3.349 \(\int \frac {\log (\frac {a}{x^2})}{a x-x^3} \, dx\)

Optimal. Leaf size=17 \[ \frac {\text {Li}_2\left (1-\frac {a}{x^2}\right )}{2 a} \]

[Out]

1/2*polylog(2,1-a/x^2)/a

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Rubi [A]  time = 0.09, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1593, 2343, 2333, 2315} \[ \frac {\text {PolyLog}\left (2,1-\frac {a}{x^2}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Log[a/x^2]/(a*x - x^3),x]

[Out]

PolyLog[2, 1 - a/x^2]/(2*a)

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)/(x_))^(q_.)*(x_)^(m_.), x_Symbol] :> Int[(e + d*
x)^q*(a + b*Log[c*x^n])^p, x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[m, q] && IntegerQ[q]

Rule 2343

Int[((a_.) + Log[(c_.)*(x_)^(n_)]*(b_.))/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Dist[1/n, Subst[Int[(a
+ b*Log[c*x])/(x*(d + e*x^(r/n))), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IntegerQ[r/n]

Rubi steps

\begin {align*} \int \frac {\log \left (\frac {a}{x^2}\right )}{a x-x^3} \, dx &=\int \frac {\log \left (\frac {a}{x^2}\right )}{x \left (a-x^2\right )} \, dx\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {\log (a x)}{\left (a-\frac {1}{x}\right ) x} \, dx,x,\frac {1}{x^2}\right )\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {\log (a x)}{-1+a x} \, dx,x,\frac {1}{x^2}\right )\right )\\ &=\frac {\text {Li}_2\left (1-\frac {a}{x^2}\right )}{2 a}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 1.24 \[ \frac {\text {Li}_2\left (-\frac {a-x^2}{x^2}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[a/x^2]/(a*x - x^3),x]

[Out]

PolyLog[2, -((a - x^2)/x^2)]/(2*a)

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fricas [A]  time = 0.40, size = 14, normalized size = 0.82 \[ \frac {{\rm Li}_2\left (-\frac {a}{x^{2}} + 1\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a/x^2)/(-x^3+a*x),x, algorithm="fricas")

[Out]

1/2*dilog(-a/x^2 + 1)/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a/x^2)/(-x^3+a*x),x, algorithm="giac")

[Out]

integrate(-log(a/x^2)/(x^3 - a*x), x)

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maple [A]  time = 0.04, size = 12, normalized size = 0.71 \[ \frac {\dilog \left (\frac {a}{x^{2}}\right )}{2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(a/x^2)/(-x^3+a*x),x)

[Out]

1/2/a*dilog(a/x^2)

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maxima [B]  time = 0.61, size = 81, normalized size = 4.76 \[ -\frac {1}{2} \, {\left (\frac {\log \left (x^{2} - a\right )}{a} - \frac {2 \, \log \relax (x)}{a}\right )} \log \left (\frac {a}{x^{2}}\right ) - \frac {\log \left (x^{2} - a\right ) \log \relax (x) - \log \relax (x)^{2}}{a} + \frac {2 \, \log \relax (x) \log \left (-\frac {x^{2}}{a} + 1\right ) + {\rm Li}_2\left (\frac {x^{2}}{a}\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(a/x^2)/(-x^3+a*x),x, algorithm="maxima")

[Out]

-1/2*(log(x^2 - a)/a - 2*log(x)/a)*log(a/x^2) - (log(x^2 - a)*log(x) - log(x)^2)/a + 1/2*(2*log(x)*log(-x^2/a
+ 1) + dilog(x^2/a))/a

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mupad [B]  time = 3.50, size = 11, normalized size = 0.65 \[ \frac {{\mathrm {Li}}_{\mathrm {2}}\left (\frac {a}{x^2}\right )}{2\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(a/x^2)/(a*x - x^3),x)

[Out]

dilog(a/x^2)/(2*a)

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sympy [C]  time = 11.31, size = 78, normalized size = 4.59 \[ - \frac {\begin {cases} i \pi \log {\relax (x )} + \frac {\operatorname {Li}_{2}\left (\frac {a}{x^{2}}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- i \pi \log {\left (\frac {1}{x} \right )} + \frac {\operatorname {Li}_{2}\left (\frac {a}{x^{2}}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- i \pi {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} + i \pi {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} + \frac {\operatorname {Li}_{2}\left (\frac {a}{x^{2}}\right )}{2} & \text {otherwise} \end {cases}}{a} - \frac {\log {\left (\frac {a}{x^{2}} \right )} \log {\left (\frac {a}{x^{2}} - 1 \right )}}{2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(a/x**2)/(-x**3+a*x),x)

[Out]

-Piecewise((I*pi*log(x) + polylog(2, a/x**2)/2, Abs(x) < 1), (-I*pi*log(1/x) + polylog(2, a/x**2)/2, 1/Abs(x)
< 1), (-I*pi*meijerg(((), (1, 1)), ((0, 0), ()), x) + I*pi*meijerg(((1, 1), ()), ((), (0, 0)), x) + polylog(2,
a/x**2)/2, True))/a - log(a/x**2)*log(a/x**2 - 1)/(2*a)

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