\(\int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+(-e^x x+2 x^2 \log (5)) \log (x)+x^2 \log ^2(x)} \, dx\) [3233]

Optimal result

Integrand size = 71, antiderivative size = 20 \[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=\log \left (-95 \left (-2+\frac {2 e^x}{x (\log (5)+\log (x))}\right )\right ) \]

[Out]

ln(-95*exp(x)/x/(1/2*ln(5)+1/2*ln(x))+190)

Rubi [F]

\[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=\int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx \]

[In]

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

[Out]

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

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

Mathematica [F]

\[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=\int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx \]

[In]

Integrate[(E^x*(1 + (1 - x)*Log[5]) + E^x*(1 - x)*Log[x])/(-(E^x*x*Log[5]) + x^2*Log[5]^2 + (-(E^x*x) + 2*x^2*
Log[5])*Log[x] + x^2*Log[x]^2),x]

[Out]

Integrate[(E^x*(1 + (1 - x)*Log[5]) + E^x*(1 - x)*Log[x])/(-(E^x*x*Log[5]) + x^2*Log[5]^2 + (-(E^x*x) + 2*x^2*
Log[5])*Log[x] + x^2*Log[x]^2), x]

Maple [A] (verified)

Time = 1.87 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35

method	result	size
risch	\(\ln \left (\ln \left (x \right )+\frac {x \ln \left (5\right )-{\mathrm e}^{x}}{x}\right )-\ln \left (\ln \left (5\right )+\ln \left (x \right )\right )\)	\(27\)
norman	\(-\ln \left (x \right )-\ln \left (\ln \left (5\right )+\ln \left (x \right )\right )+\ln \left (x \ln \left (5\right )+x \ln \left (x \right )-{\mathrm e}^{x}\right )\)	\(28\)
parallelrisch	\(-\ln \left (x \right )-\ln \left (\ln \left (5\right )+\ln \left (x \right )\right )+\ln \left (x \ln \left (5\right )+x \ln \left (x \right )-{\mathrm e}^{x}\right )\)	\(28\)

[In]

int(((1-x)*exp(x)*ln(x)+((1-x)*ln(5)+1)*exp(x))/(x^2*ln(x)^2+(-exp(x)*x+2*x^2*ln(5))*ln(x)-x*exp(x)*ln(5)+x^2*
ln(5)^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=\log \left (\frac {x \log \left (5\right ) + x \log \left (x\right ) - e^{x}}{x}\right ) - \log \left (\log \left (5\right ) + \log \left (x\right )\right ) \]

[In]

integrate(((1-x)*exp(x)*log(x)+((1-x)*log(5)+1)*exp(x))/(x^2*log(x)^2+(-exp(x)*x+2*x^2*log(5))*log(x)-x*exp(x)
*log(5)+x^2*log(5)^2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=- \log {\left (x \right )} - \log {\left (\log {\left (x \right )} + \log {\left (5 \right )} \right )} + \log {\left (- x \log {\left (x \right )} - x \log {\left (5 \right )} + e^{x} \right )} \]

[In]

integrate(((1-x)*exp(x)*ln(x)+((1-x)*ln(5)+1)*exp(x))/(x**2*ln(x)**2+(-exp(x)*x+2*x**2*ln(5))*ln(x)-x*exp(x)*l
n(5)+x**2*ln(5)**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=\log \left (-x \log \left (5\right ) - x \log \left (x\right ) + e^{x}\right ) - \log \left (x\right ) - \log \left (\log \left (5\right ) + \log \left (x\right )\right ) \]

[In]

integrate(((1-x)*exp(x)*log(x)+((1-x)*log(5)+1)*exp(x))/(x^2*log(x)^2+(-exp(x)*x+2*x^2*log(5))*log(x)-x*exp(x)
*log(5)+x^2*log(5)^2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=\log \left (-x \log \left (5\right ) - x \log \left (x\right ) + e^{x}\right ) - \log \left (x\right ) - \log \left (\log \left (5\right ) + \log \left (x\right )\right ) \]

[In]

integrate(((1-x)*exp(x)*log(x)+((1-x)*log(5)+1)*exp(x))/(x^2*log(x)^2+(-exp(x)*x+2*x^2*log(5))*log(x)-x*exp(x)
*log(5)+x^2*log(5)^2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=-\int \frac {{\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x-1\right )-1\right )+{\mathrm {e}}^x\,\ln \left (x\right )\,\left (x-1\right )}{x^2\,{\ln \left (5\right )}^2+x^2\,{\ln \left (x\right )}^2+\ln \left (x\right )\,\left (2\,x^2\,\ln \left (5\right )-x\,{\mathrm {e}}^x\right )-x\,{\mathrm {e}}^x\,\ln \left (5\right )} \,d x \]

[In]

int(-(exp(x)*(log(5)*(x - 1) - 1) + exp(x)*log(x)*(x - 1))/(x^2*log(5)^2 + x^2*log(x)^2 + log(x)*(2*x^2*log(5)
- x*exp(x)) - x*exp(x)*log(5)),x)

[Out]

-int((exp(x)*(log(5)*(x - 1) - 1) + exp(x)*log(x)*(x - 1))/(x^2*log(5)^2 + x^2*log(x)^2 + log(x)*(2*x^2*log(5)
- x*exp(x)) - x*exp(x)*log(5)), x)