3.1.81 \(\int \frac {\text {PolyLog}(3,a x^2)}{\sqrt {d x}} \, dx\) [81]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 134, 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, 327, 335, 218, 214, 211} \begin {gather*} -\frac {64 \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt [4]{a} \sqrt {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 {32 \sqrt {d x} \log \left (1-a x^2\right )}{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} \end {gather*}

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 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 327

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 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 )}{\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 \text {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 \text {Subst}\left (\int \frac {1}{d-\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )-64 \text {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*}

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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 = 68, normalized size = 0.51 \begin {gather*} -\frac {5 x \Gamma \left (\frac {5}{4}\right ) \left (-64+64 \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};a x^2\right )+16 \log \left (1-a x^2\right )+4 \text {PolyLog}\left (2,a x^2\right )-\text {PolyLog}\left (3,a x^2\right )\right )}{2 \sqrt {d x} \Gamma \left (\frac {9}{4}\right )} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.16, size = 147, normalized size = 1.10

method	result	size
meijerg	\(-\frac {\sqrt {x}\, \left (\frac {256 \sqrt {x}\, \left (-a \right )^{\frac {5}{4}}}{a}+\frac {64 \sqrt {x}\, \left (-a \right )^{\frac {5}{4}} \left (\ln \left (1-\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (a \,x^{2}\right )^{\frac {1}{4}}\right )\right )}{a \left (a \,x^{2}\right )^{\frac {1}{4}}}-\frac {64 \sqrt {x}\, \left (-a \right )^{\frac {5}{4}} \ln \left (-a \,x^{2}+1\right )}{a}-\frac {16 \sqrt {x}\, \left (-a \right )^{\frac {5}{4}} \polylog \left (2, a \,x^{2}\right )}{a}+\frac {4 \sqrt {x}\, \left (-a \right )^{\frac {5}{4}} \polylog \left (3, a \,x^{2}\right )}{a}\right )}{2 \sqrt {d x}\, \left (-a \right )^{\frac {1}{4}}}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)/a*polylog(2,a*x^2)+4*x^(1/2)*(-a)^(5/4)/a*polylog(3,a*x^2))

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Maxima [A]
time = 0.48, size = 141, normalized size = 1.05 \begin {gather*} \frac {2 \, {\left (32 \, \sqrt {d x} {\left (\log \left (d\right ) + 2\right )} - \frac {32 \, d \arctan \left (\frac {\sqrt {d x} \sqrt {a}}{\sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d}} - 4 \, \sqrt {d x} {\rm Li}_2\left (a x^{2}\right ) - 16 \, \sqrt {d x} \log \left (-a d^{2} x^{2} + d^{2}\right ) + \frac {16 \, d \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}} + \sqrt {d x} {\rm Li}_{3}(a x^{2})\right )}}{d} \end {gather*}

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]

2*(32*sqrt(d*x)*(log(d) + 2) - 32*d*arctan(sqrt(d*x)*sqrt(a)/sqrt(sqrt(a)*d))/sqrt(sqrt(a)*d) - 4*sqrt(d*x)*di
log(a*x^2) - 16*sqrt(d*x)*log(-a*d^2*x^2 + d^2) + 16*d*log((sqrt(d*x)*sqrt(a) - sqrt(sqrt(a)*d))/(sqrt(d*x)*sq
rt(a) + sqrt(sqrt(a)*d)))/sqrt(sqrt(a)*d) + sqrt(d*x)*polylog(3, a*x^2))/d

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Fricas [A]
time = 0.38, size = 169, normalized size = 1.26 \begin {gather*} \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 ) - 4 \, \sqrt {d x} {\left ({\rm Li}_2\left (a x^{2}\right ) + 4 \, \log \left (-a x^{2} + 1\right ) - 16\right )} + \sqrt {d x} {\rm polylog}\left (3, a x^{2}\right )\right )}}{d} \end {gather*}

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)) - 4*sqrt(d*x)*(dilog(a*x^2) + 4*log(-a*x^2 + 1) - 16) + sqrt(d*x)*polylog(3, a*x^2)
)/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {Li}_{3}\left (a x^{2}\right )}{\sqrt {d x}}\, dx \end {gather*}

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 = 0.00, 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)^(1/2),x, algorithm="giac")

[Out]

integrate(polylog(3, a*x^2)/sqrt(d*x), 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 )}{\sqrt {d\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x^2)/(d*x)^(1/2),x)

[Out]

int(polylog(3, a*x^2)/(d*x)^(1/2), x)

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