3.1.76 \(\int \frac {\text {PolyLog}(2,a x^2)}{(d x)^{5/2}} \, dx\) [76]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 111, normalized size of antiderivative = 1.00, 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 = {6726, 2505, 16, 335, 218, 214, 211} \begin {gather*} \frac {16 a^{3/4} \text {ArcTan}\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 {2 \text {Li}_2\left (a x^2\right )}{3 d (d x)^{3/2}}+\frac {8 \log \left (1-a x^2\right )}{9 d (d x)^{3/2}} \end {gather*}

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 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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 335

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 2505

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 6726

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]

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) \text {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) \text {Subst}\left (\int \frac {1}{d-\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{9 d^2}+\frac {(16 a) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.05, size = 62, normalized size = 0.56 \begin {gather*} \frac {x \Gamma \left (\frac {1}{4}\right ) \left (16 a x^2 \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};a x^2\right )+4 \log \left (1-a x^2\right )-3 \text {PolyLog}\left (2,a x^2\right )\right )}{18 (d x)^{5/2} \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.39, size = 116, normalized size = 1.05

method	result	size
meijerg	\(-\frac {x^{\frac {5}{2}} \left (-a \right )^{\frac {3}{4}} \left (-\frac {16 \sqrt {x}\, \left (-a \right )^{\frac {1}{4}} \left (\ln \left (1-\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (a \,x^{2}\right )^{\frac {1}{4}}\right )\right )}{9 \left (a \,x^{2}\right )^{\frac {1}{4}}}+\frac {16 \left (-a \right )^{\frac {1}{4}} \ln \left (-a \,x^{2}+1\right )}{9 x^{\frac {3}{2}} a}-\frac {4 \left (-a \right )^{\frac {1}{4}} \polylog \left (2, a \,x^{2}\right )}{3 x^{\frac {3}{2}} a}\right )}{2 \left (d x \right )^{\frac {5}{2}}}\)	\(111\)
derivativedivides	\(\frac {-\frac {2 \polylog \left (2, a \,x^{2}\right )}{3 \left (d x \right )^{\frac {3}{2}}}+\frac {8 \ln \left (\frac {-a \,d^{2} x^{2}+d^{2}}{d^{2}}\right )}{9 \left (d x \right )^{\frac {3}{2}}}+\frac {8 a \left (\frac {d^{2}}{a}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )\right )}{9 d^{2}}}{d}\)	\(116\)
default	\(\frac {-\frac {2 \polylog \left (2, a \,x^{2}\right )}{3 \left (d x \right )^{\frac {3}{2}}}+\frac {8 \ln \left (\frac {-a \,d^{2} x^{2}+d^{2}}{d^{2}}\right )}{9 \left (d x \right )^{\frac {3}{2}}}+\frac {8 a \left (\frac {d^{2}}{a}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )\right )}{9 d^{2}}}{d}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(-1/3*polylog(2,a*x^2)/(d*x)^(3/2)+4/9/(d*x)^(3/2)*ln((-a*d^2*x^2+d^2)/d^2)+4/9*a*(d^2/a)^(1/4)/d^2*(ln(((
d*x)^(1/2)+(d^2/a)^(1/4))/((d*x)^(1/2)-(d^2/a)^(1/4)))+2*arctan((d*x)^(1/2)/(d^2/a)^(1/4))))

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Maxima [A]
time = 0.47, size = 125, normalized size = 1.13 \begin {gather*} \frac {2 \, {\left (\frac {8 \, a \arctan \left (\frac {\sqrt {d x} \sqrt {a}}{\sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d} d} - \frac {4 \, a \log \left (\frac {\sqrt {d x} \sqrt {a} - \sqrt {\sqrt {a} d}}{\sqrt {d x} \sqrt {a} + \sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d} d} - \frac {3 \, {\rm Li}_2\left (a x^{2}\right ) - 4 \, \log \left (-a d^{2} x^{2} + d^{2}\right ) + 8 \, \log \left (d\right )}{\left (d x\right )^{\frac {3}{2}}}\right )}}{9 \, d} \end {gather*}

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]

2/9*(8*a*arctan(sqrt(d*x)*sqrt(a)/sqrt(sqrt(a)*d))/(sqrt(sqrt(a)*d)*d) - 4*a*log((sqrt(d*x)*sqrt(a) - sqrt(sqr
t(a)*d))/(sqrt(d*x)*sqrt(a) + sqrt(sqrt(a)*d)))/(sqrt(sqrt(a)*d)*d) - (3*dilog(a*x^2) - 4*log(-a*d^2*x^2 + d^2
) + 8*log(d))/(d*x)^(3/2))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (78) = 156\).
time = 0.45, size = 196, normalized size = 1.77 \begin {gather*} -\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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {polylog}\left (2,a\,x^2\right )}{{\left (d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x^2)/(d*x)^(5/2),x)

[Out]

int(polylog(2, a*x^2)/(d*x)^(5/2), x)

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