3.60.98 \(\int \frac {10 x+(-5+5 x) \log (-x^2)+x^3 \log ^2(-x^2)+x \log ^2(x) \log ^2(-x^2)+\log (x) (-10-2 x^2 \log ^2(-x^2))}{x^3 \log ^2(-x^2)-2 x^2 \log (x) \log ^2(-x^2)+x \log ^2(x) \log ^2(-x^2)} \, dx\)

Optimal. Leaf size=20 \[ x+\frac {5}{(-x+\log (x)) \log \left (-x^2\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 0.81, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {10 x+(-5+5 x) \log \left (-x^2\right )+x^3 \log ^2\left (-x^2\right )+x \log ^2(x) \log ^2\left (-x^2\right )+\log (x) \left (-10-2 x^2 \log ^2\left (-x^2\right )\right )}{x^3 \log ^2\left (-x^2\right )-2 x^2 \log (x) \log ^2\left (-x^2\right )+x \log ^2(x) \log ^2\left (-x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(10*x + (-5 + 5*x)*Log[-x^2] + x^3*Log[-x^2]^2 + x*Log[x]^2*Log[-x^2]^2 + Log[x]*(-10 - 2*x^2*Log[-x^2]^2)
)/(x^3*Log[-x^2]^2 - 2*x^2*Log[x]*Log[-x^2]^2 + x*Log[x]^2*Log[-x^2]^2),x]

[Out]

x + 10*Defer[Int][1/(x*(x - Log[x])*Log[-x^2]^2), x] + 5*Defer[Int][1/((x - Log[x])^2*Log[-x^2]), x] - 5*Defer
[Int][1/(x*(x - Log[x])^2*Log[-x^2]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 x+(-5+5 x) \log \left (-x^2\right )+x^3 \log ^2\left (-x^2\right )+x \log ^2(x) \log ^2\left (-x^2\right )+\log (x) \left (-10-2 x^2 \log ^2\left (-x^2\right )\right )}{x (x-\log (x))^2 \log ^2\left (-x^2\right )} \, dx\\ &=\int \left (1+\frac {10}{x (x-\log (x)) \log ^2\left (-x^2\right )}+\frac {5 (-1+x)}{x (x-\log (x))^2 \log \left (-x^2\right )}\right ) \, dx\\ &=x+5 \int \frac {-1+x}{x (x-\log (x))^2 \log \left (-x^2\right )} \, dx+10 \int \frac {1}{x (x-\log (x)) \log ^2\left (-x^2\right )} \, dx\\ &=x+5 \int \left (\frac {1}{(x-\log (x))^2 \log \left (-x^2\right )}-\frac {1}{x (x-\log (x))^2 \log \left (-x^2\right )}\right ) \, dx+10 \int \frac {1}{x (x-\log (x)) \log ^2\left (-x^2\right )} \, dx\\ &=x+5 \int \frac {1}{(x-\log (x))^2 \log \left (-x^2\right )} \, dx-5 \int \frac {1}{x (x-\log (x))^2 \log \left (-x^2\right )} \, dx+10 \int \frac {1}{x (x-\log (x)) \log ^2\left (-x^2\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.25, size = 20, normalized size = 1.00 \begin {gather*} x-\frac {5}{(x-\log (x)) \log \left (-x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x + (-5 + 5*x)*Log[-x^2] + x^3*Log[-x^2]^2 + x*Log[x]^2*Log[-x^2]^2 + Log[x]*(-10 - 2*x^2*Log[-x
^2]^2))/(x^3*Log[-x^2]^2 - 2*x^2*Log[x]*Log[-x^2]^2 + x*Log[x]^2*Log[-x^2]^2),x]

[Out]

x - 5/((x - Log[x])*Log[-x^2])

________________________________________________________________________________________

fricas [C]  time = 0.69, size = 55, normalized size = 2.75 \begin {gather*} \frac {x \log \left (-x^{2}\right )^{2} + {\left (i \, \pi x - 2 \, x^{2}\right )} \log \left (-x^{2}\right ) + 10}{{\left (i \, \pi - 2 \, x\right )} \log \left (-x^{2}\right ) + \log \left (-x^{2}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(-x^2)^2*log(x)^2+(-2*x^2*log(-x^2)^2-10)*log(x)+x^3*log(-x^2)^2+(5*x-5)*log(-x^2)+10*x)/(x*lo
g(-x^2)^2*log(x)^2-2*x^2*log(-x^2)^2*log(x)+x^3*log(-x^2)^2),x, algorithm="fricas")

[Out]

(x*log(-x^2)^2 + (I*pi*x - 2*x^2)*log(-x^2) + 10)/((I*pi - 2*x)*log(-x^2) + log(-x^2)^2)

________________________________________________________________________________________

giac [C]  time = 0.46, size = 27, normalized size = 1.35 \begin {gather*} x + \frac {5}{-i \, \pi x + i \, \pi \log \relax (x) - 2 \, x \log \relax (x) + 2 \, \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(-x^2)^2*log(x)^2+(-2*x^2*log(-x^2)^2-10)*log(x)+x^3*log(-x^2)^2+(5*x-5)*log(-x^2)+10*x)/(x*lo
g(-x^2)^2*log(x)^2-2*x^2*log(-x^2)^2*log(x)+x^3*log(-x^2)^2),x, algorithm="giac")

[Out]

x + 5/(-I*pi*x + I*pi*log(x) - 2*x*log(x) + 2*log(x)^2)

________________________________________________________________________________________

maple [C]  time = 0.30, size = 157, normalized size = 7.85

method	result	size
risch	\(x -\frac {10 i}{-\pi x \mathrm {csgn}\left (i x^{2}\right )^{3}+2 x \pi \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\ln \relax (x ) \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-2 \ln \relax (x ) \pi \mathrm {csgn}\left (i x^{2}\right )^{2}-\ln \relax (x ) \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 \ln \relax (x ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-4 i \ln \relax (x )^{2}+4 i x \ln \relax (x )-2 \pi x +2 \pi \ln \relax (x )}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*ln(-x^2)^2*ln(x)^2+(-2*x^2*ln(-x^2)^2-10)*ln(x)+x^3*ln(-x^2)^2+(5*x-5)*ln(-x^2)+10*x)/(x*ln(-x^2)^2*ln(
x)^2-2*x^2*ln(-x^2)^2*ln(x)+x^3*ln(-x^2)^2),x,method=_RETURNVERBOSE)

[Out]

x-10*I/(-Pi*x*csgn(I*x^2)^3+2*x*Pi*csgn(I*x^2)^2+Pi*x*csgn(I*x)^2*csgn(I*x^2)-2*Pi*x*csgn(I*x)*csgn(I*x^2)^2+l
n(x)*Pi*csgn(I*x^2)^3-2*ln(x)*Pi*csgn(I*x^2)^2-ln(x)*Pi*csgn(I*x)^2*csgn(I*x^2)+2*ln(x)*Pi*csgn(I*x)*csgn(I*x^
2)^2-4*I*ln(x)^2+4*I*x*ln(x)-2*Pi*x+2*Pi*ln(x))

________________________________________________________________________________________

maxima [C]  time = 0.38, size = 52, normalized size = 2.60 \begin {gather*} \frac {-i \, \pi x^{2} + 2 \, x \log \relax (x)^{2} + {\left (i \, \pi x - 2 \, x^{2}\right )} \log \relax (x) + 5}{-i \, \pi x + {\left (i \, \pi - 2 \, x\right )} \log \relax (x) + 2 \, \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*log(-x^2)^2*log(x)^2+(-2*x^2*log(-x^2)^2-10)*log(x)+x^3*log(-x^2)^2+(5*x-5)*log(-x^2)+10*x)/(x*lo
g(-x^2)^2*log(x)^2-2*x^2*log(-x^2)^2*log(x)+x^3*log(-x^2)^2),x, algorithm="maxima")

[Out]

(-I*pi*x^2 + 2*x*log(x)^2 + (I*pi*x - 2*x^2)*log(x) + 5)/(-I*pi*x + (I*pi - 2*x)*log(x) + 2*log(x)^2)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int \frac {10\,x+\ln \left (-x^2\right )\,\left (5\,x-5\right )+x^3\,{\ln \left (-x^2\right )}^2-\ln \relax (x)\,\left (2\,x^2\,{\ln \left (-x^2\right )}^2+10\right )+x\,{\ln \left (-x^2\right )}^2\,{\ln \relax (x)}^2}{x^3\,{\ln \left (-x^2\right )}^2-2\,x^2\,{\ln \left (-x^2\right )}^2\,\ln \relax (x)+x\,{\ln \left (-x^2\right )}^2\,{\ln \relax (x)}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x + log(-x^2)*(5*x - 5) + x^3*log(-x^2)^2 - log(x)*(2*x^2*log(-x^2)^2 + 10) + x*log(-x^2)^2*log(x)^2)/
(x^3*log(-x^2)^2 + x*log(-x^2)^2*log(x)^2 - 2*x^2*log(-x^2)^2*log(x)),x)

[Out]

int((10*x + log(-x^2)*(5*x - 5) + x^3*log(-x^2)^2 - log(x)*(2*x^2*log(-x^2)^2 + 10) + x*log(-x^2)^2*log(x)^2)/
(x^3*log(-x^2)^2 + x*log(-x^2)^2*log(x)^2 - 2*x^2*log(-x^2)^2*log(x)), x)

________________________________________________________________________________________

sympy [C]  time = 0.67, size = 24, normalized size = 1.20 \begin {gather*} x + \frac {5}{- i \pi x + \left (- 2 x + i \pi \right ) \log {\relax (x )} + 2 \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*ln(-x**2)**2*ln(x)**2+(-2*x**2*ln(-x**2)**2-10)*ln(x)+x**3*ln(-x**2)**2+(5*x-5)*ln(-x**2)+10*x)/(
x*ln(-x**2)**2*ln(x)**2-2*x**2*ln(-x**2)**2*ln(x)+x**3*ln(-x**2)**2),x)

[Out]

x + 5/(-I*pi*x + (-2*x + I*pi)*log(x) + 2*log(x)**2)

________________________________________________________________________________________