3.83 \(\int \frac{\text{PolyLog}(3,a x^2)}{(d x)^{5/2}} \, dx\)

Optimal. Leaf size=132 \[ -\frac{8 \text{PolyLog}\left (2,a x^2\right )}{9 d (d x)^{3/2}}-\frac{2 \text{PolyLog}\left (3,a x^2\right )}{3 d (d x)^{3/2}}+\frac{64 a^{3/4} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{27 d^{5/2}}+\frac{64 a^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{27 d^{5/2}}+\frac{32 \log \left (1-a x^2\right )}{27 d (d x)^{3/2}} \]

[Out]

(64*a^(3/4)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(27*d^(5/2)) + (64*a^(3/4)*ArcTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d
]])/(27*d^(5/2)) + (32*Log[1 - a*x^2])/(27*d*(d*x)^(3/2)) - (8*PolyLog[2, a*x^2])/(9*d*(d*x)^(3/2)) - (2*PolyL
og[3, a*x^2])/(3*d*(d*x)^(3/2))

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Rubi [A]  time = 0.0910165, antiderivative size = 132, 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, 212, 208, 205} \[ -\frac{8 \text{PolyLog}\left (2,a x^2\right )}{9 d (d x)^{3/2}}-\frac{2 \text{PolyLog}\left (3,a x^2\right )}{3 d (d x)^{3/2}}+\frac{64 a^{3/4} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{27 d^{5/2}}+\frac{64 a^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{27 d^{5/2}}+\frac{32 \log \left (1-a x^2\right )}{27 d (d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^2]/(d*x)^(5/2),x]

[Out]

(64*a^(3/4)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(27*d^(5/2)) + (64*a^(3/4)*ArcTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d
]])/(27*d^(5/2)) + (32*Log[1 - a*x^2])/(27*d*(d*x)^(3/2)) - (8*PolyLog[2, a*x^2])/(9*d*(d*x)^(3/2)) - (2*PolyL
og[3, a*x^2])/(3*d*(d*x)^(3/2))

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 212

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

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]

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]

Rubi steps

\begin{align*} \int \frac{\text{Li}_3\left (a x^2\right )}{(d x)^{5/2}} \, dx &=-\frac{2 \text{Li}_3\left (a x^2\right )}{3 d (d x)^{3/2}}+\frac{4}{3} \int \frac{\text{Li}_2\left (a x^2\right )}{(d x)^{5/2}} \, dx\\ &=-\frac{8 \text{Li}_2\left (a x^2\right )}{9 d (d x)^{3/2}}-\frac{2 \text{Li}_3\left (a x^2\right )}{3 d (d x)^{3/2}}-\frac{16}{9} \int \frac{\log \left (1-a x^2\right )}{(d x)^{5/2}} \, dx\\ &=\frac{32 \log \left (1-a x^2\right )}{27 d (d x)^{3/2}}-\frac{8 \text{Li}_2\left (a x^2\right )}{9 d (d x)^{3/2}}-\frac{2 \text{Li}_3\left (a x^2\right )}{3 d (d x)^{3/2}}+\frac{(64 a) \int \frac{x}{(d x)^{3/2} \left (1-a x^2\right )} \, dx}{27 d}\\ &=\frac{32 \log \left (1-a x^2\right )}{27 d (d x)^{3/2}}-\frac{8 \text{Li}_2\left (a x^2\right )}{9 d (d x)^{3/2}}-\frac{2 \text{Li}_3\left (a x^2\right )}{3 d (d x)^{3/2}}+\frac{(64 a) \int \frac{1}{\sqrt{d x} \left (1-a x^2\right )} \, dx}{27 d^2}\\ &=\frac{32 \log \left (1-a x^2\right )}{27 d (d x)^{3/2}}-\frac{8 \text{Li}_2\left (a x^2\right )}{9 d (d x)^{3/2}}-\frac{2 \text{Li}_3\left (a x^2\right )}{3 d (d x)^{3/2}}+\frac{(128 a) \operatorname{Subst}\left (\int \frac{1}{1-\frac{a x^4}{d^2}} \, dx,x,\sqrt{d x}\right )}{27 d^3}\\ &=\frac{32 \log \left (1-a x^2\right )}{27 d (d x)^{3/2}}-\frac{8 \text{Li}_2\left (a x^2\right )}{9 d (d x)^{3/2}}-\frac{2 \text{Li}_3\left (a x^2\right )}{3 d (d x)^{3/2}}+\frac{(64 a) \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{a} x^2} \, dx,x,\sqrt{d x}\right )}{27 d^2}+\frac{(64 a) \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{a} x^2} \, dx,x,\sqrt{d x}\right )}{27 d^2}\\ &=\frac{64 a^{3/4} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{27 d^{5/2}}+\frac{64 a^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{27 d^{5/2}}+\frac{32 \log \left (1-a x^2\right )}{27 d (d x)^{3/2}}-\frac{8 \text{Li}_2\left (a x^2\right )}{9 d (d x)^{3/2}}-\frac{2 \text{Li}_3\left (a x^2\right )}{3 d (d x)^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0935473, size = 71, normalized size = 0.54 \[ \frac{x \text{Gamma}\left (\frac{1}{4}\right ) \left (64 a x^2 \text{Hypergeometric2F1}\left (\frac{1}{4},1,\frac{5}{4},a x^2\right )-12 \text{PolyLog}\left (2,a x^2\right )-9 \text{PolyLog}\left (3,a x^2\right )+16 \log \left (1-a x^2\right )\right )}{54 \text{Gamma}\left (\frac{5}{4}\right ) (d x)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[PolyLog[3, a*x^2]/(d*x)^(5/2),x]

[Out]

(x*Gamma[1/4]*(64*a*x^2*Hypergeometric2F1[1/4, 1, 5/4, a*x^2] + 16*Log[1 - a*x^2] - 12*PolyLog[2, a*x^2] - 9*P
olyLog[3, a*x^2]))/(54*(d*x)^(5/2)*Gamma[5/4])

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Maple [A]  time = 0.194, size = 131, normalized size = 1. \begin{align*} -{\frac{1}{2}{x}^{{\frac{5}{2}}} \left ( -a \right ) ^{{\frac{3}{4}}} \left ( -{\frac{64}{27}\sqrt{x}\sqrt [4]{-a} \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 ){\frac{1}{\sqrt [4]{a{x}^{2}}}}}+{\frac{64\,\ln \left ( -a{x}^{2}+1 \right ) }{27\,a}\sqrt [4]{-a}{x}^{-{\frac{3}{2}}}}-{\frac{16\,{\it polylog} \left ( 2,a{x}^{2} \right ) }{9\,a}\sqrt [4]{-a}{x}^{-{\frac{3}{2}}}}-{\frac{4\,{\it polylog} \left ( 3,a{x}^{2} \right ) }{3\,a}\sqrt [4]{-a}{x}^{-{\frac{3}{2}}}} \right ) \left ( dx \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3,a*x^2)/(d*x)^(5/2),x)

[Out]

-1/2/(d*x)^(5/2)*x^(5/2)*(-a)^(3/4)*(-64/27*x^(1/2)*(-a)^(1/4)/(a*x^2)^(1/4)*(ln(1-(a*x^2)^(1/4))-ln(1+(a*x^2)
^(1/4))-2*arctan((a*x^2)^(1/4)))+64/27/x^(3/2)*(-a)^(1/4)/a*ln(-a*x^2+1)-16/9/x^(3/2)*(-a)^(1/4)*polylog(2,a*x
^2)/a-4/3/x^(3/2)*(-a)^(1/4)/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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.81842, size = 591, normalized size = 4.48 \begin{align*} -\frac{2 \,{\left (64 \, d^{3} x^{2} \left (\frac{a^{3}}{d^{10}}\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{d x} a d^{7} \left (\frac{a^{3}}{d^{10}}\right )^{\frac{3}{4}} - \sqrt{d^{6} \sqrt{\frac{a^{3}}{d^{10}}} + a^{2} d x} d^{7} \left (\frac{a^{3}}{d^{10}}\right )^{\frac{3}{4}}}{a^{3}}\right ) - 16 \, d^{3} x^{2} \left (\frac{a^{3}}{d^{10}}\right )^{\frac{1}{4}} \log \left (32 \, d^{3} \left (\frac{a^{3}}{d^{10}}\right )^{\frac{1}{4}} + 32 \, \sqrt{d x} a\right ) + 16 \, d^{3} x^{2} \left (\frac{a^{3}}{d^{10}}\right )^{\frac{1}{4}} \log \left (-32 \, d^{3} \left (\frac{a^{3}}{d^{10}}\right )^{\frac{1}{4}} + 32 \, \sqrt{d x} a\right ) + 12 \, \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 ) + 9 \, \sqrt{d x}{\rm polylog}\left (3, a x^{2}\right )\right )}}{27 \, d^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/(d*x)^(5/2),x, algorithm="fricas")

[Out]

-2/27*(64*d^3*x^2*(a^3/d^10)^(1/4)*arctan(-(sqrt(d*x)*a*d^7*(a^3/d^10)^(3/4) - sqrt(d^6*sqrt(a^3/d^10) + a^2*d
*x)*d^7*(a^3/d^10)^(3/4))/a^3) - 16*d^3*x^2*(a^3/d^10)^(1/4)*log(32*d^3*(a^3/d^10)^(1/4) + 32*sqrt(d*x)*a) + 1
6*d^3*x^2*(a^3/d^10)^(1/4)*log(-32*d^3*(a^3/d^10)^(1/4) + 32*sqrt(d*x)*a) + 12*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) + 9*sqrt(d*x)*polylog(3, a*x^2))/(d^3*x^2)

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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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x**2)/(d*x)**(5/2),x)

[Out]

Integral(polylog(3, a*x**2)/(d*x)**(5/2), 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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/(d*x)^(5/2),x, algorithm="giac")

[Out]

integrate(polylog(3, a*x^2)/(d*x)^(5/2), x)