3.22.65 \(\int \frac {18-12 x+2 x^2+x^2 (i \pi +\log (\frac {25}{4}))+(-9+6 x-x^2) \log (x^2)}{(3 x^2-x^3) (i \pi +\log (\frac {25}{4}))+(9 x-6 x^2+x^3) \log (x^2)} \, dx\)

Optimal. Leaf size=32 \[ \log \left (\frac {1}{3} \left (\frac {i \pi +\log \left (\frac {25}{4}\right )}{3-x}+\frac {\log \left (x^2\right )}{x}\right )\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 2.20, 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 {18-12 x+2 x^2+x^2 \left (i \pi +\log \left (\frac {25}{4}\right )\right )+\left (-9+6 x-x^2\right ) \log \left (x^2\right )}{\left (3 x^2-x^3\right ) \left (i \pi +\log \left (\frac {25}{4}\right )\right )+\left (9 x-6 x^2+x^3\right ) \log \left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(18 - 12*x + 2*x^2 + x^2*(I*Pi + Log[25/4]) + (-9 + 6*x - x^2)*Log[x^2])/((3*x^2 - x^3)*(I*Pi + Log[25/4])
+ (9*x - 6*x^2 + x^3)*Log[x^2]),x]

[Out]

-Log[x] + (6*I)*Defer[Int][1/(x*(-(x*(Pi - I*Log[25/4])) - I*(-3 + x)*Log[x^2])), x] + (2*I)*Defer[Int][(x*(Pi
- I*Log[25/4]) + I*(-3 + x)*Log[x^2])^(-1), x] + 3*(Pi - I*Log[25/4])*Defer[Int][1/((3 - x)*(x*(Pi - I*Log[25
/4]) + I*(-3 + x)*Log[x^2])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {18-12 x+x^2 \left (2+i \pi +\log \left (\frac {25}{4}\right )\right )+\left (-9+6 x-x^2\right ) \log \left (x^2\right )}{\left (3 x^2-x^3\right ) \left (i \pi +\log \left (\frac {25}{4}\right )\right )+\left (9 x-6 x^2+x^3\right ) \log \left (x^2\right )} \, dx\\ &=\int \frac {18-12 x+x^2 \left (2+i \pi +\log \left (\frac {25}{4}\right )\right )+\left (-9+6 x-x^2\right ) \log \left (x^2\right )}{(3-x) x \left (i \pi x \left (1-\frac {i \log \left (\frac {25}{4}\right )}{\pi }\right )+3 \log \left (x^2\right )-x \log \left (x^2\right )\right )} \, dx\\ &=\int \left (-\frac {1}{x}+\frac {i \left (-18-2 x^2+3 x \left (4-i \pi -\log \left (\frac {25}{4}\right )\right )\right )}{(3-x) x \left (\pi x \left (1-\frac {i \log \left (\frac {25}{4}\right )}{\pi }\right )-3 i \log \left (x^2\right )+i x \log \left (x^2\right )\right )}\right ) \, dx\\ &=-\log (x)+i \int \frac {-18-2 x^2+3 x \left (4-i \pi -\log \left (\frac {25}{4}\right )\right )}{(3-x) x \left (\pi x \left (1-\frac {i \log \left (\frac {25}{4}\right )}{\pi }\right )-3 i \log \left (x^2\right )+i x \log \left (x^2\right )\right )} \, dx\\ &=-\log (x)+i \int \frac {-18-2 x^2+3 x \left (4-i \pi -\log \left (\frac {25}{4}\right )\right )}{(3-x) x \left (x \left (\pi -i \log \left (\frac {25}{4}\right )\right )+i (-3+x) \log \left (x^2\right )\right )} \, dx\\ &=-\log (x)+i \int \left (\frac {6}{x \left (-\pi x \left (1-\frac {i \log \left (\frac {25}{4}\right )}{\pi }\right )+3 i \log \left (x^2\right )-i x \log \left (x^2\right )\right )}+\frac {2}{\pi x \left (1-\frac {i \log \left (\frac {25}{4}\right )}{\pi }\right )-3 i \log \left (x^2\right )+i x \log \left (x^2\right )}+\frac {3 \left (-i \pi -\log \left (\frac {25}{4}\right )\right )}{(3-x) \left (\pi x \left (1-\frac {i \log \left (\frac {25}{4}\right )}{\pi }\right )-3 i \log \left (x^2\right )+i x \log \left (x^2\right )\right )}\right ) \, dx\\ &=-\log (x)+2 i \int \frac {1}{\pi x \left (1-\frac {i \log \left (\frac {25}{4}\right )}{\pi }\right )-3 i \log \left (x^2\right )+i x \log \left (x^2\right )} \, dx+6 i \int \frac {1}{x \left (-\pi x \left (1-\frac {i \log \left (\frac {25}{4}\right )}{\pi }\right )+3 i \log \left (x^2\right )-i x \log \left (x^2\right )\right )} \, dx+\left (3 \left (\pi -i \log \left (\frac {25}{4}\right )\right )\right ) \int \frac {1}{(3-x) \left (\pi x \left (1-\frac {i \log \left (\frac {25}{4}\right )}{\pi }\right )-3 i \log \left (x^2\right )+i x \log \left (x^2\right )\right )} \, dx\\ &=-\log (x)+2 i \int \frac {1}{x \left (\pi -i \log \left (\frac {25}{4}\right )\right )+i (-3+x) \log \left (x^2\right )} \, dx+6 i \int \frac {1}{x \left (-x \left (\pi -i \log \left (\frac {25}{4}\right )\right )-i (-3+x) \log \left (x^2\right )\right )} \, dx+\left (3 \left (\pi -i \log \left (\frac {25}{4}\right )\right )\right ) \int \frac {1}{(3-x) \left (x \left (\pi -i \log \left (\frac {25}{4}\right )\right )+i (-3+x) \log \left (x^2\right )\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.40, size = 121, normalized size = 3.78 \begin {gather*} -i \tan ^{-1}\left (\frac {\pi x}{-x \log \left (\frac {25}{4}\right )-3 \log \left (x^2\right )+x \log \left (x^2\right )}\right )-\frac {1}{2} \log \left ((3-x)^2\right )-\log (x)+\frac {1}{2} \log \left (\pi ^2 x^2+x^2 \log ^2\left (\frac {25}{4}\right )+6 x \log \left (\frac {25}{4}\right ) \log \left (x^2\right )-2 x^2 \log \left (\frac {25}{4}\right ) \log \left (x^2\right )+9 \log ^2\left (x^2\right )-6 x \log ^2\left (x^2\right )+x^2 \log ^2\left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18 - 12*x + 2*x^2 + x^2*(I*Pi + Log[25/4]) + (-9 + 6*x - x^2)*Log[x^2])/((3*x^2 - x^3)*(I*Pi + Log[
25/4]) + (9*x - 6*x^2 + x^3)*Log[x^2]),x]

[Out]

(-I)*ArcTan[(Pi*x)/(-(x*Log[25/4]) - 3*Log[x^2] + x*Log[x^2])] - Log[(3 - x)^2]/2 - Log[x] + Log[Pi^2*x^2 + x^
2*Log[25/4]^2 + 6*x*Log[25/4]*Log[x^2] - 2*x^2*Log[25/4]*Log[x^2] + 9*Log[x^2]^2 - 6*x*Log[x^2]^2 + x^2*Log[x^
2]^2]/2

________________________________________________________________________________________

fricas [A]  time = 0.60, size = 32, normalized size = 1.00 \begin {gather*} -\frac {1}{2} \, \log \left (x^{2}\right ) + \log \left (\frac {-i \, \pi x - x \log \left (\frac {25}{4}\right ) + {\left (x - 3\right )} \log \left (x^{2}\right )}{x - 3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+6*x-9)*log(x^2)+x^2*(log(25/4)+I*pi)+2*x^2-12*x+18)/((x^3-6*x^2+9*x)*log(x^2)+(-x^3+3*x^2)*(l
og(25/4)+I*pi)),x, algorithm="fricas")

[Out]

-1/2*log(x^2) + log((-I*pi*x - x*log(25/4) + (x - 3)*log(x^2))/(x - 3))

________________________________________________________________________________________

giac [A]  time = 0.36, size = 39, normalized size = 1.22 \begin {gather*} \log \left (\pi x - 2 i \, x \log \relax (5) + 2 i \, x \log \relax (2) + i \, x \log \left (x^{2}\right ) - 3 i \, \log \left (x^{2}\right )\right ) - \log \left (x - 3\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+6*x-9)*log(x^2)+x^2*(log(25/4)+I*pi)+2*x^2-12*x+18)/((x^3-6*x^2+9*x)*log(x^2)+(-x^3+3*x^2)*(l
og(25/4)+I*pi)),x, algorithm="giac")

[Out]

log(pi*x - 2*I*x*log(5) + 2*I*x*log(2) + I*x*log(x^2) - 3*I*log(x^2)) - log(x - 3) - log(x)

________________________________________________________________________________________

maple [A]  time = 0.17, size = 33, normalized size = 1.03

method	result	size
risch	\(-\ln \relax (x )+\ln \left (\ln \left (x^{2}\right )-\frac {x \left (2 \ln \relax (5)-2 \ln \relax (2)+i \pi \right )}{x -3}\right )\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2+6*x-9)*ln(x^2)+x^2*(ln(25/4)+I*Pi)+2*x^2-12*x+18)/((x^3-6*x^2+9*x)*ln(x^2)+(-x^3+3*x^2)*(ln(25/4)+I
*Pi)),x,method=_RETURNVERBOSE)

[Out]

-ln(x)+ln(ln(x^2)-x*(2*ln(5)-2*ln(2)+I*Pi)/(x-3))

________________________________________________________________________________________

maxima [A]  time = 0.62, size = 35, normalized size = 1.09 \begin {gather*} -\log \relax (x) + \log \left (\frac {{\left (-i \, \pi - 2 \, \log \relax (5) + 2 \, \log \relax (2)\right )} x + 2 \, {\left (x - 3\right )} \log \relax (x)}{2 \, {\left (x - 3\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+6*x-9)*log(x^2)+x^2*(log(25/4)+I*pi)+2*x^2-12*x+18)/((x^3-6*x^2+9*x)*log(x^2)+(-x^3+3*x^2)*(l
og(25/4)+I*pi)),x, algorithm="maxima")

[Out]

-log(x) + log(1/2*((-I*pi - 2*log(5) + 2*log(2))*x + 2*(x - 3)*log(x))/(x - 3))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {2\,x^2-\ln \left (x^2\right )\,\left (x^2-6\,x+9\right )+x^2\,\left (\ln \left (\frac {25}{4}\right )+\Pi \,1{}\mathrm {i}\right )-12\,x+18}{\left (\ln \left (\frac {25}{4}\right )+\Pi \,1{}\mathrm {i}\right )\,\left (3\,x^2-x^3\right )+\ln \left (x^2\right )\,\left (x^3-6\,x^2+9\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(Pi*1i + log(25/4)) - log(x^2)*(x^2 - 6*x + 9) - 12*x + 2*x^2 + 18)/((Pi*1i + log(25/4))*(3*x^2 - x^3
) + log(x^2)*(9*x - 6*x^2 + x^3)),x)

[Out]

int((x^2*(Pi*1i + log(25/4)) - log(x^2)*(x^2 - 6*x + 9) - 12*x + 2*x^2 + 18)/((Pi*1i + log(25/4))*(3*x^2 - x^3
) + log(x^2)*(9*x - 6*x^2 + x^3)), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2+6*x-9)*ln(x**2)+x**2*(ln(25/4)+I*pi)+2*x**2-12*x+18)/((x**3-6*x**2+9*x)*ln(x**2)+(-x**3+3*x*
*2)*(ln(25/4)+I*pi)),x)

[Out]

Timed out

________________________________________________________________________________________