3.38.24 \(\int \frac {2 (i \pi +\log (\frac {18}{5}))+2 (i \pi +\log (\frac {18}{5})) \log (x)}{4-4 x+x^2+(4-2 x) \log (x)+\log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {2 x \left (i \pi +\log \left (\frac {18}{5}\right )\right )}{2-x+\log (x)} \]

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Rubi [F]  time = 0.19, 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 {2 \left (i \pi +\log \left (\frac {18}{5}\right )\right )+2 \left (i \pi +\log \left (\frac {18}{5}\right )\right ) \log (x)}{4-4 x+x^2+(4-2 x) \log (x)+\log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*(I*Pi + Log[18/5]) + 2*(I*Pi + Log[18/5])*Log[x])/(4 - 4*x + x^2 + (4 - 2*x)*Log[x] + Log[x]^2),x]

[Out]

-2*(I*Pi + Log[18/5])*Defer[Int][(-2 + x - Log[x])^(-2), x] + 2*(I*Pi + Log[18/5])*Defer[Int][x/(-2 + x - Log[
x])^2, x] + 2*(I*Pi + Log[18/5])*Defer[Int][(2 - x + Log[x])^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (i \pi +\log \left (\frac {18}{5}\right )\right ) (1+\log (x))}{(2-x+\log (x))^2} \, dx\\ &=\left (2 \left (i \pi +\log \left (\frac {18}{5}\right )\right )\right ) \int \frac {1+\log (x)}{(2-x+\log (x))^2} \, dx\\ &=\left (2 \left (i \pi +\log \left (\frac {18}{5}\right )\right )\right ) \int \left (\frac {-1+x}{(-2+x-\log (x))^2}+\frac {1}{2-x+\log (x)}\right ) \, dx\\ &=\left (2 \left (i \pi +\log \left (\frac {18}{5}\right )\right )\right ) \int \frac {-1+x}{(-2+x-\log (x))^2} \, dx+\left (2 \left (i \pi +\log \left (\frac {18}{5}\right )\right )\right ) \int \frac {1}{2-x+\log (x)} \, dx\\ &=\left (2 \left (i \pi +\log \left (\frac {18}{5}\right )\right )\right ) \int \left (-\frac {1}{(-2+x-\log (x))^2}+\frac {x}{(-2+x-\log (x))^2}\right ) \, dx+\left (2 \left (i \pi +\log \left (\frac {18}{5}\right )\right )\right ) \int \frac {1}{2-x+\log (x)} \, dx\\ &=-\left (\left (2 \left (i \pi +\log \left (\frac {18}{5}\right )\right )\right ) \int \frac {1}{(-2+x-\log (x))^2} \, dx\right )+\left (2 \left (i \pi +\log \left (\frac {18}{5}\right )\right )\right ) \int \frac {x}{(-2+x-\log (x))^2} \, dx+\left (2 \left (i \pi +\log \left (\frac {18}{5}\right )\right )\right ) \int \frac {1}{2-x+\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 22, normalized size = 1.00 \begin {gather*} \frac {2 x \left (i \pi +\log \left (\frac {18}{5}\right )\right )}{2-x+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*(I*Pi + Log[18/5]) + 2*(I*Pi + Log[18/5])*Log[x])/(4 - 4*x + x^2 + (4 - 2*x)*Log[x] + Log[x]^2),x
]

[Out]

(2*x*(I*Pi + Log[18/5]))/(2 - x + Log[x])

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fricas [A]  time = 0.85, size = 21, normalized size = 0.95 \begin {gather*} \frac {2 \, {\left (-i \, \pi x - x \log \left (\frac {18}{5}\right )\right )}}{x - \log \relax (x) - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(log(18/5)+I*pi)*log(x)+2*log(18/5)+2*I*pi)/(log(x)^2+(4-2*x)*log(x)+x^2-4*x+4),x, algorithm="fri
cas")

[Out]

2*(-I*pi*x - x*log(18/5))/(x - log(x) - 2)

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giac [A]  time = 0.19, size = 25, normalized size = 1.14 \begin {gather*} -\frac {2 \, {\left (i \, \pi x + x \log \left (18\right ) - x \log \relax (5)\right )}}{x - \log \relax (x) - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(log(18/5)+I*pi)*log(x)+2*log(18/5)+2*I*pi)/(log(x)^2+(4-2*x)*log(x)+x^2-4*x+4),x, algorithm="gia
c")

[Out]

-2*(I*pi*x + x*log(18) - x*log(5))/(x - log(x) - 2)

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maple [A]  time = 0.16, size = 28, normalized size = 1.27

method	result	size
risch	\(-\frac {2 \left (\ln \relax (2)+2 \ln \relax (3)-\ln \relax (5)+i \pi \right ) x}{x -\ln \relax (x )-2}\)	\(28\)
norman	\(\frac {\left (-2 i \pi -2 \ln \left (18\right )+2 \ln \relax (5)\right ) \ln \relax (x )-4 i \pi -4 \ln \left (18\right )+4 \ln \relax (5)}{x -\ln \relax (x )-2}\)	\(40\)
default	\(\frac {-2 \ln \relax (x ) \ln \left (18\right )-4 \ln \left (18\right )}{x -\ln \relax (x )-2}+\frac {-2 i \pi \ln \relax (x )-4 i \pi }{x -\ln \relax (x )-2}+\frac {2 \ln \relax (5) \ln \relax (x )+4 \ln \relax (5)}{x -\ln \relax (x )-2}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*(ln(18/5)+I*Pi)*ln(x)+2*ln(18/5)+2*I*Pi)/(ln(x)^2+(4-2*x)*ln(x)+x^2-4*x+4),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.50, size = 26, normalized size = 1.18 \begin {gather*} \frac {2 \, {\left (-i \, \pi + \log \relax (5) - 2 \, \log \relax (3) - \log \relax (2)\right )} x}{x - \log \relax (x) - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(log(18/5)+I*pi)*log(x)+2*log(18/5)+2*I*pi)/(log(x)^2+(4-2*x)*log(x)+x^2-4*x+4),x, algorithm="max
ima")

[Out]

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

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mupad [B]  time = 2.42, size = 21, normalized size = 0.95 \begin {gather*} \frac {-2\,x\,\ln \left (\frac {5}{18}\right )+\Pi \,x\,2{}\mathrm {i}}{\ln \relax (x)-x+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((Pi*2i + 2*log(18/5) + 2*log(x)*(Pi*1i + log(18/5)))/(log(x)^2 - 4*x - log(x)*(2*x - 4) + x^2 + 4),x)

[Out]

(Pi*x*2i - 2*x*log(5/18))/(log(x) - x + 2)

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sympy [A]  time = 0.20, size = 26, normalized size = 1.18 \begin {gather*} \frac {- 2 x \log {\relax (5 )} + 2 x \log {\left (18 \right )} + 2 i \pi x}{- x + \log {\relax (x )} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(ln(18/5)+I*pi)*ln(x)+2*ln(18/5)+2*I*pi)/(ln(x)**2+(4-2*x)*ln(x)+x**2-4*x+4),x)

[Out]

(-2*x*log(5) + 2*x*log(18) + 2*I*pi*x)/(-x + log(x) + 2)

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