3.84 \(\int \frac{\text{PolyLog}(3,a x^2)}{(d x)^{7/2}} \, dx\)

Optimal. Leaf size=147 \[ -\frac{8 \text{PolyLog}\left (2,a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}}-\frac{64 a^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{125 d^{7/2}}+\frac{64 a^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{125 d^{7/2}}-\frac{128 a}{125 d^3 \sqrt{d x}}+\frac{32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}} \]

[Out]

(-128*a)/(125*d^3*Sqrt[d*x]) - (64*a^(5/4)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(125*d^(7/2)) + (64*a^(5/4)*Ar
cTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(125*d^(7/2)) + (32*Log[1 - a*x^2])/(125*d*(d*x)^(5/2)) - (8*PolyLog[2, a*
x^2])/(25*d*(d*x)^(5/2)) - (2*PolyLog[3, a*x^2])/(5*d*(d*x)^(5/2))

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Rubi [A]  time = 0.106859, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ -\frac{8 \text{PolyLog}\left (2,a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{PolyLog}\left (3,a x^2\right )}{5 d (d x)^{5/2}}-\frac{64 a^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{125 d^{7/2}}+\frac{64 a^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{125 d^{7/2}}-\frac{128 a}{125 d^3 \sqrt{d x}}+\frac{32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^2]/(d*x)^(7/2),x]

[Out]

(-128*a)/(125*d^3*Sqrt[d*x]) - (64*a^(5/4)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(125*d^(7/2)) + (64*a^(5/4)*Ar
cTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(125*d^(7/2)) + (32*Log[1 - a*x^2])/(125*d*(d*x)^(5/2)) - (8*PolyLog[2, a*
x^2])/(25*d*(d*x)^(5/2)) - (2*PolyLog[3, 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}_3\left (a x^2\right )}{(d x)^{7/2}} \, dx &=-\frac{2 \text{Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac{4}{5} \int \frac{\text{Li}_2\left (a x^2\right )}{(d x)^{7/2}} \, dx\\ &=-\frac{8 \text{Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}-\frac{16}{25} \int \frac{\log \left (1-a x^2\right )}{(d x)^{7/2}} \, dx\\ &=\frac{32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}}-\frac{8 \text{Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac{(64 a) \int \frac{x}{(d x)^{5/2} \left (1-a x^2\right )} \, dx}{125 d}\\ &=\frac{32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}}-\frac{8 \text{Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac{(64 a) \int \frac{1}{(d x)^{3/2} \left (1-a x^2\right )} \, dx}{125 d^2}\\ &=-\frac{128 a}{125 d^3 \sqrt{d x}}+\frac{32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}}-\frac{8 \text{Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac{\left (64 a^2\right ) \int \frac{\sqrt{d x}}{1-a x^2} \, dx}{125 d^4}\\ &=-\frac{128 a}{125 d^3 \sqrt{d x}}+\frac{32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}}-\frac{8 \text{Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac{\left (128 a^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{a x^4}{d^2}} \, dx,x,\sqrt{d x}\right )}{125 d^5}\\ &=-\frac{128 a}{125 d^3 \sqrt{d x}}+\frac{32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}}-\frac{8 \text{Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}+\frac{\left (64 a^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{a} x^2} \, dx,x,\sqrt{d x}\right )}{125 d^3}-\frac{\left (64 a^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{a} x^2} \, dx,x,\sqrt{d x}\right )}{125 d^3}\\ &=-\frac{128 a}{125 d^3 \sqrt{d x}}-\frac{64 a^{5/4} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{125 d^{7/2}}+\frac{64 a^{5/4} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{125 d^{7/2}}+\frac{32 \log \left (1-a x^2\right )}{125 d (d x)^{5/2}}-\frac{8 \text{Li}_2\left (a x^2\right )}{25 d (d x)^{5/2}}-\frac{2 \text{Li}_3\left (a x^2\right )}{5 d (d x)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0986821, size = 79, normalized size = 0.54 \[ -\frac{x \text{Gamma}\left (-\frac{1}{4}\right ) \left (64 a^2 x^4 \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},a x^2\right )-60 \text{PolyLog}\left (2,a x^2\right )-75 \text{PolyLog}\left (3,a x^2\right )-192 a x^2+48 \log \left (1-a x^2\right )\right )}{750 \text{Gamma}\left (\frac{3}{4}\right ) (d x)^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[PolyLog[3, a*x^2]/(d*x)^(7/2),x]

[Out]

-(x*Gamma[-1/4]*(-192*a*x^2 + 64*a^2*x^4*Hypergeometric2F1[3/4, 1, 7/4, a*x^2] + 48*Log[1 - a*x^2] - 60*PolyLo
g[2, a*x^2] - 75*PolyLog[3, a*x^2]))/(750*(d*x)^(7/2)*Gamma[3/4])

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Maple [A]  time = 0.206, size = 142, normalized size = 1. \begin{align*} -{\frac{1}{2}{x}^{{\frac{7}{2}}} \left ( -a \right ) ^{{\frac{5}{4}}} \left ( -{\frac{256}{125}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt [4]{-a}}}}-{\frac{64\,a}{125}{x}^{{\frac{3}{2}}} \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}}} \left ( a{x}^{2} \right ) ^{-{\frac{3}{4}}}}+{\frac{64\,\ln \left ( -a{x}^{2}+1 \right ) }{125\,a}{x}^{-{\frac{5}{2}}}{\frac{1}{\sqrt [4]{-a}}}}-{\frac{16\,{\it polylog} \left ( 2,a{x}^{2} \right ) }{25\,a}{x}^{-{\frac{5}{2}}}{\frac{1}{\sqrt [4]{-a}}}}-{\frac{4\,{\it polylog} \left ( 3,a{x}^{2} \right ) }{5\,a}{x}^{-{\frac{5}{2}}}{\frac{1}{\sqrt [4]{-a}}}} \right ) \left ( dx \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3,a*x^2)/(d*x)^(7/2),x)

[Out]

-1/2/(d*x)^(7/2)*x^(7/2)*(-a)^(5/4)*(-256/125/x^(1/2)/(-a)^(1/4)-64/125*x^(3/2)/(-a)^(1/4)*a/(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)))+64/125/x^(5/2)/(-a)^(1/4)/a*ln(-a*x^2+1)-16/25/
x^(5/2)/(-a)^(1/4)*polylog(2,a*x^2)/a-4/5/x^(5/2)/(-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)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [C]  time = 3.00294, size = 641, normalized size = 4.36 \begin{align*} \frac{2 \,{\left (64 \, 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 ) + 16 \, d^{4} x^{3} \left (\frac{a^{5}}{d^{14}}\right )^{\frac{1}{4}} \log \left (32768 \, d^{11} \left (\frac{a^{5}}{d^{14}}\right )^{\frac{3}{4}} + 32768 \, \sqrt{d x} a^{4}\right ) - 16 \, d^{4} x^{3} \left (\frac{a^{5}}{d^{14}}\right )^{\frac{1}{4}} \log \left (-32768 \, d^{11} \left (\frac{a^{5}}{d^{14}}\right )^{\frac{3}{4}} + 32768 \, \sqrt{d x} a^{4}\right ) - 16 \,{\left (4 \, a x^{2} - \log \left (-a x^{2} + 1\right )\right )} \sqrt{d x} - 20 \, \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 ) - 25 \, \sqrt{d x}{\rm polylog}\left (3, a x^{2}\right )\right )}}{125 \, d^{4} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/(d*x)^(7/2),x, algorithm="fricas")

[Out]

2/125*(64*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) + 16*d^4*x^3*(a^5/d^14)^(1/4)*log(32768*d^11*(a^5/d^14)^(3/4) + 32768*sqr
t(d*x)*a^4) - 16*d^4*x^3*(a^5/d^14)^(1/4)*log(-32768*d^11*(a^5/d^14)^(3/4) + 32768*sqrt(d*x)*a^4) - 16*(4*a*x^
2 - log(-a*x^2 + 1))*sqrt(d*x) - 20*sqrt(d*x)*\%iint(a, x, -log(-a*x^2 + 1)/a, -2*log(-a*x^2 + 1)/x) - 25*sqrt(
d*x)*polylog(3, a*x^2))/(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(3,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}_{3}(a x^{2})}{\left (d x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/(d*x)^(7/2),x, algorithm="giac")

[Out]

integrate(polylog(3, a*x^2)/(d*x)^(7/2), x)