3.242 \(\int \frac{x \log (\frac{x^2}{c})}{c-x^2} \, dx\)

Optimal. Leaf size=16 \[ \frac{1}{2} \text{PolyLog}\left (2,1-\frac{x^2}{c}\right ) \]

[Out]

PolyLog[2, 1 - x^2/c]/2

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Rubi [A]  time = 0.0429158, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2336, 2315} \[ \frac{1}{2} \text{PolyLog}\left (2,1-\frac{x^2}{c}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x*Log[x^2/c])/(c - x^2),x]

[Out]

PolyLog[2, 1 - x^2/c]/2

Rule 2336

Int[((a_.) + Log[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_))^(q_.), x_Symbol] :>
Dist[f^m/n, Subst[Int[(d + e*x)^q*(a + b*Log[c*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}
, x] && EqQ[m, r - 1] && IGtQ[p, 0] && (IntegerQ[m] || GtQ[f, 0]) && EqQ[r, n]

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]

Rubi steps

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

Mathematica [A]  time = 0.0039653, size = 17, normalized size = 1.06 \[ \frac{1}{2} \text{PolyLog}\left (2,\frac{c-x^2}{c}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Log[x^2/c])/(c - x^2),x]

[Out]

PolyLog[2, (c - x^2)/c]/2

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Maple [C]  time = 0.139, size = 52, normalized size = 3.3 \begin{align*} -{\frac{1}{2}\sum _{{\it \_alpha}={\it RootOf} \left ({{\it \_Z}}^{2}-c \right ) }\ln \left ( x-{\it \_alpha} \right ) \ln \left ({\frac{{x}^{2}}{c}} \right ) -2\,{\it dilog} \left ({\frac{x}{{\it \_alpha}}} \right ) -2\,\ln \left ( x-{\it \_alpha} \right ) \ln \left ({\frac{x}{{\it \_alpha}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sum(ln(x-_alpha)*ln(x^2/c)-2*dilog(x/_alpha)-2*ln(x-_alpha)*ln(x/_alpha),_alpha=RootOf(_Z^2-c))

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Maxima [B]  time = 1.13157, size = 78, normalized size = 4.88 \begin{align*} -\frac{1}{2} \, \log \left (x^{2} - c\right ) \log \left (\frac{x^{2}}{c}\right ) + \frac{1}{2} \, \log \left (x^{2} - c\right ) \log \left (\frac{x^{2} - c}{c} + 1\right ) + \frac{1}{2} \,{\rm Li}_2\left (-\frac{x^{2} - c}{c}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.29892, size = 31, normalized size = 1.94 \begin{align*} \frac{1}{2} \,{\rm Li}_2\left (-\frac{x^{2}}{c} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*dilog(-x^2/c + 1)

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Sympy [A]  time = 6.5004, size = 102, normalized size = 6.38 \begin{align*} \begin{cases} \log{\left (c \right )} \log{\left (x \right )} + i \pi \log{\left (x \right )} - \frac{\operatorname{Li}_{2}\left (\frac{x^{2}}{c}\right )}{2} & \text{for}\: \left |{x}\right | < 1 \\- \log{\left (c \right )} \log{\left (\frac{1}{x} \right )} - i \pi \log{\left (\frac{1}{x} \right )} - \frac{\operatorname{Li}_{2}\left (\frac{x^{2}}{c}\right )}{2} & \text{for}\: \frac{1}{\left |{x}\right |} < 1 \\-{G_{2, 2}^{2, 0}\left (\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle |{x} \right )} \log{\left (c \right )} - i \pi{G_{2, 2}^{2, 0}\left (\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle |{x} \right )} +{G_{2, 2}^{0, 2}\left (\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle |{x} \right )} \log{\left (c \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{x^{2}}{c}\right )}{2} & \text{otherwise} \end{cases} - \frac{\log{\left (\frac{x^{2}}{c} \right )} \log{\left (- c + x^{2} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*ln(x**2/c)/(-x**2+c),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x \log \left (\frac{x^{2}}{c}\right )}{x^{2} - c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-x*log(x^2/c)/(x^2 - c), x)