3.132 \(\int x \text{PolyLog}(3,c (a+b x)) \, dx\)

Optimal. Leaf size=198 \[ -\frac{\left (a^2-b^2 x^2\right ) \text{PolyLog}(3,c (a+b x))}{2 b^2}+\frac{3 a^2 \text{PolyLog}(2,c (a+b x))}{4 b^2}-\frac{1}{4} x^2 \text{PolyLog}(2,c (a+b x))+\frac{a x \text{PolyLog}(2,c (a+b x))}{2 b}+\frac{(1-a c)^2 \log (-a c-b c x+1)}{8 b^2 c^2}-\frac{3 a (-a c-b c x+1) \log (-a c-b c x+1)}{4 b^2 c}-\frac{1}{8} x^2 \log (-a c-b c x+1)+\frac{x (1-a c)}{8 b c}-\frac{3 a x}{4 b}+\frac{x^2}{16} \]

[Out]

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

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Rubi [A]  time = 0.2869, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 12, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.091, Rules used = {6599, 6595, 2444, 2389, 2295, 2421, 2393, 2391, 6598, 43, 2416, 2395} \[ -\frac{\left (a^2-b^2 x^2\right ) \text{PolyLog}(3,c (a+b x))}{2 b^2}+\frac{3 a^2 \text{PolyLog}(2,c (a+b x))}{4 b^2}-\frac{1}{4} x^2 \text{PolyLog}(2,c (a+b x))+\frac{a x \text{PolyLog}(2,c (a+b x))}{2 b}+\frac{(1-a c)^2 \log (-a c-b c x+1)}{8 b^2 c^2}-\frac{3 a (-a c-b c x+1) \log (-a c-b c x+1)}{4 b^2 c}-\frac{1}{8} x^2 \log (-a c-b c x+1)+\frac{x (1-a c)}{8 b c}-\frac{3 a x}{4 b}+\frac{x^2}{16} \]

Antiderivative was successfully verified.

[In]

Int[x*PolyLog[3, c*(a + b*x)],x]

[Out]

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

Rule 6599

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

Rule 6595

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

Rule 2444

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2421

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*(a + b*Log[c*
ExpandToSum[v, x]^n])^p, x] /; FreeQ[{a, b, c, n, p, q}, x] && BinomialQ[u, x] && LinearQ[v, x] &&  !(Binomial
MatchQ[u, x] && LinearMatchQ[v, x])

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 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 6598

Int[((d_.) + (e_.)*(x_))^(m_.)*PolyLog[2, (c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> Simp[((d + e*x)^(m + 1)*Po
lyLog[2, c*(a + b*x)])/(e*(m + 1)), x] + Dist[b/(e*(m + 1)), Int[((d + e*x)^(m + 1)*Log[1 - a*c - b*c*x])/(a +
b*x), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[m, -1]

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 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

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

Mathematica [A]  time = 0.0466226, size = 198, normalized size = 1. \[ \frac{4 c^2 \left (3 a^2+2 a b x-b^2 x^2\right ) \text{PolyLog}(2,c (a+b x))-8 c^2 \left (a^2-b^2 x^2\right ) \text{PolyLog}(3,c (a+b x))+14 a^2 c^2 \log (-a c-b c x+1)-15 a^2 c^2-2 b^2 c^2 x^2 \log (-a c-b c x+1)-14 a b c^2 x+12 a b c^2 x \log (-a c-b c x+1)-16 a c \log (-a c-b c x+1)+2 \log (-a c-b c x+1)+2 a c+b^2 c^2 x^2+2 b c x}{16 b^2 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*PolyLog[3, c*(a + b*x)],x]

[Out]

(2*a*c - 15*a^2*c^2 + 2*b*c*x - 14*a*b*c^2*x + b^2*c^2*x^2 + 2*Log[1 - a*c - b*c*x] - 16*a*c*Log[1 - a*c - b*c
*x] + 14*a^2*c^2*Log[1 - a*c - b*c*x] + 12*a*b*c^2*x*Log[1 - a*c - b*c*x] - 2*b^2*c^2*x^2*Log[1 - a*c - b*c*x]
+ 4*c^2*(3*a^2 + 2*a*b*x - b^2*x^2)*PolyLog[2, c*(a + b*x)] - 8*c^2*(a^2 - b^2*x^2)*PolyLog[3, c*(a + b*x)])/
(16*b^2*c^2)

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Maple [F]  time = 0.006, size = 0, normalized size = 0. \begin{align*} \int x{\it polylog} \left ( 3,c \left ( bx+a \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*polylog(3,c*(b*x+a)),x)

[Out]

int(x*polylog(3,c*(b*x+a)),x)

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Maxima [A]  time = 1.01089, size = 261, normalized size = 1.32 \begin{align*} -\frac{3 \,{\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) +{\rm Li}_2\left (-b c x - a c + 1\right )\right )} a^{2}}{4 \, b^{2}} - \frac{a^{2}{\rm Li}_{3}(b c x + a c)}{2 \, b^{2}} + \frac{8 \, b^{2} c^{2} x^{2}{\rm Li}_{3}(b c x + a c) + b^{2} c^{2} x^{2} - 2 \,{\left (7 \, a b c^{2} - b c\right )} x - 4 \,{\left (b^{2} c^{2} x^{2} - 2 \, a b c^{2} x\right )}{\rm Li}_2\left (b c x + a c\right ) - 2 \,{\left (b^{2} c^{2} x^{2} - 6 \, a b c^{2} x - 7 \, a^{2} c^{2} + 8 \, a c - 1\right )} \log \left (-b c x - a c + 1\right )}{16 \, b^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*polylog(3,c*(b*x+a)),x, algorithm="maxima")

[Out]

-3/4*(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*a^2/b^2 - 1/2*a^2*polylog(3, b*c*x + a
*c)/b^2 + 1/16*(8*b^2*c^2*x^2*polylog(3, b*c*x + a*c) + b^2*c^2*x^2 - 2*(7*a*b*c^2 - b*c)*x - 4*(b^2*c^2*x^2 -
2*a*b*c^2*x)*dilog(b*c*x + a*c) - 2*(b^2*c^2*x^2 - 6*a*b*c^2*x - 7*a^2*c^2 + 8*a*c - 1)*log(-b*c*x - a*c + 1)
)/(b^2*c^2)

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Fricas [C]  time = 2.51134, size = 510, normalized size = 2.58 \begin{align*} \frac{b^{2} c^{2} x^{2} - 2 \,{\left (7 \, a b c^{2} - b c\right )} x - 4 \,{\left (b^{2} c^{2} x^{2} - 2 \, a b c^{2} x - 3 \, a^{2} c^{2}\right )}{\rm \%iint}\left (a, b, c, x, -\frac{\log \left (-b c x - a c + 1\right )}{b x + a}, -\frac{x \log \left (-b c x - a c + 1\right )}{b x + a}, -\frac{\log \left (-b c x - a c + 1\right )}{c}, -\frac{b \log \left (-b c x - a c + 1\right )}{b x + a}\right ) - 2 \,{\left (b^{2} c^{2} x^{2} - 6 \, a b c^{2} x - 7 \, a^{2} c^{2} + 8 \, a c - 1\right )} \log \left (-b c x - a c + 1\right ) + 8 \,{\left (b^{2} c^{2} x^{2} - a^{2} c^{2}\right )}{\rm polylog}\left (3, b c x + a c\right )}{16 \, b^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*polylog(3,c*(b*x+a)),x, algorithm="fricas")

[Out]

1/16*(b^2*c^2*x^2 - 2*(7*a*b*c^2 - b*c)*x - 4*(b^2*c^2*x^2 - 2*a*b*c^2*x - 3*a^2*c^2)*\%iint(a, b, c, x, -log(-
b*c*x - a*c + 1)/(b*x + a), -x*log(-b*c*x - a*c + 1)/(b*x + a), -log(-b*c*x - a*c + 1)/c, -b*log(-b*c*x - a*c
+ 1)/(b*x + a)) - 2*(b^2*c^2*x^2 - 6*a*b*c^2*x - 7*a^2*c^2 + 8*a*c - 1)*log(-b*c*x - a*c + 1) + 8*(b^2*c^2*x^2
- a^2*c^2)*polylog(3, b*c*x + a*c))/(b^2*c^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{Li}_{3}\left (a c + b c x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\rm Li}_{3}({\left (b x + a\right )} c)\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*polylog(3,c*(b*x+a)),x, algorithm="giac")

[Out]

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