3.241 \(\int \frac {x \log (c x^2)}{1-c x^2} \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2336, 2315} \[ \frac {\text {PolyLog}\left (2,1-c x^2\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(x*Log[c*x^2])/(1 - c*x^2),x]

[Out]

PolyLog[2, 1 - c*x^2]/(2*c)

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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 17, normalized size = 1.00 \[ \frac {\text {Li}_2\left (1-c x^2\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Log[c*x^2])/(1 - c*x^2),x]

[Out]

PolyLog[2, 1 - c*x^2]/(2*c)

________________________________________________________________________________________

fricas [A]  time = 0.38, size = 14, normalized size = 0.82 \[ \frac {{\rm Li}_2\left (-c x^{2} + 1\right )}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(c*x^2)/(-c*x^2+1),x, algorithm="fricas")

[Out]

1/2*dilog(-c*x^2 + 1)/c

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x \log \left (c x^{2}\right )}{c x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(c*x^2)/(-c*x^2+1),x, algorithm="giac")

[Out]

integrate(-x*log(c*x^2)/(c*x^2 - 1), x)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 12, normalized size = 0.71 \[ \frac {\dilog \left (c \,x^{2}\right )}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*ln(c*x^2)/(-c*x^2+1),x)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.49, size = 76, normalized size = 4.47 \[ -\frac {\log \left (c x^{2} - 1\right ) \log \left (c x^{2}\right )}{2 \, c} + \frac {\log \left (c x^{2} - 1\right ) \log \relax (x)}{c} + \frac {\log \left (c x^{2} - 1\right ) \log \left (c x^{2}\right ) - 2 \, \log \left (c x^{2} - 1\right ) \log \relax (x) + {\rm Li}_2\left (-c x^{2} + 1\right )}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(c*x^2)/(-c*x^2+1),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.38, size = 11, normalized size = 0.65 \[ \frac {{\mathrm {Li}}_{\mathrm {2}}\left (c\,x^2\right )}{2\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*log(c*x^2))/(c*x^2 - 1),x)

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 8.25, size = 78, normalized size = 4.59 \[ \frac {\begin {cases} i \pi \log {\relax (x )} - \frac {\operatorname {Li}_{2}\left (c x^{2}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- i \pi \log {\left (\frac {1}{x} \right )} - \frac {\operatorname {Li}_{2}\left (c x^{2}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- i \pi {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} + i \pi {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} - \frac {\operatorname {Li}_{2}\left (c x^{2}\right )}{2} & \text {otherwise} \end {cases}}{c} - \frac {\log {\left (c x^{2} \right )} \log {\left (c x^{2} - 1 \right )}}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*ln(c*x**2)/(-c*x**2+1),x)

[Out]

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

________________________________________________________________________________________