3.30 \(\int \frac {\text {Li}_2(a x^2)}{x^4} \, dx\)

Optimal. Leaf size=56 \[ \frac {4}{9} a^{3/2} \tanh ^{-1}\left (\sqrt {a} x\right )-\frac {\text {Li}_2\left (a x^2\right )}{3 x^3}+\frac {2 \log \left (1-a x^2\right )}{9 x^3}-\frac {4 a}{9 x} \]

[Out]

-4/9*a/x+4/9*a^(3/2)*arctanh(x*a^(1/2))+2/9*ln(-a*x^2+1)/x^3-1/3*polylog(2,a*x^2)/x^3

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Rubi [A]  time = 0.03, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6591, 2455, 325, 206} \[ -\frac {\text {PolyLog}\left (2,a x^2\right )}{3 x^3}+\frac {4}{9} a^{3/2} \tanh ^{-1}\left (\sqrt {a} x\right )+\frac {2 \log \left (1-a x^2\right )}{9 x^3}-\frac {4 a}{9 x} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^2]/x^4,x]

[Out]

(-4*a)/(9*x) + (4*a^(3/2)*ArcTanh[Sqrt[a]*x])/9 + (2*Log[1 - a*x^2])/(9*x^3) - PolyLog[2, a*x^2]/(3*x^3)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2455

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

Rule 6591

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 \frac {\text {Li}_2\left (a x^2\right )}{x^4} \, dx &=-\frac {\text {Li}_2\left (a x^2\right )}{3 x^3}-\frac {2}{3} \int \frac {\log \left (1-a x^2\right )}{x^4} \, dx\\ &=\frac {2 \log \left (1-a x^2\right )}{9 x^3}-\frac {\text {Li}_2\left (a x^2\right )}{3 x^3}+\frac {1}{9} (4 a) \int \frac {1}{x^2 \left (1-a x^2\right )} \, dx\\ &=-\frac {4 a}{9 x}+\frac {2 \log \left (1-a x^2\right )}{9 x^3}-\frac {\text {Li}_2\left (a x^2\right )}{3 x^3}+\frac {1}{9} \left (4 a^2\right ) \int \frac {1}{1-a x^2} \, dx\\ &=-\frac {4 a}{9 x}+\frac {4}{9} a^{3/2} \tanh ^{-1}\left (\sqrt {a} x\right )+\frac {2 \log \left (1-a x^2\right )}{9 x^3}-\frac {\text {Li}_2\left (a x^2\right )}{3 x^3}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 47, normalized size = 0.84 \[ -\frac {4 a x^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};a x^2\right )+3 \text {Li}_2\left (a x^2\right )-2 \log \left (1-a x^2\right )}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x^2]/x^4,x]

[Out]

-1/9*(4*a*x^2*Hypergeometric2F1[-1/2, 1, 1/2, a*x^2] - 2*Log[1 - a*x^2] + 3*PolyLog[2, a*x^2])/x^3

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fricas [A]  time = 0.60, size = 114, normalized size = 2.04 \[ \left [\frac {2 \, a^{\frac {3}{2}} x^{3} \log \left (\frac {a x^{2} + 2 \, \sqrt {a} x + 1}{a x^{2} - 1}\right ) - 4 \, a x^{2} - 3 \, {\rm Li}_2\left (a x^{2}\right ) + 2 \, \log \left (-a x^{2} + 1\right )}{9 \, x^{3}}, -\frac {4 \, \sqrt {-a} a x^{3} \arctan \left (\sqrt {-a} x\right ) + 4 \, a x^{2} + 3 \, {\rm Li}_2\left (a x^{2}\right ) - 2 \, \log \left (-a x^{2} + 1\right )}{9 \, x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/x^4,x, algorithm="fricas")

[Out]

[1/9*(2*a^(3/2)*x^3*log((a*x^2 + 2*sqrt(a)*x + 1)/(a*x^2 - 1)) - 4*a*x^2 - 3*dilog(a*x^2) + 2*log(-a*x^2 + 1))
/x^3, -1/9*(4*sqrt(-a)*a*x^3*arctan(sqrt(-a)*x) + 4*a*x^2 + 3*dilog(a*x^2) - 2*log(-a*x^2 + 1))/x^3]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\left (a x^{2}\right )}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/x^4,x, algorithm="giac")

[Out]

integrate(dilog(a*x^2)/x^4, x)

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maple [A]  time = 0.01, size = 45, normalized size = 0.80 \[ -\frac {4 a}{9 x}+\frac {4 a^{\frac {3}{2}} \arctanh \left (x \sqrt {a}\right )}{9}+\frac {2 \ln \left (-a \,x^{2}+1\right )}{9 x^{3}}-\frac {\polylog \left (2, a \,x^{2}\right )}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,a*x^2)/x^4,x)

[Out]

-4/9*a/x+4/9*a^(3/2)*arctanh(x*a^(1/2))+2/9*ln(-a*x^2+1)/x^3-1/3*polylog(2,a*x^2)/x^3

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maxima [A]  time = 0.42, size = 57, normalized size = 1.02 \[ -\frac {2}{9} \, a^{\frac {3}{2}} \log \left (\frac {a x - \sqrt {a}}{a x + \sqrt {a}}\right ) - \frac {4 \, a x^{2} + 3 \, {\rm Li}_2\left (a x^{2}\right ) - 2 \, \log \left (-a x^{2} + 1\right )}{9 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/x^4,x, algorithm="maxima")

[Out]

-2/9*a^(3/2)*log((a*x - sqrt(a))/(a*x + sqrt(a))) - 1/9*(4*a*x^2 + 3*dilog(a*x^2) - 2*log(-a*x^2 + 1))/x^3

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mupad [B]  time = 0.33, size = 47, normalized size = 0.84 \[ \frac {2\,\ln \left (1-a\,x^2\right )}{9\,x^3}-\frac {4\,a}{9\,x}-\frac {\mathrm {polylog}\left (2,a\,x^2\right )}{3\,x^3}-\frac {a^{3/2}\,\mathrm {atan}\left (\sqrt {a}\,x\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x^2)/x^4,x)

[Out]

(2*log(1 - a*x^2))/(9*x^3) - polylog(2, a*x^2)/(3*x^3) - (4*a)/(9*x) - (a^(3/2)*atan(a^(1/2)*x*1i)*4i)/9

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sympy [A]  time = 133.63, size = 275, normalized size = 4.91 \[ \begin {cases} - \frac {\pi ^{2}}{18 x^{3}} & \text {for}\: a = \frac {1}{x^{2}} \\0 & \text {for}\: a = 0 \\- \frac {4 a^{2} x^{5} \sqrt {\frac {1}{a}} \log {\left (x - \sqrt {\frac {1}{a}} \right )}}{9 x^{5} - \frac {9 x^{3}}{a}} - \frac {2 a^{2} x^{5} \sqrt {\frac {1}{a}} \operatorname {Li}_{1}\left (a x^{2}\right )}{9 x^{5} - \frac {9 x^{3}}{a}} - \frac {4 a x^{4}}{9 x^{5} - \frac {9 x^{3}}{a}} + \frac {4 a x^{3} \sqrt {\frac {1}{a}} \log {\left (x - \sqrt {\frac {1}{a}} \right )}}{9 x^{5} - \frac {9 x^{3}}{a}} + \frac {2 a x^{3} \sqrt {\frac {1}{a}} \operatorname {Li}_{1}\left (a x^{2}\right )}{9 x^{5} - \frac {9 x^{3}}{a}} - \frac {2 x^{2} \operatorname {Li}_{1}\left (a x^{2}\right )}{9 x^{5} - \frac {9 x^{3}}{a}} - \frac {3 x^{2} \operatorname {Li}_{2}\left (a x^{2}\right )}{9 x^{5} - \frac {9 x^{3}}{a}} + \frac {4 x^{2}}{9 x^{5} - \frac {9 x^{3}}{a}} + \frac {2 \operatorname {Li}_{1}\left (a x^{2}\right )}{9 a x^{5} - 9 x^{3}} + \frac {3 \operatorname {Li}_{2}\left (a x^{2}\right )}{9 a x^{5} - 9 x^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x**2)/x**4,x)

[Out]

Piecewise((-pi**2/(18*x**3), Eq(a, x**(-2))), (0, Eq(a, 0)), (-4*a**2*x**5*sqrt(1/a)*log(x - sqrt(1/a))/(9*x**
5 - 9*x**3/a) - 2*a**2*x**5*sqrt(1/a)*polylog(1, a*x**2)/(9*x**5 - 9*x**3/a) - 4*a*x**4/(9*x**5 - 9*x**3/a) +
4*a*x**3*sqrt(1/a)*log(x - sqrt(1/a))/(9*x**5 - 9*x**3/a) + 2*a*x**3*sqrt(1/a)*polylog(1, a*x**2)/(9*x**5 - 9*
x**3/a) - 2*x**2*polylog(1, a*x**2)/(9*x**5 - 9*x**3/a) - 3*x**2*polylog(2, a*x**2)/(9*x**5 - 9*x**3/a) + 4*x*
*2/(9*x**5 - 9*x**3/a) + 2*polylog(1, a*x**2)/(9*a*x**5 - 9*x**3) + 3*polylog(2, a*x**2)/(9*a*x**5 - 9*x**3),
True))

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