3.14 \(\int \frac{(a+b \log (c x^n))^2 \log (1+e x)}{x} \, dx\)

Optimal. Leaf size=55 \[ 2 b n \text{PolyLog}(3,-e x) \left (a+b \log \left (c x^n\right )\right )-\text{PolyLog}(2,-e x) \left (a+b \log \left (c x^n\right )\right )^2-2 b^2 n^2 \text{PolyLog}(4,-e x) \]

[Out]

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

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Rubi [A]  time = 0.060267, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {2374, 2383, 6589} \[ 2 b n \text{PolyLog}(3,-e x) \left (a+b \log \left (c x^n\right )\right )-\text{PolyLog}(2,-e x) \left (a+b \log \left (c x^n\right )\right )^2-2 b^2 n^2 \text{PolyLog}(4,-e x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*Log[c*x^n])^2*PolyLog[2, -(e*x)]) + 2*b*n*(a + b*Log[c*x^n])*PolyLog[3, -(e*x)] - 2*b^2*n^2*PolyLog[4
, -(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 )^2 \log (1+e x)}{x} \, dx &=-\left (a+b \log \left (c x^n\right )\right )^2 \text{Li}_2(-e x)+(2 b n) \int \frac{\left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(-e x)}{x} \, dx\\ &=-\left (a+b \log \left (c x^n\right )\right )^2 \text{Li}_2(-e x)+2 b n \left (a+b \log \left (c x^n\right )\right ) \text{Li}_3(-e x)-\left (2 b^2 n^2\right ) \int \frac{\text{Li}_3(-e x)}{x} \, dx\\ &=-\left (a+b \log \left (c x^n\right )\right )^2 \text{Li}_2(-e x)+2 b n \left (a+b \log \left (c x^n\right )\right ) \text{Li}_3(-e x)-2 b^2 n^2 \text{Li}_4(-e x)\\ \end{align*}

Mathematica [A]  time = 0.0753697, size = 53, normalized size = 0.96 \[ 2 b n \left (\text{PolyLog}(3,-e x) \left (a+b \log \left (c x^n\right )\right )-b n \text{PolyLog}(4,-e x)\right )-\text{PolyLog}(2,-e x) \left (a+b \log \left (c x^n\right )\right )^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.175, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) ^{2}\ln \left ( ex+1 \right ) }{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a+b*ln(c*x^n))^2*ln(e*x+1)/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 )}^{2} \log \left (e x + 1\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))^2*log(e*x+1)/x,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \log \left (c x^{n}\right )^{2} \log \left (e x + 1\right ) + 2 \, a b \log \left (c x^{n}\right ) \log \left (e x + 1\right ) + a^{2} \log \left (e x + 1\right )}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))^2*log(e*x+1)/x,x, algorithm="fricas")

[Out]

integral((b^2*log(c*x^n)^2*log(e*x + 1) + 2*a*b*log(c*x^n)*log(e*x + 1) + a^2*log(e*x + 1))/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))**2*ln(e*x+1)/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 )}^{2} \log \left (e x + 1\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))^2*log(e*x+1)/x,x, algorithm="giac")

[Out]

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