3.1.85 \(\int \frac {\text {PolyLog}(3,a x^2)}{(d x)^{9/2}} \, dx\) [85]

Optimal. Leaf size=147 \[ -\frac {128 a}{1029 d^3 (d x)^{3/2}}+\frac {64 a^{7/4} \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{343 d^{9/2}}+\frac {64 a^{7/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{343 d^{9/2}}+\frac {32 \log \left (1-a x^2\right )}{343 d (d x)^{7/2}}-\frac {8 \text {PolyLog}\left (2,a x^2\right )}{49 d (d x)^{7/2}}-\frac {2 \text {PolyLog}\left (3,a x^2\right )}{7 d (d x)^{7/2}} \]

[Out]

-128/1029*a/d^3/(d*x)^(3/2)+64/343*a^(7/4)*arctan(a^(1/4)*(d*x)^(1/2)/d^(1/2))/d^(9/2)+64/343*a^(7/4)*arctanh(
a^(1/4)*(d*x)^(1/2)/d^(1/2))/d^(9/2)+32/343*ln(-a*x^2+1)/d/(d*x)^(7/2)-8/49*polylog(2,a*x^2)/d/(d*x)^(7/2)-2/7
*polylog(3,a*x^2)/d/(d*x)^(7/2)

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Rubi [A]
time = 0.07, antiderivative size = 147, normalized size of antiderivative = 1.00, 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 = {6726, 2505, 16, 331, 335, 218, 214, 211} \begin {gather*} \frac {64 a^{7/4} \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{343 d^{9/2}}+\frac {64 a^{7/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{343 d^{9/2}}-\frac {128 a}{1029 d^3 (d x)^{3/2}}-\frac {8 \text {Li}_2\left (a x^2\right )}{49 d (d x)^{7/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{7 d (d x)^{7/2}}+\frac {32 \log \left (1-a x^2\right )}{343 d (d x)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^2]/(d*x)^(9/2),x]

[Out]

(-128*a)/(1029*d^3*(d*x)^(3/2)) + (64*a^(7/4)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(343*d^(9/2)) + (64*a^(7/4)
*ArcTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(343*d^(9/2)) + (32*Log[1 - a*x^2])/(343*d*(d*x)^(7/2)) - (8*PolyLog[2,
a*x^2])/(49*d*(d*x)^(7/2)) - (2*PolyLog[3, a*x^2])/(7*d*(d*x)^(7/2))

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 331

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 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}_3\left (a x^2\right )}{(d x)^{9/2}} \, dx &=-\frac {2 \text {Li}_3\left (a x^2\right )}{7 d (d x)^{7/2}}+\frac {4}{7} \int \frac {\text {Li}_2\left (a x^2\right )}{(d x)^{9/2}} \, dx\\ &=-\frac {8 \text {Li}_2\left (a x^2\right )}{49 d (d x)^{7/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{7 d (d x)^{7/2}}-\frac {16}{49} \int \frac {\log \left (1-a x^2\right )}{(d x)^{9/2}} \, dx\\ &=\frac {32 \log \left (1-a x^2\right )}{343 d (d x)^{7/2}}-\frac {8 \text {Li}_2\left (a x^2\right )}{49 d (d x)^{7/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{7 d (d x)^{7/2}}+\frac {(64 a) \int \frac {x}{(d x)^{7/2} \left (1-a x^2\right )} \, dx}{343 d}\\ &=\frac {32 \log \left (1-a x^2\right )}{343 d (d x)^{7/2}}-\frac {8 \text {Li}_2\left (a x^2\right )}{49 d (d x)^{7/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{7 d (d x)^{7/2}}+\frac {(64 a) \int \frac {1}{(d x)^{5/2} \left (1-a x^2\right )} \, dx}{343 d^2}\\ &=-\frac {128 a}{1029 d^3 (d x)^{3/2}}+\frac {32 \log \left (1-a x^2\right )}{343 d (d x)^{7/2}}-\frac {8 \text {Li}_2\left (a x^2\right )}{49 d (d x)^{7/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{7 d (d x)^{7/2}}+\frac {\left (64 a^2\right ) \int \frac {1}{\sqrt {d x} \left (1-a x^2\right )} \, dx}{343 d^4}\\ &=-\frac {128 a}{1029 d^3 (d x)^{3/2}}+\frac {32 \log \left (1-a x^2\right )}{343 d (d x)^{7/2}}-\frac {8 \text {Li}_2\left (a x^2\right )}{49 d (d x)^{7/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{7 d (d x)^{7/2}}+\frac {\left (128 a^2\right ) \text {Subst}\left (\int \frac {1}{1-\frac {a x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{343 d^5}\\ &=-\frac {128 a}{1029 d^3 (d x)^{3/2}}+\frac {32 \log \left (1-a x^2\right )}{343 d (d x)^{7/2}}-\frac {8 \text {Li}_2\left (a x^2\right )}{49 d (d x)^{7/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{7 d (d x)^{7/2}}+\frac {\left (64 a^2\right ) \text {Subst}\left (\int \frac {1}{d-\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{343 d^4}+\frac {\left (64 a^2\right ) \text {Subst}\left (\int \frac {1}{d+\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{343 d^4}\\ &=-\frac {128 a}{1029 d^3 (d x)^{3/2}}+\frac {64 a^{7/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{343 d^{9/2}}+\frac {64 a^{7/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{343 d^{9/2}}+\frac {32 \log \left (1-a x^2\right )}{343 d (d x)^{7/2}}-\frac {8 \text {Li}_2\left (a x^2\right )}{49 d (d x)^{7/2}}-\frac {2 \text {Li}_3\left (a x^2\right )}{7 d (d x)^{7/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.07, size = 84, normalized size = 0.57 \begin {gather*} -\frac {\sqrt {d x} \Gamma \left (-\frac {3}{4}\right ) \left (-64 a x^2+192 a^2 x^4 \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};a x^2\right )+48 \log \left (1-a x^2\right )-84 \text {PolyLog}\left (2,a x^2\right )-147 \text {PolyLog}\left (3,a x^2\right )\right )}{686 d^5 x^4 \Gamma \left (\frac {1}{4}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[3, a*x^2]/(d*x)^(9/2),x]

[Out]

-1/686*(Sqrt[d*x]*Gamma[-3/4]*(-64*a*x^2 + 192*a^2*x^4*Hypergeometric2F1[1/4, 1, 5/4, a*x^2] + 48*Log[1 - a*x^
2] - 84*PolyLog[2, a*x^2] - 147*PolyLog[3, a*x^2]))/(d^5*x^4*Gamma[1/4])

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Maple [A]
time = 0.16, size = 142, normalized size = 0.97

method	result	size
meijerg	\(-\frac {x^{\frac {9}{2}} \left (-a \right )^{\frac {7}{4}} \left (-\frac {256}{1029 x^{\frac {3}{2}} \left (-a \right )^{\frac {3}{4}}}-\frac {64 \sqrt {x}\, a \left (\ln \left (1-\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (a \,x^{2}\right )^{\frac {1}{4}}\right )\right )}{343 \left (-a \right )^{\frac {3}{4}} \left (a \,x^{2}\right )^{\frac {1}{4}}}+\frac {64 \ln \left (-a \,x^{2}+1\right )}{343 x^{\frac {7}{2}} \left (-a \right )^{\frac {3}{4}} a}-\frac {16 \polylog \left (2, a \,x^{2}\right )}{49 x^{\frac {7}{2}} \left (-a \right )^{\frac {3}{4}} a}-\frac {4 \polylog \left (3, a \,x^{2}\right )}{7 x^{\frac {7}{2}} \left (-a \right )^{\frac {3}{4}} a}\right )}{2 \left (d x \right )^{\frac {9}{2}}}\)	\(142\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3,a*x^2)/(d*x)^(9/2),x,method=_RETURNVERBOSE)

[Out]

-1/2/(d*x)^(9/2)*x^(9/2)*(-a)^(7/4)*(-256/1029/x^(3/2)/(-a)^(3/4)-64/343*x^(1/2)/(-a)^(3/4)*a/(a*x^2)^(1/4)*(l
n(1-(a*x^2)^(1/4))-ln(1+(a*x^2)^(1/4))-2*arctan((a*x^2)^(1/4)))+64/343/x^(7/2)/(-a)^(3/4)/a*ln(-a*x^2+1)-16/49
/x^(7/2)/(-a)^(3/4)/a*polylog(2,a*x^2)-4/7/x^(7/2)/(-a)^(3/4)/a*polylog(3,a*x^2))

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Maxima [A]
time = 0.48, size = 168, normalized size = 1.14 \begin {gather*} \frac {2 \, {\left (\frac {48 \, {\left (\frac {2 \, a^{2} \arctan \left (\frac {\sqrt {d x} \sqrt {a}}{\sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d} d} - \frac {a^{2} \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}\right )}}{d^{2}} - \frac {64 \, a d^{2} x^{2} + 84 \, d^{2} {\rm Li}_2\left (a x^{2}\right ) - 48 \, d^{2} \log \left (-a d^{2} x^{2} + d^{2}\right ) + 96 \, d^{2} \log \left (d\right ) + 147 \, d^{2} {\rm Li}_{3}(a x^{2})}{\left (d x\right )^{\frac {7}{2}} d^{2}}\right )}}{1029 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/(d*x)^(9/2),x, algorithm="maxima")

[Out]

2/1029*(48*(2*a^2*arctan(sqrt(d*x)*sqrt(a)/sqrt(sqrt(a)*d))/(sqrt(sqrt(a)*d)*d) - a^2*log((sqrt(d*x)*sqrt(a) -
sqrt(sqrt(a)*d))/(sqrt(d*x)*sqrt(a) + sqrt(sqrt(a)*d)))/(sqrt(sqrt(a)*d)*d))/d^2 - (64*a*d^2*x^2 + 84*d^2*dil
og(a*x^2) - 48*d^2*log(-a*d^2*x^2 + d^2) + 96*d^2*log(d) + 147*d^2*polylog(3, a*x^2))/((d*x)^(7/2)*d^2))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (106) = 212\).
time = 0.41, size = 223, normalized size = 1.52 \begin {gather*} -\frac {2 \, {\left (192 \, d^{5} x^{4} \left (\frac {a^{7}}{d^{18}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {d x} a^{2} d^{13} \left (\frac {a^{7}}{d^{18}}\right )^{\frac {3}{4}} - \sqrt {d^{10} \sqrt {\frac {a^{7}}{d^{18}}} + a^{4} d x} d^{13} \left (\frac {a^{7}}{d^{18}}\right )^{\frac {3}{4}}}{a^{7}}\right ) - 48 \, d^{5} x^{4} \left (\frac {a^{7}}{d^{18}}\right )^{\frac {1}{4}} \log \left (32 \, d^{5} \left (\frac {a^{7}}{d^{18}}\right )^{\frac {1}{4}} + 32 \, \sqrt {d x} a^{2}\right ) + 48 \, d^{5} x^{4} \left (\frac {a^{7}}{d^{18}}\right )^{\frac {1}{4}} \log \left (-32 \, d^{5} \left (\frac {a^{7}}{d^{18}}\right )^{\frac {1}{4}} + 32 \, \sqrt {d x} a^{2}\right ) + 4 \, {\left (16 \, a x^{2} + 21 \, {\rm Li}_2\left (a x^{2}\right ) - 12 \, \log \left (-a x^{2} + 1\right )\right )} \sqrt {d x} + 147 \, \sqrt {d x} {\rm polylog}\left (3, a x^{2}\right )\right )}}{1029 \, d^{5} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(3,a*x^2)/(d*x)^(9/2),x, algorithm="fricas")

[Out]

-2/1029*(192*d^5*x^4*(a^7/d^18)^(1/4)*arctan(-(sqrt(d*x)*a^2*d^13*(a^7/d^18)^(3/4) - sqrt(d^10*sqrt(a^7/d^18)
+ a^4*d*x)*d^13*(a^7/d^18)^(3/4))/a^7) - 48*d^5*x^4*(a^7/d^18)^(1/4)*log(32*d^5*(a^7/d^18)^(1/4) + 32*sqrt(d*x
)*a^2) + 48*d^5*x^4*(a^7/d^18)^(1/4)*log(-32*d^5*(a^7/d^18)^(1/4) + 32*sqrt(d*x)*a^2) + 4*(16*a*x^2 + 21*dilog
(a*x^2) - 12*log(-a*x^2 + 1))*sqrt(d*x) + 147*sqrt(d*x)*polylog(3, a*x^2))/(d^5*x^4)

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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(3,a*x**2)/(d*x)**(9/2),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(3,a*x^2)/(d*x)^(9/2),x, algorithm="giac")

[Out]

integrate(polylog(3, a*x^2)/(d*x)^(9/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {polylog}\left (3,a\,x^2\right )}{{\left (d\,x\right )}^{9/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x^2)/(d*x)^(9/2),x)

[Out]

int(polylog(3, a*x^2)/(d*x)^(9/2), x)

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