\(\int \frac {-3+5 e^x-5 x+(-5+5 x) \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))} \, dx\) [1548]

Optimal result

Integrand size = 43, antiderivative size = 26 \[ \int \frac {-3+5 e^x-5 x+(-5+5 x) \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))} \, dx=-2+x+\log \left (\frac {1}{3+5 \left (x-x \left (-e^x+\log (x \log (2))\right )\right )}\right ) \]

[Out]

x-2+ln(1/(3+5*x-5*x*(ln(x*ln(2))-exp(x))))

Rubi [F]

\[ \int \frac {-3+5 e^x-5 x+(-5+5 x) \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))} \, dx=\int \frac {-3+5 e^x-5 x+(-5+5 x) \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))} \, dx \]

[In]

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

[Out]

-Log[x] + 8*Defer[Int][(3 + 5*x + 5*E^x*x - 5*x*Log[x*Log[2]])^(-1), x] + 3*Defer[Int][1/(x*(3 + 5*x + 5*E^x*x
- 5*x*Log[x*Log[2]])), x] + 5*Defer[Int][x/(3 + 5*x + 5*E^x*x - 5*x*Log[x*Log[2]]), x] + 5*Defer[Int][(x*Log[
x*Log[2]])/(-3 - 5*x - 5*E^x*x + 5*x*Log[x*Log[2]]), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{x}-\frac {-3-8 x-5 x^2+5 x^2 \log (x \log (2))}{x \left (3+5 x+5 e^x x-5 x \log (x \log (2))\right )}\right ) \, dx \\ & = -\log (x)-\int \frac {-3-8 x-5 x^2+5 x^2 \log (x \log (2))}{x \left (3+5 x+5 e^x x-5 x \log (x \log (2))\right )} \, dx \\ & = -\log (x)-\int \left (-\frac {8}{3+5 x+5 e^x x-5 x \log (x \log (2))}-\frac {3}{x \left (3+5 x+5 e^x x-5 x \log (x \log (2))\right )}-\frac {5 x}{3+5 x+5 e^x x-5 x \log (x \log (2))}-\frac {5 x \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))}\right ) \, dx \\ & = -\log (x)+3 \int \frac {1}{x \left (3+5 x+5 e^x x-5 x \log (x \log (2))\right )} \, dx+5 \int \frac {x}{3+5 x+5 e^x x-5 x \log (x \log (2))} \, dx+5 \int \frac {x \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))} \, dx+8 \int \frac {1}{3+5 x+5 e^x x-5 x \log (x \log (2))} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-3+5 e^x-5 x+(-5+5 x) \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))} \, dx=x-\log \left (3+5 x+5 e^x x-5 x \log (x \log (2))\right ) \]

[In]

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

[Out]

x - Log[3 + 5*x + 5*E^x*x - 5*x*Log[x*Log[2]]]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81

method	result	size
parallelrisch	\(-\ln \left ({\mathrm e}^{x} x -x \ln \left (x \ln \left (2\right )\right )+x +\frac {3}{5}\right )+x\)	\(21\)
norman	\(x -\ln \left (5 \,{\mathrm e}^{x} x -5 x \ln \left (x \ln \left (2\right )\right )+5 x +3\right )\)	\(24\)
risch	\(x -\ln \left (x \right )-\ln \left (\ln \left (x \ln \left (2\right )\right )-\frac {5 \,{\mathrm e}^{x} x +5 x +3}{5 x}\right )\)	\(31\)

[In]

int(((5*x-5)*ln(x*ln(2))+5*exp(x)-5*x-3)/(5*x*ln(x*ln(2))-5*exp(x)*x-5*x-3),x,method=_RETURNVERBOSE)

[Out]

-ln(exp(x)*x-x*ln(x*ln(2))+x+3/5)+x

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {-3+5 e^x-5 x+(-5+5 x) \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))} \, dx=x - \log \left (x \log \left (2\right )\right ) - \log \left (-\frac {5 \, x e^{x} - 5 \, x \log \left (x \log \left (2\right )\right ) + 5 \, x + 3}{x}\right ) \]

[In]

integrate(((5*x-5)*log(x*log(2))+5*exp(x)-5*x-3)/(5*x*log(x*log(2))-5*exp(x)*x-5*x-3),x, algorithm="fricas")

[Out]

x - log(x*log(2)) - log(-(5*x*e^x - 5*x*log(x*log(2)) + 5*x + 3)/x)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {-3+5 e^x-5 x+(-5+5 x) \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))} \, dx=x - \log {\left (x \right )} - \log {\left (e^{x} + \frac {- 5 x \log {\left (x \log {\left (2 \right )} \right )} + 5 x + 3}{5 x} \right )} \]

[In]

integrate(((5*x-5)*ln(x*ln(2))+5*exp(x)-5*x-3)/(5*x*ln(x*ln(2))-5*exp(x)*x-5*x-3),x)

[Out]

x - log(x) - log(exp(x) + (-5*x*log(x*log(2)) + 5*x + 3)/(5*x))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {-3+5 e^x-5 x+(-5+5 x) \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))} \, dx=x - \log \left (x\right ) - \log \left (-\frac {5 \, x {\left (\log \left (\log \left (2\right )\right ) - 1\right )} - 5 \, x e^{x} + 5 \, x \log \left (x\right ) - 3}{5 \, x}\right ) \]

[In]

integrate(((5*x-5)*log(x*log(2))+5*exp(x)-5*x-3)/(5*x*log(x*log(2))-5*exp(x)*x-5*x-3),x, algorithm="maxima")

[Out]

x - log(x) - log(-1/5*(5*x*(log(log(2)) - 1) - 5*x*e^x + 5*x*log(x) - 3)/x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {-3+5 e^x-5 x+(-5+5 x) \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))} \, dx=x - \log \left (-5 \, x e^{x} + 5 \, x \log \left (x \log \left (2\right )\right ) - 5 \, x - 3\right ) \]

[In]

integrate(((5*x-5)*log(x*log(2))+5*exp(x)-5*x-3)/(5*x*log(x*log(2))-5*exp(x)*x-5*x-3),x, algorithm="giac")

[Out]

x - log(-5*x*e^x + 5*x*log(x*log(2)) - 5*x - 3)

Mupad [B] (verification not implemented)

Time = 8.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {-3+5 e^x-5 x+(-5+5 x) \log (x \log (2))}{-3-5 x-5 e^x x+5 x \log (x \log (2))} \, dx=x-\ln \left (5\,x+5\,x\,{\mathrm {e}}^x-5\,x\,\left (\ln \left (\ln \left (2\right )\right )+\ln \left (x\right )\right )+3\right ) \]

[In]

int((5*x - 5*exp(x) - log(x*log(2))*(5*x - 5) + 3)/(5*x - 5*x*log(x*log(2)) + 5*x*exp(x) + 3),x)

[Out]

x - log(5*x + 5*x*exp(x) - 5*x*(log(log(2)) + log(x)) + 3)