3.77 \(\int \frac{\text{PolyLog}(2,a x^2)}{(d x)^{7/2}} \, dx\)

Optimal. Leaf size=126 \[ -\frac{2 \text{PolyLog}\left (2,a x^2\right )}{5 d (d x)^{5/2}}-\frac{16 a^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{25 d^{7/2}}+\frac{16 a^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{25 d^{7/2}}-\frac{32 a}{25 d^3 \sqrt{d x}}+\frac{8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}} \]

[Out]

(-32*a)/(25*d^3*Sqrt[d*x]) - (16*a^(5/4)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(25*d^(7/2)) + (16*a^(5/4)*ArcTa
nh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(25*d^(7/2)) + (8*Log[1 - a*x^2])/(25*d*(d*x)^(5/2)) - (2*PolyLog[2, a*x^2])/
(5*d*(d*x)^(5/2))

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Rubi [A]  time = 0.0858323, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.533, Rules used = {6591, 2455, 16, 325, 329, 298, 205, 208} \[ -\frac{2 \text{PolyLog}\left (2,a x^2\right )}{5 d (d x)^{5/2}}-\frac{16 a^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{25 d^{7/2}}+\frac{16 a^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{25 d^{7/2}}-\frac{32 a}{25 d^3 \sqrt{d x}}+\frac{8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^2]/(d*x)^(7/2),x]

[Out]

(-32*a)/(25*d^3*Sqrt[d*x]) - (16*a^(5/4)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(25*d^(7/2)) + (16*a^(5/4)*ArcTa
nh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(25*d^(7/2)) + (8*Log[1 - a*x^2])/(25*d*(d*x)^(5/2)) - (2*PolyLog[2, a*x^2])/
(5*d*(d*x)^(5/2))

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 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 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)^{7/2}} \, dx &=-\frac{2 \text{Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}-\frac{4}{5} \int \frac{\log \left (1-a x^2\right )}{(d x)^{7/2}} \, dx\\ &=\frac{8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac{(16 a) \int \frac{x}{(d x)^{5/2} \left (1-a x^2\right )} \, dx}{25 d}\\ &=\frac{8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac{(16 a) \int \frac{1}{(d x)^{3/2} \left (1-a x^2\right )} \, dx}{25 d^2}\\ &=-\frac{32 a}{25 d^3 \sqrt{d x}}+\frac{8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac{\left (16 a^2\right ) \int \frac{\sqrt{d x}}{1-a x^2} \, dx}{25 d^4}\\ &=-\frac{32 a}{25 d^3 \sqrt{d x}}+\frac{8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac{\left (32 a^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{a x^4}{d^2}} \, dx,x,\sqrt{d x}\right )}{25 d^5}\\ &=-\frac{32 a}{25 d^3 \sqrt{d x}}+\frac{8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac{\left (16 a^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{a} x^2} \, dx,x,\sqrt{d x}\right )}{25 d^3}-\frac{\left (16 a^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{a} x^2} \, dx,x,\sqrt{d x}\right )}{25 d^3}\\ &=-\frac{32 a}{25 d^3 \sqrt{d x}}-\frac{16 a^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{25 d^{7/2}}+\frac{16 a^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{25 d^{7/2}}+\frac{8 \log \left (1-a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{Li}_2\left (a x^2\right )}{5 d (d x)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0765584, size = 70, normalized size = 0.56 \[ -\frac{x \text{Gamma}\left (-\frac{1}{4}\right ) \left (16 a^2 x^4 \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},a x^2\right )-15 \text{PolyLog}\left (2,a x^2\right )-48 a x^2+12 \log \left (1-a x^2\right )\right )}{150 \text{Gamma}\left (\frac{3}{4}\right ) (d x)^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[PolyLog[2, a*x^2]/(d*x)^(7/2),x]

[Out]

-(x*Gamma[-1/4]*(-48*a*x^2 + 16*a^2*x^4*Hypergeometric2F1[3/4, 1, 7/4, a*x^2] + 12*Log[1 - a*x^2] - 15*PolyLog
[2, a*x^2]))/(150*(d*x)^(7/2)*Gamma[3/4])

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Maple [A]  time = 0.056, size = 140, normalized size = 1.1 \begin{align*} -{\frac{2\,{\it polylog} \left ( 2,a{x}^{2} \right ) }{5\,d} \left ( dx \right ) ^{-{\frac{5}{2}}}}+{\frac{8}{25\,d}\ln \left ({\frac{-a{d}^{2}{x}^{2}+{d}^{2}}{{d}^{2}}} \right ) \left ( dx \right ) ^{-{\frac{5}{2}}}}-{\frac{32\,a}{25\,{d}^{3}}{\frac{1}{\sqrt{dx}}}}-{\frac{16\,a}{25\,{d}^{3}}\arctan \left ({\sqrt{dx}{\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{a}}}}}} \right ){\frac{1}{\sqrt [4]{{\frac{{d}^{2}}{a}}}}}}+{\frac{8\,a}{25\,{d}^{3}}\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)^(7/2),x)

[Out]

-2/5*polylog(2,a*x^2)/d/(d*x)^(5/2)+8/25/d/(d*x)^(5/2)*ln((-a*d^2*x^2+d^2)/d^2)-32/25*a/d^3/(d*x)^(1/2)-16/25/
d^3*a/(d^2/a)^(1/4)*arctan((d*x)^(1/2)/(d^2/a)^(1/4))+8/25/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)))

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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)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.75536, size = 512, normalized size = 4.06 \begin{align*} \frac{2 \,{\left (16 \, d^{4} x^{3} \left (\frac{a^{5}}{d^{14}}\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{d x} a^{4} d^{3} \left (\frac{a^{5}}{d^{14}}\right )^{\frac{1}{4}} - \sqrt{a^{5} d^{8} \sqrt{\frac{a^{5}}{d^{14}}} + a^{8} d x} d^{3} \left (\frac{a^{5}}{d^{14}}\right )^{\frac{1}{4}}}{a^{5}}\right ) + 4 \, d^{4} x^{3} \left (\frac{a^{5}}{d^{14}}\right )^{\frac{1}{4}} \log \left (512 \, d^{11} \left (\frac{a^{5}}{d^{14}}\right )^{\frac{3}{4}} + 512 \, \sqrt{d x} a^{4}\right ) - 4 \, d^{4} x^{3} \left (\frac{a^{5}}{d^{14}}\right )^{\frac{1}{4}} \log \left (-512 \, d^{11} \left (\frac{a^{5}}{d^{14}}\right )^{\frac{3}{4}} + 512 \, \sqrt{d x} a^{4}\right ) -{\left (16 \, a x^{2} + 5 \,{\rm Li}_2\left (a x^{2}\right ) - 4 \, \log \left (-a x^{2} + 1\right )\right )} \sqrt{d x}\right )}}{25 \, d^{4} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/(d*x)^(7/2),x, algorithm="fricas")

[Out]

2/25*(16*d^4*x^3*(a^5/d^14)^(1/4)*arctan(-(sqrt(d*x)*a^4*d^3*(a^5/d^14)^(1/4) - sqrt(a^5*d^8*sqrt(a^5/d^14) +
a^8*d*x)*d^3*(a^5/d^14)^(1/4))/a^5) + 4*d^4*x^3*(a^5/d^14)^(1/4)*log(512*d^11*(a^5/d^14)^(3/4) + 512*sqrt(d*x)
*a^4) - 4*d^4*x^3*(a^5/d^14)^(1/4)*log(-512*d^11*(a^5/d^14)^(3/4) + 512*sqrt(d*x)*a^4) - (16*a*x^2 + 5*dilog(a
*x^2) - 4*log(-a*x^2 + 1))*sqrt(d*x))/(d^4*x^3)

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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)**(7/2),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{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,a*x^2)/(d*x)^(7/2),x, algorithm="giac")

[Out]

integrate(dilog(a*x^2)/(d*x)^(7/2), x)