3.31 \(\int \frac {\text {Li}_2(a x^2)}{x^6} \, dx\)

Optimal. Leaf size=66 \[ \frac {4}{25} a^{5/2} \tanh ^{-1}\left (\sqrt {a} x\right )-\frac {4 a^2}{25 x}-\frac {\text {Li}_2\left (a x^2\right )}{5 x^5}-\frac {4 a}{75 x^3}+\frac {2 \log \left (1-a x^2\right )}{25 x^5} \]

[Out]

-4/75*a/x^3-4/25*a^2/x+4/25*a^(5/2)*arctanh(x*a^(1/2))+2/25*ln(-a*x^2+1)/x^5-1/5*polylog(2,a*x^2)/x^5

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Rubi [A]  time = 0.04, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6591, 2455, 325, 206} \[ -\frac {\text {PolyLog}\left (2,a x^2\right )}{5 x^5}-\frac {4 a^2}{25 x}+\frac {4}{25} a^{5/2} \tanh ^{-1}\left (\sqrt {a} x\right )-\frac {4 a}{75 x^3}+\frac {2 \log \left (1-a x^2\right )}{25 x^5} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^2]/x^6,x]

[Out]

(-4*a)/(75*x^3) - (4*a^2)/(25*x) + (4*a^(5/2)*ArcTanh[Sqrt[a]*x])/25 + (2*Log[1 - a*x^2])/(25*x^5) - PolyLog[2
, a*x^2]/(5*x^5)

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^6} \, dx &=-\frac {\text {Li}_2\left (a x^2\right )}{5 x^5}-\frac {2}{5} \int \frac {\log \left (1-a x^2\right )}{x^6} \, dx\\ &=\frac {2 \log \left (1-a x^2\right )}{25 x^5}-\frac {\text {Li}_2\left (a x^2\right )}{5 x^5}+\frac {1}{25} (4 a) \int \frac {1}{x^4 \left (1-a x^2\right )} \, dx\\ &=-\frac {4 a}{75 x^3}+\frac {2 \log \left (1-a x^2\right )}{25 x^5}-\frac {\text {Li}_2\left (a x^2\right )}{5 x^5}+\frac {1}{25} \left (4 a^2\right ) \int \frac {1}{x^2 \left (1-a x^2\right )} \, dx\\ &=-\frac {4 a}{75 x^3}-\frac {4 a^2}{25 x}+\frac {2 \log \left (1-a x^2\right )}{25 x^5}-\frac {\text {Li}_2\left (a x^2\right )}{5 x^5}+\frac {1}{25} \left (4 a^3\right ) \int \frac {1}{1-a x^2} \, dx\\ &=-\frac {4 a}{75 x^3}-\frac {4 a^2}{25 x}+\frac {4}{25} a^{5/2} \tanh ^{-1}\left (\sqrt {a} x\right )+\frac {2 \log \left (1-a x^2\right )}{25 x^5}-\frac {\text {Li}_2\left (a x^2\right )}{5 x^5}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 47, normalized size = 0.71 \[ -\frac {4 a x^2 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};a x^2\right )+15 \text {Li}_2\left (a x^2\right )-6 \log \left (1-a x^2\right )}{75 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, a*x^2]/x^6,x]

[Out]

-1/75*(4*a*x^2*Hypergeometric2F1[-3/2, 1, -1/2, a*x^2] - 6*Log[1 - a*x^2] + 15*PolyLog[2, a*x^2])/x^5

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fricas [A]  time = 0.52, size = 132, normalized size = 2.00 \[ \left [\frac {6 \, a^{\frac {5}{2}} x^{5} \log \left (\frac {a x^{2} + 2 \, \sqrt {a} x + 1}{a x^{2} - 1}\right ) - 12 \, a^{2} x^{4} - 4 \, a x^{2} - 15 \, {\rm Li}_2\left (a x^{2}\right ) + 6 \, \log \left (-a x^{2} + 1\right )}{75 \, x^{5}}, -\frac {12 \, \sqrt {-a} a^{2} x^{5} \arctan \left (\sqrt {-a} x\right ) + 12 \, a^{2} x^{4} + 4 \, a x^{2} + 15 \, {\rm Li}_2\left (a x^{2}\right ) - 6 \, \log \left (-a x^{2} + 1\right )}{75 \, x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/x^6,x, algorithm="fricas")

[Out]

[1/75*(6*a^(5/2)*x^5*log((a*x^2 + 2*sqrt(a)*x + 1)/(a*x^2 - 1)) - 12*a^2*x^4 - 4*a*x^2 - 15*dilog(a*x^2) + 6*l
og(-a*x^2 + 1))/x^5, -1/75*(12*sqrt(-a)*a^2*x^5*arctan(sqrt(-a)*x) + 12*a^2*x^4 + 4*a*x^2 + 15*dilog(a*x^2) -
6*log(-a*x^2 + 1))/x^5]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\left (a x^{2}\right )}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/x^6,x, algorithm="giac")

[Out]

integrate(dilog(a*x^2)/x^6, x)

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maple [A]  time = 0.01, size = 53, normalized size = 0.80 \[ -\frac {4 a}{75 x^{3}}-\frac {4 a^{2}}{25 x}+\frac {4 a^{\frac {5}{2}} \arctanh \left (x \sqrt {a}\right )}{25}+\frac {2 \ln \left (-a \,x^{2}+1\right )}{25 x^{5}}-\frac {\polylog \left (2, a \,x^{2}\right )}{5 x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,a*x^2)/x^6,x)

[Out]

-4/75*a/x^3-4/25*a^2/x+4/25*a^(5/2)*arctanh(x*a^(1/2))+2/25*ln(-a*x^2+1)/x^5-1/5*polylog(2,a*x^2)/x^5

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maxima [A]  time = 0.41, size = 65, normalized size = 0.98 \[ -\frac {2}{25} \, a^{\frac {5}{2}} \log \left (\frac {a x - \sqrt {a}}{a x + \sqrt {a}}\right ) - \frac {12 \, a^{2} x^{4} + 4 \, a x^{2} + 15 \, {\rm Li}_2\left (a x^{2}\right ) - 6 \, \log \left (-a x^{2} + 1\right )}{75 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/x^6,x, algorithm="maxima")

[Out]

-2/25*a^(5/2)*log((a*x - sqrt(a))/(a*x + sqrt(a))) - 1/75*(12*a^2*x^4 + 4*a*x^2 + 15*dilog(a*x^2) - 6*log(-a*x
^2 + 1))/x^5

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mupad [B]  time = 0.34, size = 58, normalized size = 0.88 \[ \frac {2\,\ln \left (1-a\,x^2\right )}{25\,x^5}-\frac {4\,a^2\,x^2+\frac {4\,a}{3}}{25\,x^3}-\frac {\mathrm {polylog}\left (2,a\,x^2\right )}{5\,x^5}-\frac {a^{5/2}\,\mathrm {atan}\left (\sqrt {a}\,x\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{25} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x^2)/x^6,x)

[Out]

(2*log(1 - a*x^2))/(25*x^5) - polylog(2, a*x^2)/(5*x^5) - ((4*a)/3 + 4*a^2*x^2)/(25*x^3) - (a^(5/2)*atan(a^(1/
2)*x*1i)*4i)/25

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x**2)/x**6,x)

[Out]

Timed out

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