3.3.42 \(\int \frac {x \log (\frac {x^2}{c})}{c-x^2} \, dx\) [242]

Optimal. Leaf size=16 \[ \frac {1}{2} \text {Li}_2\left (1-\frac {x^2}{c}\right ) \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 16, normalized size of antiderivative = 1.00, 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 = {2374, 2352} \begin {gather*} \frac {1}{2} \text {PolyLog}\left (2,1-\frac {x^2}{c}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*Log[x^2/c])/(c - x^2),x]

[Out]

PolyLog[2, 1 - x^2/c]/2

Rule 2352

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

Rule 2374

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]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 17, normalized size = 1.06 \begin {gather*} \frac {1}{2} \text {Li}_2\left (\frac {c-x^2}{c}\right ) \end {gather*}

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] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 53, normalized size = 3.31

method	result	size
default	\(\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{2}-c \right )}{\sum }\left (-\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x^{2}}{c}\right )+2 \dilog \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )+2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )\right )\right )}{2}\)	\(53\)
risch	\(\frac {\left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{2}-c \right )}{\sum }\left (-\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x^{2}}{c}\right )+2 \dilog \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )+2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )\right )\right )}{2}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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] Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (13) = 26\).
time = 0.29, size = 58, normalized size = 3.62 \begin {gather*} -\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 {gather*}

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 = 0.37, size = 13, normalized size = 0.81 \begin {gather*} \frac {1}{2} \, {\rm Li}_2\left (-\frac {x^{2}}{c} + 1\right ) \end {gather*}

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 = 3.21, size = 117, normalized size = 7.31 \begin {gather*} \begin {cases} - \frac {\operatorname {Li}_{2}\left (\frac {x^{2}}{c}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-polylog(2, x**2/c)/2, (Abs(x) < 1) & (1/Abs(x) < 1)), (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)*lo
g(-c + x**2)/2

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

[Out]

integrate(-x*log(x^2/c)/(x^2 - c), x)

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Mupad [B]
time = 3.34, size = 10, normalized size = 0.62 \begin {gather*} \frac {{\mathrm {Li}}_{\mathrm {2}}\left (\frac {x^2}{c}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*log(x^2/c))/(c - x^2),x)

[Out]

dilog(x^2/c)/2

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