3.219 \(\int \frac {(a+b \log (c x^n)) \text {Li}_3(e x)}{x^3} \, dx\)

Optimal. Leaf size=238 \[ \frac {1}{8} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{8} e^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {\text {Li}_2(e x) \left (a+b \log \left (c x^n\right )\right )}{4 x^2}-\frac {\text {Li}_3(e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{8 x}+\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{8 x^2}-\frac {1}{8} b e^2 n \text {Li}_2(e x)-\frac {1}{16} b e^2 n \log ^2(x)+\frac {3}{16} b e^2 n \log (x)-\frac {3}{16} b e^2 n \log (1-e x)-\frac {b n \text {Li}_2(e x)}{4 x^2}-\frac {b n \text {Li}_3(e x)}{4 x^2}+\frac {3 b n \log (1-e x)}{16 x^2}-\frac {5 b e n}{16 x} \]

[Out]

-5/16*b*e*n/x+3/16*b*e^2*n*ln(x)-1/16*b*e^2*n*ln(x)^2-1/8*e*(a+b*ln(c*x^n))/x+1/8*e^2*ln(x)*(a+b*ln(c*x^n))-3/
16*b*e^2*n*ln(-e*x+1)+3/16*b*n*ln(-e*x+1)/x^2-1/8*e^2*(a+b*ln(c*x^n))*ln(-e*x+1)+1/8*(a+b*ln(c*x^n))*ln(-e*x+1
)/x^2-1/8*b*e^2*n*polylog(2,e*x)-1/4*b*n*polylog(2,e*x)/x^2-1/4*(a+b*ln(c*x^n))*polylog(2,e*x)/x^2-1/4*b*n*pol
ylog(3,e*x)/x^2-1/2*(a+b*ln(c*x^n))*polylog(3,e*x)/x^2

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Rubi [A]  time = 0.22, antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2385, 2395, 44, 2376, 2301, 2391, 6591} \[ -\frac {\text {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )}{4 x^2}-\frac {\text {PolyLog}(3,e x) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {1}{8} b e^2 n \text {PolyLog}(2,e x)-\frac {b n \text {PolyLog}(2,e x)}{4 x^2}-\frac {b n \text {PolyLog}(3,e x)}{4 x^2}+\frac {1}{8} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{8} e^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {e \left (a+b \log \left (c x^n\right )\right )}{8 x}+\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{8 x^2}-\frac {1}{16} b e^2 n \log ^2(x)+\frac {3}{16} b e^2 n \log (x)-\frac {3}{16} b e^2 n \log (1-e x)+\frac {3 b n \log (1-e x)}{16 x^2}-\frac {5 b e n}{16 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b*e*n)/(16*x) + (3*b*e^2*n*Log[x])/16 - (b*e^2*n*Log[x]^2)/16 - (e*(a + b*Log[c*x^n]))/(8*x) + (e^2*Log[x]
*(a + b*Log[c*x^n]))/8 - (3*b*e^2*n*Log[1 - e*x])/16 + (3*b*n*Log[1 - e*x])/(16*x^2) - (e^2*(a + b*Log[c*x^n])
*Log[1 - e*x])/8 + ((a + b*Log[c*x^n])*Log[1 - e*x])/(8*x^2) - (b*e^2*n*PolyLog[2, e*x])/8 - (b*n*PolyLog[2, e
*x])/(4*x^2) - ((a + b*Log[c*x^n])*PolyLog[2, e*x])/(4*x^2) - (b*n*PolyLog[3, e*x])/(4*x^2) - ((a + b*Log[c*x^
n])*PolyLog[3, e*x])/(2*x^2)

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

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 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 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]

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 6591

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

Rubi steps

\begin {align*} \int \frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)}{x^3} \, dx &=-\frac {b n \text {Li}_3(e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)}{2 x^2}+\frac {1}{2} \int \frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)}{x^3} \, dx+\frac {1}{4} (b n) \int \frac {\text {Li}_2(e x)}{x^3} \, dx\\ &=-\frac {b n \text {Li}_2(e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)}{4 x^2}-\frac {b n \text {Li}_3(e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)}{2 x^2}-\frac {1}{4} \int \frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{x^3} \, dx-2 \left (\frac {1}{8} (b n) \int \frac {\log (1-e x)}{x^3} \, dx\right )\\ &=-\frac {e \left (a+b \log \left (c x^n\right )\right )}{8 x}+\frac {1}{8} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{8} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 x^2}-\frac {b n \text {Li}_2(e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)}{4 x^2}-\frac {b n \text {Li}_3(e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)}{2 x^2}+\frac {1}{4} (b n) \int \left (\frac {e}{2 x^2}-\frac {e^2 \log (x)}{2 x}-\frac {\log (1-e x)}{2 x^3}+\frac {e^2 \log (1-e x)}{2 x}\right ) \, dx-2 \left (-\frac {b n \log (1-e x)}{16 x^2}-\frac {1}{16} (b e n) \int \frac {1}{x^2 (1-e x)} \, dx\right )\\ &=-\frac {b e n}{8 x}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{8 x}+\frac {1}{8} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{8} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 x^2}-\frac {b n \text {Li}_2(e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)}{4 x^2}-\frac {b n \text {Li}_3(e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)}{2 x^2}-\frac {1}{8} (b n) \int \frac {\log (1-e x)}{x^3} \, dx-2 \left (-\frac {b n \log (1-e x)}{16 x^2}-\frac {1}{16} (b e n) \int \left (\frac {1}{x^2}+\frac {e}{x}-\frac {e^2}{-1+e x}\right ) \, dx\right )-\frac {1}{8} \left (b e^2 n\right ) \int \frac {\log (x)}{x} \, dx+\frac {1}{8} \left (b e^2 n\right ) \int \frac {\log (1-e x)}{x} \, dx\\ &=-\frac {b e n}{8 x}-\frac {1}{16} b e^2 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{8 x}+\frac {1}{8} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1-e x)}{16 x^2}-\frac {1}{8} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 x^2}-2 \left (\frac {b e n}{16 x}-\frac {1}{16} b e^2 n \log (x)+\frac {1}{16} b e^2 n \log (1-e x)-\frac {b n \log (1-e x)}{16 x^2}\right )-\frac {1}{8} b e^2 n \text {Li}_2(e x)-\frac {b n \text {Li}_2(e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)}{4 x^2}-\frac {b n \text {Li}_3(e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)}{2 x^2}+\frac {1}{16} (b e n) \int \frac {1}{x^2 (1-e x)} \, dx\\ &=-\frac {b e n}{8 x}-\frac {1}{16} b e^2 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{8 x}+\frac {1}{8} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1-e x)}{16 x^2}-\frac {1}{8} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 x^2}-2 \left (\frac {b e n}{16 x}-\frac {1}{16} b e^2 n \log (x)+\frac {1}{16} b e^2 n \log (1-e x)-\frac {b n \log (1-e x)}{16 x^2}\right )-\frac {1}{8} b e^2 n \text {Li}_2(e x)-\frac {b n \text {Li}_2(e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)}{4 x^2}-\frac {b n \text {Li}_3(e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)}{2 x^2}+\frac {1}{16} (b e n) \int \left (\frac {1}{x^2}+\frac {e}{x}-\frac {e^2}{-1+e x}\right ) \, dx\\ &=-\frac {3 b e n}{16 x}+\frac {1}{16} b e^2 n \log (x)-\frac {1}{16} b e^2 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{8 x}+\frac {1}{8} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{16} b e^2 n \log (1-e x)+\frac {b n \log (1-e x)}{16 x^2}-\frac {1}{8} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 x^2}-2 \left (\frac {b e n}{16 x}-\frac {1}{16} b e^2 n \log (x)+\frac {1}{16} b e^2 n \log (1-e x)-\frac {b n \log (1-e x)}{16 x^2}\right )-\frac {1}{8} b e^2 n \text {Li}_2(e x)-\frac {b n \text {Li}_2(e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)}{4 x^2}-\frac {b n \text {Li}_3(e x)}{4 x^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)}{2 x^2}\\ \end {align*}

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Mathematica [F]  time = 0.13, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)}{x^3} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [C]  time = 0.55, size = 221, normalized size = 0.93 \[ \frac {b e^{2} n x^{2} \log \relax (x)^{2} - {\left (5 \, b e n + 2 \, a e\right )} x - 2 \, {\left (b e^{2} n x^{2} + 2 \, b n + 2 \, a\right )} {\rm Li}_2\left (e x\right ) - {\left ({\left (3 \, b e^{2} n + 2 \, a e^{2}\right )} x^{2} - 3 \, b n - 2 \, a\right )} \log \left (-e x + 1\right ) - 2 \, {\left (b e x + 2 \, b {\rm Li}_2\left (e x\right ) + {\left (b e^{2} x^{2} - b\right )} \log \left (-e x + 1\right )\right )} \log \relax (c) + {\left (2 \, b e^{2} x^{2} \log \relax (c) - 2 \, b e n x + {\left (3 \, b e^{2} n + 2 \, a e^{2}\right )} x^{2} - 4 \, b n {\rm Li}_2\left (e x\right ) - 2 \, {\left (b e^{2} n x^{2} - b n\right )} \log \left (-e x + 1\right )\right )} \log \relax (x) - 4 \, {\left (2 \, b n \log \relax (x) + b n + 2 \, b \log \relax (c) + 2 \, a\right )} {\rm polylog}\left (3, e x\right )}{16 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(b*e^2*n*x^2*log(x)^2 - (5*b*e*n + 2*a*e)*x - 2*(b*e^2*n*x^2 + 2*b*n + 2*a)*dilog(e*x) - ((3*b*e^2*n + 2*
a*e^2)*x^2 - 3*b*n - 2*a)*log(-e*x + 1) - 2*(b*e*x + 2*b*dilog(e*x) + (b*e^2*x^2 - b)*log(-e*x + 1))*log(c) +
(2*b*e^2*x^2*log(c) - 2*b*e*n*x + (3*b*e^2*n + 2*a*e^2)*x^2 - 4*b*n*dilog(e*x) - 2*(b*e^2*n*x^2 - b*n)*log(-e*
x + 1))*log(x) - 4*(2*b*n*log(x) + b*n + 2*b*log(c) + 2*a)*polylog(3, e*x))/x^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} {\rm Li}_{3}(e x)}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.46, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) \polylog \left (3, e x \right )}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{8} \, {\left (e^{2} \log \relax (x) - \frac {e x + {\left (e^{2} x^{2} - 1\right )} \log \left (-e x + 1\right ) + 2 \, {\rm Li}_2\left (e x\right ) + 4 \, {\rm Li}_{3}(e x)}{x^{2}}\right )} a - \frac {1}{16} \, b {\left (\frac {4 \, {\left (n + \log \relax (c) + \log \left (x^{n}\right )\right )} {\rm Li}_2\left (e x\right ) - {\left (2 \, e^{2} n x^{2} \log \relax (x) + 3 \, n + 2 \, \log \relax (c)\right )} \log \left (-e x + 1\right ) - 2 \, {\left (e^{2} x^{2} \log \relax (x) - e x - {\left (e^{2} x^{2} - 1\right )} \log \left (-e x + 1\right )\right )} \log \left (x^{n}\right ) + 4 \, {\left (n + 2 \, \log \relax (c) + 2 \, \log \left (x^{n}\right )\right )} {\rm Li}_{3}(e x)}{x^{2}} + 16 \, \int -\frac {2 \, e^{2} n x - 5 \, e n - 2 \, e \log \relax (c) - 2 \, {\left (2 \, e^{3} n x^{2} - e^{2} n x\right )} \log \relax (x)}{16 \, {\left (e x^{3} - x^{2}\right )}}\,{d x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(e^2*log(x) - (e*x + (e^2*x^2 - 1)*log(-e*x + 1) + 2*dilog(e*x) + 4*polylog(3, e*x))/x^2)*a - 1/16*b*((4*(
n + log(c) + log(x^n))*dilog(e*x) - (2*e^2*n*x^2*log(x) + 3*n + 2*log(c))*log(-e*x + 1) - 2*(e^2*x^2*log(x) -
e*x - (e^2*x^2 - 1)*log(-e*x + 1))*log(x^n) + 4*(n + 2*log(c) + 2*log(x^n))*polylog(3, e*x))/x^2 + 16*integrat
e(-1/16*(2*e^2*n*x - 5*e*n - 2*e*log(c) - 2*(2*e^3*n*x^2 - e^2*n*x)*log(x))/(e*x^3 - x^2), x))

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \log {\left (c x^{n} \right )}\right ) \operatorname {Li}_{3}\left (e x\right )}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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