\(\int \frac {2-2 x-45 x \log (4)}{-2 x^2+(-12 x-45 x^2) \log (4)+2 x \log (x)} \, dx\) [6691]

Optimal result

Integrand size = 36, antiderivative size = 24 \[ \int \frac {2-2 x-45 x \log (4)}{-2 x^2+\left (-12 x-45 x^2\right ) \log (4)+2 x \log (x)} \, dx=\log \left (4-x+16 \left (x-\frac {-x+\log (x)}{24 \log (4)}\right )\right ) \]

[Out]

ln(4+15*x-1/3*(ln(x)-x)/ln(2))

Rubi [F]

\[ \int \frac {2-2 x-45 x \log (4)}{-2 x^2+\left (-12 x-45 x^2\right ) \log (4)+2 x \log (x)} \, dx=\int \frac {2-2 x-45 x \log (4)}{-2 x^2+\left (-12 x-45 x^2\right ) \log (4)+2 x \log (x)} \, dx \]

[In]

Int[(2 - 2*x - 45*x*Log[4])/(-2*x^2 + (-12*x - 45*x^2)*Log[4] + 2*x*Log[x]),x]

[Out]

(2 + 45*Log[4])*Defer[Int][(12*Log[4] + 2*x*(1 + (45*Log[4])/2) - 2*Log[x])^(-1), x] + 2*Defer[Int][1/(x*(-12*
Log[4] - 2*x*(1 + (45*Log[4])/2) + 2*Log[x])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {2+x (-2-45 \log (4))}{-2 x^2+\left (-12 x-45 x^2\right ) \log (4)+2 x \log (x)} \, dx \\ & = \int \left (\frac {2+45 \log (4)}{12 \log (4)+2 x \left (1+\frac {45 \log (4)}{2}\right )-2 \log (x)}+\frac {2}{x \left (-12 \log (4)-2 x \left (1+\frac {45 \log (4)}{2}\right )+2 \log (x)\right )}\right ) \, dx \\ & = 2 \int \frac {1}{x \left (-12 \log (4)-2 x \left (1+\frac {45 \log (4)}{2}\right )+2 \log (x)\right )} \, dx+(2+45 \log (4)) \int \frac {1}{12 \log (4)+2 x \left (1+\frac {45 \log (4)}{2}\right )-2 \log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {2-2 x-45 x \log (4)}{-2 x^2+\left (-12 x-45 x^2\right ) \log (4)+2 x \log (x)} \, dx=\log (2 x+12 \log (4)+45 x \log (4)-2 \log (x)) \]

[In]

Integrate[(2 - 2*x - 45*x*Log[4])/(-2*x^2 + (-12*x - 45*x^2)*Log[4] + 2*x*Log[x]),x]

[Out]

Log[2*x + 12*Log[4] + 45*x*Log[4] - 2*Log[x]]

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71

method	result	size
default	\(\ln \left (45 x \ln \left (2\right )+12 \ln \left (2\right )+x -\ln \left (x \right )\right )\)	\(17\)
norman	\(\ln \left (45 x \ln \left (2\right )+12 \ln \left (2\right )+x -\ln \left (x \right )\right )\)	\(17\)
risch	\(\ln \left (-45 x \ln \left (2\right )+\ln \left (x \right )-12 \ln \left (2\right )-x \right )\)	\(17\)
parallelrisch	\(\ln \left (\frac {45 x \ln \left (2\right )+12 \ln \left (2\right )+x -\ln \left (x \right )}{45 \ln \left (2\right )+1}\right )\)	\(26\)

[In]

int((-90*x*ln(2)-2*x+2)/(2*x*ln(x)+2*(-45*x^2-12*x)*ln(2)-2*x^2),x,method=_RETURNVERBOSE)

[Out]

ln(45*x*ln(2)+12*ln(2)+x-ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {2-2 x-45 x \log (4)}{-2 x^2+\left (-12 x-45 x^2\right ) \log (4)+2 x \log (x)} \, dx=\log \left (-3 \, {\left (15 \, x + 4\right )} \log \left (2\right ) - x + \log \left (x\right )\right ) \]

[In]

integrate((-90*x*log(2)-2*x+2)/(2*x*log(x)+2*(-45*x^2-12*x)*log(2)-2*x^2),x, algorithm="fricas")

[Out]

log(-3*(15*x + 4)*log(2) - x + log(x))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {2-2 x-45 x \log (4)}{-2 x^2+\left (-12 x-45 x^2\right ) \log (4)+2 x \log (x)} \, dx=\log {\left (- 45 x \log {\left (2 \right )} - x + \log {\left (x \right )} - 12 \log {\left (2 \right )} \right )} \]

[In]

integrate((-90*x*ln(2)-2*x+2)/(2*x*ln(x)+2*(-45*x**2-12*x)*ln(2)-2*x**2),x)

[Out]

log(-45*x*log(2) - x + log(x) - 12*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {2-2 x-45 x \log (4)}{-2 x^2+\left (-12 x-45 x^2\right ) \log (4)+2 x \log (x)} \, dx=\log \left (-x {\left (45 \, \log \left (2\right ) + 1\right )} - 12 \, \log \left (2\right ) + \log \left (x\right )\right ) \]

[In]

integrate((-90*x*log(2)-2*x+2)/(2*x*log(x)+2*(-45*x^2-12*x)*log(2)-2*x^2),x, algorithm="maxima")

[Out]

log(-x*(45*log(2) + 1) - 12*log(2) + log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {2-2 x-45 x \log (4)}{-2 x^2+\left (-12 x-45 x^2\right ) \log (4)+2 x \log (x)} \, dx=\log \left (-45 \, x \log \left (2\right ) - x - 12 \, \log \left (2\right ) + \log \left (x\right )\right ) \]

[In]

integrate((-90*x*log(2)-2*x+2)/(2*x*log(x)+2*(-45*x^2-12*x)*log(2)-2*x^2),x, algorithm="giac")

[Out]

log(-45*x*log(2) - x - 12*log(2) + log(x))

Mupad [B] (verification not implemented)

Time = 14.06 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {2-2 x-45 x \log (4)}{-2 x^2+\left (-12 x-45 x^2\right ) \log (4)+2 x \log (x)} \, dx=\ln \left (x+12\,\ln \left (2\right )-\ln \left (x\right )+45\,x\,\ln \left (2\right )\right ) \]

[In]

int((2*x + 90*x*log(2) - 2)/(2*log(2)*(12*x + 45*x^2) - 2*x*log(x) + 2*x^2),x)

[Out]

log(x + 12*log(2) - log(x) + 45*x*log(2))