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

Optimal. Leaf size=115 \[ \frac {2 \sqrt {d x} \text {Li}_2\left (a x^2\right )}{d}+\frac {8 \sqrt {d x} \log \left (1-a x^2\right )}{d}+\frac {16 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt [4]{a} \sqrt {d}}+\frac {16 \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt [4]{a} \sqrt {d}}-\frac {32 \sqrt {d x}}{d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 115, 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, 212, 208, 205} \[ \frac {2 \sqrt {d x} \text {PolyLog}\left (2,a x^2\right )}{d}+\frac {8 \sqrt {d x} \log \left (1-a x^2\right )}{d}+\frac {16 \tan ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt [4]{a} \sqrt {d}}+\frac {16 \tanh ^{-1}\left (\frac {\sqrt [4]{a} \sqrt {d x}}{\sqrt {d}}\right )}{\sqrt [4]{a} \sqrt {d}}-\frac {32 \sqrt {d x}}{d} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, a*x^2]/Sqrt[d*x],x]

[Out]

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

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Mathematica [C]  time = 0.07, size = 57, normalized size = 0.50 \[ \frac {5 x \Gamma \left (\frac {5}{4}\right ) \left (16 \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};a x^2\right )+\text {Li}_2\left (a x^2\right )+4 \log \left (1-a x^2\right )-16\right )}{2 \Gamma \left (\frac {9}{4}\right ) \sqrt {d x}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(5*x*Gamma[5/4]*(-16 + 16*Hypergeometric2F1[1/4, 1, 5/4, a*x^2] + 4*Log[1 - a*x^2] + PolyLog[2, a*x^2]))/(2*Sq
rt[d*x]*Gamma[9/4])

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fricas [A]  time = 0.59, size = 156, normalized size = 1.36 \[ -\frac {2 \, {\left (16 \, d \left (\frac {1}{a d^{2}}\right )^{\frac {1}{4}} \arctan \left (\sqrt {d^{2} \sqrt {\frac {1}{a d^{2}}} + d x} a d \left (\frac {1}{a d^{2}}\right )^{\frac {3}{4}} - \sqrt {d x} a d \left (\frac {1}{a d^{2}}\right )^{\frac {3}{4}}\right ) - 4 \, d \left (\frac {1}{a d^{2}}\right )^{\frac {1}{4}} \log \left (d \left (\frac {1}{a d^{2}}\right )^{\frac {1}{4}} + \sqrt {d x}\right ) + 4 \, d \left (\frac {1}{a d^{2}}\right )^{\frac {1}{4}} \log \left (-d \left (\frac {1}{a d^{2}}\right )^{\frac {1}{4}} + \sqrt {d x}\right ) - \sqrt {d x} {\left ({\rm Li}_2\left (a x^{2}\right ) + 4 \, \log \left (-a x^{2} + 1\right ) - 16\right )}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(dilog(a*x^2)/sqrt(d*x), x)

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maple [A]  time = 0.02, size = 137, normalized size = 1.19 \[ \frac {2 \polylog \left (2, a \,x^{2}\right ) \sqrt {d x}}{d}+\frac {8 \sqrt {d x}\, \ln \left (\frac {-a \,d^{2} x^{2}+d^{2}}{d^{2}}\right )}{d}+\frac {16 \left (\frac {d^{2}}{a}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {d x}}{\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )}{d}+\frac {8 \left (\frac {d^{2}}{a}\right )^{\frac {1}{4}} \ln \left (\frac {\sqrt {d x}+\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}{\sqrt {d x}-\left (\frac {d^{2}}{a}\right )^{\frac {1}{4}}}\right )}{d}-\frac {32 \sqrt {d x}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.36, size = 128, normalized size = 1.11 \[ -\frac {2 \, {\left (8 \, \sqrt {d x} {\left (\log \relax (d) + 2\right )} - \frac {8 \, d \arctan \left (\frac {\sqrt {d x} \sqrt {a}}{\sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d}} - \sqrt {d x} {\rm Li}_2\left (a x^{2}\right ) - 4 \, \sqrt {d x} \log \left (-a d^{2} x^{2} + d^{2}\right ) + \frac {4 \, d \log \left (\frac {\sqrt {d x} \sqrt {a} - \sqrt {\sqrt {a} d}}{\sqrt {d x} \sqrt {a} + \sqrt {\sqrt {a} d}}\right )}{\sqrt {\sqrt {a} d}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(8*sqrt(d*x)*(log(d) + 2) - 8*d*arctan(sqrt(d*x)*sqrt(a)/sqrt(sqrt(a)*d))/sqrt(sqrt(a)*d) - sqrt(d*x)*dilog
(a*x^2) - 4*sqrt(d*x)*log(-a*d^2*x^2 + d^2) + 4*d*log((sqrt(d*x)*sqrt(a) - sqrt(sqrt(a)*d))/(sqrt(d*x)*sqrt(a)
+ sqrt(sqrt(a)*d)))/sqrt(sqrt(a)*d))/d

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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