3.75 \(\int \frac{\text{PolyLog}(2,a x^2)}{(d x)^{3/2}} \, dx\)

Optimal. Leaf size=103 \[ -\frac{2 \text{PolyLog}\left (2,a x^2\right )}{d \sqrt{d x}}-\frac{16 \sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{16 \sqrt [4]{a} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{8 \log \left (1-a x^2\right )}{d \sqrt{d x}} \]

[Out]

(-16*a^(1/4)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/d^(3/2) + (16*a^(1/4)*ArcTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/
d^(3/2) + (8*Log[1 - a*x^2])/(d*Sqrt[d*x]) - (2*PolyLog[2, a*x^2])/(d*Sqrt[d*x])

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Rubi [A]  time = 0.0738395, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {6591, 2455, 16, 329, 298, 205, 208} \[ -\frac{2 \text{PolyLog}\left (2,a x^2\right )}{d \sqrt{d x}}-\frac{16 \sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{16 \sqrt [4]{a} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{8 \log \left (1-a x^2\right )}{d \sqrt{d x}} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^2]/(d*x)^(3/2),x]

[Out]

(-16*a^(1/4)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/d^(3/2) + (16*a^(1/4)*ArcTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/
d^(3/2) + (8*Log[1 - a*x^2])/(d*Sqrt[d*x]) - (2*PolyLog[2, a*x^2])/(d*Sqrt[d*x])

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]

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 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\text{Li}_2\left (a x^2\right )}{(d x)^{3/2}} \, dx &=-\frac{2 \text{Li}_2\left (a x^2\right )}{d \sqrt{d x}}-4 \int \frac{\log \left (1-a x^2\right )}{(d x)^{3/2}} \, dx\\ &=\frac{8 \log \left (1-a x^2\right )}{d \sqrt{d x}}-\frac{2 \text{Li}_2\left (a x^2\right )}{d \sqrt{d x}}+\frac{(16 a) \int \frac{x}{\sqrt{d x} \left (1-a x^2\right )} \, dx}{d}\\ &=\frac{8 \log \left (1-a x^2\right )}{d \sqrt{d x}}-\frac{2 \text{Li}_2\left (a x^2\right )}{d \sqrt{d x}}+\frac{(16 a) \int \frac{\sqrt{d x}}{1-a x^2} \, dx}{d^2}\\ &=\frac{8 \log \left (1-a x^2\right )}{d \sqrt{d x}}-\frac{2 \text{Li}_2\left (a x^2\right )}{d \sqrt{d x}}+\frac{(32 a) \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{a x^4}{d^2}} \, dx,x,\sqrt{d x}\right )}{d^3}\\ &=\frac{8 \log \left (1-a x^2\right )}{d \sqrt{d x}}-\frac{2 \text{Li}_2\left (a x^2\right )}{d \sqrt{d x}}+\frac{\left (16 \sqrt{a}\right ) \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{a} x^2} \, dx,x,\sqrt{d x}\right )}{d}-\frac{\left (16 \sqrt{a}\right ) \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{a} x^2} \, dx,x,\sqrt{d x}\right )}{d}\\ &=-\frac{16 \sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{16 \sqrt [4]{a} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{8 \log \left (1-a x^2\right )}{d \sqrt{d x}}-\frac{2 \text{Li}_2\left (a x^2\right )}{d \sqrt{d x}}\\ \end{align*}

Mathematica [C]  time = 0.0700504, size = 62, normalized size = 0.6 \[ \frac{x \text{Gamma}\left (\frac{3}{4}\right ) \left (16 a x^2 \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},a x^2\right )-3 \text{PolyLog}\left (2,a x^2\right )+12 \log \left (1-a x^2\right )\right )}{2 \text{Gamma}\left (\frac{7}{4}\right ) (d x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[PolyLog[2, a*x^2]/(d*x)^(3/2),x]

[Out]

(x*Gamma[3/4]*(16*a*x^2*Hypergeometric2F1[3/4, 1, 7/4, a*x^2] + 12*Log[1 - a*x^2] - 3*PolyLog[2, a*x^2]))/(2*(
d*x)^(3/2)*Gamma[7/4])

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Maple [A]  time = 0.052, size = 127, normalized size = 1.2 \begin{align*} -2\,{\frac{{\it polylog} \left ( 2,a{x}^{2} \right ) }{d\sqrt{dx}}}+8\,{\frac{1}{d\sqrt{dx}}\ln \left ({\frac{-a{d}^{2}{x}^{2}+{d}^{2}}{{d}^{2}}} \right ) }-16\,{\frac{1}{d}\arctan \left ({\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{a}}}}}} \right ){\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{a}}}}}}+8\,{\frac{1}{d}\ln \left ({ \left ( \sqrt{dx}+\sqrt [4]{{\frac{{d}^{2}}{a}}} \right ) \left ( \sqrt{dx}-\sqrt [4]{{\frac{{d}^{2}}{a}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{a}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,a*x^2)/(d*x)^(3/2),x)

[Out]

-2*polylog(2,a*x^2)/d/(d*x)^(1/2)+8/d/(d*x)^(1/2)*ln((-a*d^2*x^2+d^2)/d^2)-16/d/(d^2/a)^(1/4)*arctan((d*x)^(1/
2)/(d^2/a)^(1/4))+8/d/(d^2/a)^(1/4)*ln(((d*x)^(1/2)+(d^2/a)^(1/4))/((d*x)^(1/2)-(d^2/a)^(1/4)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/(d*x)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.66427, size = 425, normalized size = 4.13 \begin{align*} \frac{2 \,{\left (16 \, d^{2} x \left (\frac{a}{d^{6}}\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{d x} a d \left (\frac{a}{d^{6}}\right )^{\frac{1}{4}} - \sqrt{a d^{4} \sqrt{\frac{a}{d^{6}}} + a^{2} d x} d \left (\frac{a}{d^{6}}\right )^{\frac{1}{4}}}{a}\right ) + 4 \, d^{2} x \left (\frac{a}{d^{6}}\right )^{\frac{1}{4}} \log \left (512 \, d^{5} \left (\frac{a}{d^{6}}\right )^{\frac{3}{4}} + 512 \, \sqrt{d x} a\right ) - 4 \, d^{2} x \left (\frac{a}{d^{6}}\right )^{\frac{1}{4}} \log \left (-512 \, d^{5} \left (\frac{a}{d^{6}}\right )^{\frac{3}{4}} + 512 \, \sqrt{d x} a\right ) - \sqrt{d x}{\left ({\rm Li}_2\left (a x^{2}\right ) - 4 \, \log \left (-a x^{2} + 1\right )\right )}\right )}}{d^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/(d*x)^(3/2),x, algorithm="fricas")

[Out]

2*(16*d^2*x*(a/d^6)^(1/4)*arctan(-(sqrt(d*x)*a*d*(a/d^6)^(1/4) - sqrt(a*d^4*sqrt(a/d^6) + a^2*d*x)*d*(a/d^6)^(
1/4))/a) + 4*d^2*x*(a/d^6)^(1/4)*log(512*d^5*(a/d^6)^(3/4) + 512*sqrt(d*x)*a) - 4*d^2*x*(a/d^6)^(1/4)*log(-512
*d^5*(a/d^6)^(3/4) + 512*sqrt(d*x)*a) - sqrt(d*x)*(dilog(a*x^2) - 4*log(-a*x^2 + 1)))/(d^2*x)

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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(polylog(2,a*x**2)/(d*x)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm Li}_2\left (a x^{2}\right )}{\left (d x\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/(d*x)^(3/2),x, algorithm="giac")

[Out]

integrate(dilog(a*x^2)/(d*x)^(3/2), x)