3.1.72 \(\int (d x)^{3/2} \text {PolyLog}(2,a x^2) \, dx\) [72]

Optimal. Leaf size=140 \[ -\frac {32 d \sqrt {d x}}{25 a}-\frac {32 (d x)^{5/2}}{125 d}+\frac {16 d^{3/2} \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{25 a^{5/4}}+\frac {16 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{25 a^{5/4}}+\frac {8 (d x)^{5/2} \log \left (1-a x^2\right )}{25 d}+\frac {2 (d x)^{5/2} \text {PolyLog}\left (2,a x^2\right )}{5 d} \]

[Out]

-32/125*(d*x)^(5/2)/d+16/25*d^(3/2)*arctan(a^(1/4)*(d*x)^(1/2)/d^(1/2))/a^(5/4)+16/25*d^(3/2)*arctanh(a^(1/4)*
(d*x)^(1/2)/d^(1/2))/a^(5/4)+8/25*(d*x)^(5/2)*ln(-a*x^2+1)/d+2/5*(d*x)^(5/2)*polylog(2,a*x^2)/d-32/25*d*(d*x)^
(1/2)/a

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Rubi [A]
time = 0.07, antiderivative size = 140, 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 {16 d^{3/2} \text {ArcTan}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{25 a^{5/4}}+\frac {16 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{25 a^{5/4}}+\frac {2 (d x)^{5/2} \text {Li}_2\left (a x^2\right )}{5 d}+\frac {8 (d x)^{5/2} \log \left (1-a x^2\right )}{25 d}-\frac {32 d \sqrt {d x}}{25 a}-\frac {32 (d x)^{5/2}}{125 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^(3/2)*PolyLog[2, a*x^2],x]

[Out]

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

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Mathematica [A]
time = 0.07, size = 101, normalized size = 0.72 \begin {gather*} \frac {2 (d x)^{3/2} \left (\frac {40 \text {ArcTan}\left (\sqrt [4]{a} \sqrt {x}\right )+40 \tanh ^{-1}\left (\sqrt [4]{a} \sqrt {x}\right )+4 \sqrt [4]{a} \sqrt {x} \left (-20-4 a x^2+5 a x^2 \log \left (1-a x^2\right )\right )}{a^{5/4}}+25 x^{5/2} \text {PolyLog}\left (2,a x^2\right )\right )}{125 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*x)^(3/2)*((40*ArcTan[a^(1/4)*Sqrt[x]] + 40*ArcTanh[a^(1/4)*Sqrt[x]] + 4*a^(1/4)*Sqrt[x]*(-20 - 4*a*x^2 +
5*a*x^2*Log[1 - a*x^2]))/a^(5/4) + 25*x^(5/2)*PolyLog[2, a*x^2]))/(125*x^(3/2))

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Maple [A]
time = 0.42, size = 145, normalized size = 1.04

method	result	size
meijerg	\(-\frac {\left (d x \right )^{\frac {3}{2}} \left (-\frac {4 \sqrt {x}\, \left (-a \right )^{\frac {9}{4}} \left (144 a \,x^{2}+720\right )}{1125 a^{2}}-\frac {16 \sqrt {x}\, \left (-a \right )^{\frac {9}{4}} \left (\ln \left (1-\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-\ln \left (1+\left (a \,x^{2}\right )^{\frac {1}{4}}\right )-2 \arctan \left (\left (a \,x^{2}\right )^{\frac {1}{4}}\right )\right )}{25 a^{2} \left (a \,x^{2}\right )^{\frac {1}{4}}}+\frac {16 x^{\frac {5}{2}} \left (-a \right )^{\frac {9}{4}} \ln \left (-a \,x^{2}+1\right )}{25 a}+\frac {4 x^{\frac {5}{2}} \left (-a \right )^{\frac {9}{4}} \polylog \left (2, a \,x^{2}\right )}{5 a}\right )}{2 x^{\frac {3}{2}} \left (-a \right )^{\frac {5}{4}}}\)	\(135\)
derivativedivides	\(\frac {\frac {2 \left (d x \right )^{\frac {5}{2}} \polylog \left (2, a \,x^{2}\right )}{5}+\frac {8 \left (d x \right )^{\frac {5}{2}} \ln \left (\frac {-a \,d^{2} x^{2}+d^{2}}{d^{2}}\right )}{25}+\frac {32 a \left (-\frac {\frac {a \left (d x \right )^{\frac {5}{2}}}{5}+d^{2} \sqrt {d x}}{a^{2}}+\frac {d^{2} \left (\frac {d^{2}}{a}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )\right )}{4 a^{2}}\right )}{25}}{d}\)	\(145\)
default	\(\frac {\frac {2 \left (d x \right )^{\frac {5}{2}} \polylog \left (2, a \,x^{2}\right )}{5}+\frac {8 \left (d x \right )^{\frac {5}{2}} \ln \left (\frac {-a \,d^{2} x^{2}+d^{2}}{d^{2}}\right )}{25}+\frac {32 a \left (-\frac {\frac {a \left (d x \right )^{\frac {5}{2}}}{5}+d^{2} \sqrt {d x}}{a^{2}}+\frac {d^{2} \left (\frac {d^{2}}{a}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )\right )}{4 a^{2}}\right )}{25}}{d}\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(1/5*(d*x)^(5/2)*polylog(2,a*x^2)+4/25*(d*x)^(5/2)*ln((-a*d^2*x^2+d^2)/d^2)+16/25*a*(-1/a^2*(1/5*a*(d*x)^(
5/2)+d^2*(d*x)^(1/2))+1/4*d^2/a^2*(d^2/a)^(1/4)*(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 = 160, normalized size = 1.14 \begin {gather*} \frac {2 \, {\left (\frac {25 \, \left (d x\right )^{\frac {5}{2}} a {\rm Li}_2\left (a x^{2}\right ) + 20 \, \left (d x\right )^{\frac {5}{2}} a \log \left (-a d^{2} x^{2} + d^{2}\right ) - 8 \, \left (d x\right )^{\frac {5}{2}} {\left (5 \, a \log \left (d\right ) + 2 \, a\right )} - 80 \, \sqrt {d x} d^{2}}{a} + \frac {20 \, {\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 )}}{125 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/125*((25*(d*x)^(5/2)*a*dilog(a*x^2) + 20*(d*x)^(5/2)*a*log(-a*d^2*x^2 + d^2) - 8*(d*x)^(5/2)*(5*a*log(d) + 2
*a) - 80*sqrt(d*x)*d^2)/a + 20*(2*d^3*arctan(sqrt(d*x)*sqrt(a)/sqrt(sqrt(a)*d))/sqrt(sqrt(a)*d) - d^3*log((sqr
t(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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Fricas [A]
time = 0.43, size = 194, normalized size = 1.39 \begin {gather*} -\frac {2 \, {\left (80 \, 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 ) - 20 \, a \left (\frac {d^{6}}{a^{5}}\right )^{\frac {1}{4}} \log \left (8 \, \sqrt {d x} d + 8 \, a \left (\frac {d^{6}}{a^{5}}\right )^{\frac {1}{4}}\right ) + 20 \, a \left (\frac {d^{6}}{a^{5}}\right )^{\frac {1}{4}} \log \left (8 \, \sqrt {d x} d - 8 \, a \left (\frac {d^{6}}{a^{5}}\right )^{\frac {1}{4}}\right ) - {\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 )}}{125 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/125*(80*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) - 20*a*(d^6/a^5)^(1/4)*log(8*sqrt(d*x)*d + 8*a*(d^6/a^5)^(1/4)) + 20*a*(d^6/a^5)^(1/4)*log
(8*sqrt(d*x)*d - 8*a*(d^6/a^5)^(1/4)) - (25*a*d*x^2*dilog(a*x^2) + 20*a*d*x^2*log(-a*x^2 + 1) - 16*a*d*x^2 - 8
0*d)*sqrt(d*x))/a

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

[Out]

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

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

[Out]

int(polylog(2, a*x^2)*(d*x)^(3/2), x)

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