3.213 \(\int \frac{(a+b \log (c x^n)) \text{PolyLog}(2,e x)}{x^3} \, dx\)

Optimal. Leaf size=202 \[ -\frac{\text{PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac{1}{4} b e^2 n \text{PolyLog}(2,e x)-\frac{b n \text{PolyLog}(2,e x)}{4 x^2}+\frac{1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{4} e^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac{e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac{\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{4 x^2}-\frac{1}{8} b e^2 n \log ^2(x)+\frac{1}{4} b e^2 n \log (x)-\frac{1}{4} b e^2 n \log (1-e x)+\frac{b n \log (1-e x)}{4 x^2}-\frac{b e n}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.15729, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {2385, 2395, 44, 2376, 2301, 2391} \[ -\frac{\text{PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac{1}{4} b e^2 n \text{PolyLog}(2,e x)-\frac{b n \text{PolyLog}(2,e x)}{4 x^2}+\frac{1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{4} e^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac{e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac{\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{4 x^2}-\frac{1}{8} b e^2 n \log ^2(x)+\frac{1}{4} b e^2 n \log (x)-\frac{1}{4} b e^2 n \log (1-e x)+\frac{b n \log (1-e x)}{4 x^2}-\frac{b e n}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Log[c*x^n])*PolyLog[2, e*x])/x^3,x]

[Out]

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

Rule 2385

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.)*PolyLog[k_, (e_.)*(x_)^(q_.)], x_Symbol] :> -Simp
[(b*n*(d*x)^(m + 1)*PolyLog[k, e*x^q])/(d*(m + 1)^2), x] + (-Dist[q/(m + 1), Int[(d*x)^m*PolyLog[k - 1, e*x^q]
*(a + b*Log[c*x^n]), x], x] + Dist[(b*n*q)/(m + 1)^2, Int[(d*x)^m*PolyLog[k - 1, e*x^q], x], x] + Simp[((d*x)^
(m + 1)*PolyLog[k, e*x^q]*(a + b*Log[c*x^n]))/(d*(m + 1)), x]) /; FreeQ[{a, b, c, d, e, m, n, q}, x] && IGtQ[k
, 0]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2376

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((g_.)*(x_))^(q_.), x_Sym
bol] :> With[{u = IntHide[(g*x)^q*Log[d*(e + f*x^m)^r], x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[Dist
[1/x, u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] &
& RationalQ[q])) && NeQ[q, -1]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)}{x^3} \, dx &=-\frac{b n \text{Li}_2(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)}{2 x^2}-\frac{1}{2} \int \frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{x^3} \, dx-\frac{1}{4} (b n) \int \frac{\log (1-e x)}{x^3} \, dx\\ &=-\frac{e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac{1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac{b n \log (1-e x)}{8 x^2}-\frac{1}{4} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 x^2}-\frac{b n \text{Li}_2(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)}{2 x^2}+\frac{1}{2} (b n) \int \left (\frac{e}{2 x^2}-\frac{e^2 \log (x)}{2 x}-\frac{\log (1-e x)}{2 x^3}+\frac{e^2 \log (1-e x)}{2 x}\right ) \, dx+\frac{1}{8} (b e n) \int \frac{1}{x^2 (1-e x)} \, dx\\ &=-\frac{b e n}{4 x}-\frac{e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac{1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac{b n \log (1-e x)}{8 x^2}-\frac{1}{4} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 x^2}-\frac{b n \text{Li}_2(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)}{2 x^2}-\frac{1}{4} (b n) \int \frac{\log (1-e x)}{x^3} \, dx+\frac{1}{8} (b e n) \int \left (\frac{1}{x^2}+\frac{e}{x}-\frac{e^2}{-1+e x}\right ) \, dx-\frac{1}{4} \left (b e^2 n\right ) \int \frac{\log (x)}{x} \, dx+\frac{1}{4} \left (b e^2 n\right ) \int \frac{\log (1-e x)}{x} \, dx\\ &=-\frac{3 b e n}{8 x}+\frac{1}{8} b e^2 n \log (x)-\frac{1}{8} b e^2 n \log ^2(x)-\frac{e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac{1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{8} b e^2 n \log (1-e x)+\frac{b n \log (1-e x)}{4 x^2}-\frac{1}{4} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 x^2}-\frac{1}{4} b e^2 n \text{Li}_2(e x)-\frac{b n \text{Li}_2(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)}{2 x^2}+\frac{1}{8} (b e n) \int \frac{1}{x^2 (1-e x)} \, dx\\ &=-\frac{3 b e n}{8 x}+\frac{1}{8} b e^2 n \log (x)-\frac{1}{8} b e^2 n \log ^2(x)-\frac{e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac{1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{8} b e^2 n \log (1-e x)+\frac{b n \log (1-e x)}{4 x^2}-\frac{1}{4} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 x^2}-\frac{1}{4} b e^2 n \text{Li}_2(e x)-\frac{b n \text{Li}_2(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)}{2 x^2}+\frac{1}{8} (b e n) \int \left (\frac{1}{x^2}+\frac{e}{x}-\frac{e^2}{-1+e x}\right ) \, dx\\ &=-\frac{b e n}{2 x}+\frac{1}{4} b e^2 n \log (x)-\frac{1}{8} b e^2 n \log ^2(x)-\frac{e \left (a+b \log \left (c x^n\right )\right )}{4 x}+\frac{1}{4} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{4} b e^2 n \log (1-e x)+\frac{b n \log (1-e x)}{4 x^2}-\frac{1}{4} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 x^2}-\frac{1}{4} b e^2 n \text{Li}_2(e x)-\frac{b n \text{Li}_2(e x)}{4 x^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.188619, size = 163, normalized size = 0.81 \[ \frac{\left (-2 \text{PolyLog}(2,e x)+e^2 x^2 \log (x)-e^2 x^2 \log (1-e x)-e x+\log (1-e x)\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{4 x^2}+\frac{b n \left (-2 \left (e^2 x^2+2 \log (x)+1\right ) \text{PolyLog}(2,e x)+e^2 x^2 \log ^2(x)-2 e^2 x^2 \log (1-e x)-4 e x+2 \log (1-e x)-2 (e x-1) \log (x) ((e x+1) \log (1-e x)-e x)\right )}{8 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Log[c*x^n])*PolyLog[2, e*x])/x^3,x]

[Out]

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

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Maple [F]  time = 0.194, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\ln \left ( c{x}^{n} \right ) \right ){\it polylog} \left ( 2,ex \right ) }{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \,{\left (e^{2} \log \left (x\right ) - \frac{e x +{\left (e^{2} x^{2} - 1\right )} \log \left (-e x + 1\right ) + 2 \,{\rm Li}_2\left (e x\right )}{x^{2}}\right )} a - \frac{1}{4} \, b{\left (\frac{{\left (n + 2 \, \log \left (c\right ) + 2 \, \log \left (x^{n}\right )\right )}{\rm Li}_2\left (e x\right ) -{\left (e^{2} n x^{2} \log \left (x\right ) + n + \log \left (c\right )\right )} \log \left (-e x + 1\right ) -{\left (e^{2} x^{2} \log \left (x\right ) - e x -{\left (e^{2} x^{2} - 1\right )} \log \left (-e x + 1\right )\right )} \log \left (x^{n}\right )}{x^{2}} + 4 \, \int -\frac{e^{2} n x - 2 \, e n - e \log \left (c\right ) -{\left (2 \, e^{3} n x^{2} - e^{2} n x\right )} \log \left (x\right )}{4 \,{\left (e x^{3} - x^{2}\right )}}\,{d x}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*polylog(2,e*x)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.961958, size = 451, normalized size = 2.23 \begin{align*} \frac{b e^{2} n x^{2} \log \left (x\right )^{2} - 2 \,{\left (2 \, b e n + a e\right )} x - 2 \,{\left (b e^{2} n x^{2} + b n + 2 \, a\right )}{\rm Li}_2\left (e x\right ) - 2 \,{\left ({\left (b e^{2} n + a e^{2}\right )} x^{2} - b n - a\right )} \log \left (-e x + 1\right ) - 2 \,{\left (b e x + 2 \, b{\rm Li}_2\left (e x\right ) +{\left (b e^{2} x^{2} - b\right )} \log \left (-e x + 1\right )\right )} \log \left (c\right ) + 2 \,{\left (b e^{2} x^{2} \log \left (c\right ) - b e n x +{\left (b e^{2} n + a e^{2}\right )} x^{2} - 2 \, b n{\rm Li}_2\left (e x\right ) -{\left (b e^{2} n x^{2} - b n\right )} \log \left (-e x + 1\right )\right )} \log \left (x\right )}{8 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*polylog(2,e*x)/x^3,x, algorithm="fricas")

[Out]

1/8*(b*e^2*n*x^2*log(x)^2 - 2*(2*b*e*n + a*e)*x - 2*(b*e^2*n*x^2 + b*n + 2*a)*dilog(e*x) - 2*((b*e^2*n + a*e^2
)*x^2 - b*n - a)*log(-e*x + 1) - 2*(b*e*x + 2*b*dilog(e*x) + (b*e^2*x^2 - b)*log(-e*x + 1))*log(c) + 2*(b*e^2*
x^2*log(c) - b*e*n*x + (b*e^2*n + a*e^2)*x^2 - 2*b*n*dilog(e*x) - (b*e^2*n*x^2 - b*n)*log(-e*x + 1))*log(x))/x
^2

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (c x^{n}\right ) + a\right )}{\rm Li}_2\left (e x\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*polylog(2,e*x)/x^3,x, algorithm="giac")

[Out]

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