3.79.20 \(\int \frac {x-\log (-130 e^{-4-x})+4 \log ^2(-130 e^{-4-x})}{1+(-8-4 x) \log (-130 e^{-4-x})+(16+16 x+4 x^2) \log ^2(-130 e^{-4-x})} \, dx\)

Optimal. Leaf size=23 \[ \frac {x}{4+2 x-\frac {1}{\log \left (-130 e^{-4-x}\right )}} \]

________________________________________________________________________________________

Rubi [F]  time = 1.31, 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 {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{1+(-8-4 x) \log \left (-130 e^{-4-x}\right )+\left (16+16 x+4 x^2\right ) \log ^2\left (-130 e^{-4-x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x - Log[-130*E^(-4 - x)] + 4*Log[-130*E^(-4 - x)]^2)/(1 + (-8 - 4*x)*Log[-130*E^(-4 - x)] + (16 + 16*x +
4*x^2)*Log[-130*E^(-4 - x)]^2),x]

[Out]

-(2 + x)^(-1) - Defer[Int][1/((2 + x)*(-1 + 4*Log[-130*E^(-4 - x)] + 2*x*Log[-130*E^(-4 - x)])), x]/2 + Defer[
Int][x/(1 - 2*(2 + x)*Log[-130*E^(-4 - x)])^2, x] + Defer[Int][1/((2 + x)^2*(1 - 2*(2 + x)*Log[-130*E^(-4 - x)
])^2), x] - Defer[Int][1/((2 + x)*(1 - 2*(2 + x)*Log[-130*E^(-4 - x)])^2), x]/2 + 2*Defer[Int][1/((2 + x)^2*(-
1 + 2*(2 + x)*Log[-130*E^(-4 - x)])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x-\log \left (-130 e^{-4-x}\right )+4 \log ^2\left (-130 e^{-4-x}\right )}{\left (1-2 (2+x) \log \left (-130 e^{-4-x}\right )\right )^2} \, dx\\ &=\int \left (\frac {1}{(2+x)^2}+\frac {x \left (7+8 x+2 x^2\right )}{2 (2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2}+\frac {2-x}{2 (2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )}\right ) \, dx\\ &=-\frac {1}{2+x}+\frac {1}{2} \int \frac {x \left (7+8 x+2 x^2\right )}{(2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2} \, dx+\frac {1}{2} \int \frac {2-x}{(2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )} \, dx\\ &=-\frac {1}{2+x}+\frac {1}{2} \int \frac {x \left (7+8 x+2 x^2\right )}{(2+x)^2 \left (1-2 (2+x) \log \left (-130 e^{-4-x}\right )\right )^2} \, dx+\frac {1}{2} \int \frac {-2+x}{(2+x)^2 \left (1-2 (2+x) \log \left (-130 e^{-4-x}\right )\right )} \, dx\\ &=-\frac {1}{2+x}+\frac {1}{2} \int \left (\frac {2 x}{\left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2}+\frac {2}{(2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2}-\frac {1}{(2+x) \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2}\right ) \, dx+\frac {1}{2} \int \left (\frac {4}{(2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )}-\frac {1}{(2+x) \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )}\right ) \, dx\\ &=-\frac {1}{2+x}-\frac {1}{2} \int \frac {1}{(2+x) \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2} \, dx-\frac {1}{2} \int \frac {1}{(2+x) \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )} \, dx+2 \int \frac {1}{(2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )} \, dx+\int \frac {x}{\left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2} \, dx+\int \frac {1}{(2+x)^2 \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )^2} \, dx\\ &=-\frac {1}{2+x}-\frac {1}{2} \int \frac {1}{(2+x) \left (-1+4 \log \left (-130 e^{-4-x}\right )+2 x \log \left (-130 e^{-4-x}\right )\right )} \, dx-\frac {1}{2} \int \frac {1}{(2+x) \left (1-2 (2+x) \log \left (-130 e^{-4-x}\right )\right )^2} \, dx+2 \int \frac {1}{(2+x)^2 \left (-1+2 (2+x) \log \left (-130 e^{-4-x}\right )\right )} \, dx+\int \frac {x}{\left (1-2 (2+x) \log \left (-130 e^{-4-x}\right )\right )^2} \, dx+\int \frac {1}{(2+x)^2 \left (1-2 (2+x) \log \left (-130 e^{-4-x}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.37, size = 34, normalized size = 1.48 \begin {gather*} \frac {1-4 \log \left (-130 e^{-4-x}\right )}{-2+4 (2+x) \log \left (-130 e^{-4-x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x - Log[-130*E^(-4 - x)] + 4*Log[-130*E^(-4 - x)]^2)/(1 + (-8 - 4*x)*Log[-130*E^(-4 - x)] + (16 + 1
6*x + 4*x^2)*Log[-130*E^(-4 - x)]^2),x]

[Out]

(1 - 4*Log[-130*E^(-4 - x)])/(-2 + 4*(2 + x)*Log[-130*E^(-4 - x)])

________________________________________________________________________________________

fricas [C]  time = 0.70, size = 37, normalized size = 1.61 \begin {gather*} \frac {-4 i \, \pi + 4 \, x - 4 \, \log \left (130\right ) + 17}{2 \, {\left (2 \, {\left (i \, \pi + \log \left (130\right )\right )} {\left (x + 2\right )} - 2 \, x^{2} - 12 \, x - 17\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(-130/exp(4+x))^2-log(-130/exp(4+x))+x)/((4*x^2+16*x+16)*log(-130/exp(4+x))^2+(-4*x-8)*log(-13
0/exp(4+x))+1),x, algorithm="fricas")

[Out]

1/2*(-4*I*pi + 4*x - 4*log(130) + 17)/(2*(I*pi + log(130))*(x + 2) - 2*x^2 - 12*x - 17)

________________________________________________________________________________________

giac [C]  time = 0.23, size = 41, normalized size = 1.78 \begin {gather*} \frac {4 i \, \pi - 4 \, x + 4 \, \log \left (130\right ) - 17}{-8 i \, \pi - 4 i \, \pi x + 4 \, x^{2} - 4 \, x \log \left (130\right ) + 24 \, x - 8 \, \log \left (130\right ) + 34} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(-130/exp(4+x))^2-log(-130/exp(4+x))+x)/((4*x^2+16*x+16)*log(-130/exp(4+x))^2+(-4*x-8)*log(-13
0/exp(4+x))+1),x, algorithm="giac")

[Out]

(4*I*pi - 4*x + 4*log(130) - 17)/(-8*I*pi - 4*I*pi*x + 4*x^2 - 4*x*log(130) + 24*x - 8*log(130) + 34)

________________________________________________________________________________________

maple [A]  time = 0.51, size = 42, normalized size = 1.83

method	result	size
norman	\(\frac {\frac {1}{2}-2 \ln \left (-130 \,{\mathrm e}^{-x -4}\right )}{2 \ln \left (-130 \,{\mathrm e}^{-x -4}\right ) x +4 \ln \left (-130 \,{\mathrm e}^{-x -4}\right )-1}\)	\(42\)
default	\(\frac {-\frac {1}{4}+\ln \left (-130 \,{\mathrm e}^{-x -4}\right )}{x^{2}+x \left (\ln \left ({\mathrm e}^{4+x}\right )-4-x \right )-x \left (\ln \left (-130 \,{\mathrm e}^{-x -4}\right )+\ln \left ({\mathrm e}^{4+x}\right )\right )+4 x +\frac {1}{2}-2 \ln \left (-130 \,{\mathrm e}^{-x -4}\right )}\)	\(64\)
risch	\(-\frac {1}{2+x}-\frac {i x}{2 \left (2+x \right ) \left (-2 x \pi \mathrm {csgn}\left (i {\mathrm e}^{-x -4}\right )^{2}+2 x \pi \mathrm {csgn}\left (i {\mathrm e}^{-x -4}\right )^{3}-4 \pi \mathrm {csgn}\left (i {\mathrm e}^{-x -4}\right )^{2}+4 \pi \mathrm {csgn}\left (i {\mathrm e}^{-x -4}\right )^{3}+2 \pi x +4 \pi +4 i \ln \left ({\mathrm e}^{4+x}\right )-4 i \ln \left (13\right )-4 i \ln \relax (2)-4 i \ln \relax (5)-2 i x \ln \left (13\right )-2 i \ln \relax (5) x -2 i x \ln \relax (2)+2 i x \ln \left ({\mathrm e}^{4+x}\right )+i\right )}\)	\(142\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*ln(-130/exp(4+x))^2-ln(-130/exp(4+x))+x)/((4*x^2+16*x+16)*ln(-130/exp(4+x))^2+(-4*x-8)*ln(-130/exp(4+x)
)+1),x,method=_RETURNVERBOSE)

[Out]

(1/2-2*ln(-130/exp(4+x)))/(2*ln(-130/exp(4+x))*x+4*ln(-130/exp(4+x))-1)

________________________________________________________________________________________

maxima [B]  time = 0.50, size = 60, normalized size = 2.61 \begin {gather*} -\frac {4 \, \log \left (13\right ) + 4 \, \log \relax (5) + 4 \, \log \relax (2) - 4 \, \log \left (-e^{x}\right ) - 17}{2 \, {\left (2 \, x {\left (\log \left (13\right ) + \log \relax (5) + \log \relax (2) - 4\right )} - 2 \, {\left (x + 2\right )} \log \left (-e^{x}\right ) + 4 \, \log \left (13\right ) + 4 \, \log \relax (5) + 4 \, \log \relax (2) - 17\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*log(-130/exp(4+x))^2-log(-130/exp(4+x))+x)/((4*x^2+16*x+16)*log(-130/exp(4+x))^2+(-4*x-8)*log(-13
0/exp(4+x))+1),x, algorithm="maxima")

[Out]

-1/2*(4*log(13) + 4*log(5) + 4*log(2) - 4*log(-e^x) - 17)/(2*x*(log(13) + log(5) + log(2) - 4) - 2*(x + 2)*log
(-e^x) + 4*log(13) + 4*log(5) + 4*log(2) - 17)

________________________________________________________________________________________

mupad [B]  time = 5.92, size = 43, normalized size = 1.87 \begin {gather*} \frac {2\,x-2\,\ln \left (130\right )+\frac {17}{2}-\pi \,2{}\mathrm {i}}{-2\,x^2+\left (2\,\ln \left (130\right )-12+\pi \,2{}\mathrm {i}\right )\,x+\pi \,4{}\mathrm {i}+4\,\ln \left (130\right )-17} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - log(-130*exp(- x - 4)) + 4*log(-130*exp(- x - 4))^2)/(log(-130*exp(- x - 4))^2*(16*x + 4*x^2 + 16) -
log(-130*exp(- x - 4))*(4*x + 8) + 1),x)

[Out]

(2*x - pi*2i - 2*log(130) + 17/2)/(pi*4i + 4*log(130) + x*(pi*2i + 2*log(130) - 12) - 2*x^2 - 17)

________________________________________________________________________________________

sympy [C]  time = 1.81, size = 44, normalized size = 1.91 \begin {gather*} \frac {4 x - 4 \log {\left (130 \right )} + 17 - 4 i \pi }{- 4 x^{2} + x \left (-24 + 4 \log {\left (130 \right )} + 4 i \pi \right ) - 34 + 8 \log {\left (130 \right )} + 8 i \pi } \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*ln(-130/exp(4+x))**2-ln(-130/exp(4+x))+x)/((4*x**2+16*x+16)*ln(-130/exp(4+x))**2+(-4*x-8)*ln(-130
/exp(4+x))+1),x)

[Out]

(4*x - 4*log(130) + 17 - 4*I*pi)/(-4*x**2 + x*(-24 + 4*log(130) + 4*I*pi) - 34 + 8*log(130) + 8*I*pi)

________________________________________________________________________________________