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

Optimal. Leaf size=111 \[ -\frac{2 \text{PolyLog}\left (2,a x^2\right )}{3 d (d x)^{3/2}}+\frac{16 a^{3/4} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{9 d^{5/2}}+\frac{16 a^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{9 d^{5/2}}+\frac{8 \log \left (1-a x^2\right )}{9 d (d x)^{3/2}} \]

[Out]

(16*a^(3/4)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(9*d^(5/2)) + (16*a^(3/4)*ArcTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]
])/(9*d^(5/2)) + (8*Log[1 - a*x^2])/(9*d*(d*x)^(3/2)) - (2*PolyLog[2, a*x^2])/(3*d*(d*x)^(3/2))

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Rubi [A]  time = 0.0720639, antiderivative size = 111, 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, 212, 208, 205} \[ -\frac{2 \text{PolyLog}\left (2,a x^2\right )}{3 d (d x)^{3/2}}+\frac{16 a^{3/4} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{9 d^{5/2}}+\frac{16 a^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{9 d^{5/2}}+\frac{8 \log \left (1-a x^2\right )}{9 d (d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^2]/(d*x)^(5/2),x]

[Out]

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

Mathematica [C]  time = 0.0673089, size = 62, normalized size = 0.56 \[ \frac{x \text{Gamma}\left (\frac{1}{4}\right ) \left (16 a x^2 \text{Hypergeometric2F1}\left (\frac{1}{4},1,\frac{5}{4},a x^2\right )-3 \text{PolyLog}\left (2,a x^2\right )+4 \log \left (1-a x^2\right )\right )}{18 \text{Gamma}\left (\frac{5}{4}\right ) (d x)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*Gamma[1/4]*(16*a*x^2*Hypergeometric2F1[1/4, 1, 5/4, a*x^2] + 4*Log[1 - a*x^2] - 3*PolyLog[2, a*x^2]))/(18*(
d*x)^(5/2)*Gamma[5/4])

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Maple [A]  time = 0.054, size = 129, normalized size = 1.2 \begin{align*} -{\frac{2\,{\it polylog} \left ( 2,a{x}^{2} \right ) }{3\,d} \left ( dx \right ) ^{-{\frac{3}{2}}}}+{\frac{8}{9\,d}\ln \left ({\frac{-a{d}^{2}{x}^{2}+{d}^{2}}{{d}^{2}}} \right ) \left ( dx \right ) ^{-{\frac{3}{2}}}}+{\frac{8\,a}{9\,{d}^{3}}\sqrt [4]{{\frac{{d}^{2}}{a}}}\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{16\,a}{9\,{d}^{3}}\sqrt [4]{{\frac{{d}^{2}}{a}}}\arctan \left ({\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{a}}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*polylog(2,a*x^2)/d/(d*x)^(3/2)+8/9/d/(d*x)^(3/2)*ln((-a*d^2*x^2+d^2)/d^2)+8/9/d^3*a*(d^2/a)^(1/4)*ln(((d*
x)^(1/2)+(d^2/a)^(1/4))/((d*x)^(1/2)-(d^2/a)^(1/4)))+16/9/d^3*a*(d^2/a)^(1/4)*arctan((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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.73533, size = 470, normalized size = 4.23 \begin{align*} -\frac{2 \,{\left (16 \, 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 ) - 4 \, d^{3} x^{2} \left (\frac{a^{3}}{d^{10}}\right )^{\frac{1}{4}} \log \left (8 \, d^{3} \left (\frac{a^{3}}{d^{10}}\right )^{\frac{1}{4}} + 8 \, \sqrt{d x} a\right ) + 4 \, d^{3} x^{2} \left (\frac{a^{3}}{d^{10}}\right )^{\frac{1}{4}} \log \left (-8 \, d^{3} \left (\frac{a^{3}}{d^{10}}\right )^{\frac{1}{4}} + 8 \, \sqrt{d x} a\right ) + \sqrt{d x}{\left (3 \,{\rm Li}_2\left (a x^{2}\right ) - 4 \, \log \left (-a x^{2} + 1\right )\right )}\right )}}{9 \, d^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*(16*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) - 4*d^3*x^2*(a^3/d^10)^(1/4)*log(8*d^3*(a^3/d^10)^(1/4) + 8*sqrt(d*x)*a) + 4*d^3
*x^2*(a^3/d^10)^(1/4)*log(-8*d^3*(a^3/d^10)^(1/4) + 8*sqrt(d*x)*a) + sqrt(d*x)*(3*dilog(a*x^2) - 4*log(-a*x^2
+ 1)))/(d^3*x^2)

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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)**(5/2),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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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