3.3.19 \(\int \frac {(a+b \log (c x^n)) \text {Li}_3(e x)}{x^3} \, dx\) [219]

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

[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 = 0.16, 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 = {2432, 2442, 46, 2423, 2338, 2438, 6726} \begin {gather*} -\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} \end {gather*}

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 46

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] && Lt
Q[m + n + 2, 0])

Rule 2338

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 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]

Rule 6726

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 = 0.09, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)}{x^3} \, dx \end {gather*}

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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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \ln \left (c \,x^{n}\right )\right ) \polylog \left (3, e x \right )}{x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 217, normalized size = 0.91 \begin {gather*} \frac {b n x^{2} e^{2} \log \left (x\right )^{2} - {\left (5 \, b n + 2 \, a\right )} x e - 2 \, {\left (b n x^{2} e^{2} + 2 \, b n + 2 \, a\right )} {\rm Li}_2\left (x e\right ) - {\left ({\left (3 \, b n + 2 \, a\right )} x^{2} e^{2} - 3 \, b n - 2 \, a\right )} \log \left (-x e + 1\right ) - 2 \, {\left (b x e + 2 \, b {\rm Li}_2\left (x e\right ) + {\left (b x^{2} e^{2} - b\right )} \log \left (-x e + 1\right )\right )} \log \left (c\right ) + {\left (2 \, b x^{2} e^{2} \log \left (c\right ) + {\left (3 \, b n + 2 \, a\right )} x^{2} e^{2} - 2 \, b n x e - 4 \, b n {\rm Li}_2\left (x e\right ) - 2 \, {\left (b n x^{2} e^{2} - b n\right )} \log \left (-x e + 1\right )\right )} \log \left (x\right ) - 4 \, {\left (2 \, b n \log \left (x\right ) + b n + 2 \, b \log \left (c\right ) + 2 \, a\right )} {\rm polylog}\left (3, x e\right )}{16 \, x^{2}} \end {gather*}

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

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

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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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((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, x*e)/x^3, x)

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

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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