\(\int (-128+32 x-32 (i \pi +\log (-\frac {-2-\log (4)}{-1+\log (4)}))) \, dx\) [7892]

Optimal result

Integrand size = 28, antiderivative size = 27 \[ \int \left (-128+32 x-32 \left (i \pi +\log \left (-\frac {-2-\log (4)}{-1+\log (4)}\right )\right )\right ) \, dx=16 \left (4+i \pi -x+\log \left (1-\frac {3}{1-\log (4)}\right )\right )^2 \]

[Out]

4*(4-x+ln(3/(1-2*ln(2))-1))*(16-4*x+4*ln(3/(1-2*ln(2))-1))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15, number of steps used = 1, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \left (-128+32 x-32 \left (i \pi +\log \left (-\frac {-2-\log (4)}{-1+\log (4)}\right )\right )\right ) \, dx=16 x^2-32 x \left (4+i \pi +\log \left (-\frac {2+\log (4)}{1-\log (4)}\right )\right ) \]

[In]

Int[-128 + 32*x - 32*(I*Pi + Log[-((-2 - Log[4])/(-1 + Log[4]))]),x]

[Out]

16*x^2 - 32*x*(4 + I*Pi + Log[-((2 + Log[4])/(1 - Log[4]))])

Rubi steps \begin{align*} \text {integral}& = 16 x^2-32 x \left (4+i \pi +\log \left (-\frac {2+\log (4)}{1-\log (4)}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \left (-128+32 x-32 \left (i \pi +\log \left (-\frac {-2-\log (4)}{-1+\log (4)}\right )\right )\right ) \, dx=-128 x+16 x^2-32 x \left (i \pi +\log \left (-\frac {-2-\log (4)}{-1+\log (4)}\right )\right ) \]

[In]

Integrate[-128 + 32*x - 32*(I*Pi + Log[-((-2 - Log[4])/(-1 + Log[4]))]),x]

[Out]

-128*x + 16*x^2 - 32*x*(I*Pi + Log[-((-2 - Log[4])/(-1 + Log[4]))])

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00

method	result	size
gosper	\(-16 x \left (-x +2 \ln \left (\frac {-2 \ln \left (2\right )-2}{2 \ln \left (2\right )-1}\right )+8\right )\)	\(27\)
default	\(-32 x \ln \left (\frac {-2 \ln \left (2\right )-2}{2 \ln \left (2\right )-1}\right )+16 x^{2}-128 x\)	\(29\)
parallelrisch	\(16 x^{2}+\left (-32 \ln \left (\frac {-2 \ln \left (2\right )-2}{2 \ln \left (2\right )-1}\right )-128\right ) x\)	\(29\)
parts	\(-32 x \ln \left (\frac {-2 \ln \left (2\right )-2}{2 \ln \left (2\right )-1}\right )+16 x^{2}-128 x\)	\(29\)
norman	\(\left (-32 \ln \left (2\right )-32 \ln \left (1+\ln \left (2\right )\right )+32 \ln \left (2 \ln \left (2\right )-1\right )-32 i \pi -128\right ) x +16 x^{2}\)	\(35\)
risch	\(-32 i \pi x -32 x \ln \left (2\right )-32 \ln \left (1+\ln \left (2\right )\right ) x +32 \ln \left (2 \ln \left (2\right )-1\right ) x +16 x^{2}-128 x\)	\(38\)

[In]

int(-32*ln((-2*ln(2)-2)/(2*ln(2)-1))+32*x-128,x,method=_RETURNVERBOSE)

[Out]

-16*x*(-x+2*ln((-2*ln(2)-2)/(2*ln(2)-1))+8)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \left (-128+32 x-32 \left (i \pi +\log \left (-\frac {-2-\log (4)}{-1+\log (4)}\right )\right )\right ) \, dx=16 \, x^{2} - 32 \, x \log \left (-\frac {2 \, {\left (\log \left (2\right ) + 1\right )}}{2 \, \log \left (2\right ) - 1}\right ) - 128 \, x \]

[In]

integrate(-32*log((-2*log(2)-2)/(2*log(2)-1))+32*x-128,x, algorithm="fricas")

[Out]

16*x^2 - 32*x*log(-2*(log(2) + 1)/(2*log(2) - 1)) - 128*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \left (-128+32 x-32 \left (i \pi +\log \left (-\frac {-2-\log (4)}{-1+\log (4)}\right )\right )\right ) \, dx=16 x^{2} + x \left (-128 - 32 \log {\left (2 \log {\left (2 \right )} + 2 \right )} + 32 \log {\left (-1 + 2 \log {\left (2 \right )} \right )} - 32 i \pi \right ) \]

[In]

integrate(-32*ln((-2*ln(2)-2)/(2*ln(2)-1))+32*x-128,x)

[Out]

16*x**2 + x*(-128 - 32*log(2*log(2) + 2) + 32*log(-1 + 2*log(2)) - 32*I*pi)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \left (-128+32 x-32 \left (i \pi +\log \left (-\frac {-2-\log (4)}{-1+\log (4)}\right )\right )\right ) \, dx=16 \, x^{2} - 32 \, x \log \left (-\frac {2 \, {\left (\log \left (2\right ) + 1\right )}}{2 \, \log \left (2\right ) - 1}\right ) - 128 \, x \]

[In]

integrate(-32*log((-2*log(2)-2)/(2*log(2)-1))+32*x-128,x, algorithm="maxima")

[Out]

16*x^2 - 32*x*log(-2*(log(2) + 1)/(2*log(2) - 1)) - 128*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \left (-128+32 x-32 \left (i \pi +\log \left (-\frac {-2-\log (4)}{-1+\log (4)}\right )\right )\right ) \, dx=16 \, x^{2} - 32 \, x \log \left (-\frac {2 \, {\left (\log \left (2\right ) + 1\right )}}{2 \, \log \left (2\right ) - 1}\right ) - 128 \, x \]

[In]

integrate(-32*log((-2*log(2)-2)/(2*log(2)-1))+32*x-128,x, algorithm="giac")

[Out]

16*x^2 - 32*x*log(-2*(log(2) + 1)/(2*log(2) - 1)) - 128*x

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \left (-128+32 x-32 \left (i \pi +\log \left (-\frac {-2-\log (4)}{-1+\log (4)}\right )\right )\right ) \, dx=16\,x^2-x\,\left (32\,\ln \left (-\frac {\ln \left (4\right )+2}{\ln \left (4\right )-1}\right )+128\right ) \]

[In]

int(32*x - 32*log(-(2*log(2) + 2)/(2*log(2) - 1)) - 128,x)

[Out]

16*x^2 - x*(32*log(-(log(4) + 2)/(log(4) - 1)) + 128)