 + e*x)*Log[1 - e*x]) + (-1 - e*x + e*x*Log[x])*PolyLog[2, 
e*x]))/e
 

3.3.10.3 Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.22, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2828, 25, 2817, 2009, 2836, 2732}

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.

\(\displaystyle \int \operatorname {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right ) \, dx\)

\(\Big \downarrow \) 2828

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2817

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 2836

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

\(\Big \downarrow \) 2732

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

input

Int[(a + b*Log[c*x^n])*PolyLog[2, e*x],x]
 

output

-(x*(a + b*Log[c*x^n])) - ((1 - e*x)*(a + b*Log[c*x^n])*Log[1 - e*x])/e + 
(b*n*(-1 + e*x + (1 - e*x)*Log[1 - e*x]))/e - b*n*x*PolyLog[2, e*x] + x*(a 
 + b*Log[c*x^n])*PolyLog[2, e*x] - b*n*(-2*x - ((1 - e*x)*Log[1 - e*x])/e 
+ PolyLog[2, e*x]/e)
 

3.3.10.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

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

rule 2732

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x 
] /; FreeQ[{c, n}, x]
 

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]))
 

rule 2828

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*PolyLog[k_, (e_.)*(x_)^(q_.)], x_ 
Symbol] :> Simp[(-b)*n*x*PolyLog[k, e*x^q], x] + (Simp[x*PolyLog[k, e*x^q]* 
(a + b*Log[c*x^n]), x] - Simp[q   Int[PolyLog[k - 1, e*x^q]*(a + b*Log[c*x^ 
n]), x], x] + Simp[b*n*q   Int[PolyLog[k - 1, e*x^q], x], x]) /; FreeQ[{a, 
b, c, e, n, q}, x] && IGtQ[k, 0]
 

rule 2836

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

3.3.10.4 Maple [A] (verified)

Time = 4.88 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.59

method	result	size
parallelrisch	\(\frac {x \ln \left (c \,x^{n}\right ) \operatorname {Li}_{2}\left (e x \right ) b e n +b \ln \left (-e x +1\right ) \ln \left (c \,x^{n}\right ) x e n -x \,\operatorname {Li}_{2}\left (e x \right ) b e \,n^{2}-2 x \ln \left (-e x +1\right ) b e \,n^{2}-x \ln \left (c \,x^{n}\right ) b e n +x \,\operatorname {Li}_{2}\left (e x \right ) a e n +x \ln \left (-e x +1\right ) a e n +3 x b e \,n^{2}+2 \ln \left (e x -1\right ) b \,n^{2}-x a e n -b \ln \left (-e x +1\right ) \ln \left (c \,x^{n}\right ) n -b \,n^{2} \operatorname {Li}_{2}\left (e x \right )-\ln \left (e x -1\right ) a n}{e n}\)	\(169\)

input

int((a+b*ln(c*x^n))*polylog(2,e*x),x,method=_RETURNVERBOSE)
 

output

(x*ln(c*x^n)*polylog(2,e*x)*b*e*n+b*ln(-e*x+1)*ln(c*x^n)*x*e*n-x*polylog(2 
,e*x)*b*e*n^2-2*x*ln(-e*x+1)*b*e*n^2-x*ln(c*x^n)*b*e*n+x*polylog(2,e*x)*a* 
e*n+x*ln(-e*x+1)*a*e*n+3*x*b*e*n^2+2*ln(e*x-1)*b*n^2-x*a*e*n-b*ln(-e*x+1)* 
ln(c*x^n)*n-b*n^2*polylog(2,e*x)-ln(e*x-1)*a*n)/e/n
 

3.3.10.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.29 \[ \int \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\frac {{\left (3 \, b e n - a e\right )} x - {\left (b n + {\left (b e n - a e\right )} x\right )} {\rm Li}_2\left (e x\right ) + {\left (2 \, b n - {\left (2 \, b e n - a e\right )} x - a\right )} \log \left (-e x + 1\right ) + {\left (b e x {\rm Li}_2\left (e x\right ) - b e x + {\left (b e x - b\right )} \log \left (-e x + 1\right )\right )} \log \left (c\right ) + {\left (b e n x {\rm Li}_2\left (e x\right ) - b e n x + {\left (b e n x - b n\right )} \log \left (-e x + 1\right )\right )} \log \left (x\right )}{e} \]

input

integrate((a+b*log(c*x^n))*polylog(2,e*x),x, algorithm="fricas")
 

output

((3*b*e*n - a*e)*x - (b*n + (b*e*n - a*e)*x)*dilog(e*x) + (2*b*n - (2*b*e* 
n - a*e)*x - a)*log(-e*x + 1) + (b*e*x*dilog(e*x) - b*e*x + (b*e*x - b)*lo 
g(-e*x + 1))*log(c) + (b*e*n*x*dilog(e*x) - b*e*n*x + (b*e*n*x - b*n)*log( 
-e*x + 1))*log(x))/e
 

3.3.10.6 Sympy [A] (verification not implemented)

Time = 6.03 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.28 \[ \int \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\begin {cases} - a x \operatorname {Li}_{1}\left (e x\right ) + a x \operatorname {Li}_{2}\left (e x\right ) - a x + \frac {a \operatorname {Li}_{1}\left (e x\right )}{e} + 2 b n x \operatorname {Li}_{1}\left (e x\right ) - b n x \operatorname {Li}_{2}\left (e x\right ) + 3 b n x - b x \log {\left (c x^{n} \right )} \operatorname {Li}_{1}\left (e x\right ) + b x \log {\left (c x^{n} \right )} \operatorname {Li}_{2}\left (e x\right ) - b x \log {\left (c x^{n} \right )} - \frac {2 b n \operatorname {Li}_{1}\left (e x\right )}{e} - \frac {b n \operatorname {Li}_{2}\left (e x\right )}{e} + \frac {b \log {\left (c x^{n} \right )} \operatorname {Li}_{1}\left (e x\right )}{e} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]

input

integrate((a+b*ln(c*x**n))*polylog(2,e*x),x)
 

output

Piecewise((-a*x*polylog(1, e*x) + a*x*polylog(2, e*x) - a*x + a*polylog(1, 
 e*x)/e + 2*b*n*x*polylog(1, e*x) - b*n*x*polylog(2, e*x) + 3*b*n*x - b*x* 
log(c*x**n)*polylog(1, e*x) + b*x*log(c*x**n)*polylog(2, e*x) - b*x*log(c* 
x**n) - 2*b*n*polylog(1, e*x)/e - b*n*polylog(2, e*x)/e + b*log(c*x**n)*po 
lylog(1, e*x)/e, Ne(e, 0)), (0, True))
 

3.3.10.7 Maxima [F]

\[ \int \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} {\rm Li}_2\left (e x\right ) \,d x } \]

input

integrate((a+b*log(c*x^n))*polylog(2,e*x),x, algorithm="maxima")
 

output

b*(((e*x*log(x^n) - (e*n - e*log(c))*x)*dilog(e*x) - ((2*e*n - e*log(c))*x 
 - n*log(x))*log(-e*x + 1) - (e*x - (e*x - 1)*log(-e*x + 1))*log(x^n))/e - 
 integrate(-((3*e*n - e*log(c))*x - n*log(x) - n)/(e*x - 1), x)) + (e*x*di 
log(e*x) - e*x + (e*x - 1)*log(-e*x + 1))*a/e
 

3.3.10.8 Giac [F]

\[ \int \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} {\rm Li}_2\left (e x\right ) \,d x } \]

input

integrate((a+b*log(c*x^n))*polylog(2,e*x),x, algorithm="giac")
 

output

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

3.3.10.9 Mupad [F(-1)]

Timed out. \[ \int \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}(2,e x) \, dx=\int \mathrm {polylog}\left (2,e\,x\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]

input

int(polylog(2, e*x)*(a + b*log(c*x^n)),x)
 

output

int(polylog(2, e*x)*(a + b*log(c*x^n)), x)