3.208 \(\int x^2 (a+b \log (c x^n)) \text {Li}_2(e x) \, dx\)

Optimal. Leaf size=217 \[ -\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{9 e^2}+\frac {1}{3} x^3 \text {Li}_2(e x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{9} x^3 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{18 e}-\frac {1}{27} x^3 \left (a+b \log \left (c x^n\right )\right )-\frac {b n \text {Li}_2(e x)}{9 e^3}+\frac {2 b n \log (1-e x)}{27 e^3}+\frac {5 b n x}{27 e^2}-\frac {1}{9} b n x^3 \text {Li}_2(e x)-\frac {2}{27} b n x^3 \log (1-e x)+\frac {7 b n x^2}{108 e}+\frac {1}{27} b n x^3 \]

[Out]

5/27*b*n*x/e^2+7/108*b*n*x^2/e+1/27*b*n*x^3-1/9*x*(a+b*ln(c*x^n))/e^2-1/18*x^2*(a+b*ln(c*x^n))/e-1/27*x^3*(a+b
*ln(c*x^n))+2/27*b*n*ln(-e*x+1)/e^3-2/27*b*n*x^3*ln(-e*x+1)-1/9*(a+b*ln(c*x^n))*ln(-e*x+1)/e^3+1/9*x^3*(a+b*ln
(c*x^n))*ln(-e*x+1)-1/9*b*n*polylog(2,e*x)/e^3-1/9*b*n*x^3*polylog(2,e*x)+1/3*x^3*(a+b*ln(c*x^n))*polylog(2,e*
x)

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Rubi [A]  time = 0.18, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2385, 2395, 43, 2376, 2391} \[ \frac {1}{3} x^3 \text {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )-\frac {b n \text {PolyLog}(2,e x)}{9 e^3}-\frac {1}{9} b n x^3 \text {PolyLog}(2,e x)-\frac {x \left (a+b \log \left (c x^n\right )\right )}{9 e^2}-\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{9 e^3}+\frac {1}{9} x^3 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{18 e}-\frac {1}{27} x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {5 b n x}{27 e^2}+\frac {2 b n \log (1-e x)}{27 e^3}+\frac {7 b n x^2}{108 e}-\frac {2}{27} b n x^3 \log (1-e x)+\frac {1}{27} b n x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b*n*x)/(27*e^2) + (7*b*n*x^2)/(108*e) + (b*n*x^3)/27 - (x*(a + b*Log[c*x^n]))/(9*e^2) - (x^2*(a + b*Log[c*x
^n]))/(18*e) - (x^3*(a + b*Log[c*x^n]))/27 + (2*b*n*Log[1 - e*x])/(27*e^3) - (2*b*n*x^3*Log[1 - e*x])/27 - ((a
+ b*Log[c*x^n])*Log[1 - e*x])/(9*e^3) + (x^3*(a + b*Log[c*x^n])*Log[1 - e*x])/9 - (b*n*PolyLog[2, e*x])/(9*e^
3) - (b*n*x^3*PolyLog[2, e*x])/9 + (x^3*(a + b*Log[c*x^n])*PolyLog[2, e*x])/3

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 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]

Rubi steps

\begin {align*} \int x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x) \, dx &=-\frac {1}{9} b n x^3 \text {Li}_2(e x)+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)+\frac {1}{3} \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x) \, dx-\frac {1}{9} (b n) \int x^2 \log (1-e x) \, dx\\ &=-\frac {x \left (a+b \log \left (c x^n\right )\right )}{9 e^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{18 e}-\frac {1}{27} x^3 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{27} b n x^3 \log (1-e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{9 e^3}+\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac {1}{9} b n x^3 \text {Li}_2(e x)+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{3} (b n) \int \left (-\frac {1}{3 e^2}-\frac {x}{6 e}-\frac {x^2}{9}-\frac {\log (1-e x)}{3 e^3 x}+\frac {1}{3} x^2 \log (1-e x)\right ) \, dx-\frac {1}{27} (b e n) \int \frac {x^3}{1-e x} \, dx\\ &=\frac {b n x}{9 e^2}+\frac {b n x^2}{36 e}+\frac {1}{81} b n x^3-\frac {x \left (a+b \log \left (c x^n\right )\right )}{9 e^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{18 e}-\frac {1}{27} x^3 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{27} b n x^3 \log (1-e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{9 e^3}+\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac {1}{9} b n x^3 \text {Li}_2(e x)+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{9} (b n) \int x^2 \log (1-e x) \, dx+\frac {(b n) \int \frac {\log (1-e x)}{x} \, dx}{9 e^3}-\frac {1}{27} (b e n) \int \left (-\frac {1}{e^3}-\frac {x}{e^2}-\frac {x^2}{e}-\frac {1}{e^3 (-1+e x)}\right ) \, dx\\ &=\frac {4 b n x}{27 e^2}+\frac {5 b n x^2}{108 e}+\frac {2}{81} b n x^3-\frac {x \left (a+b \log \left (c x^n\right )\right )}{9 e^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{18 e}-\frac {1}{27} x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1-e x)}{27 e^3}-\frac {2}{27} b n x^3 \log (1-e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{9 e^3}+\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac {b n \text {Li}_2(e x)}{9 e^3}-\frac {1}{9} b n x^3 \text {Li}_2(e x)+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{27} (b e n) \int \frac {x^3}{1-e x} \, dx\\ &=\frac {4 b n x}{27 e^2}+\frac {5 b n x^2}{108 e}+\frac {2}{81} b n x^3-\frac {x \left (a+b \log \left (c x^n\right )\right )}{9 e^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{18 e}-\frac {1}{27} x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1-e x)}{27 e^3}-\frac {2}{27} b n x^3 \log (1-e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{9 e^3}+\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac {b n \text {Li}_2(e x)}{9 e^3}-\frac {1}{9} b n x^3 \text {Li}_2(e x)+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{27} (b e n) \int \left (-\frac {1}{e^3}-\frac {x}{e^2}-\frac {x^2}{e}-\frac {1}{e^3 (-1+e x)}\right ) \, dx\\ &=\frac {5 b n x}{27 e^2}+\frac {7 b n x^2}{108 e}+\frac {1}{27} b n x^3-\frac {x \left (a+b \log \left (c x^n\right )\right )}{9 e^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{18 e}-\frac {1}{27} x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {2 b n \log (1-e x)}{27 e^3}-\frac {2}{27} b n x^3 \log (1-e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{9 e^3}+\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)-\frac {b n \text {Li}_2(e x)}{9 e^3}-\frac {1}{9} b n x^3 \text {Li}_2(e x)+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 196, normalized size = 0.90 \[ \frac {\left (18 e^3 x^3 \text {Li}_2(e x)+6 \left (e^3 x^3-1\right ) \log (1-e x)-e x \left (2 e^2 x^2+3 e x+6\right )\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{54 e^3}+\frac {b n \left (12 \text {Li}_2(e x) \left (-e^3 x^3+3 e^3 x^3 \log (x)-1\right )+4 e^3 x^3-8 e^3 x^3 \log (1-e x)+7 e^2 x^2+2 \log (x) \left (6 \left (e^3 x^3-1\right ) \log (1-e x)-e x \left (2 e^2 x^2+3 e x+6\right )\right )+20 e x+8 \log (1-e x)\right )}{108 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a - b*n*Log[x] + b*Log[c*x^n])*(-(e*x*(6 + 3*e*x + 2*e^2*x^2)) + 6*(-1 + e^3*x^3)*Log[1 - e*x] + 18*e^3*x^3*
PolyLog[2, e*x]))/(54*e^3) + (b*n*(20*e*x + 7*e^2*x^2 + 4*e^3*x^3 + 8*Log[1 - e*x] - 8*e^3*x^3*Log[1 - e*x] +
2*Log[x]*(-(e*x*(6 + 3*e*x + 2*e^2*x^2)) + 6*(-1 + e^3*x^3)*Log[1 - e*x]) + 12*(-1 - e^3*x^3 + 3*e^3*x^3*Log[x
])*PolyLog[2, e*x]))/(108*e^3)

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fricas [A]  time = 0.69, size = 247, normalized size = 1.14 \[ \frac {4 \, {\left (b e^{3} n - a e^{3}\right )} x^{3} + {\left (7 \, b e^{2} n - 6 \, a e^{2}\right )} x^{2} + 4 \, {\left (5 \, b e n - 3 \, a e\right )} x - 12 \, {\left ({\left (b e^{3} n - 3 \, a e^{3}\right )} x^{3} + b n\right )} {\rm Li}_2\left (e x\right ) - 4 \, {\left ({\left (2 \, b e^{3} n - 3 \, a e^{3}\right )} x^{3} - 2 \, b n + 3 \, a\right )} \log \left (-e x + 1\right ) + 2 \, {\left (18 \, b e^{3} x^{3} {\rm Li}_2\left (e x\right ) - 2 \, b e^{3} x^{3} - 3 \, b e^{2} x^{2} - 6 \, b e x + 6 \, {\left (b e^{3} x^{3} - b\right )} \log \left (-e x + 1\right )\right )} \log \relax (c) + 2 \, {\left (18 \, b e^{3} n x^{3} {\rm Li}_2\left (e x\right ) - 2 \, b e^{3} n x^{3} - 3 \, b e^{2} n x^{2} - 6 \, b e n x + 6 \, {\left (b e^{3} n x^{3} - b n\right )} \log \left (-e x + 1\right )\right )} \log \relax (x)}{108 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/108*(4*(b*e^3*n - a*e^3)*x^3 + (7*b*e^2*n - 6*a*e^2)*x^2 + 4*(5*b*e*n - 3*a*e)*x - 12*((b*e^3*n - 3*a*e^3)*x
^3 + b*n)*dilog(e*x) - 4*((2*b*e^3*n - 3*a*e^3)*x^3 - 2*b*n + 3*a)*log(-e*x + 1) + 2*(18*b*e^3*x^3*dilog(e*x)
- 2*b*e^3*x^3 - 3*b*e^2*x^2 - 6*b*e*x + 6*(b*e^3*x^3 - b)*log(-e*x + 1))*log(c) + 2*(18*b*e^3*n*x^3*dilog(e*x)
- 2*b*e^3*n*x^3 - 3*b*e^2*n*x^2 - 6*b*e*n*x + 6*(b*e^3*n*x^3 - b*n)*log(-e*x + 1))*log(x))/e^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} {\rm Li}_2\left (e x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \left (b \ln \left (c \,x^{n}\right )+a \right ) x^{2} \polylog \left (2, e x \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{54} \, b {\left (\frac {6 \, {\left (3 \, e^{3} x^{3} \log \left (x^{n}\right ) - {\left (e^{3} n - 3 \, e^{3} \log \relax (c)\right )} x^{3}\right )} {\rm Li}_2\left (e x\right ) - 2 \, {\left ({\left (2 \, e^{3} n - 3 \, e^{3} \log \relax (c)\right )} x^{3} - 3 \, n \log \relax (x)\right )} \log \left (-e x + 1\right ) - {\left (2 \, e^{3} x^{3} + 3 \, e^{2} x^{2} + 6 \, e x - 6 \, {\left (e^{3} x^{3} - 1\right )} \log \left (-e x + 1\right )\right )} \log \left (x^{n}\right )}{e^{3}} - 54 \, \int -\frac {e^{2} n x^{2} + 6 \, {\left (e^{3} n - e^{3} \log \relax (c)\right )} x^{3} + 3 \, e n x - 6 \, n \log \relax (x) - 6 \, n}{54 \, {\left (e^{3} x - e^{2}\right )}}\,{d x}\right )} + \frac {{\left (18 \, e^{3} x^{3} {\rm Li}_2\left (e x\right ) - 2 \, e^{3} x^{3} - 3 \, e^{2} x^{2} - 6 \, e x + 6 \, {\left (e^{3} x^{3} - 1\right )} \log \left (-e x + 1\right )\right )} a}{54 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/54*b*((6*(3*e^3*x^3*log(x^n) - (e^3*n - 3*e^3*log(c))*x^3)*dilog(e*x) - 2*((2*e^3*n - 3*e^3*log(c))*x^3 - 3*
n*log(x))*log(-e*x + 1) - (2*e^3*x^3 + 3*e^2*x^2 + 6*e*x - 6*(e^3*x^3 - 1)*log(-e*x + 1))*log(x^n))/e^3 - 54*i
ntegrate(-1/54*(e^2*n*x^2 + 6*(e^3*n - e^3*log(c))*x^3 + 3*e*n*x - 6*n*log(x) - 6*n)/(e^3*x - e^2), x)) + 1/54
*(18*e^3*x^3*dilog(e*x) - 2*e^3*x^3 - 3*e^2*x^2 - 6*e*x + 6*(e^3*x^3 - 1)*log(-e*x + 1))*a/e^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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