3.1.82 \(\int \frac {\text {PolyLog}(3,a x^2)}{(d x)^{3/2}} \, dx\) [82]

Optimal. Leaf size=122 \[ -\frac {64 \sqrt [4]{a} \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {64 \sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {32 \log \left (1-a x^2\right )}{d \sqrt {d x}}-\frac {8 \text {PolyLog}\left (2,a x^2\right )}{d \sqrt {d x}}-\frac {2 \text {PolyLog}\left (3,a x^2\right )}{d \sqrt {d x}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {6726, 2505, 16, 335, 304, 211, 214} \begin {gather*} -\frac {64 \sqrt [4]{a} \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {64 \sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {8 \text {Li}_2\left (a x^2\right )}{d \sqrt {d x}}-\frac {2 \text {Li}_3\left (a x^2\right )}{d \sqrt {d x}}+\frac {32 \log \left (1-a x^2\right )}{d \sqrt {d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[3, a*x^2]/(d*x)^(3/2),x]

[Out]

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

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 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[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}_3\left (a x^2\right )}{(d x)^{3/2}} \, dx &=-\frac {2 \text {Li}_3\left (a x^2\right )}{d \sqrt {d x}}+4 \int \frac {\text {Li}_2\left (a x^2\right )}{(d x)^{3/2}} \, dx\\ &=-\frac {8 \text {Li}_2\left (a x^2\right )}{d \sqrt {d x}}-\frac {2 \text {Li}_3\left (a x^2\right )}{d \sqrt {d x}}-16 \int \frac {\log \left (1-a x^2\right )}{(d x)^{3/2}} \, dx\\ &=\frac {32 \log \left (1-a x^2\right )}{d \sqrt {d x}}-\frac {8 \text {Li}_2\left (a x^2\right )}{d \sqrt {d x}}-\frac {2 \text {Li}_3\left (a x^2\right )}{d \sqrt {d x}}+\frac {(64 a) \int \frac {x}{\sqrt {d x} \left (1-a x^2\right )} \, dx}{d}\\ &=\frac {32 \log \left (1-a x^2\right )}{d \sqrt {d x}}-\frac {8 \text {Li}_2\left (a x^2\right )}{d \sqrt {d x}}-\frac {2 \text {Li}_3\left (a x^2\right )}{d \sqrt {d x}}+\frac {(64 a) \int \frac {\sqrt {d x}}{1-a x^2} \, dx}{d^2}\\ &=\frac {32 \log \left (1-a x^2\right )}{d \sqrt {d x}}-\frac {8 \text {Li}_2\left (a x^2\right )}{d \sqrt {d x}}-\frac {2 \text {Li}_3\left (a x^2\right )}{d \sqrt {d x}}+\frac {(128 a) \text {Subst}\left (\int \frac {x^2}{1-\frac {a x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{d^3}\\ &=\frac {32 \log \left (1-a x^2\right )}{d \sqrt {d x}}-\frac {8 \text {Li}_2\left (a x^2\right )}{d \sqrt {d x}}-\frac {2 \text {Li}_3\left (a x^2\right )}{d \sqrt {d x}}+\frac {\left (64 \sqrt {a}\right ) \text {Subst}\left (\int \frac {1}{d-\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{d}-\frac {\left (64 \sqrt {a}\right ) \text {Subst}\left (\int \frac {1}{d+\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{d}\\ &=-\frac {64 \sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {64 \sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {32 \log \left (1-a x^2\right )}{d \sqrt {d x}}-\frac {8 \text {Li}_2\left (a x^2\right )}{d \sqrt {d x}}-\frac {2 \text {Li}_3\left (a x^2\right )}{d \sqrt {d x}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.06, size = 71, normalized size = 0.58 \begin {gather*} \frac {x \Gamma \left (\frac {3}{4}\right ) \left (64 a x^2 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};a x^2\right )+48 \log \left (1-a x^2\right )-12 \text {PolyLog}\left (2,a x^2\right )-3 \text {PolyLog}\left (3,a x^2\right )\right )}{2 (d x)^{3/2} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Gamma[3/4]*(64*a*x^2*Hypergeometric2F1[3/4, 1, 7/4, a*x^2] + 48*Log[1 - a*x^2] - 12*PolyLog[2, a*x^2] - 3*P
olyLog[3, a*x^2]))/(2*(d*x)^(3/2)*Gamma[7/4])

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Maple [A]
time = 0.14, size = 131, normalized size = 1.07

method	result	size
meijerg	\(-\frac {x^{\frac {3}{2}} \left (-a \right )^{\frac {1}{4}} \left (-\frac {64 x^{\frac {3}{2}} \left (-a \right )^{\frac {3}{4}} \left (\ln \left (1-\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (a \,x^{2}\right )^{\frac {1}{4}}\right )+2 \arctan \left (\left (a \,x^{2}\right )^{\frac {1}{4}}\right )\right )}{\left (a \,x^{2}\right )^{\frac {3}{4}}}+\frac {64 \left (-a \right )^{\frac {3}{4}} \ln \left (-a \,x^{2}+1\right )}{\sqrt {x}\, a}-\frac {16 \left (-a \right )^{\frac {3}{4}} \polylog \left (2, a \,x^{2}\right )}{\sqrt {x}\, a}-\frac {4 \left (-a \right )^{\frac {3}{4}} \polylog \left (3, a \,x^{2}\right )}{\sqrt {x}\, a}\right )}{2 \left (d x \right )^{\frac {3}{2}}}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(d*x)^(3/2)*x^(3/2)*(-a)^(1/4)*(-64*x^(3/2)*(-a)^(3/4)/(a*x^2)^(3/4)*(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)^(3/4)/a*ln(-a*x^2+1)-16/x^(1/2)*(-a)^(3/4)/a*polylog(2,a*x^2)-4/
x^(1/2)*(-a)^(3/4)/a*polylog(3,a*x^2))

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Maxima [A]
time = 0.47, size = 132, normalized size = 1.08 \begin {gather*} -\frac {2 \, {\left (16 \, a {\left (\frac {2 \, \arctan \left (\frac {\sqrt {d x} \sqrt {a}}{\sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d} \sqrt {a}} + \frac {\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 {a}}\right )} + \frac {4 \, {\rm Li}_2\left (a x^{2}\right ) - 16 \, \log \left (-a d^{2} x^{2} + d^{2}\right ) + 32 \, \log \left (d\right ) + {\rm Li}_{3}(a x^{2})}{\sqrt {d x}}\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(16*a*(2*arctan(sqrt(d*x)*sqrt(a)/sqrt(sqrt(a)*d))/(sqrt(sqrt(a)*d)*sqrt(a)) + log((sqrt(d*x)*sqrt(a) - sqr
t(sqrt(a)*d))/(sqrt(d*x)*sqrt(a) + sqrt(sqrt(a)*d)))/(sqrt(sqrt(a)*d)*sqrt(a))) + (4*dilog(a*x^2) - 16*log(-a*
d^2*x^2 + d^2) + 32*log(d) + polylog(3, a*x^2))/sqrt(d*x))/d

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Fricas [A]
time = 0.44, size = 184, normalized size = 1.51 \begin {gather*} \frac {2 \, {\left (64 \, d^{2} x \left (\frac {a}{d^{6}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {d x} a d \left (\frac {a}{d^{6}}\right )^{\frac {1}{4}} - \sqrt {a d^{4} \sqrt {\frac {a}{d^{6}}} + a^{2} d x} d \left (\frac {a}{d^{6}}\right )^{\frac {1}{4}}}{a}\right ) + 16 \, d^{2} x \left (\frac {a}{d^{6}}\right )^{\frac {1}{4}} \log \left (32768 \, d^{5} \left (\frac {a}{d^{6}}\right )^{\frac {3}{4}} + 32768 \, \sqrt {d x} a\right ) - 16 \, d^{2} x \left (\frac {a}{d^{6}}\right )^{\frac {1}{4}} \log \left (-32768 \, d^{5} \left (\frac {a}{d^{6}}\right )^{\frac {3}{4}} + 32768 \, \sqrt {d x} a\right ) - 4 \, \sqrt {d x} {\left ({\rm Li}_2\left (a x^{2}\right ) - 4 \, \log \left (-a x^{2} + 1\right )\right )} - \sqrt {d x} {\rm polylog}\left (3, a x^{2}\right )\right )}}{d^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(64*d^2*x*(a/d^6)^(1/4)*arctan(-(sqrt(d*x)*a*d*(a/d^6)^(1/4) - sqrt(a*d^4*sqrt(a/d^6) + a^2*d*x)*d*(a/d^6)^(
1/4))/a) + 16*d^2*x*(a/d^6)^(1/4)*log(32768*d^5*(a/d^6)^(3/4) + 32768*sqrt(d*x)*a) - 16*d^2*x*(a/d^6)^(1/4)*lo
g(-32768*d^5*(a/d^6)^(3/4) + 32768*sqrt(d*x)*a) - 4*sqrt(d*x)*(dilog(a*x^2) - 4*log(-a*x^2 + 1)) - sqrt(d*x)*p
olylog(3, a*x^2))/(d^2*x)

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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 )}{\left (d x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(3, a*x^2)/(d*x)^(3/2),x)

[Out]

int(polylog(3, a*x^2)/(d*x)^(3/2), x)

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