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

Optimal. Leaf size=260 \[ \frac{a^3 \text{PolyLog}(2,c (a+b x))}{3 b^3}+\frac{1}{3} x^3 \text{PolyLog}(2,c (a+b x))-\frac{a^2 (-a c-b c x+1) \log (-a c-b c x+1)}{3 b^3 c}-\frac{a^2 x}{3 b^2}-\frac{x (1-a c)^2}{9 b^2 c^2}+\frac{a (1-a c)^2 \log (-a c-b c x+1)}{6 b^3 c^2}-\frac{(1-a c)^3 \log (-a c-b c x+1)}{9 b^3 c^3}+\frac{a x (1-a c)}{6 b^2 c}-\frac{x^2 (1-a c)}{18 b c}-\frac{a x^2 \log (-a c-b c x+1)}{6 b}+\frac{1}{9} x^3 \log (-a c-b c x+1)+\frac{a x^2}{12 b}-\frac{x^3}{27} \]

[Out]

-(a^2*x)/(3*b^2) + (a*(1 - a*c)*x)/(6*b^2*c) - ((1 - a*c)^2*x)/(9*b^2*c^2) + (a*x^2)/(12*b) - ((1 - a*c)*x^2)/
(18*b*c) - x^3/27 + (a*(1 - a*c)^2*Log[1 - a*c - b*c*x])/(6*b^3*c^2) - ((1 - a*c)^3*Log[1 - a*c - b*c*x])/(9*b
^3*c^3) - (a*x^2*Log[1 - a*c - b*c*x])/(6*b) + (x^3*Log[1 - a*c - b*c*x])/9 - (a^2*(1 - a*c - b*c*x)*Log[1 - a
*c - b*c*x])/(3*b^3*c) + (a^3*PolyLog[2, c*(a + b*x)])/(3*b^3) + (x^3*PolyLog[2, c*(a + b*x)])/3

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Rubi [A]  time = 0.320951, antiderivative size = 260, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {6598, 43, 2416, 2389, 2295, 2395, 2393, 2391} \[ \frac{a^3 \text{PolyLog}(2,c (a+b x))}{3 b^3}+\frac{1}{3} x^3 \text{PolyLog}(2,c (a+b x))-\frac{a^2 (-a c-b c x+1) \log (-a c-b c x+1)}{3 b^3 c}-\frac{a^2 x}{3 b^2}-\frac{x (1-a c)^2}{9 b^2 c^2}+\frac{a (1-a c)^2 \log (-a c-b c x+1)}{6 b^3 c^2}-\frac{(1-a c)^3 \log (-a c-b c x+1)}{9 b^3 c^3}+\frac{a x (1-a c)}{6 b^2 c}-\frac{x^2 (1-a c)}{18 b c}-\frac{a x^2 \log (-a c-b c x+1)}{6 b}+\frac{1}{9} x^3 \log (-a c-b c x+1)+\frac{a x^2}{12 b}-\frac{x^3}{27} \]

Antiderivative was successfully verified.

[In]

Int[x^2*PolyLog[2, c*(a + b*x)],x]

[Out]

-(a^2*x)/(3*b^2) + (a*(1 - a*c)*x)/(6*b^2*c) - ((1 - a*c)^2*x)/(9*b^2*c^2) + (a*x^2)/(12*b) - ((1 - a*c)*x^2)/
(18*b*c) - x^3/27 + (a*(1 - a*c)^2*Log[1 - a*c - b*c*x])/(6*b^3*c^2) - ((1 - a*c)^3*Log[1 - a*c - b*c*x])/(9*b
^3*c^3) - (a*x^2*Log[1 - a*c - b*c*x])/(6*b) + (x^3*Log[1 - a*c - b*c*x])/9 - (a^2*(1 - a*c - b*c*x)*Log[1 - a
*c - b*c*x])/(3*b^3*c) + (a^3*PolyLog[2, c*(a + b*x)])/(3*b^3) + (x^3*PolyLog[2, c*(a + b*x)])/3

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^2 \text{Li}_2(c (a+b x)) \, dx &=\frac{1}{3} x^3 \text{Li}_2(c (a+b x))+\frac{1}{3} b \int \frac{x^3 \log (1-a c-b c x)}{a+b x} \, dx\\ &=\frac{1}{3} x^3 \text{Li}_2(c (a+b x))+\frac{1}{3} b \int \left (\frac{a^2 \log (1-a c-b c x)}{b^3}-\frac{a x \log (1-a c-b c x)}{b^2}+\frac{x^2 \log (1-a c-b c x)}{b}-\frac{a^3 \log (1-a c-b c x)}{b^3 (a+b x)}\right ) \, dx\\ &=\frac{1}{3} x^3 \text{Li}_2(c (a+b x))+\frac{1}{3} \int x^2 \log (1-a c-b c x) \, dx+\frac{a^2 \int \log (1-a c-b c x) \, dx}{3 b^2}-\frac{a^3 \int \frac{\log (1-a c-b c x)}{a+b x} \, dx}{3 b^2}-\frac{a \int x \log (1-a c-b c x) \, dx}{3 b}\\ &=-\frac{a x^2 \log (1-a c-b c x)}{6 b}+\frac{1}{9} x^3 \log (1-a c-b c x)+\frac{1}{3} x^3 \text{Li}_2(c (a+b x))-\frac{a^3 \operatorname{Subst}\left (\int \frac{\log (1-c x)}{x} \, dx,x,a+b x\right )}{3 b^3}-\frac{a^2 \operatorname{Subst}(\int \log (x) \, dx,x,1-a c-b c x)}{3 b^3 c}-\frac{1}{6} (a c) \int \frac{x^2}{1-a c-b c x} \, dx+\frac{1}{9} (b c) \int \frac{x^3}{1-a c-b c x} \, dx\\ &=-\frac{a^2 x}{3 b^2}-\frac{a x^2 \log (1-a c-b c x)}{6 b}+\frac{1}{9} x^3 \log (1-a c-b c x)-\frac{a^2 (1-a c-b c x) \log (1-a c-b c x)}{3 b^3 c}+\frac{a^3 \text{Li}_2(c (a+b x))}{3 b^3}+\frac{1}{3} x^3 \text{Li}_2(c (a+b x))-\frac{1}{6} (a 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{1}{9} (b c) \int \left (-\frac{(-1+a c)^2}{b^3 c^3}+\frac{(-1+a c) x}{b^2 c^2}-\frac{x^2}{b c}+\frac{(-1+a c)^3}{b^3 c^3 (-1+a c+b c x)}\right ) \, dx\\ &=-\frac{a^2 x}{3 b^2}+\frac{a (1-a c) x}{6 b^2 c}-\frac{(1-a c)^2 x}{9 b^2 c^2}+\frac{a x^2}{12 b}-\frac{(1-a c) x^2}{18 b c}-\frac{x^3}{27}+\frac{a (1-a c)^2 \log (1-a c-b c x)}{6 b^3 c^2}-\frac{(1-a c)^3 \log (1-a c-b c x)}{9 b^3 c^3}-\frac{a x^2 \log (1-a c-b c x)}{6 b}+\frac{1}{9} x^3 \log (1-a c-b c x)-\frac{a^2 (1-a c-b c x) \log (1-a c-b c x)}{3 b^3 c}+\frac{a^3 \text{Li}_2(c (a+b x))}{3 b^3}+\frac{1}{3} x^3 \text{Li}_2(c (a+b x))\\ \end{align*}

Mathematica [A]  time = 0.208213, size = 144, normalized size = 0.55 \[ \frac{36 c^3 \left (a^3+b^3 x^3\right ) \text{PolyLog}(2,c (a+b x))-b c x \left (66 a^2 c^2-3 a c (5 b c x+14)+4 b^2 c^2 x^2+6 b c x+12\right )+6 \left (6 a^2 c^2 (b c x-3)+11 a^3 c^3+a \left (9 c-3 b^2 c^3 x^2\right )+2 b^3 c^3 x^3-2\right ) \log (-a c-b c x+1)}{108 b^3 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*c*x*(12 + 66*a^2*c^2 + 6*b*c*x + 4*b^2*c^2*x^2 - 3*a*c*(14 + 5*b*c*x))) + 6*(-2 + 11*a^3*c^3 + 2*b^3*c^3*
x^3 + 6*a^2*c^2*(-3 + b*c*x) + a*(9*c - 3*b^2*c^3*x^2))*Log[1 - a*c - b*c*x] + 36*c^3*(a^3 + b^3*x^3)*PolyLog[
2, c*(a + b*x)])/(108*b^3*c^3)

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Maple [A]  time = 0.008, size = 269, normalized size = 1. \begin{align*} -{\frac{31\,a}{36\,{b}^{3}{c}^{2}}}-{\frac{\ln \left ( -xbc-ac+1 \right ) }{9\,{b}^{3}{c}^{3}}}-{\frac{85\,{a}^{3}}{108\,{b}^{3}}}+{\frac{{a}^{3}{\it dilog} \left ( -xbc-ac+1 \right ) }{3\,{b}^{3}}}-{\frac{{x}^{3}}{27}}+{\frac{11}{54\,{b}^{3}{c}^{3}}}+{\frac{11\,\ln \left ( -xbc-ac+1 \right ){a}^{3}}{18\,{b}^{3}}}-{\frac{\ln \left ( -xbc-ac+1 \right ){a}^{2}}{c{b}^{3}}}+{\frac{\ln \left ( -xbc-ac+1 \right ) a}{2\,{b}^{3}{c}^{2}}}-{\frac{{x}^{2}}{18\,bc}}-{\frac{a{x}^{2}\ln \left ( -xbc-ac+1 \right ) }{6\,b}}+{\frac{\ln \left ( -xbc-ac+1 \right ) x{a}^{2}}{3\,{b}^{2}}}+{\frac{7\,ax}{18\,{b}^{2}c}}-{\frac{11\,{a}^{2}x}{18\,{b}^{2}}}+{\frac{5\,a{x}^{2}}{36\,b}}-{\frac{x}{9\,{b}^{2}{c}^{2}}}+{\frac{{\it polylog} \left ( 2,xbc+ac \right ){x}^{3}}{3}}+{\frac{{x}^{3}\ln \left ( -xbc-ac+1 \right ) }{9}}+{\frac{13\,{a}^{2}}{9\,c{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-31/36/b^3/c^2*a-1/9/b^3/c^3*ln(-b*c*x-a*c+1)-85/108/b^3*a^3+1/3/b^3*a^3*dilog(-b*c*x-a*c+1)-1/27*x^3+11/54/b^
3/c^3+11/18/b^3*ln(-b*c*x-a*c+1)*a^3-1/b^3/c*ln(-b*c*x-a*c+1)*a^2+1/2/b^3/c^2*ln(-b*c*x-a*c+1)*a-1/18/b/c*x^2-
1/6*a*x^2*ln(-b*c*x-a*c+1)/b+1/3/b^2*ln(-b*c*x-a*c+1)*x*a^2+7/18/b^2/c*x*a-11/18*a^2*x/b^2+5/36*a*x^2/b-1/9/b^
2/c^2*x+1/3*polylog(2,b*c*x+a*c)*x^3+1/9*x^3*ln(-b*c*x-a*c+1)+13/9/b^3/c*a^2

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Maxima [A]  time = 0.994565, size = 270, normalized size = 1.04 \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^{3}}{3 \, b^{3}} + \frac{36 \, b^{3} c^{3} x^{3}{\rm Li}_2\left (b c x + a c\right ) - 4 \, b^{3} c^{3} x^{3} + 3 \,{\left (5 \, a b^{2} c^{3} - 2 \, b^{2} c^{2}\right )} x^{2} - 6 \,{\left (11 \, a^{2} b c^{3} - 7 \, a b c^{2} + 2 \, b c\right )} x + 6 \,{\left (2 \, b^{3} c^{3} x^{3} - 3 \, a b^{2} c^{3} x^{2} + 6 \, a^{2} b c^{3} x + 11 \, a^{3} c^{3} - 18 \, a^{2} c^{2} + 9 \, a c - 2\right )} \log \left (-b c x - a c + 1\right )}{108 \, b^{3} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*a^3/b^3 + 1/108*(36*b^3*c^3*x^3*dilog(
b*c*x + a*c) - 4*b^3*c^3*x^3 + 3*(5*a*b^2*c^3 - 2*b^2*c^2)*x^2 - 6*(11*a^2*b*c^3 - 7*a*b*c^2 + 2*b*c)*x + 6*(2
*b^3*c^3*x^3 - 3*a*b^2*c^3*x^2 + 6*a^2*b*c^3*x + 11*a^3*c^3 - 18*a^2*c^2 + 9*a*c - 2)*log(-b*c*x - a*c + 1))/(
b^3*c^3)

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Fricas [A]  time = 2.39335, size = 366, normalized size = 1.41 \begin{align*} -\frac{4 \, b^{3} c^{3} x^{3} - 3 \,{\left (5 \, a b^{2} c^{3} - 2 \, b^{2} c^{2}\right )} x^{2} + 6 \,{\left (11 \, a^{2} b c^{3} - 7 \, a b c^{2} + 2 \, b c\right )} x - 36 \,{\left (b^{3} c^{3} x^{3} + a^{3} c^{3}\right )}{\rm Li}_2\left (b c x + a c\right ) - 6 \,{\left (2 \, b^{3} c^{3} x^{3} - 3 \, a b^{2} c^{3} x^{2} + 6 \, a^{2} b c^{3} x + 11 \, a^{3} c^{3} - 18 \, a^{2} c^{2} + 9 \, a c - 2\right )} \log \left (-b c x - a c + 1\right )}{108 \, b^{3} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/108*(4*b^3*c^3*x^3 - 3*(5*a*b^2*c^3 - 2*b^2*c^2)*x^2 + 6*(11*a^2*b*c^3 - 7*a*b*c^2 + 2*b*c)*x - 36*(b^3*c^3
*x^3 + a^3*c^3)*dilog(b*c*x + a*c) - 6*(2*b^3*c^3*x^3 - 3*a*b^2*c^3*x^2 + 6*a^2*b*c^3*x + 11*a^3*c^3 - 18*a^2*
c^2 + 9*a*c - 2)*log(-b*c*x - a*c + 1))/(b^3*c^3)

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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**2*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^{2}{\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^2*polylog(2,c*(b*x+a)),x, algorithm="giac")

[Out]

integrate(x^2*dilog((b*x + a)*c), x)