3.73 \(\int \sqrt {d x} \text {Li}_2(a x^2) \, dx\)

Optimal. Leaf size=125 \[ -\frac {16 \sqrt {d} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{9 a^{3/4}}+\frac {16 \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{9 a^{3/4}}+\frac {2 (d x)^{3/2} \text {Li}_2\left (a x^2\right )}{3 d}+\frac {8 (d x)^{3/2} \log \left (1-a x^2\right )}{9 d}-\frac {32 (d x)^{3/2}}{27 d} \]

[Out]

-32/27*(d*x)^(3/2)/d+8/9*(d*x)^(3/2)*ln(-a*x^2+1)/d+2/3*(d*x)^(3/2)*polylog(2,a*x^2)/d-16/9*arctan(a^(1/4)*(d*
x)^(1/2)/d^(1/2))*d^(1/2)/a^(3/4)+16/9*arctanh(a^(1/4)*(d*x)^(1/2)/d^(1/2))*d^(1/2)/a^(3/4)

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Rubi [A]  time = 0.09, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 8, 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, 298, 205, 208} \[ \frac {2 (d x)^{3/2} \text {PolyLog}\left (2,a x^2\right )}{3 d}-\frac {16 \sqrt {d} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{9 a^{3/4}}+\frac {16 \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{9 a^{3/4}}+\frac {8 (d x)^{3/2} \log \left (1-a x^2\right )}{9 d}-\frac {32 (d x)^{3/2}}{27 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*x]*PolyLog[2, a*x^2],x]

[Out]

(-32*(d*x)^(3/2))/(27*d) - (16*Sqrt[d]*ArcTan[(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(9*a^(3/4)) + (16*Sqrt[d]*ArcTanh[
(a^(1/4)*Sqrt[d*x])/Sqrt[d]])/(9*a^(3/4)) + (8*(d*x)^(3/2)*Log[1 - a*x^2])/(9*d) + (2*(d*x)^(3/2)*PolyLog[2, a
*x^2])/(3*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 298

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 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 \sqrt {d x} \text {Li}_2\left (a x^2\right ) \, dx &=\frac {2 (d x)^{3/2} \text {Li}_2\left (a x^2\right )}{3 d}+\frac {4}{3} \int \sqrt {d x} \log \left (1-a x^2\right ) \, dx\\ &=\frac {8 (d x)^{3/2} \log \left (1-a x^2\right )}{9 d}+\frac {2 (d x)^{3/2} \text {Li}_2\left (a x^2\right )}{3 d}+\frac {(16 a) \int \frac {x (d x)^{3/2}}{1-a x^2} \, dx}{9 d}\\ &=\frac {8 (d x)^{3/2} \log \left (1-a x^2\right )}{9 d}+\frac {2 (d x)^{3/2} \text {Li}_2\left (a x^2\right )}{3 d}+\frac {(16 a) \int \frac {(d x)^{5/2}}{1-a x^2} \, dx}{9 d^2}\\ &=-\frac {32 (d x)^{3/2}}{27 d}+\frac {8 (d x)^{3/2} \log \left (1-a x^2\right )}{9 d}+\frac {2 (d x)^{3/2} \text {Li}_2\left (a x^2\right )}{3 d}+\frac {16}{9} \int \frac {\sqrt {d x}}{1-a x^2} \, dx\\ &=-\frac {32 (d x)^{3/2}}{27 d}+\frac {8 (d x)^{3/2} \log \left (1-a x^2\right )}{9 d}+\frac {2 (d x)^{3/2} \text {Li}_2\left (a x^2\right )}{3 d}+\frac {32 \operatorname {Subst}\left (\int \frac {x^2}{1-\frac {a x^4}{d^2}} \, dx,x,\sqrt {d x}\right )}{9 d}\\ &=-\frac {32 (d x)^{3/2}}{27 d}+\frac {8 (d x)^{3/2} \log \left (1-a x^2\right )}{9 d}+\frac {2 (d x)^{3/2} \text {Li}_2\left (a x^2\right )}{3 d}+\frac {(16 d) \operatorname {Subst}\left (\int \frac {1}{d-\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{9 \sqrt {a}}-\frac {(16 d) \operatorname {Subst}\left (\int \frac {1}{d+\sqrt {a} x^2} \, dx,x,\sqrt {d x}\right )}{9 \sqrt {a}}\\ &=-\frac {32 (d x)^{3/2}}{27 d}-\frac {16 \sqrt {d} \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{9 a^{3/4}}+\frac {16 \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{9 a^{3/4}}+\frac {8 (d x)^{3/2} \log \left (1-a x^2\right )}{9 d}+\frac {2 (d x)^{3/2} \text {Li}_2\left (a x^2\right )}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 91, normalized size = 0.73 \[ \frac {2 \sqrt {d x} \left (\frac {4 \left (a^{3/4} x^{3/2} \left (3 \log \left (1-a x^2\right )-4\right )-6 \tan ^{-1}\left (\sqrt [4]{a} \sqrt {x}\right )+6 \tanh ^{-1}\left (\sqrt [4]{a} \sqrt {x}\right )\right )}{a^{3/4}}+9 x^{3/2} \text {Li}_2\left (a x^2\right )\right )}{27 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*x]*PolyLog[2, a*x^2],x]

[Out]

(2*Sqrt[d*x]*((4*(-6*ArcTan[a^(1/4)*Sqrt[x]] + 6*ArcTanh[a^(1/4)*Sqrt[x]] + a^(3/4)*x^(3/2)*(-4 + 3*Log[1 - a*
x^2])))/a^(3/4) + 9*x^(3/2)*PolyLog[2, a*x^2]))/(27*Sqrt[x])

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fricas [A]  time = 0.74, size = 172, normalized size = 1.38 \[ \frac {2}{27} \, \sqrt {d x} {\left (9 \, x {\rm Li}_2\left (a x^{2}\right ) + 12 \, x \log \left (-a x^{2} + 1\right ) - 16 \, x\right )} + \frac {32}{9} \, \left (\frac {d^{2}}{a^{3}}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {d x} a d \left (\frac {d^{2}}{a^{3}}\right )^{\frac {1}{4}} - \sqrt {d^{3} x + a d^{2} \sqrt {\frac {d^{2}}{a^{3}}}} a \left (\frac {d^{2}}{a^{3}}\right )^{\frac {1}{4}}}{d^{2}}\right ) + \frac {8}{9} \, \left (\frac {d^{2}}{a^{3}}\right )^{\frac {1}{4}} \log \left (512 \, a^{2} \left (\frac {d^{2}}{a^{3}}\right )^{\frac {3}{4}} + 512 \, \sqrt {d x} d\right ) - \frac {8}{9} \, \left (\frac {d^{2}}{a^{3}}\right )^{\frac {1}{4}} \log \left (-512 \, a^{2} \left (\frac {d^{2}}{a^{3}}\right )^{\frac {3}{4}} + 512 \, \sqrt {d x} d\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(1/2)*polylog(2,a*x^2),x, algorithm="fricas")

[Out]

2/27*sqrt(d*x)*(9*x*dilog(a*x^2) + 12*x*log(-a*x^2 + 1) - 16*x) + 32/9*(d^2/a^3)^(1/4)*arctan(-(sqrt(d*x)*a*d*
(d^2/a^3)^(1/4) - sqrt(d^3*x + a*d^2*sqrt(d^2/a^3))*a*(d^2/a^3)^(1/4))/d^2) + 8/9*(d^2/a^3)^(1/4)*log(512*a^2*
(d^2/a^3)^(3/4) + 512*sqrt(d*x)*d) - 8/9*(d^2/a^3)^(1/4)*log(-512*a^2*(d^2/a^3)^(3/4) + 512*sqrt(d*x)*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {d x} {\rm Li}_2\left (a x^{2}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(1/2)*polylog(2,a*x^2),x, algorithm="giac")

[Out]

integrate(sqrt(d*x)*dilog(a*x^2), x)

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maple [A]  time = 0.01, size = 139, normalized size = 1.11 \[ \frac {2 \left (d x \right )^{\frac {3}{2}} \polylog \left (2, a \,x^{2}\right )}{3 d}+\frac {8 \left (d x \right )^{\frac {3}{2}} \ln \left (\frac {-a \,d^{2} x^{2}+d^{2}}{d^{2}}\right )}{9 d}-\frac {32 \left (d x \right )^{\frac {3}{2}}}{27 d}-\frac {16 d \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )}{9 a \left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}+\frac {8 d \ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )}{9 a \left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(1/2)*polylog(2,a*x^2),x)

[Out]

2/3*(d*x)^(3/2)*polylog(2,a*x^2)/d+8/9/d*(d*x)^(3/2)*ln((-a*d^2*x^2+d^2)/d^2)-32/27*(d*x)^(3/2)/d-16/9*d/a/(d^
2/a)^(1/4)*arctan((d*x)^(1/2)/(d^2/a)^(1/4))+8/9*d/a/(d^2/a)^(1/4)*ln(((d*x)^(1/2)+(d^2/a)^(1/4))/((d*x)^(1/2)
-(d^2/a)^(1/4)))

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maxima [A]  time = 1.06, size = 139, normalized size = 1.11 \[ -\frac {2 \, {\left (12 \, d^{2} {\left (\frac {2 \, \arctan \left (\frac {\sqrt {d x} \sqrt {a}}{\sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d} \sqrt {a}} + \frac {\log \left (\frac {\sqrt {d x} \sqrt {a} - \sqrt {\sqrt {a} d}}{\sqrt {d x} \sqrt {a} + \sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d} \sqrt {a}}\right )} + 8 \, \left (d x\right )^{\frac {3}{2}} {\left (3 \, \log \relax (d) + 2\right )} - 9 \, \left (d x\right )^{\frac {3}{2}} {\rm Li}_2\left (a x^{2}\right ) - 12 \, \left (d x\right )^{\frac {3}{2}} \log \left (-a d^{2} x^{2} + d^{2}\right )\right )}}{27 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(1/2)*polylog(2,a*x^2),x, algorithm="maxima")

[Out]

-2/27*(12*d^2*(2*arctan(sqrt(d*x)*sqrt(a)/sqrt(sqrt(a)*d))/(sqrt(sqrt(a)*d)*sqrt(a)) + log((sqrt(d*x)*sqrt(a)
- sqrt(sqrt(a)*d))/(sqrt(d*x)*sqrt(a) + sqrt(sqrt(a)*d)))/(sqrt(sqrt(a)*d)*sqrt(a))) + 8*(d*x)^(3/2)*(3*log(d)
+ 2) - 9*(d*x)^(3/2)*dilog(a*x^2) - 12*(d*x)^(3/2)*log(-a*d^2*x^2 + d^2))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, a*x^2)*(d*x)^(1/2),x)

[Out]

int(polylog(2, a*x^2)*(d*x)^(1/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(1/2)*polylog(2,a*x**2),x)

[Out]

Timed out

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