3.124 \(\int x^2 \text {Li}_2(c (a+b x)) \, dx\)

Optimal. Leaf size=260 \[ \frac {a^3 \text {Li}_2(c (a+b x))}{3 b^3}-\frac {a^2 (-a c-b c x+1) \log (-a c-b c x+1)}{3 b^3 c}-\frac {a^2 x}{3 b^2}-\frac {(1-a c)^3 \log (-a c-b c x+1)}{9 b^3 c^3}+\frac {a (1-a c)^2 \log (-a c-b c x+1)}{6 b^3 c^2}-\frac {x (1-a c)^2}{9 b^2 c^2}+\frac {a x (1-a c)}{6 b^2 c}+\frac {1}{3} x^3 \text {Li}_2(c (a+b x))+\frac {1}{9} x^3 \log (-a c-b c x+1)-\frac {x^2 (1-a c)}{18 b c}-\frac {a x^2 \log (-a c-b c x+1)}{6 b}+\frac {a x^2}{12 b}-\frac {x^3}{27} \]

[Out]

-1/3*a^2*x/b^2+1/6*a*(-a*c+1)*x/b^2/c-1/9*(-a*c+1)^2*x/b^2/c^2+1/12*a*x^2/b-1/18*(-a*c+1)*x^2/b/c-1/27*x^3+1/6
*a*(-a*c+1)^2*ln(-b*c*x-a*c+1)/b^3/c^2-1/9*(-a*c+1)^3*ln(-b*c*x-a*c+1)/b^3/c^3-1/6*a*x^2*ln(-b*c*x-a*c+1)/b+1/
9*x^3*ln(-b*c*x-a*c+1)-1/3*a^2*(-b*c*x-a*c+1)*ln(-b*c*x-a*c+1)/b^3/c+1/3*a^3*polylog(2,c*(b*x+a))/b^3+1/3*x^3*
polylog(2,c*(b*x+a))

________________________________________________________________________________________

Rubi [A]  time = 0.32, antiderivative size = 260, normalized size of antiderivative = 1.00, 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 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 2295

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

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 = 0.20, size = 144, normalized size = 0.55 \[ \frac {36 c^3 \left (a^3+b^3 x^3\right ) \text {Li}_2(c (a+b x))-b c x \left (66 a^2 c^2-3 a c (5 b c x+14)+4 b^2 c^2 x^2+6 b c x+12\right )+6 \left (11 a^3 c^3+6 a^2 c^2 (b c x-3)+a \left (9 c-3 b^2 c^3 x^2\right )+2 b^3 c^3 x^3-2\right ) \log (-a c-b c x+1)}{108 b^3 c^3} \]

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)

________________________________________________________________________________________

fricas [A]  time = 0.79, size = 165, normalized size = 0.63 \[ -\frac {4 \, b^{3} c^{3} x^{3} - 3 \, {\left (5 \, a b^{2} c^{3} - 2 \, b^{2} c^{2}\right )} x^{2} + 6 \, {\left (11 \, a^{2} b c^{3} - 7 \, a b c^{2} + 2 \, b c\right )} x - 36 \, {\left (b^{3} c^{3} x^{3} + a^{3} c^{3}\right )} {\rm Li}_2\left (b c x + a c\right ) - 6 \, {\left (2 \, b^{3} c^{3} x^{3} - 3 \, a b^{2} c^{3} x^{2} + 6 \, a^{2} b c^{3} x + 11 \, a^{3} c^{3} - 18 \, a^{2} c^{2} + 9 \, a c - 2\right )} \log \left (-b c x - a c + 1\right )}{108 \, b^{3} c^{3}} \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} {\rm Li}_2\left ({\left (b x + a\right )} c\right )\,{d x} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 269, normalized size = 1.03 \[ \frac {11}{54 b^{3} c^{3}}-\frac {\ln \left (-b c x -a c +1\right )}{9 b^{3} c^{3}}-\frac {85 a^{3}}{108 b^{3}}+\frac {13 a^{2}}{9 b^{3} c}-\frac {x}{9 b^{2} c^{2}}-\frac {31 a}{36 b^{3} c^{2}}+\frac {\ln \left (-b c x -a c +1\right ) x \,a^{2}}{3 b^{2}}-\frac {a \,x^{2} \ln \left (-b c x -a c +1\right )}{6 b}+\frac {7 x a}{18 b^{2} c}+\frac {5 a \,x^{2}}{36 b}+\frac {\polylog \left (2, b c x +a c \right ) x^{3}}{3}+\frac {x^{3} \ln \left (-b c x -a c +1\right )}{9}-\frac {11 a^{2} x}{18 b^{2}}+\frac {11 \ln \left (-b c x -a c +1\right ) a^{3}}{18 b^{3}}-\frac {\ln \left (-b c x -a c +1\right ) a^{2}}{b^{3} c}-\frac {x^{2}}{18 b c}+\frac {\ln \left (-b c x -a c +1\right ) a}{2 b^{3} c^{2}}-\frac {x^{3}}{27}+\frac {\dilog \left (-b c x -a c +1\right ) a^{3}}{3 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.31, size = 200, normalized size = 0.77 \[ -\frac {{\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )} a^{3}}{3 \, b^{3}} + \frac {36 \, b^{3} c^{3} x^{3} {\rm Li}_2\left (b c x + a c\right ) - 4 \, b^{3} c^{3} x^{3} + 3 \, {\left (5 \, a b^{2} c^{3} - 2 \, b^{2} c^{2}\right )} x^{2} - 6 \, {\left (11 \, a^{2} b c^{3} - 7 \, a b c^{2} + 2 \, b c\right )} x + 6 \, {\left (2 \, b^{3} c^{3} x^{3} - 3 \, a b^{2} c^{3} x^{2} + 6 \, a^{2} b c^{3} x + 11 \, a^{3} c^{3} - 18 \, a^{2} c^{2} + 9 \, a c - 2\right )} \log \left (-b c x - a c + 1\right )}{108 \, b^{3} c^{3}} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,\mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*polylog(2, c*(a + b*x)),x)

[Out]

int(x^2*polylog(2, c*(a + b*x)), x)

________________________________________________________________________________________

sympy [A]  time = 8.80, size = 236, normalized size = 0.91 \[ \begin {cases} 0 & \text {for}\: c = 0 \wedge \left (b = 0 \vee c = 0\right ) \\\frac {x^{3} \operatorname {Li}_{2}\left (a c\right )}{3} & \text {for}\: b = 0 \\- \frac {11 a^{3} \operatorname {Li}_{1}\left (a c + b c x\right )}{18 b^{3}} + \frac {a^{3} \operatorname {Li}_{2}\left (a c + b c x\right )}{3 b^{3}} - \frac {a^{2} x \operatorname {Li}_{1}\left (a c + b c x\right )}{3 b^{2}} - \frac {11 a^{2} x}{18 b^{2}} + \frac {a^{2} \operatorname {Li}_{1}\left (a c + b c x\right )}{b^{3} c} + \frac {a x^{2} \operatorname {Li}_{1}\left (a c + b c x\right )}{6 b} + \frac {5 a x^{2}}{36 b} + \frac {7 a x}{18 b^{2} c} - \frac {a \operatorname {Li}_{1}\left (a c + b c x\right )}{2 b^{3} c^{2}} - \frac {x^{3} \operatorname {Li}_{1}\left (a c + b c x\right )}{9} + \frac {x^{3} \operatorname {Li}_{2}\left (a c + b c x\right )}{3} - \frac {x^{3}}{27} - \frac {x^{2}}{18 b c} - \frac {x}{9 b^{2} c^{2}} + \frac {\operatorname {Li}_{1}\left (a c + b c x\right )}{9 b^{3} c^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((0, Eq(c, 0) & (Eq(b, 0) | Eq(c, 0))), (x**3*polylog(2, a*c)/3, Eq(b, 0)), (-11*a**3*polylog(1, a*c
+ b*c*x)/(18*b**3) + a**3*polylog(2, a*c + b*c*x)/(3*b**3) - a**2*x*polylog(1, a*c + b*c*x)/(3*b**2) - 11*a**2
*x/(18*b**2) + a**2*polylog(1, a*c + b*c*x)/(b**3*c) + a*x**2*polylog(1, a*c + b*c*x)/(6*b) + 5*a*x**2/(36*b)
+ 7*a*x/(18*b**2*c) - a*polylog(1, a*c + b*c*x)/(2*b**3*c**2) - x**3*polylog(1, a*c + b*c*x)/9 + x**3*polylog(
2, a*c + b*c*x)/3 - x**3/27 - x**2/(18*b*c) - x/(9*b**2*c**2) + polylog(1, a*c + b*c*x)/(9*b**3*c**3), True))

________________________________________________________________________________________