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

Optimal. Leaf size=185 \[ \frac{1}{2} x^2 \text{PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac{b n \text{PolyLog}(2,e x)}{4 e^2}-\frac{1}{4} b n x^2 \text{PolyLog}(2,e x)-\frac{\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{4 e^2}+\frac{1}{4} x^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac{x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac{1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac{b n \log (1-e x)}{4 e^2}-\frac{1}{4} b n x^2 \log (1-e x)+\frac{b n x}{2 e}+\frac{3}{16} b n x^2 \]

[Out]

(b*n*x)/(2*e) + (3*b*n*x^2)/16 - (x*(a + b*Log[c*x^n]))/(4*e) - (x^2*(a + b*Log[c*x^n]))/8 + (b*n*Log[1 - e*x]
)/(4*e^2) - (b*n*x^2*Log[1 - e*x])/4 - ((a + b*Log[c*x^n])*Log[1 - e*x])/(4*e^2) + (x^2*(a + b*Log[c*x^n])*Log
[1 - e*x])/4 - (b*n*PolyLog[2, e*x])/(4*e^2) - (b*n*x^2*PolyLog[2, e*x])/4 + (x^2*(a + b*Log[c*x^n])*PolyLog[2
, e*x])/2

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Rubi [A]  time = 0.131546, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {2385, 2395, 43, 2376, 2391} \[ \frac{1}{2} x^2 \text{PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac{b n \text{PolyLog}(2,e x)}{4 e^2}-\frac{1}{4} b n x^2 \text{PolyLog}(2,e x)-\frac{\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{4 e^2}+\frac{1}{4} x^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac{x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac{1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac{b n \log (1-e x)}{4 e^2}-\frac{1}{4} b n x^2 \log (1-e x)+\frac{b n x}{2 e}+\frac{3}{16} b n x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*n*x)/(2*e) + (3*b*n*x^2)/16 - (x*(a + b*Log[c*x^n]))/(4*e) - (x^2*(a + b*Log[c*x^n]))/8 + (b*n*Log[1 - e*x]
)/(4*e^2) - (b*n*x^2*Log[1 - e*x])/4 - ((a + b*Log[c*x^n])*Log[1 - e*x])/(4*e^2) + (x^2*(a + b*Log[c*x^n])*Log
[1 - e*x])/4 - (b*n*PolyLog[2, e*x])/(4*e^2) - (b*n*x^2*PolyLog[2, e*x])/4 + (x^2*(a + b*Log[c*x^n])*PolyLog[2
, e*x])/2

Rule 2385

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.)*PolyLog[k_, (e_.)*(x_)^(q_.)], x_Symbol] :> -Simp
[(b*n*(d*x)^(m + 1)*PolyLog[k, e*x^q])/(d*(m + 1)^2), x] + (-Dist[q/(m + 1), Int[(d*x)^m*PolyLog[k - 1, e*x^q]
*(a + b*Log[c*x^n]), x], x] + Dist[(b*n*q)/(m + 1)^2, Int[(d*x)^m*PolyLog[k - 1, e*x^q], x], x] + Simp[((d*x)^
(m + 1)*PolyLog[k, e*x^q]*(a + b*Log[c*x^n]))/(d*(m + 1)), x]) /; FreeQ[{a, b, c, d, e, m, n, q}, x] && IGtQ[k
, 0]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

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 2376

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((g_.)*(x_))^(q_.), x_Sym
bol] :> With[{u = IntHide[(g*x)^q*Log[d*(e + f*x^m)^r], x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[Dist
[1/x, u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] &
& RationalQ[q])) && NeQ[q, -1]

Rule 2391

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

Rubi steps

\begin{align*} \int x \left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x) \, dx &=-\frac{1}{4} b n x^2 \text{Li}_2(e x)+\frac{1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)+\frac{1}{2} \int x \left (a+b \log \left (c x^n\right )\right ) \log (1-e x) \, dx-\frac{1}{4} (b n) \int x \log (1-e x) \, dx\\ &=-\frac{x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac{1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac{1}{8} b n x^2 \log (1-e x)-\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 e^2}+\frac{1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac{1}{4} b n x^2 \text{Li}_2(e x)+\frac{1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)-\frac{1}{2} (b n) \int \left (-\frac{1}{2 e}-\frac{x}{4}-\frac{\log (1-e x)}{2 e^2 x}+\frac{1}{2} x \log (1-e x)\right ) \, dx-\frac{1}{8} (b e n) \int \frac{x^2}{1-e x} \, dx\\ &=\frac{b n x}{4 e}+\frac{1}{16} b n x^2-\frac{x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac{1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac{1}{8} b n x^2 \log (1-e x)-\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 e^2}+\frac{1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac{1}{4} b n x^2 \text{Li}_2(e x)+\frac{1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)-\frac{1}{4} (b n) \int x \log (1-e x) \, dx+\frac{(b n) \int \frac{\log (1-e x)}{x} \, dx}{4 e^2}-\frac{1}{8} (b e n) \int \left (-\frac{1}{e^2}-\frac{x}{e}-\frac{1}{e^2 (-1+e x)}\right ) \, dx\\ &=\frac{3 b n x}{8 e}+\frac{1}{8} b n x^2-\frac{x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac{1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac{b n \log (1-e x)}{8 e^2}-\frac{1}{4} b n x^2 \log (1-e x)-\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 e^2}+\frac{1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac{b n \text{Li}_2(e x)}{4 e^2}-\frac{1}{4} b n x^2 \text{Li}_2(e x)+\frac{1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)-\frac{1}{8} (b e n) \int \frac{x^2}{1-e x} \, dx\\ &=\frac{3 b n x}{8 e}+\frac{1}{8} b n x^2-\frac{x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac{1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac{b n \log (1-e x)}{8 e^2}-\frac{1}{4} b n x^2 \log (1-e x)-\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 e^2}+\frac{1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac{b n \text{Li}_2(e x)}{4 e^2}-\frac{1}{4} b n x^2 \text{Li}_2(e x)+\frac{1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)-\frac{1}{8} (b e n) \int \left (-\frac{1}{e^2}-\frac{x}{e}-\frac{1}{e^2 (-1+e x)}\right ) \, dx\\ &=\frac{b n x}{2 e}+\frac{3}{16} b n x^2-\frac{x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac{1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac{b n \log (1-e x)}{4 e^2}-\frac{1}{4} b n x^2 \log (1-e x)-\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 e^2}+\frac{1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac{b n \text{Li}_2(e x)}{4 e^2}-\frac{1}{4} b n x^2 \text{Li}_2(e x)+\frac{1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text{Li}_2(e x)\\ \end{align*}

Mathematica [A]  time = 0.323954, size = 168, normalized size = 0.91 \[ \frac{\left (4 e^2 x^2 \text{PolyLog}(2,e x)+2 \left (e^2 x^2-1\right ) \log (1-e x)-e x (e x+2)\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{8 e^2}+\frac{b n \left (\left (-4 e^2 x^2+8 e^2 x^2 \log (x)-4\right ) \text{PolyLog}(2,e x)+3 e^2 x^2-4 e^2 x^2 \log (1-e x)+\log (x) \left (4 \left (e^2 x^2-1\right ) \log (1-e x)-2 e x (e x+2)\right )+8 e x+4 \log (1-e x)\right )}{16 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - b*n*Log[x] + b*Log[c*x^n])*(-(e*x*(2 + e*x)) + 2*(-1 + e^2*x^2)*Log[1 - e*x] + 4*e^2*x^2*PolyLog[2, e*x]
))/(8*e^2) + (b*n*(8*e*x + 3*e^2*x^2 + 4*Log[1 - e*x] - 4*e^2*x^2*Log[1 - e*x] + Log[x]*(-2*e*x*(2 + e*x) + 4*
(-1 + e^2*x^2)*Log[1 - e*x]) + (-4 - 4*e^2*x^2 + 8*e^2*x^2*Log[x])*PolyLog[2, e*x]))/(16*e^2)

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Maple [F]  time = 0.181, size = 0, normalized size = 0. \begin{align*} \int x \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(x*(a+b*ln(c*x^n))*polylog(2,e*x),x)

[Out]

int(x*(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*} \frac{1}{8} \, b{\left (\frac{2 \,{\left (2 \, e^{2} x^{2} \log \left (x^{n}\right ) -{\left (e^{2} n - 2 \, e^{2} \log \left (c\right )\right )} x^{2}\right )}{\rm Li}_2\left (e x\right ) - 2 \,{\left ({\left (e^{2} n - e^{2} \log \left (c\right )\right )} x^{2} - n \log \left (x\right )\right )} \log \left (-e x + 1\right ) -{\left (e^{2} x^{2} + 2 \, e x - 2 \,{\left (e^{2} x^{2} - 1\right )} \log \left (-e x + 1\right )\right )} \log \left (x^{n}\right )}{e^{2}} - 8 \, \int -\frac{e n x +{\left (3 \, e^{2} n - 2 \, e^{2} \log \left (c\right )\right )} x^{2} - 2 \, n \log \left (x\right ) - 2 \, n}{8 \,{\left (e^{2} x - e\right )}}\,{d x}\right )} + \frac{{\left (4 \, e^{2} x^{2}{\rm Li}_2\left (e x\right ) - e^{2} x^{2} - 2 \, e x + 2 \,{\left (e^{2} x^{2} - 1\right )} \log \left (-e x + 1\right )\right )} a}{8 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b*((2*(2*e^2*x^2*log(x^n) - (e^2*n - 2*e^2*log(c))*x^2)*dilog(e*x) - 2*((e^2*n - e^2*log(c))*x^2 - n*log(x
))*log(-e*x + 1) - (e^2*x^2 + 2*e*x - 2*(e^2*x^2 - 1)*log(-e*x + 1))*log(x^n))/e^2 - 8*integrate(-1/8*(e*n*x +
(3*e^2*n - 2*e^2*log(c))*x^2 - 2*n*log(x) - 2*n)/(e^2*x - e), x)) + 1/8*(4*e^2*x^2*dilog(e*x) - e^2*x^2 - 2*e
*x + 2*(e^2*x^2 - 1)*log(-e*x + 1))*a/e^2

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Fricas [A]  time = 0.981664, size = 477, normalized size = 2.58 \begin{align*} \frac{{\left (3 \, b e^{2} n - 2 \, a e^{2}\right )} x^{2} + 4 \,{\left (2 \, b e n - a e\right )} x - 4 \,{\left ({\left (b e^{2} n - 2 \, a e^{2}\right )} x^{2} + b n\right )}{\rm Li}_2\left (e x\right ) - 4 \,{\left ({\left (b e^{2} n - a e^{2}\right )} x^{2} - b n + a\right )} \log \left (-e x + 1\right ) + 2 \,{\left (4 \, b e^{2} x^{2}{\rm Li}_2\left (e x\right ) - b e^{2} x^{2} - 2 \, b e x + 2 \,{\left (b e^{2} x^{2} - b\right )} \log \left (-e x + 1\right )\right )} \log \left (c\right ) + 2 \,{\left (4 \, b e^{2} n x^{2}{\rm Li}_2\left (e x\right ) - b e^{2} n x^{2} - 2 \, b e n x + 2 \,{\left (b e^{2} n x^{2} - b n\right )} \log \left (-e x + 1\right )\right )} \log \left (x\right )}{16 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((3*b*e^2*n - 2*a*e^2)*x^2 + 4*(2*b*e*n - a*e)*x - 4*((b*e^2*n - 2*a*e^2)*x^2 + b*n)*dilog(e*x) - 4*((b*e
^2*n - a*e^2)*x^2 - b*n + a)*log(-e*x + 1) + 2*(4*b*e^2*x^2*dilog(e*x) - b*e^2*x^2 - 2*b*e*x + 2*(b*e^2*x^2 -
b)*log(-e*x + 1))*log(c) + 2*(4*b*e^2*n*x^2*dilog(e*x) - b*e^2*n*x^2 - 2*b*e*n*x + 2*(b*e^2*n*x^2 - b*n)*log(-
e*x + 1))*log(x))/e^2

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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