3.102.95 \(\int \frac {-2-x+2 \log (-5 x+x (i \pi +\log (3)))}{(4+4 x+x^2) \log ^2(-5 x+x (i \pi +\log (3)))} \, dx\)

Optimal. Leaf size=21 \[ \frac {x}{(2+x) \log (x (-5+i \pi +\log (3)))} \]

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Rubi [F]  time = 0.20, 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 {-2-x+2 \log (-5 x+x (i \pi +\log (3)))}{\left (4+4 x+x^2\right ) \log ^2(-5 x+x (i \pi +\log (3)))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2 - x + 2*Log[-5*x + x*(I*Pi + Log[3])])/((4 + 4*x + x^2)*Log[-5*x + x*(I*Pi + Log[3])]^2),x]

[Out]

Defer[Int][1/((-2 - x)*Log[x*(-5 + I*Pi + Log[3])]^2), x] + 2*Defer[Int][1/((2 + x)^2*Log[x*(-5 + I*Pi + Log[3
])]), x]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 21, normalized size = 1.00 \begin {gather*} \frac {x}{(2+x) \log (x (-5+i \pi +\log (3)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - x + 2*Log[-5*x + x*(I*Pi + Log[3])])/((4 + 4*x + x^2)*Log[-5*x + x*(I*Pi + Log[3])]^2),x]

[Out]

x/((2 + x)*Log[x*(-5 + I*Pi + Log[3])])

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fricas [A]  time = 1.04, size = 22, normalized size = 1.05 \begin {gather*} \frac {x}{{\left (x + 2\right )} \log \left ({\left (i \, \pi - 5\right )} x + x \log \relax (3)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x*(log(3)+I*pi)-5*x)-x-2)/(x^2+4*x+4)/log(x*(log(3)+I*pi)-5*x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 35, normalized size = 1.67 \begin {gather*} \frac {x}{x \log \left (i \, \pi x + x \log \relax (3) - 5 \, x\right ) + 2 \, \log \left (i \, \pi x + x \log \relax (3) - 5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x*(log(3)+I*pi)-5*x)-x-2)/(x^2+4*x+4)/log(x*(log(3)+I*pi)-5*x)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 1.43, size = 24, normalized size = 1.14

method	result	size
norman	\(\frac {x}{\left (2+x \right ) \ln \left (x \left (\ln \relax (3)+i \pi \right )-5 x \right )}\)	\(24\)
risch	\(\frac {x}{\left (2+x \right ) \ln \left (x \left (\ln \relax (3)+i \pi \right )-5 x \right )}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*ln(x*(ln(3)+I*Pi)-5*x)-x-2)/(x^2+4*x+4)/ln(x*(ln(3)+I*Pi)-5*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.49, size = 31, normalized size = 1.48 \begin {gather*} \frac {x}{x \log \left (i \, \pi + \log \relax (3) - 5\right ) + {\left (x + 2\right )} \log \relax (x) + 2 \, \log \left (i \, \pi + \log \relax (3) - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*log(x*(log(3)+I*pi)-5*x)-x-2)/(x^2+4*x+4)/log(x*(log(3)+I*pi)-5*x)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.82, size = 20, normalized size = 0.95 \begin {gather*} \frac {x}{\ln \left (x\,\left (\ln \relax (3)-5+\Pi \,1{}\mathrm {i}\right )\right )\,\left (x+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x - 2*log(x*(Pi*1i + log(3)) - 5*x) + 2)/(log(x*(Pi*1i + log(3)) - 5*x)^2*(4*x + x^2 + 4)),x)

[Out]

x/(log(x*(Pi*1i + log(3) - 5))*(x + 2))

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sympy [A]  time = 0.13, size = 19, normalized size = 0.90 \begin {gather*} \frac {x}{\left (x + 2\right ) \log {\left (- 5 x + x \log {\relax (3 )} + i \pi x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*ln(x*(ln(3)+I*pi)-5*x)-x-2)/(x**2+4*x+4)/ln(x*(ln(3)+I*pi)-5*x)**2,x)

[Out]

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

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