\(\int \frac {10 e^x x+e^x (20 x+10 x^2) \log (x)}{4+5 e^x x^2 \log (x)} \, dx\) [2613]

Optimal result

Integrand size = 37, antiderivative size = 19 \[ \int \frac {10 e^x x+e^x \left (20 x+10 x^2\right ) \log (x)}{4+5 e^x x^2 \log (x)} \, dx=-5+\log \left (\left (-1-\frac {5}{4} e^x x^2 \log (x)\right )^2\right ) \]

[Out]

ln((-1-5/4*x^2*exp(x)*ln(x))^2)-5

Rubi [F]

\[ \int \frac {10 e^x x+e^x \left (20 x+10 x^2\right ) \log (x)}{4+5 e^x x^2 \log (x)} \, dx=\int \frac {10 e^x x+e^x \left (20 x+10 x^2\right ) \log (x)}{4+5 e^x x^2 \log (x)} \, dx \]

[In]

Int[(10*E^x*x + E^x*(20*x + 10*x^2)*Log[x])/(4 + 5*E^x*x^2*Log[x]),x]

[Out]

2*x - 8*Defer[Int][(4 + 5*E^x*x^2*Log[x])^(-1), x] + 10*Defer[Int][(E^x*x)/(4 + 5*E^x*x^2*Log[x]), x] + 20*Def
er[Int][(E^x*x*Log[x])/(4 + 5*E^x*x^2*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {10 e^x x (1+(2+x) \log (x))}{4+5 e^x x^2 \log (x)} \, dx \\ & = 10 \int \frac {e^x x (1+(2+x) \log (x))}{4+5 e^x x^2 \log (x)} \, dx \\ & = 10 \int \left (\frac {e^x x}{4+5 e^x x^2 \log (x)}+\frac {2 e^x x \log (x)}{4+5 e^x x^2 \log (x)}+\frac {e^x x^2 \log (x)}{4+5 e^x x^2 \log (x)}\right ) \, dx \\ & = 10 \int \frac {e^x x}{4+5 e^x x^2 \log (x)} \, dx+10 \int \frac {e^x x^2 \log (x)}{4+5 e^x x^2 \log (x)} \, dx+20 \int \frac {e^x x \log (x)}{4+5 e^x x^2 \log (x)} \, dx \\ & = 10 \int \frac {e^x x}{4+5 e^x x^2 \log (x)} \, dx+10 \int \left (\frac {1}{5}-\frac {4}{5 \left (4+5 e^x x^2 \log (x)\right )}\right ) \, dx+20 \int \frac {e^x x \log (x)}{4+5 e^x x^2 \log (x)} \, dx \\ & = 2 x-8 \int \frac {1}{4+5 e^x x^2 \log (x)} \, dx+10 \int \frac {e^x x}{4+5 e^x x^2 \log (x)} \, dx+20 \int \frac {e^x x \log (x)}{4+5 e^x x^2 \log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {10 e^x x+e^x \left (20 x+10 x^2\right ) \log (x)}{4+5 e^x x^2 \log (x)} \, dx=2 \log \left (4+5 e^x x^2 \log (x)\right ) \]

[In]

Integrate[(10*E^x*x + E^x*(20*x + 10*x^2)*Log[x])/(4 + 5*E^x*x^2*Log[x]),x]

[Out]

2*Log[4 + 5*E^x*x^2*Log[x]]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
parallelrisch	\(2 \ln \left (x^{2} {\mathrm e}^{x} \ln \left (x \right )+\frac {4}{5}\right )\)	\(14\)
norman	\(2 \ln \left (5 x^{2} {\mathrm e}^{x} \ln \left (x \right )+4\right )\)	\(15\)
risch	\(2 x +4 \ln \left (x \right )+2 \ln \left (\ln \left (x \right )+\frac {4 \,{\mathrm e}^{-x}}{5 x^{2}}\right )\)	\(24\)

[In]

int(((10*x^2+20*x)*exp(x)*ln(x)+10*exp(x)*x)/(5*x^2*exp(x)*ln(x)+4),x,method=_RETURNVERBOSE)

[Out]

2*ln(x^2*exp(x)*ln(x)+4/5)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {10 e^x x+e^x \left (20 x+10 x^2\right ) \log (x)}{4+5 e^x x^2 \log (x)} \, dx=2 \, x + 4 \, \log \left (x\right ) + 2 \, \log \left (\frac {{\left (5 \, x^{2} e^{x} \log \left (x\right ) + 4\right )} e^{\left (-x\right )}}{x^{2}}\right ) \]

[In]

integrate(((10*x^2+20*x)*exp(x)*log(x)+10*exp(x)*x)/(5*x^2*exp(x)*log(x)+4),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {10 e^x x+e^x \left (20 x+10 x^2\right ) \log (x)}{4+5 e^x x^2 \log (x)} \, dx=4 \log {\left (x \right )} + 2 \log {\left (e^{x} + \frac {4}{5 x^{2} \log {\left (x \right )}} \right )} + 2 \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate(((10*x**2+20*x)*exp(x)*ln(x)+10*exp(x)*x)/(5*x**2*exp(x)*ln(x)+4),x)

[Out]

4*log(x) + 2*log(exp(x) + 4/(5*x**2*log(x))) + 2*log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {10 e^x x+e^x \left (20 x+10 x^2\right ) \log (x)}{4+5 e^x x^2 \log (x)} \, dx=4 \, \log \left (x\right ) + 2 \, \log \left (\frac {5 \, x^{2} e^{x} \log \left (x\right ) + 4}{5 \, x^{2} \log \left (x\right )}\right ) + 2 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((10*x^2+20*x)*exp(x)*log(x)+10*exp(x)*x)/(5*x^2*exp(x)*log(x)+4),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {10 e^x x+e^x \left (20 x+10 x^2\right ) \log (x)}{4+5 e^x x^2 \log (x)} \, dx=2 \, \log \left (5 \, x^{2} e^{x} \log \left (x\right ) + 4\right ) \]

[In]

integrate(((10*x^2+20*x)*exp(x)*log(x)+10*exp(x)*x)/(5*x^2*exp(x)*log(x)+4),x, algorithm="giac")

[Out]

2*log(5*x^2*e^x*log(x) + 4)

Mupad [B] (verification not implemented)

Time = 8.69 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {10 e^x x+e^x \left (20 x+10 x^2\right ) \log (x)}{4+5 e^x x^2 \log (x)} \, dx=2\,\ln \left (5\,x^2\,{\mathrm {e}}^x\,\ln \left (x\right )+4\right ) \]

[In]

int((10*x*exp(x) + exp(x)*log(x)*(20*x + 10*x^2))/(5*x^2*exp(x)*log(x) + 4),x)

[Out]

2*log(5*x^2*exp(x)*log(x) + 4)