3.210 \(\int (a+b \log (c x^n)) \text{PolyLog}(2,e x) \, dx\)

Optimal. Leaf size=106 \[ x \text{PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-b n x \text{PolyLog}(2,e x)-\frac{b n \text{PolyLog}(2,e x)}{e}-\frac{(1-e x) \log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{e}-x \left (a+b \log \left (c x^n\right )\right )+\frac{2 b n (1-e x) \log (1-e x)}{e}+3 b n x \]

[Out]

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

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Rubi [A]  time = 0.114032, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {2381, 2389, 2295, 2370, 2411, 43, 2351, 2315} \[ x \text{PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-b n x \text{PolyLog}(2,e x)-\frac{b n \text{PolyLog}(2,e x)}{e}-\frac{(1-e x) \log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{e}-x \left (a+b \log \left (c x^n\right )\right )+\frac{2 b n (1-e x) \log (1-e x)}{e}+3 b n x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.0739469, size = 113, normalized size = 1.07 \[ \left (x \text{PolyLog}(2,e x)+\left (x-\frac{1}{e}\right ) \log (1-e x)-x\right ) \left (a+b \left (\log \left (c x^n\right )-n \log (x)\right )\right )+\frac{b n ((-e x+e x \log (x)-1) \text{PolyLog}(2,e x)+3 e x-2 e x \log (1-e x)+2 \log (1-e x)+\log (x) ((e x-1) \log (1-e x)-e x))}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.151, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\ln \left ( c{x}^{n} \right ) \right ){\it polylog} \left ( 2,ex \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a+b*ln(c*x^n))*polylog(2,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 (e n - e \log \left (c\right )\right )} x\right )}{\rm Li}_2\left (e x\right ) -{\left ({\left (2 \, 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 )}{e} - \int -\frac{{\left (3 \, 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 +{\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(2,e*x),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.897499, size = 329, normalized size = 3.1 \begin{align*} \frac{{\left (3 \, b e n - a e\right )} x -{\left (b n +{\left (b e n - a e\right )} x\right )}{\rm Li}_2\left (e x\right ) +{\left (2 \, b n -{\left (2 \, b e n - a e\right )} x - a\right )} \log \left (-e x + 1\right ) +{\left (b e x{\rm Li}_2\left (e x\right ) - 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{\rm Li}_2\left (e x\right ) - b e n x +{\left (b e n x - b n\right )} \log \left (-e x + 1\right )\right )} \log \left (x\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*b*e*n - a*e)*x - (b*n + (b*e*n - a*e)*x)*dilog(e*x) + (2*b*n - (2*b*e*n - a*e)*x - a)*log(-e*x + 1) + (b*e
*x*dilog(e*x) - b*e*x + (b*e*x - b)*log(-e*x + 1))*log(c) + (b*e*n*x*dilog(e*x) - b*e*n*x + (b*e*n*x - b*n)*lo
g(-e*x + 1))*log(x))/e

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Sympy [A]  time = 99.2439, size = 172, normalized size = 1.62 \begin{align*} \begin{cases} - a x \operatorname{Li}_{1}\left (e x\right ) + a x \operatorname{Li}_{2}\left (e x\right ) - a x + \frac{a \operatorname{Li}_{1}\left (e x\right )}{e} - b n x \log{\left (x \right )} \operatorname{Li}_{1}\left (e x\right ) + b n x \log{\left (x \right )} \operatorname{Li}_{2}\left (e x\right ) - b n x \log{\left (x \right )} + 2 b n x \operatorname{Li}_{1}\left (e x\right ) - b n x \operatorname{Li}_{2}\left (e x\right ) + 3 b n x - b x \log{\left (c \right )} \operatorname{Li}_{1}\left (e x\right ) + b x \log{\left (c \right )} \operatorname{Li}_{2}\left (e x\right ) - b x \log{\left (c \right )} + \frac{b n \log{\left (x \right )} \operatorname{Li}_{1}\left (e x\right )}{e} - \frac{2 b n \operatorname{Li}_{1}\left (e x\right )}{e} - \frac{b n \operatorname{Li}_{2}\left (e x\right )}{e} + \frac{b \log{\left (c \right )} \operatorname{Li}_{1}\left (e x\right )}{e} & \text{for}\: e \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))*polylog(2,e*x),x)

[Out]

Piecewise((-a*x*polylog(1, e*x) + a*x*polylog(2, e*x) - a*x + a*polylog(1, e*x)/e - b*n*x*log(x)*polylog(1, e*
x) + b*n*x*log(x)*polylog(2, e*x) - b*n*x*log(x) + 2*b*n*x*polylog(1, e*x) - b*n*x*polylog(2, e*x) + 3*b*n*x -
b*x*log(c)*polylog(1, e*x) + b*x*log(c)*polylog(2, e*x) - b*x*log(c) + b*n*log(x)*polylog(1, e*x)/e - 2*b*n*p
olylog(1, e*x)/e - b*n*polylog(2, e*x)/e + b*log(c)*polylog(1, e*x)/e, Ne(e, 0)), (0, True))

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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}_2\left (e x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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