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

Optimal. Leaf size=104 \[ \frac{6 b^2 n^2 \text{PolyLog}\left (k+3,e x^q\right ) \left (a+b \log \left (c x^n\right )\right )}{q^3}-\frac{3 b n \text{PolyLog}\left (k+2,e x^q\right ) \left (a+b \log \left (c x^n\right )\right )^2}{q^2}+\frac{\text{PolyLog}\left (k+1,e x^q\right ) \left (a+b \log \left (c x^n\right )\right )^3}{q}-\frac{6 b^3 n^3 \text{PolyLog}\left (k+4,e x^q\right )}{q^4} \]

[Out]

((a + b*Log[c*x^n])^3*PolyLog[1 + k, e*x^q])/q - (3*b*n*(a + b*Log[c*x^n])^2*PolyLog[2 + k, e*x^q])/q^2 + (6*b
^2*n^2*(a + b*Log[c*x^n])*PolyLog[3 + k, e*x^q])/q^3 - (6*b^3*n^3*PolyLog[4 + k, e*x^q])/q^4

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Rubi [A]  time = 0.112095, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2383, 6589} \[ \frac{6 b^2 n^2 \text{PolyLog}\left (k+3,e x^q\right ) \left (a+b \log \left (c x^n\right )\right )}{q^3}-\frac{3 b n \text{PolyLog}\left (k+2,e x^q\right ) \left (a+b \log \left (c x^n\right )\right )^2}{q^2}+\frac{\text{PolyLog}\left (k+1,e x^q\right ) \left (a+b \log \left (c x^n\right )\right )^3}{q}-\frac{6 b^3 n^3 \text{PolyLog}\left (k+4,e x^q\right )}{q^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*Log[c*x^n])^3*PolyLog[1 + k, e*x^q])/q - (3*b*n*(a + b*Log[c*x^n])^2*PolyLog[2 + k, e*x^q])/q^2 + (6*b
^2*n^2*(a + b*Log[c*x^n])*PolyLog[3 + k, e*x^q])/q^3 - (6*b^3*n^3*PolyLog[4 + k, e*x^q])/q^4

Rule 2383

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

Rule 6589

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

Rubi steps

\begin{align*} \int \frac{\left (a+b \log \left (c x^n\right )\right )^3 \text{Li}_k\left (e x^q\right )}{x} \, dx &=\frac{\left (a+b \log \left (c x^n\right )\right )^3 \text{Li}_{1+k}\left (e x^q\right )}{q}-\frac{(3 b n) \int \frac{\left (a+b \log \left (c x^n\right )\right )^2 \text{Li}_{1+k}\left (e x^q\right )}{x} \, dx}{q}\\ &=\frac{\left (a+b \log \left (c x^n\right )\right )^3 \text{Li}_{1+k}\left (e x^q\right )}{q}-\frac{3 b n \left (a+b \log \left (c x^n\right )\right )^2 \text{Li}_{2+k}\left (e x^q\right )}{q^2}+\frac{\left (6 b^2 n^2\right ) \int \frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_{2+k}\left (e x^q\right )}{x} \, dx}{q^2}\\ &=\frac{\left (a+b \log \left (c x^n\right )\right )^3 \text{Li}_{1+k}\left (e x^q\right )}{q}-\frac{3 b n \left (a+b \log \left (c x^n\right )\right )^2 \text{Li}_{2+k}\left (e x^q\right )}{q^2}+\frac{6 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \text{Li}_{3+k}\left (e x^q\right )}{q^3}-\frac{\left (6 b^3 n^3\right ) \int \frac{\text{Li}_{3+k}\left (e x^q\right )}{x} \, dx}{q^3}\\ &=\frac{\left (a+b \log \left (c x^n\right )\right )^3 \text{Li}_{1+k}\left (e x^q\right )}{q}-\frac{3 b n \left (a+b \log \left (c x^n\right )\right )^2 \text{Li}_{2+k}\left (e x^q\right )}{q^2}+\frac{6 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \text{Li}_{3+k}\left (e x^q\right )}{q^3}-\frac{6 b^3 n^3 \text{Li}_{4+k}\left (e x^q\right )}{q^4}\\ \end{align*}

Mathematica [A]  time = 0.0485369, size = 99, normalized size = 0.95 \[ \frac{q^3 \text{PolyLog}\left (k+1,e x^q\right ) \left (a+b \log \left (c x^n\right )\right )^3-3 b n \left (q^2 \text{PolyLog}\left (k+2,e x^q\right ) \left (a+b \log \left (c x^n\right )\right )^2+2 b n \left (b n \text{PolyLog}\left (k+4,e x^q\right )-q \text{PolyLog}\left (k+3,e x^q\right ) \left (a+b \log \left (c x^n\right )\right )\right )\right )}{q^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(q^3*(a + b*Log[c*x^n])^3*PolyLog[1 + k, e*x^q] - 3*b*n*(q^2*(a + b*Log[c*x^n])^2*PolyLog[2 + k, e*x^q] + 2*b*
n*(-(q*(a + b*Log[c*x^n])*PolyLog[3 + k, e*x^q]) + b*n*PolyLog[4 + k, e*x^q])))/q^4

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Maple [F]  time = 0.027, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) ^{3}{\it polylog} \left ( k,e{x}^{q} \right ) }{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{3} \log \left (c x^{n}\right )^{3} + 3 \, a b^{2} \log \left (c x^{n}\right )^{2} + 3 \, a^{2} b \log \left (c x^{n}\right ) + a^{3}\right )}{\rm polylog}\left (k, e x^{q}\right )}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^3*log(c*x^n)^3 + 3*a*b^2*log(c*x^n)^2 + 3*a^2*b*log(c*x^n) + a^3)*polylog(k, e*x^q)/x, x)

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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))**3*polylog(k,e*x**q)/x,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 )}^{3}{\rm Li}_{k}(e x^{q})}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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