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

Optimal. Leaf size=134 \[ -\frac{8 \sqrt{d x} \text{PolyLog}\left (2,a x^2\right )}{d}+\frac{2 \sqrt{d x} \text{PolyLog}\left (3,a x^2\right )}{d}-\frac{32 \sqrt{d x} \log \left (1-a x^2\right )}{d}-\frac{64 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{\sqrt [4]{a} \sqrt{d}}-\frac{64 \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{\sqrt [4]{a} \sqrt{d}}+\frac{128 \sqrt{d x}}{d} \]

[Out]

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

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Rubi [A]  time = 0.101164, antiderivative size = 134, 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, 321, 329, 212, 208, 205} \[ -\frac{8 \sqrt{d x} \text{PolyLog}\left (2,a x^2\right )}{d}+\frac{2 \sqrt{d x} \text{PolyLog}\left (3,a x^2\right )}{d}-\frac{32 \sqrt{d x} \log \left (1-a x^2\right )}{d}-\frac{64 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{\sqrt [4]{a} \sqrt{d}}-\frac{64 \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{\sqrt [4]{a} \sqrt{d}}+\frac{128 \sqrt{d x}}{d} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^2]/Sqrt[d*x],x]

[Out]

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

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 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 )}{\sqrt{d x}} \, dx &=\frac{2 \sqrt{d x} \text{Li}_3\left (a x^2\right )}{d}-4 \int \frac{\text{Li}_2\left (a x^2\right )}{\sqrt{d x}} \, dx\\ &=-\frac{8 \sqrt{d x} \text{Li}_2\left (a x^2\right )}{d}+\frac{2 \sqrt{d x} \text{Li}_3\left (a x^2\right )}{d}-16 \int \frac{\log \left (1-a x^2\right )}{\sqrt{d x}} \, dx\\ &=-\frac{32 \sqrt{d x} \log \left (1-a x^2\right )}{d}-\frac{8 \sqrt{d x} \text{Li}_2\left (a x^2\right )}{d}+\frac{2 \sqrt{d x} \text{Li}_3\left (a x^2\right )}{d}-\frac{(64 a) \int \frac{x \sqrt{d x}}{1-a x^2} \, dx}{d}\\ &=-\frac{32 \sqrt{d x} \log \left (1-a x^2\right )}{d}-\frac{8 \sqrt{d x} \text{Li}_2\left (a x^2\right )}{d}+\frac{2 \sqrt{d x} \text{Li}_3\left (a x^2\right )}{d}-\frac{(64 a) \int \frac{(d x)^{3/2}}{1-a x^2} \, dx}{d^2}\\ &=\frac{128 \sqrt{d x}}{d}-\frac{32 \sqrt{d x} \log \left (1-a x^2\right )}{d}-\frac{8 \sqrt{d x} \text{Li}_2\left (a x^2\right )}{d}+\frac{2 \sqrt{d x} \text{Li}_3\left (a x^2\right )}{d}-64 \int \frac{1}{\sqrt{d x} \left (1-a x^2\right )} \, dx\\ &=\frac{128 \sqrt{d x}}{d}-\frac{32 \sqrt{d x} \log \left (1-a x^2\right )}{d}-\frac{8 \sqrt{d x} \text{Li}_2\left (a x^2\right )}{d}+\frac{2 \sqrt{d x} \text{Li}_3\left (a x^2\right )}{d}-\frac{128 \operatorname{Subst}\left (\int \frac{1}{1-\frac{a x^4}{d^2}} \, dx,x,\sqrt{d x}\right )}{d}\\ &=\frac{128 \sqrt{d x}}{d}-\frac{32 \sqrt{d x} \log \left (1-a x^2\right )}{d}-\frac{8 \sqrt{d x} \text{Li}_2\left (a x^2\right )}{d}+\frac{2 \sqrt{d x} \text{Li}_3\left (a x^2\right )}{d}-64 \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{a} x^2} \, dx,x,\sqrt{d x}\right )-64 \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{a} x^2} \, dx,x,\sqrt{d x}\right )\\ &=\frac{128 \sqrt{d x}}{d}-\frac{64 \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{\sqrt [4]{a} \sqrt{d}}-\frac{64 \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{\sqrt [4]{a} \sqrt{d}}-\frac{32 \sqrt{d x} \log \left (1-a x^2\right )}{d}-\frac{8 \sqrt{d x} \text{Li}_2\left (a x^2\right )}{d}+\frac{2 \sqrt{d x} \text{Li}_3\left (a x^2\right )}{d}\\ \end{align*}

Mathematica [C]  time = 0.0861366, size = 68, normalized size = 0.51 \[ -\frac{5 x \text{Gamma}\left (\frac{5}{4}\right ) \left (64 \text{Hypergeometric2F1}\left (\frac{1}{4},1,\frac{5}{4},a x^2\right )+4 \text{PolyLog}\left (2,a x^2\right )-\text{PolyLog}\left (3,a x^2\right )+16 \log \left (1-a x^2\right )-64\right )}{2 \text{Gamma}\left (\frac{9}{4}\right ) \sqrt{d x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.191, size = 147, normalized size = 1.1 \begin{align*} -{\frac{1}{2}\sqrt{x} \left ( 256\,{\frac{\sqrt{x} \left ( -a \right ) ^{5/4}}{a}}+64\,{\frac{\sqrt{x} \left ( -a \right ) ^{5/4} \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 ) }{a\sqrt [4]{a{x}^{2}}}}-64\,{\frac{\sqrt{x} \left ( -a \right ) ^{5/4}\ln \left ( -a{x}^{2}+1 \right ) }{a}}-16\,{\frac{\sqrt{x} \left ( -a \right ) ^{5/4}{\it polylog} \left ( 2,a{x}^{2} \right ) }{a}}+4\,{\frac{\sqrt{x} \left ( -a \right ) ^{5/4}{\it polylog} \left ( 3,a{x}^{2} \right ) }{a}} \right ){\frac{1}{\sqrt{dx}}}{\frac{1}{\sqrt [4]{-a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3,a*x^2)/(d*x)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.8615, size = 521, normalized size = 3.89 \begin{align*} \frac{2 \,{\left (64 \, d \left (\frac{1}{a d^{2}}\right )^{\frac{1}{4}} \arctan \left (\sqrt{d^{2} \sqrt{\frac{1}{a d^{2}}} + d x} a d \left (\frac{1}{a d^{2}}\right )^{\frac{3}{4}} - \sqrt{d x} a d \left (\frac{1}{a d^{2}}\right )^{\frac{3}{4}}\right ) - 16 \, d \left (\frac{1}{a d^{2}}\right )^{\frac{1}{4}} \log \left (d \left (\frac{1}{a d^{2}}\right )^{\frac{1}{4}} + \sqrt{d x}\right ) + 16 \, d \left (\frac{1}{a d^{2}}\right )^{\frac{1}{4}} \log \left (-d \left (\frac{1}{a d^{2}}\right )^{\frac{1}{4}} + \sqrt{d x}\right ) - 16 \, \sqrt{d x}{\left (\log \left (-a x^{2} + 1\right ) - 4\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 ) + \sqrt{d x}{\rm polylog}\left (3, a x^{2}\right )\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/(d*x)^(1/2),x, algorithm="fricas")

[Out]

2*(64*d*(1/(a*d^2))^(1/4)*arctan(sqrt(d^2*sqrt(1/(a*d^2)) + d*x)*a*d*(1/(a*d^2))^(3/4) - sqrt(d*x)*a*d*(1/(a*d
^2))^(3/4)) - 16*d*(1/(a*d^2))^(1/4)*log(d*(1/(a*d^2))^(1/4) + sqrt(d*x)) + 16*d*(1/(a*d^2))^(1/4)*log(-d*(1/(
a*d^2))^(1/4) + sqrt(d*x)) - 16*sqrt(d*x)*(log(-a*x^2 + 1) - 4) - 4*sqrt(d*x)*\%iint(a, x, -log(-a*x^2 + 1)/a,
-2*log(-a*x^2 + 1)/x) + sqrt(d*x)*polylog(3, a*x^2))/d

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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 )}{\sqrt{d x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x**2)/(d*x)**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/(d*x)^(1/2),x, algorithm="giac")

[Out]

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