3.3.9 \(\int x (a+b \log (c x^n)) \text {Li}_2(e x) \, dx\) [209]

Optimal. Leaf size=185 \[ \frac {b n x}{2 e}+\frac {3}{16} b n x^2-\frac {x \left (a+b \log \left (c x^n\right )\right )}{4 e}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1-e x)}{4 e^2}-\frac {1}{4} b n x^2 \log (1-e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{4 e^2}+\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac {b n \text {Li}_2(e x)}{4 e^2}-\frac {1}{4} b n x^2 \text {Li}_2(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x) \]

[Out]

1/2*b*n*x/e+3/16*b*n*x^2-1/4*x*(a+b*ln(c*x^n))/e-1/8*x^2*(a+b*ln(c*x^n))+1/4*b*n*ln(-e*x+1)/e^2-1/4*b*n*x^2*ln
(-e*x+1)-1/4*(a+b*ln(c*x^n))*ln(-e*x+1)/e^2+1/4*x^2*(a+b*ln(c*x^n))*ln(-e*x+1)-1/4*b*n*polylog(2,e*x)/e^2-1/4*
b*n*x^2*polylog(2,e*x)+1/2*x^2*(a+b*ln(c*x^n))*polylog(2,e*x)

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Rubi [A]
time = 0.10, antiderivative size = 185, normalized size of antiderivative = 1.00, 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 = {2432, 2442, 45, 2423, 2438} \begin {gather*} \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 {x \left (a+b \log \left (c x^n\right )\right )}{4 e}+\frac {1}{4} x^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\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 \end {gather*}

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 45

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 2423

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 2432

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 2438

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]

Rule 2442

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]

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*}

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Mathematica [A]
time = 0.24, size = 168, normalized size = 0.91 \begin {gather*} \frac {\left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \left (-e x (2+e x)+2 \left (-1+e^2 x^2\right ) \log (1-e x)+4 e^2 x^2 \text {Li}_2(e x)\right )}{8 e^2}+\frac {b n \left (8 e x+3 e^2 x^2+4 \log (1-e x)-4 e^2 x^2 \log (1-e x)+\log (x) \left (-2 e x (2+e x)+4 \left (-1+e^2 x^2\right ) \log (1-e x)\right )+\left (-4-4 e^2 x^2+8 e^2 x^2 \log (x)\right ) \text {Li}_2(e x)\right )}{16 e^2} \end {gather*}

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 = 0.04, size = 0, normalized size = 0.00 \[\int x \left (a +b \ln \left (c \,x^{n}\right )\right ) \polylog \left (2, e x \right )\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [A]
time = 0.37, size = 196, normalized size = 1.06 \begin {gather*} \frac {1}{16} \, {\left ({\left (3 \, b n - 2 \, a\right )} x^{2} e^{2} + 4 \, {\left (2 \, b n - a\right )} x e - 4 \, {\left ({\left (b n - 2 \, a\right )} x^{2} e^{2} + b n\right )} {\rm Li}_2\left (x e\right ) - 4 \, {\left ({\left (b n - a\right )} x^{2} e^{2} - b n + a\right )} \log \left (-x e + 1\right ) + 2 \, {\left (4 \, b x^{2} {\rm Li}_2\left (x e\right ) e^{2} - b x^{2} e^{2} - 2 \, b x e + 2 \, {\left (b x^{2} e^{2} - b\right )} \log \left (-x e + 1\right )\right )} \log \left (c\right ) + 2 \, {\left (4 \, b n x^{2} {\rm Li}_2\left (x e\right ) e^{2} - b n x^{2} e^{2} - 2 \, b n x e + 2 \, {\left (b n x^{2} e^{2} - b n\right )} \log \left (-x e + 1\right )\right )} \log \left (x\right )\right )} e^{\left (-2\right )} \end {gather*}

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

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Sympy [A]
time = 24.94, size = 206, normalized size = 1.11 \begin {gather*} \begin {cases} - \frac {a x^{2} \operatorname {Li}_{1}\left (e x\right )}{4} + \frac {a x^{2} \operatorname {Li}_{2}\left (e x\right )}{2} - \frac {a x^{2}}{8} - \frac {a x}{4 e} + \frac {a \operatorname {Li}_{1}\left (e x\right )}{4 e^{2}} + \frac {b n x^{2} \operatorname {Li}_{1}\left (e x\right )}{4} - \frac {b n x^{2} \operatorname {Li}_{2}\left (e x\right )}{4} + \frac {3 b n x^{2}}{16} - \frac {b x^{2} \log {\left (c x^{n} \right )} \operatorname {Li}_{1}\left (e x\right )}{4} + \frac {b x^{2} \log {\left (c x^{n} \right )} \operatorname {Li}_{2}\left (e x\right )}{2} - \frac {b x^{2} \log {\left (c x^{n} \right )}}{8} + \frac {b n x}{2 e} - \frac {b x \log {\left (c x^{n} \right )}}{4 e} - \frac {b n \operatorname {Li}_{1}\left (e x\right )}{4 e^{2}} - \frac {b n \operatorname {Li}_{2}\left (e x\right )}{4 e^{2}} + \frac {b \log {\left (c x^{n} \right )} \operatorname {Li}_{1}\left (e x\right )}{4 e^{2}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

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]

Piecewise((-a*x**2*polylog(1, e*x)/4 + a*x**2*polylog(2, e*x)/2 - a*x**2/8 - a*x/(4*e) + a*polylog(1, e*x)/(4*
e**2) + b*n*x**2*polylog(1, e*x)/4 - b*n*x**2*polylog(2, e*x)/4 + 3*b*n*x**2/16 - b*x**2*log(c*x**n)*polylog(1
, e*x)/4 + b*x**2*log(c*x**n)*polylog(2, e*x)/2 - b*x**2*log(c*x**n)/8 + b*n*x/(2*e) - b*x*log(c*x**n)/(4*e) -
b*n*polylog(1, e*x)/(4*e**2) - b*n*polylog(2, e*x)/(4*e**2) + b*log(c*x**n)*polylog(1, e*x)/(4*e**2), Ne(e, 0
)), (0, True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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(x*e), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,\mathrm {polylog}\left (2,e\,x\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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