3.125 \(\int x \text{PolyLog}(2,c (a+b x)) \, dx\)

Optimal. Leaf size=152 \[ -\frac{a^2 \text{PolyLog}(2,c (a+b x))}{2 b^2}+\frac{1}{2} x^2 \text{PolyLog}(2,c (a+b x))-\frac{(1-a c)^2 \log (-a c-b c x+1)}{4 b^2 c^2}+\frac{a (-a c-b c x+1) \log (-a c-b c x+1)}{2 b^2 c}+\frac{1}{4} x^2 \log (-a c-b c x+1)-\frac{x (1-a c)}{4 b c}+\frac{a x}{2 b}-\frac{x^2}{8} \]

[Out]

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

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Rubi [A]  time = 0.169219, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.727, Rules used = {6598, 43, 2416, 2389, 2295, 2395, 2393, 2391} \[ -\frac{a^2 \text{PolyLog}(2,c (a+b x))}{2 b^2}+\frac{1}{2} x^2 \text{PolyLog}(2,c (a+b x))-\frac{(1-a c)^2 \log (-a c-b c x+1)}{4 b^2 c^2}+\frac{a (-a c-b c x+1) \log (-a c-b c x+1)}{2 b^2 c}+\frac{1}{4} x^2 \log (-a c-b c x+1)-\frac{x (1-a c)}{4 b c}+\frac{a x}{2 b}-\frac{x^2}{8} \]

Antiderivative was successfully verified.

[In]

Int[x*PolyLog[2, c*(a + b*x)],x]

[Out]

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

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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.100904, size = 96, normalized size = 0.63 \[ \frac{-4 c^2 \left (a^2-b^2 x^2\right ) \text{PolyLog}(2,c (a+b x))+\left (-6 a^2 c^2-4 a c (b c x-2)+2 b^2 c^2 x^2-2\right ) \log (-a c-b c x+1)-b c x (-6 a c+b c x+2)}{8 b^2 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*PolyLog[2, c*(a + b*x)],x]

[Out]

(-(b*c*x*(2 - 6*a*c + b*c*x)) + (-2 - 6*a^2*c^2 + 2*b^2*c^2*x^2 - 4*a*c*(-2 + b*c*x))*Log[1 - a*c - b*c*x] - 4
*c^2*(a^2 - b^2*x^2)*PolyLog[2, c*(a + b*x)])/(8*b^2*c^2)

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Maple [A]  time = 0.004, size = 177, normalized size = 1.2 \begin{align*} -{\frac{{\it polylog} \left ( 2,xbc+ac \right ){a}^{2}}{2\,{b}^{2}}}+{\frac{{\it polylog} \left ( 2,xbc+ac \right ){x}^{2}}{2}}-{\frac{\ln \left ( -xbc-ac+1 \right ) xa}{2\,b}}-{\frac{3\,\ln \left ( -xbc-ac+1 \right ){a}^{2}}{4\,{b}^{2}}}+{\frac{3\,ax}{4\,b}}+{\frac{7\,{a}^{2}}{8\,{b}^{2}}}+{\frac{\ln \left ( -xbc-ac+1 \right ) a}{{b}^{2}c}}-{\frac{5\,a}{4\,{b}^{2}c}}+{\frac{{x}^{2}\ln \left ( -xbc-ac+1 \right ) }{4}}-{\frac{\ln \left ( -xbc-ac+1 \right ) }{4\,{b}^{2}{c}^{2}}}-{\frac{{x}^{2}}{8}}-{\frac{x}{4\,bc}}+{\frac{3}{8\,{b}^{2}{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*polylog(2,c*(b*x+a)),x)

[Out]

-1/2/b^2*polylog(2,b*c*x+a*c)*a^2+1/2*polylog(2,b*c*x+a*c)*x^2-1/2/b*ln(-b*c*x-a*c+1)*x*a-3/4/b^2*ln(-b*c*x-a*
c+1)*a^2+3/4*a*x/b+7/8/b^2*a^2+1/b^2/c*ln(-b*c*x-a*c+1)*a-5/4/b^2/c*a+1/4*x^2*ln(-b*c*x-a*c+1)-1/4/b^2/c^2*ln(
-b*c*x-a*c+1)-1/8*x^2-1/4/b/c*x+3/8/b^2/c^2

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Maxima [A]  time = 0.992845, size = 196, normalized size = 1.29 \begin{align*} \frac{{\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}}{2 \, b^{2}} + \frac{4 \, b^{2} c^{2} x^{2}{\rm Li}_2\left (b c x + a c\right ) - b^{2} c^{2} x^{2} + 2 \,{\left (3 \, a b c^{2} - b c\right )} x + 2 \,{\left (b^{2} c^{2} x^{2} - 2 \, a b c^{2} x - 3 \, a^{2} c^{2} + 4 \, a c - 1\right )} \log \left (-b c x - a c + 1\right )}{8 \, b^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*a^2/b^2 + 1/8*(4*b^2*c^2*x^2*dilog(b*c*
x + a*c) - b^2*c^2*x^2 + 2*(3*a*b*c^2 - b*c)*x + 2*(b^2*c^2*x^2 - 2*a*b*c^2*x - 3*a^2*c^2 + 4*a*c - 1)*log(-b*
c*x - a*c + 1))/(b^2*c^2)

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Fricas [A]  time = 2.348, size = 242, normalized size = 1.59 \begin{align*} -\frac{b^{2} c^{2} x^{2} - 2 \,{\left (3 \, a b c^{2} - b c\right )} x - 4 \,{\left (b^{2} c^{2} x^{2} - a^{2} c^{2}\right )}{\rm Li}_2\left (b c x + a c\right ) - 2 \,{\left (b^{2} c^{2} x^{2} - 2 \, a b c^{2} x - 3 \, a^{2} c^{2} + 4 \, a c - 1\right )} \log \left (-b c x - a c + 1\right )}{8 \, b^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(b^2*c^2*x^2 - 2*(3*a*b*c^2 - b*c)*x - 4*(b^2*c^2*x^2 - a^2*c^2)*dilog(b*c*x + a*c) - 2*(b^2*c^2*x^2 - 2*
a*b*c^2*x - 3*a^2*c^2 + 4*a*c - 1)*log(-b*c*x - a*c + 1))/(b^2*c^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\rm Li}_2\left ({\left (b x + a\right )} c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*dilog((b*x + a)*c), x)