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

Optimal. Leaf size=122 \[ -\frac{8 \text{PolyLog}\left (2,a x^2\right )}{d \sqrt{d x}}-\frac{2 \text{PolyLog}\left (3,a x^2\right )}{d \sqrt{d x}}-\frac{64 \sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{64 \sqrt [4]{a} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{32 \log \left (1-a x^2\right )}{d \sqrt{d x}} \]

[Out]

(-64*a^(1/4)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/d^(3/2) + (64*a^(1/4)*ArcTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/
d^(3/2) + (32*Log[1 - a*x^2])/(d*Sqrt[d*x]) - (8*PolyLog[2, a*x^2])/(d*Sqrt[d*x]) - (2*PolyLog[3, a*x^2])/(d*S
qrt[d*x])

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Rubi [A]  time = 0.0932373, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{8 \text{PolyLog}\left (2,a x^2\right )}{d \sqrt{d x}}-\frac{2 \text{PolyLog}\left (3,a x^2\right )}{d \sqrt{d x}}-\frac{64 \sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{64 \sqrt [4]{a} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{32 \log \left (1-a x^2\right )}{d \sqrt{d x}} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^2]/(d*x)^(3/2),x]

[Out]

(-64*a^(1/4)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/d^(3/2) + (64*a^(1/4)*ArcTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/
d^(3/2) + (32*Log[1 - a*x^2])/(d*Sqrt[d*x]) - (8*PolyLog[2, a*x^2])/(d*Sqrt[d*x]) - (2*PolyLog[3, a*x^2])/(d*S
qrt[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}_3\left (a x^2\right )}{(d x)^{3/2}} \, dx &=-\frac{2 \text{Li}_3\left (a x^2\right )}{d \sqrt{d x}}+4 \int \frac{\text{Li}_2\left (a x^2\right )}{(d x)^{3/2}} \, dx\\ &=-\frac{8 \text{Li}_2\left (a x^2\right )}{d \sqrt{d x}}-\frac{2 \text{Li}_3\left (a x^2\right )}{d \sqrt{d x}}-16 \int \frac{\log \left (1-a x^2\right )}{(d x)^{3/2}} \, dx\\ &=\frac{32 \log \left (1-a x^2\right )}{d \sqrt{d x}}-\frac{8 \text{Li}_2\left (a x^2\right )}{d \sqrt{d x}}-\frac{2 \text{Li}_3\left (a x^2\right )}{d \sqrt{d x}}+\frac{(64 a) \int \frac{x}{\sqrt{d x} \left (1-a x^2\right )} \, dx}{d}\\ &=\frac{32 \log \left (1-a x^2\right )}{d \sqrt{d x}}-\frac{8 \text{Li}_2\left (a x^2\right )}{d \sqrt{d x}}-\frac{2 \text{Li}_3\left (a x^2\right )}{d \sqrt{d x}}+\frac{(64 a) \int \frac{\sqrt{d x}}{1-a x^2} \, dx}{d^2}\\ &=\frac{32 \log \left (1-a x^2\right )}{d \sqrt{d x}}-\frac{8 \text{Li}_2\left (a x^2\right )}{d \sqrt{d x}}-\frac{2 \text{Li}_3\left (a x^2\right )}{d \sqrt{d x}}+\frac{(128 a) \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{a x^4}{d^2}} \, dx,x,\sqrt{d x}\right )}{d^3}\\ &=\frac{32 \log \left (1-a x^2\right )}{d \sqrt{d x}}-\frac{8 \text{Li}_2\left (a x^2\right )}{d \sqrt{d x}}-\frac{2 \text{Li}_3\left (a x^2\right )}{d \sqrt{d x}}+\frac{\left (64 \sqrt{a}\right ) \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{a} x^2} \, dx,x,\sqrt{d x}\right )}{d}-\frac{\left (64 \sqrt{a}\right ) \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{a} x^2} \, dx,x,\sqrt{d x}\right )}{d}\\ &=-\frac{64 \sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{64 \sqrt [4]{a} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{32 \log \left (1-a x^2\right )}{d \sqrt{d x}}-\frac{8 \text{Li}_2\left (a x^2\right )}{d \sqrt{d x}}-\frac{2 \text{Li}_3\left (a x^2\right )}{d \sqrt{d x}}\\ \end{align*}

Mathematica [C]  time = 0.0897261, size = 71, normalized size = 0.58 \[ \frac{x \text{Gamma}\left (\frac{3}{4}\right ) \left (64 a x^2 \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},a x^2\right )-12 \text{PolyLog}\left (2,a x^2\right )-3 \text{PolyLog}\left (3,a x^2\right )+48 \log \left (1-a x^2\right )\right )}{2 \text{Gamma}\left (\frac{7}{4}\right ) (d x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*Gamma[3/4]*(64*a*x^2*Hypergeometric2F1[3/4, 1, 7/4, a*x^2] + 48*Log[1 - a*x^2] - 12*PolyLog[2, a*x^2] - 3*P
olyLog[3, a*x^2]))/(2*(d*x)^(3/2)*Gamma[7/4])

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Maple [A]  time = 0.191, size = 131, normalized size = 1.1 \begin{align*} -{\frac{1}{2}{x}^{{\frac{3}{2}}}\sqrt [4]{-a} \left ( -64\,{\frac{{x}^{3/2} \left ( -a \right ) ^{3/4} \left ( \ln \left ( 1-\sqrt [4]{a{x}^{2}} \right ) -\ln \left ( 1+\sqrt [4]{a{x}^{2}} \right ) +2\,\arctan \left ( \sqrt [4]{a{x}^{2}} \right ) \right ) }{ \left ( a{x}^{2} \right ) ^{3/4}}}+64\,{\frac{ \left ( -a \right ) ^{3/4}\ln \left ( -a{x}^{2}+1 \right ) }{\sqrt{x}a}}-16\,{\frac{ \left ( -a \right ) ^{3/4}{\it polylog} \left ( 2,a{x}^{2} \right ) }{\sqrt{x}a}}-4\,{\frac{ \left ( -a \right ) ^{3/4}{\it polylog} \left ( 3,a{x}^{2} \right ) }{\sqrt{x}a}} \right ) \left ( dx \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(d*x)^(3/2)*x^(3/2)*(-a)^(1/4)*(-64*x^(3/2)*(-a)^(3/4)/(a*x^2)^(3/4)*(ln(1-(a*x^2)^(1/4))-ln(1+(a*x^2)^(1
/4))+2*arctan((a*x^2)^(1/4)))+64/x^(1/2)*(-a)^(3/4)/a*ln(-a*x^2+1)-16/x^(1/2)*(-a)^(3/4)*polylog(2,a*x^2)/a-4/
x^(1/2)*(-a)^(3/4)/a*polylog(3,a*x^2))

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 3.0072, size = 549, normalized size = 4.5 \begin{align*} \frac{2 \,{\left (64 \, 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 ) + 16 \, d^{2} x \left (\frac{a}{d^{6}}\right )^{\frac{1}{4}} \log \left (32768 \, d^{5} \left (\frac{a}{d^{6}}\right )^{\frac{3}{4}} + 32768 \, \sqrt{d x} a\right ) - 16 \, d^{2} x \left (\frac{a}{d^{6}}\right )^{\frac{1}{4}} \log \left (-32768 \, d^{5} \left (\frac{a}{d^{6}}\right )^{\frac{3}{4}} + 32768 \, \sqrt{d x} a\right ) - 4 \, \sqrt{d x}{\rm \%iint}\left (a, x, -\frac{\log \left (-a x^{2} + 1\right )}{a}, -\frac{2 \, \log \left (-a x^{2} + 1\right )}{x}\right ) + 16 \, \sqrt{d x} \log \left (-a x^{2} + 1\right ) - \sqrt{d x}{\rm polylog}\left (3, a x^{2}\right )\right )}}{d^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(64*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) + 16*d^2*x*(a/d^6)^(1/4)*log(32768*d^5*(a/d^6)^(3/4) + 32768*sqrt(d*x)*a) - 16*d^2*x*(a/d^6)^(1/4)*lo
g(-32768*d^5*(a/d^6)^(3/4) + 32768*sqrt(d*x)*a) - 4*sqrt(d*x)*\%iint(a, x, -log(-a*x^2 + 1)/a, -2*log(-a*x^2 +
1)/x) + 16*sqrt(d*x)*log(-a*x^2 + 1) - sqrt(d*x)*polylog(3, a*x^2))/(d^2*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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