3.2.99 \(\int \frac {(a+b \log (c x^n))^3 \text {Li}_k(e x^q)}{x} \, dx\) [199]

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

[Out]

(a+b*ln(c*x^n))^3*polylog(1+k,e*x^q)/q-3*b*n*(a+b*ln(c*x^n))^2*polylog(2+k,e*x^q)/q^2+6*b^2*n^2*(a+b*ln(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 = 0.07, antiderivative size = 104, normalized size of antiderivative = 1.00, 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 = {2430, 6724} \begin {gather*} \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} \end {gather*}

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 2430

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*PolyLog[k_, (e_.)*(x_)^(q_.)])/(x_), x_Symbol] :> Simp[PolyLo
g[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 6724

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*}

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Mathematica [A]
time = 0.04, size = 99, normalized size = 0.95 \begin {gather*} \frac {q^3 \left (a+b \log \left (c x^n\right )\right )^3 \text {Li}_{1+k}\left (e x^q\right )-3 b n \left (q^2 \left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_{2+k}\left (e x^q\right )+2 b n \left (-q \left (a+b \log \left (c x^n\right )\right ) \text {Li}_{3+k}\left (e x^q\right )+b n \text {Li}_{4+k}\left (e x^q\right )\right )\right )}{q^4} \end {gather*}

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 = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \ln \left (c \,x^{n}\right )\right )^{3} \polylog \left (k , e \,x^{q}\right )}{x}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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, x^q*e)/x, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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, x^q*e)/x, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \log {\left (c x^{n} \right )}\right )^{3} \operatorname {Li}_{k}\left (e x^{q}\right )}{x}\, dx \end {gather*}

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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, x^q*e)/x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {polylog}\left (k,e\,x^q\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^3}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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