3.79 \(\int (d x)^{3/2} \text {Li}_3(a x^2) \, dx\)

Optimal. Leaf size=161 \[ -\frac {64 d^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{125 a^{5/4}}-\frac {64 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{125 a^{5/4}}-\frac {8 (d x)^{5/2} \text {Li}_2\left (a x^2\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^2\right )}{5 d}-\frac {32 (d x)^{5/2} \log \left (1-a x^2\right )}{125 d}+\frac {128 d \sqrt {d x}}{125 a}+\frac {128 (d x)^{5/2}}{625 d} \]

[Out]

128/625*(d*x)^(5/2)/d-64/125*d^(3/2)*arctan(a^(1/4)*(d*x)^(1/2)/d^(1/2))/a^(5/4)-64/125*d^(3/2)*arctanh(a^(1/4
)*(d*x)^(1/2)/d^(1/2))/a^(5/4)-32/125*(d*x)^(5/2)*ln(-a*x^2+1)/d-8/25*(d*x)^(5/2)*polylog(2,a*x^2)/d+2/5*(d*x)
^(5/2)*polylog(3,a*x^2)/d+128/125*d*(d*x)^(1/2)/a

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Rubi [A]  time = 0.12, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ -\frac {8 (d x)^{5/2} \text {PolyLog}\left (2,a x^2\right )}{25 d}+\frac {2 (d x)^{5/2} \text {PolyLog}\left (3,a x^2\right )}{5 d}-\frac {64 d^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{125 a^{5/4}}-\frac {64 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{125 a^{5/4}}-\frac {32 (d x)^{5/2} \log \left (1-a x^2\right )}{125 d}+\frac {128 d \sqrt {d x}}{125 a}+\frac {128 (d x)^{5/2}}{625 d} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(3/2)*PolyLog[3, a*x^2],x]

[Out]

(128*d*Sqrt[d*x])/(125*a) + (128*(d*x)^(5/2))/(625*d) - (64*d^(3/2)*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(125*
a^(5/4)) - (64*d^(3/2)*ArcTanh[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(125*a^(5/4)) - (32*(d*x)^(5/2)*Log[1 - a*x^2])/(
125*d) - (8*(d*x)^(5/2)*PolyLog[2, a*x^2])/(25*d) + (2*(d*x)^(5/2)*PolyLog[3, a*x^2])/(5*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 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]

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 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 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 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 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]

Rubi steps

\begin {align*} \int (d x)^{3/2} \text {Li}_3\left (a x^2\right ) \, dx &=\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^2\right )}{5 d}-\frac {4}{5} \int (d x)^{3/2} \text {Li}_2\left (a x^2\right ) \, dx\\ &=-\frac {8 (d x)^{5/2} \text {Li}_2\left (a x^2\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^2\right )}{5 d}-\frac {16}{25} \int (d x)^{3/2} \log \left (1-a x^2\right ) \, dx\\ &=-\frac {32 (d x)^{5/2} \log \left (1-a x^2\right )}{125 d}-\frac {8 (d x)^{5/2} \text {Li}_2\left (a x^2\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^2\right )}{5 d}-\frac {(64 a) \int \frac {x (d x)^{5/2}}{1-a x^2} \, dx}{125 d}\\ &=-\frac {32 (d x)^{5/2} \log \left (1-a x^2\right )}{125 d}-\frac {8 (d x)^{5/2} \text {Li}_2\left (a x^2\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^2\right )}{5 d}-\frac {(64 a) \int \frac {(d x)^{7/2}}{1-a x^2} \, dx}{125 d^2}\\ &=\frac {128 (d x)^{5/2}}{625 d}-\frac {32 (d x)^{5/2} \log \left (1-a x^2\right )}{125 d}-\frac {8 (d x)^{5/2} \text {Li}_2\left (a x^2\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^2\right )}{5 d}-\frac {64}{125} \int \frac {(d x)^{3/2}}{1-a x^2} \, dx\\ &=\frac {128 d \sqrt {d x}}{125 a}+\frac {128 (d x)^{5/2}}{625 d}-\frac {32 (d x)^{5/2} \log \left (1-a x^2\right )}{125 d}-\frac {8 (d x)^{5/2} \text {Li}_2\left (a x^2\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^2\right )}{5 d}-\frac {\left (64 d^2\right ) \int \frac {1}{\sqrt {d x} \left (1-a x^2\right )} \, dx}{125 a}\\ &=\frac {128 d \sqrt {d x}}{125 a}+\frac {128 (d x)^{5/2}}{625 d}-\frac {32 (d x)^{5/2} \log \left (1-a x^2\right )}{125 d}-\frac {8 (d x)^{5/2} \text {Li}_2\left (a x^2\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^2\right )}{5 d}-\frac {(128 d) \operatorname {Subst}\left (\int \frac {1}{1-\frac {a x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{125 a}\\ &=\frac {128 d \sqrt {d x}}{125 a}+\frac {128 (d x)^{5/2}}{625 d}-\frac {32 (d x)^{5/2} \log \left (1-a x^2\right )}{125 d}-\frac {8 (d x)^{5/2} \text {Li}_2\left (a x^2\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^2\right )}{5 d}-\frac {\left (64 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{d-\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{125 a}-\frac {\left (64 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{d+\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{125 a}\\ &=\frac {128 d \sqrt {d x}}{125 a}+\frac {128 (d x)^{5/2}}{625 d}-\frac {64 d^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{125 a^{5/4}}-\frac {64 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{125 a^{5/4}}-\frac {32 (d x)^{5/2} \log \left (1-a x^2\right )}{125 d}-\frac {8 (d x)^{5/2} \text {Li}_2\left (a x^2\right )}{25 d}+\frac {2 (d x)^{5/2} \text {Li}_3\left (a x^2\right )}{5 d}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 89, normalized size = 0.55 \[ -\frac {9 d \Gamma \left (\frac {9}{4}\right ) \sqrt {d x} \left (320 \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};a x^2\right )+100 a x^2 \text {Li}_2\left (a x^2\right )-125 a x^2 \text {Li}_3\left (a x^2\right )-64 a x^2+80 a x^2 \log \left (1-a x^2\right )-320\right )}{1250 a \Gamma \left (\frac {13}{4}\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-9*d*Sqrt[d*x]*Gamma[9/4]*(-320 - 64*a*x^2 + 320*Hypergeometric2F1[1/4, 1, 5/4, a*x^2] + 80*a*x^2*Log[1 - a*x
^2] + 100*a*x^2*PolyLog[2, a*x^2] - 125*a*x^2*PolyLog[3, a*x^2]))/(1250*a*Gamma[13/4])

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fricas [C]  time = 0.91, size = 213, normalized size = 1.32 \[ \frac {2 \, {\left (125 \, \sqrt {d x} a d x^{2} {\rm polylog}\left (3, a x^{2}\right ) + 320 \, a \left (\frac {d^{6}}{a^{5}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {d x} a^{4} d \left (\frac {d^{6}}{a^{5}}\right )^{\frac {3}{4}} - \sqrt {d^{3} x + a^{2} \sqrt {\frac {d^{6}}{a^{5}}}} a^{4} \left (\frac {d^{6}}{a^{5}}\right )^{\frac {3}{4}}}{d^{6}}\right ) - 80 \, a \left (\frac {d^{6}}{a^{5}}\right )^{\frac {1}{4}} \log \left (32 \, \sqrt {d x} d + 32 \, a \left (\frac {d^{6}}{a^{5}}\right )^{\frac {1}{4}}\right ) + 80 \, a \left (\frac {d^{6}}{a^{5}}\right )^{\frac {1}{4}} \log \left (32 \, \sqrt {d x} d - 32 \, a \left (\frac {d^{6}}{a^{5}}\right )^{\frac {1}{4}}\right ) - 4 \, {\left (25 \, a d x^{2} {\rm Li}_2\left (a x^{2}\right ) + 20 \, a d x^{2} \log \left (-a x^{2} + 1\right ) - 16 \, a d x^{2} - 80 \, d\right )} \sqrt {d x}\right )}}{625 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/625*(125*sqrt(d*x)*a*d*x^2*polylog(3, a*x^2) + 320*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) - 80*a*(d^6/a^5)^(1/4)*log(32*sqrt(d*x)*d + 32*a
*(d^6/a^5)^(1/4)) + 80*a*(d^6/a^5)^(1/4)*log(32*sqrt(d*x)*d - 32*a*(d^6/a^5)^(1/4)) - 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 - 80*d)*sqrt(d*x))/a

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d x\right )^{\frac {3}{2}} {\rm Li}_{3}(a x^{2})\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.16, size = 155, normalized size = 0.96 \[ -\frac {\left (d x \right )^{\frac {3}{2}} \left (\frac {4 \sqrt {x}\, \left (-a \right )^{\frac {9}{4}} \left (576 a \,x^{2}+2880\right )}{5625 a^{2}}+\frac {64 \sqrt {x}\, \left (-a \right )^{\frac {9}{4}} \left (\ln \left (1-\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (a \,x^{2}\right )^{\frac {1}{4}}\right )\right )}{125 a^{2} \left (a \,x^{2}\right )^{\frac {1}{4}}}-\frac {64 x^{\frac {5}{2}} \left (-a \right )^{\frac {9}{4}} \ln \left (-a \,x^{2}+1\right )}{125 a}-\frac {16 x^{\frac {5}{2}} \left (-a \right )^{\frac {9}{4}} \polylog \left (2, a \,x^{2}\right )}{25 a}+\frac {4 x^{\frac {5}{2}} \left (-a \right )^{\frac {9}{4}} \polylog \left (3, a \,x^{2}\right )}{5 a}\right )}{2 x^{\frac {3}{2}} \left (-a \right )^{\frac {5}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(d*x)^(3/2)/x^(3/2)/(-a)^(5/4)*(4/5625*x^(1/2)*(-a)^(9/4)*(576*a*x^2+2880)/a^2+64/125*x^(1/2)*(-a)^(9/4)/
a^2/(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)))-64/125*x^(5/2)*(-a)^(9/4)/
a*ln(-a*x^2+1)-16/25*x^(5/2)*(-a)^(9/4)*polylog(2,a*x^2)/a+4/5*x^(5/2)*(-a)^(9/4)/a*polylog(3,a*x^2))

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maxima [A]  time = 1.35, size = 175, normalized size = 1.09 \[ -\frac {2 \, {\left (\frac {100 \, \left (d x\right )^{\frac {5}{2}} a {\rm Li}_2\left (a x^{2}\right ) + 80 \, \left (d x\right )^{\frac {5}{2}} a \log \left (-a d^{2} x^{2} + d^{2}\right ) - 125 \, \left (d x\right )^{\frac {5}{2}} a {\rm Li}_{3}(a x^{2}) - 32 \, \left (d x\right )^{\frac {5}{2}} {\left (5 \, a \log \relax (d) + 2 \, a\right )} - 320 \, \sqrt {d x} d^{2}}{a} + \frac {80 \, {\left (\frac {2 \, d^{3} \arctan \left (\frac {\sqrt {d x} \sqrt {a}}{\sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d}} - \frac {d^{3} \log \left (\frac {\sqrt {d x} \sqrt {a} - \sqrt {\sqrt {a} d}}{\sqrt {d x} \sqrt {a} + \sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d}}\right )}}{a}\right )}}{625 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/625*((100*(d*x)^(5/2)*a*dilog(a*x^2) + 80*(d*x)^(5/2)*a*log(-a*d^2*x^2 + d^2) - 125*(d*x)^(5/2)*a*polylog(3
, a*x^2) - 32*(d*x)^(5/2)*(5*a*log(d) + 2*a) - 320*sqrt(d*x)*d^2)/a + 80*(2*d^3*arctan(sqrt(d*x)*sqrt(a)/sqrt(
sqrt(a)*d))/sqrt(sqrt(a)*d) - d^3*log((sqrt(d*x)*sqrt(a) - sqrt(sqrt(a)*d))/(sqrt(d*x)*sqrt(a) + sqrt(sqrt(a)*
d)))/sqrt(sqrt(a)*d))/a)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {polylog}\left (3,a\,x^2\right )\,{\left (d\,x\right )}^{3/2} \,d x \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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