3.3.15 \(\int x (a+b \log (c x^n)) \text {Li}_3(e x) \, dx\) [215]

Optimal. Leaf size=221 \[ -\frac {5 b n x}{16 e}-\frac {1}{8} b n x^2+\frac {x \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {3 b n \log (1-e x)}{16 e^2}+\frac {3}{16} b n x^2 \log (1-e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 e^2}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+\frac {b n \text {Li}_2(e x)}{8 e^2}+\frac {1}{4} b n x^2 \text {Li}_2(e x)-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2432, 2442, 45, 2423, 2438, 6726} \begin {gather*} -\frac {1}{4} x^2 \text {PolyLog}(2,e x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} x^2 \text {PolyLog}(3,e x) \left (a+b \log \left (c x^n\right )\right )+\frac {b n \text {PolyLog}(2,e x)}{8 e^2}+\frac {1}{4} b n x^2 \text {PolyLog}(2,e x)-\frac {1}{4} b n x^2 \text {PolyLog}(3,e x)+\frac {\log (1-e x) \left (a+b \log \left (c x^n\right )\right )}{8 e^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{8 e}-\frac {1}{8} x^2 \log (1-e x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {3 b n \log (1-e x)}{16 e^2}+\frac {3}{16} b n x^2 \log (1-e x)-\frac {5 b n x}{16 e}-\frac {1}{8} b n x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 x \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x) \, dx &=-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)-\frac {1}{2} \int x \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x) \, dx+\frac {1}{4} (b n) \int x \text {Li}_2(e x) \, dx\\ &=\frac {1}{4} b n x^2 \text {Li}_2(e x)-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)-\frac {1}{4} \int x \left (a+b \log \left (c x^n\right )\right ) \log (1-e x) \, dx+2 \left (\frac {1}{8} (b n) \int x \log (1-e x) \, dx\right )\\ &=\frac {x \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 e^2}-\frac {1}{8} 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}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)+\frac {1}{4} (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+2 \left (\frac {1}{16} b n x^2 \log (1-e x)+\frac {1}{16} (b e n) \int \frac {x^2}{1-e x} \, dx\right )\\ &=-\frac {b n x}{8 e}-\frac {1}{32} b n x^2+\frac {x \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 e^2}-\frac {1}{8} 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}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)+\frac {1}{8} (b n) \int x \log (1-e x) \, dx-\frac {(b n) \int \frac {\log (1-e x)}{x} \, dx}{8 e^2}+2 \left (\frac {1}{16} b n x^2 \log (1-e x)+\frac {1}{16} (b e n) \int \left (-\frac {1}{e^2}-\frac {x}{e}-\frac {1}{e^2 (-1+e x)}\right ) \, dx\right )\\ &=-\frac {b n x}{8 e}-\frac {1}{32} b n x^2+\frac {x \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{16} b n x^2 \log (1-e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 e^2}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+2 \left (-\frac {b n x}{16 e}-\frac {1}{32} b n x^2-\frac {b n \log (1-e x)}{16 e^2}+\frac {1}{16} b n x^2 \log (1-e x)\right )+\frac {b n \text {Li}_2(e x)}{8 e^2}+\frac {1}{4} b n x^2 \text {Li}_2(e x)-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)+\frac {1}{16} (b e n) \int \frac {x^2}{1-e x} \, dx\\ &=-\frac {b n x}{8 e}-\frac {1}{32} b n x^2+\frac {x \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{16} b n x^2 \log (1-e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 e^2}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+2 \left (-\frac {b n x}{16 e}-\frac {1}{32} b n x^2-\frac {b n \log (1-e x)}{16 e^2}+\frac {1}{16} b n x^2 \log (1-e x)\right )+\frac {b n \text {Li}_2(e x)}{8 e^2}+\frac {1}{4} b n x^2 \text {Li}_2(e x)-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)+\frac {1}{16} (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}{16 e}-\frac {1}{16} b n x^2+\frac {x \left (a+b \log \left (c x^n\right )\right )}{8 e}+\frac {1}{16} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {b n \log (1-e x)}{16 e^2}+\frac {1}{16} b n x^2 \log (1-e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1-e x)}{8 e^2}-\frac {1}{8} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1-e x)+2 \left (-\frac {b n x}{16 e}-\frac {1}{32} b n x^2-\frac {b n \log (1-e x)}{16 e^2}+\frac {1}{16} b n x^2 \log (1-e x)\right )+\frac {b n \text {Li}_2(e x)}{8 e^2}+\frac {1}{4} b n x^2 \text {Li}_2(e x)-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2(e x)-\frac {1}{4} b n x^2 \text {Li}_3(e x)+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3(e x)\\ \end {align*}

________________________________________________________________________________________

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

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int x \left (a +b \ln \left (c \,x^{n}\right )\right ) \polylog \left (3, e x \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(3,e*x),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 241, normalized size = 1.09 \begin {gather*} -\frac {1}{16} \, {\left ({\left (2 \, b n - a\right )} x^{2} e^{2} + {\left (5 \, b n - 2 \, a\right )} x e - 2 \, {\left (2 \, {\left (b n - a\right )} x^{2} e^{2} + b n\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 ) + {\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 ) + {\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 ) - 4 \, {\left (2 \, b n x^{2} e^{2} \log \left (x\right ) + 2 \, b x^{2} e^{2} \log \left (c\right ) - {\left (b n - 2 \, a\right )} x^{2} e^{2}\right )} {\rm polylog}\left (3, x e\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(3,e*x),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(3,e*x),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

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

[Out]

\text{Hanged}

________________________________________________________________________________________