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

Optimal. Leaf size=131 \[ -x \text{PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )+x \text{PolyLog}(3,e x) \left (a+b \log \left (c x^n\right )\right )+2 b n x \text{PolyLog}(2,e x)-b n x \text{PolyLog}(3,e x)+\frac{b n \text{PolyLog}(2,e x)}{e}+\frac{(1-e x) \log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{e}+x \left (a+b \log \left (c x^n\right )\right )-\frac{3 b n (1-e x) \log (1-e x)}{e}-4 b n x \]

[Out]

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

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Rubi [A]  time = 0.127397, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.562, Rules used = {2381, 2389, 2295, 2370, 2411, 43, 2351, 2315, 6586} \[ -x \text{PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )+x \text{PolyLog}(3,e x) \left (a+b \log \left (c x^n\right )\right )+2 b n x \text{PolyLog}(2,e x)-b n x \text{PolyLog}(3,e x)+\frac{b n \text{PolyLog}(2,e x)}{e}+\frac{(1-e x) \log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{e}+x \left (a+b \log \left (c x^n\right )\right )-\frac{3 b n (1-e x) \log (1-e x)}{e}-4 b n x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2381

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

Rule 2389

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

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2370

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> With[
{u = IntHide[Log[d*(e + f*x^m)^r], x]}, Dist[(a + b*Log[c*x^n])^p, u, x] - Dist[b*n*p, Int[Dist[(a + b*Log[c*x
^n])^(p - 1)/x, u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, r, m, n}, x] && IGtQ[p, 0] && RationalQ[m] && (EqQ[
p, 1] || (FractionQ[m] && IntegerQ[1/m]) || (EqQ[r, 1] && EqQ[m, 1] && EqQ[d*e, 1]))

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 6586

Int[PolyLog[n_, (a_.)*((b_.)*(x_)^(p_.))^(q_.)], x_Symbol] :> Simp[x*PolyLog[n, a*(b*x^p)^q], x] - Dist[p*q, I
nt[PolyLog[n - 1, a*(b*x^p)^q], x], x] /; FreeQ[{a, b, p, q}, x] && GtQ[n, 0]

Rubi steps

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

Mathematica [F]  time = 0.0799283, size = 0, normalized size = 0. \[ \int \left (a+b \log \left (c x^n\right )\right ) \text{PolyLog}(3,e x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.273, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\ln \left ( c{x}^{n} \right ) \right ){\it polylog} \left ( 3,ex \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 0.98276, size = 464, normalized size = 3.54 \begin{align*} -\frac{{\left (4 \, b e n - a e\right )} x +{\left (b e n x \log \left (x\right ) + b e x \log \left (c\right ) - b n -{\left (2 \, b e n - a e\right )} x\right )}{\rm \%iint}\left (e, x, -\frac{\log \left (-e x + 1\right )}{e}, -\frac{\log \left (-e x + 1\right )}{x}\right ) +{\left (3 \, b n -{\left (3 \, b e n - a e\right )} x - a\right )} \log \left (-e x + 1\right ) -{\left (b e x -{\left (b e x - b\right )} \log \left (-e x + 1\right )\right )} \log \left (c\right ) -{\left (b e n x -{\left (b e n x - b n\right )} \log \left (-e x + 1\right )\right )} \log \left (x\right ) -{\left (b e n x \log \left (x\right ) + b e x \log \left (c\right ) -{\left (b e n - a e\right )} x\right )}{\rm polylog}\left (3, e x\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((4*b*e*n - a*e)*x + (b*e*n*x*log(x) + b*e*x*log(c) - b*n - (2*b*e*n - a*e)*x)*\%iint(e, x, -log(-e*x + 1)/e,
-log(-e*x + 1)/x) + (3*b*n - (3*b*e*n - a*e)*x - a)*log(-e*x + 1) - (b*e*x - (b*e*x - b)*log(-e*x + 1))*log(c)
- (b*e*n*x - (b*e*n*x - b*n)*log(-e*x + 1))*log(x) - (b*e*n*x*log(x) + b*e*x*log(c) - (b*e*n - a*e)*x)*polylo
g(3, e*x))/e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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