3.21 \(\int \frac {(a+b \log (c x^n))^3 \log (1+e x)}{x} \, dx\)

Optimal. Leaf size=81 \[ -6 b^2 n^2 \text {Li}_4(-e x) \left (a+b \log \left (c x^n\right )\right )+3 b n \text {Li}_3(-e x) \left (a+b \log \left (c x^n\right )\right )^2-\text {Li}_2(-e x) \left (a+b \log \left (c x^n\right )\right )^3+6 b^3 n^3 \text {Li}_5(-e x) \]

[Out]

-(a+b*ln(c*x^n))^3*polylog(2,-e*x)+3*b*n*(a+b*ln(c*x^n))^2*polylog(3,-e*x)-6*b^2*n^2*(a+b*ln(c*x^n))*polylog(4
,-e*x)+6*b^3*n^3*polylog(5,-e*x)

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Rubi [A]  time = 0.10, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2374, 2383, 6589} \[ -6 b^2 n^2 \text {PolyLog}(4,-e x) \left (a+b \log \left (c x^n\right )\right )+3 b n \text {PolyLog}(3,-e x) \left (a+b \log \left (c x^n\right )\right )^2-\text {PolyLog}(2,-e x) \left (a+b \log \left (c x^n\right )\right )^3+6 b^3 n^3 \text {PolyLog}(5,-e x) \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Log[c*x^n])^3*Log[1 + e*x])/x,x]

[Out]

-((a + b*Log[c*x^n])^3*PolyLog[2, -(e*x)]) + 3*b*n*(a + b*Log[c*x^n])^2*PolyLog[3, -(e*x)] - 6*b^2*n^2*(a + b*
Log[c*x^n])*PolyLog[4, -(e*x)] + 6*b^3*n^3*PolyLog[5, -(e*x)]

Rule 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

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 \log (1+e x)}{x} \, dx &=-\left (a+b \log \left (c x^n\right )\right )^3 \text {Li}_2(-e x)+(3 b n) \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2(-e x)}{x} \, dx\\ &=-\left (a+b \log \left (c x^n\right )\right )^3 \text {Li}_2(-e x)+3 b n \left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_3(-e x)-\left (6 b^2 n^2\right ) \int \frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(-e x)}{x} \, dx\\ &=-\left (a+b \log \left (c x^n\right )\right )^3 \text {Li}_2(-e x)+3 b n \left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_3(-e x)-6 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_4(-e x)+\left (6 b^3 n^3\right ) \int \frac {\text {Li}_4(-e x)}{x} \, dx\\ &=-\left (a+b \log \left (c x^n\right )\right )^3 \text {Li}_2(-e x)+3 b n \left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_3(-e x)-6 b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_4(-e x)+6 b^3 n^3 \text {Li}_5(-e x)\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 77, normalized size = 0.95 \[ 3 b n \left (\text {Li}_3(-e x) \left (a+b \log \left (c x^n\right )\right )^2+2 b n \left (b n \text {Li}_5(-e x)-\text {Li}_4(-e x) \left (a+b \log \left (c x^n\right )\right )\right )\right )-\text {Li}_2(-e x) \left (a+b \log \left (c x^n\right )\right )^3 \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Log[c*x^n])^3*Log[1 + e*x])/x,x]

[Out]

-((a + b*Log[c*x^n])^3*PolyLog[2, -(e*x)]) + 3*b*n*((a + b*Log[c*x^n])^2*PolyLog[3, -(e*x)] + 2*b*n*(-((a + b*
Log[c*x^n])*PolyLog[4, -(e*x)]) + b*n*PolyLog[5, -(e*x)]))

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \log \left (c x^{n}\right )^{3} \log \left (e x + 1\right ) + 3 \, a b^{2} \log \left (c x^{n}\right )^{2} \log \left (e x + 1\right ) + 3 \, a^{2} b \log \left (c x^{n}\right ) \log \left (e x + 1\right ) + a^{3} \log \left (e x + 1\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))^3*log(e*x+1)/x,x, algorithm="fricas")

[Out]

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, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left (e x + 1\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))^3*log(e*x+1)/x,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right )^{3} \ln \left (e x +1\right )}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left (e x + 1\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))^3*log(e*x+1)/x,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (e\,x+1\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^3}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))**3*ln(e*x+1)/x,x)

[Out]

Timed out

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