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

Optimal. Leaf size=140 \[ \frac{2 (d x)^{5/2} \text{PolyLog}\left (2,a x^2\right )}{5 d}+\frac{16 d^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{25 a^{5/4}}+\frac{16 d^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{25 a^{5/4}}+\frac{8 (d x)^{5/2} \log \left (1-a x^2\right )}{25 d}-\frac{32 d \sqrt{d x}}{25 a}-\frac{32 (d x)^{5/2}}{125 d} \]

[Out]

(-32*d*Sqrt[d*x])/(25*a) - (32*(d*x)^(5/2))/(125*d) + (16*d^(3/2)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(25*a^(
5/4)) + (16*d^(3/2)*ArcTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(25*a^(5/4)) + (8*(d*x)^(5/2)*Log[1 - a*x^2])/(25*d)
+ (2*(d*x)^(5/2)*PolyLog[2, a*x^2])/(5*d)

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Rubi [A]  time = 0.10671, antiderivative size = 140, 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} \[ \frac{2 (d x)^{5/2} \text{PolyLog}\left (2,a x^2\right )}{5 d}+\frac{16 d^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{25 a^{5/4}}+\frac{16 d^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{a} \sqrt{d x}}{\sqrt{d}}\right )}{25 a^{5/4}}+\frac{8 (d x)^{5/2} \log \left (1-a x^2\right )}{25 d}-\frac{32 d \sqrt{d x}}{25 a}-\frac{32 (d x)^{5/2}}{125 d} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(3/2)*PolyLog[2, a*x^2],x]

[Out]

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

Mathematica [A]  time = 0.104193, size = 101, normalized size = 0.72 \[ \frac{2 (d x)^{3/2} \left (25 x^{5/2} \text{PolyLog}\left (2,a x^2\right )+\frac{4 \sqrt [4]{a} \sqrt{x} \left (-4 a x^2+5 a x^2 \log \left (1-a x^2\right )-20\right )+40 \tan ^{-1}\left (\sqrt [4]{a} \sqrt{x}\right )+40 \tanh ^{-1}\left (\sqrt [4]{a} \sqrt{x}\right )}{a^{5/4}}\right )}{125 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*x)^(3/2)*((40*ArcTan[a^(1/4)*Sqrt[x]] + 40*ArcTanh[a^(1/4)*Sqrt[x]] + 4*a^(1/4)*Sqrt[x]*(-20 - 4*a*x^2 +
5*a*x^2*Log[1 - a*x^2]))/a^(5/4) + 25*x^(5/2)*PolyLog[2, a*x^2]))/(125*x^(3/2))

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Maple [A]  time = 0.214, size = 150, 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} \left ( dx \right ) ^{{\frac{5}{2}}}\ln \left ({\frac{-a{d}^{2}{x}^{2}+{d}^{2}}{{d}^{2}}} \right ) }-{\frac{32}{125\,d} \left ( dx \right ) ^{{\frac{5}{2}}}}-{\frac{32\,d}{25\,a}\sqrt{dx}}+{\frac{8\,d}{25\,a}\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\,d}{25\,a}\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((d*x)^(3/2)*polylog(2,a*x^2),x)

[Out]

2/5*(d*x)^(5/2)*polylog(2,a*x^2)/d+8/25/d*(d*x)^(5/2)*ln((-a*d^2*x^2+d^2)/d^2)-32/125*(d*x)^(5/2)/d-32/25*d*(d
*x)^(1/2)/a+8/25*d/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)))+16/25*d/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((d*x)^(3/2)*polylog(2,a*x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.84643, size = 471, normalized size = 3.36 \begin{align*} -\frac{2 \,{\left (80 \, a \left (\frac{d^{6}}{a^{5}}\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{d x} a^{4} d \left (\frac{d^{6}}{a^{5}}\right )^{\frac{3}{4}} - \sqrt{d^{3} x + a^{2} \sqrt{\frac{d^{6}}{a^{5}}}} a^{4} \left (\frac{d^{6}}{a^{5}}\right )^{\frac{3}{4}}}{d^{6}}\right ) - 20 \, a \left (\frac{d^{6}}{a^{5}}\right )^{\frac{1}{4}} \log \left (8 \, \sqrt{d x} d + 8 \, a \left (\frac{d^{6}}{a^{5}}\right )^{\frac{1}{4}}\right ) + 20 \, a \left (\frac{d^{6}}{a^{5}}\right )^{\frac{1}{4}} \log \left (8 \, \sqrt{d x} d - 8 \, a \left (\frac{d^{6}}{a^{5}}\right )^{\frac{1}{4}}\right ) -{\left (25 \, a d x^{2}{\rm Li}_2\left (a x^{2}\right ) + 20 \, a d x^{2} \log \left (-a x^{2} + 1\right ) - 16 \, a d x^{2} - 80 \, d\right )} \sqrt{d x}\right )}}{125 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/125*(80*a*(d^6/a^5)^(1/4)*arctan(-(sqrt(d*x)*a^4*d*(d^6/a^5)^(3/4) - sqrt(d^3*x + a^2*sqrt(d^6/a^5))*a^4*(d
^6/a^5)^(3/4))/d^6) - 20*a*(d^6/a^5)^(1/4)*log(8*sqrt(d*x)*d + 8*a*(d^6/a^5)^(1/4)) + 20*a*(d^6/a^5)^(1/4)*log
(8*sqrt(d*x)*d - 8*a*(d^6/a^5)^(1/4)) - (25*a*d*x^2*dilog(a*x^2) + 20*a*d*x^2*log(-a*x^2 + 1) - 16*a*d*x^2 - 8
0*d)*sqrt(d*x))/a

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{\frac{3}{2}}{\rm Li}_2\left (a x^{2}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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