\(\int \frac {2358774 e^{2 x}+2358774 e^x x^2}{5 x^2+e^{2 x} (5+x)+e^x (10 x+x^2)} \, dx\) [7569]

Optimal result

Integrand size = 45, antiderivative size = 18 \[ \int \frac {2358774 e^{2 x}+2358774 e^x x^2}{5 x^2+e^{2 x} (5+x)+e^x \left (10 x+x^2\right )} \, dx=2358774 \log \left (5+\frac {x}{1+e^{-x} x}\right ) \]

[Out]

2358774*ln(5+x/(x/exp(x)+1))

Rubi [F]

\[ \int \frac {2358774 e^{2 x}+2358774 e^x x^2}{5 x^2+e^{2 x} (5+x)+e^x \left (10 x+x^2\right )} \, dx=\int \frac {2358774 e^{2 x}+2358774 e^x x^2}{5 x^2+e^{2 x} (5+x)+e^x \left (10 x+x^2\right )} \, dx \]

[In]

Int[(2358774*E^(2*x) + 2358774*E^x*x^2)/(5*x^2 + E^(2*x)*(5 + x) + E^x*(10*x + x^2)),x]

[Out]

-2358774*Log[1 + E^x/x] + 11793870*Defer[Int][E^x/(5*E^x + 5*x + E^x*x), x] - 11793870*Defer[Int][E^x/(x*(5*E^
x + 5*x + E^x*x)), x] + 2358774*Defer[Int][(E^x*x)/(5*E^x + 5*x + E^x*x), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {2358774 e^x \left (e^x+x^2\right )}{5 x^2+e^{2 x} (5+x)+e^x \left (10 x+x^2\right )} \, dx \\ & = 2358774 \int \frac {e^x \left (e^x+x^2\right )}{5 x^2+e^{2 x} (5+x)+e^x \left (10 x+x^2\right )} \, dx \\ & = 2358774 \int \left (-\frac {e^x (-1+x)}{x \left (e^x+x\right )}+\frac {e^x \left (-5+5 x+x^2\right )}{x \left (5 e^x+5 x+e^x x\right )}\right ) \, dx \\ & = -\left (2358774 \int \frac {e^x (-1+x)}{x \left (e^x+x\right )} \, dx\right )+2358774 \int \frac {e^x \left (-5+5 x+x^2\right )}{x \left (5 e^x+5 x+e^x x\right )} \, dx \\ & = 2358774 \int \left (\frac {5 e^x}{5 e^x+5 x+e^x x}-\frac {5 e^x}{x \left (5 e^x+5 x+e^x x\right )}+\frac {e^x x}{5 e^x+5 x+e^x x}\right ) \, dx-2358774 \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\frac {e^x}{x}\right ) \\ & = -2358774 \log \left (1+\frac {e^x}{x}\right )+2358774 \int \frac {e^x x}{5 e^x+5 x+e^x x} \, dx+11793870 \int \frac {e^x}{5 e^x+5 x+e^x x} \, dx-11793870 \int \frac {e^x}{x \left (5 e^x+5 x+e^x x\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 4.81 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {2358774 e^{2 x}+2358774 e^x x^2}{5 x^2+e^{2 x} (5+x)+e^x \left (10 x+x^2\right )} \, dx=2358774 \left (-\log \left (e^x+x\right )+\log \left (5 e^x+5 x+e^x x\right )\right ) \]

[In]

Integrate[(2358774*E^(2*x) + 2358774*E^x*x^2)/(5*x^2 + E^(2*x)*(5 + x) + E^x*(10*x + x^2)),x]

[Out]

2358774*(-Log[E^x + x] + Log[5*E^x + 5*x + E^x*x])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33

method	result	size
norman	\(-2358774 \ln \left ({\mathrm e}^{x}+x \right )+2358774 \ln \left ({\mathrm e}^{x} x +5 x +5 \,{\mathrm e}^{x}\right )\)	\(24\)
parallelrisch	\(-2358774 \ln \left ({\mathrm e}^{x}+x \right )+2358774 \ln \left ({\mathrm e}^{x} x +5 x +5 \,{\mathrm e}^{x}\right )\)	\(24\)
risch	\(2358774 \ln \left (5+x \right )+2358774 \ln \left ({\mathrm e}^{x}+\frac {5 x}{5+x}\right )-2358774 \ln \left ({\mathrm e}^{x}+x \right )\)	\(29\)

[In]

int((2358774*exp(x)^2+2358774*exp(x)*x^2)/((5+x)*exp(x)^2+(x^2+10*x)*exp(x)+5*x^2),x,method=_RETURNVERBOSE)

[Out]

-2358774*ln(exp(x)+x)+2358774*ln(exp(x)*x+5*x+5*exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.83 \[ \int \frac {2358774 e^{2 x}+2358774 e^x x^2}{5 x^2+e^{2 x} (5+x)+e^x \left (10 x+x^2\right )} \, dx=-2358774 \, \log \left (x + e^{x}\right ) + 2358774 \, \log \left (x + 5\right ) + 2358774 \, \log \left (\frac {{\left (x + 5\right )} e^{x} + 5 \, x}{x + 5}\right ) \]

[In]

integrate((2358774*exp(x)^2+2358774*exp(x)*x^2)/((5+x)*exp(x)^2+(x^2+10*x)*exp(x)+5*x^2),x, algorithm="fricas"
)

[Out]

-2358774*log(x + e^x) + 2358774*log(x + 5) + 2358774*log(((x + 5)*e^x + 5*x)/(x + 5))

Sympy [F(-2)]

Exception generated. \[ \int \frac {2358774 e^{2 x}+2358774 e^x x^2}{5 x^2+e^{2 x} (5+x)+e^x \left (10 x+x^2\right )} \, dx=\text {Exception raised: PolynomialError} \]

[In]

integrate((2358774*exp(x)**2+2358774*exp(x)*x**2)/((5+x)*exp(x)**2+(x**2+10*x)*exp(x)+5*x**2),x)

[Out]

Exception raised: PolynomialError >> 1/(x**2 + 10*x + 25) contains an element of the set of generators.

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.83 \[ \int \frac {2358774 e^{2 x}+2358774 e^x x^2}{5 x^2+e^{2 x} (5+x)+e^x \left (10 x+x^2\right )} \, dx=-2358774 \, \log \left (x + e^{x}\right ) + 2358774 \, \log \left (x + 5\right ) + 2358774 \, \log \left (\frac {{\left (x + 5\right )} e^{x} + 5 \, x}{x + 5}\right ) \]

[In]

integrate((2358774*exp(x)^2+2358774*exp(x)*x^2)/((5+x)*exp(x)^2+(x^2+10*x)*exp(x)+5*x^2),x, algorithm="maxima"
)

[Out]

-2358774*log(x + e^x) + 2358774*log(x + 5) + 2358774*log(((x + 5)*e^x + 5*x)/(x + 5))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {2358774 e^{2 x}+2358774 e^x x^2}{5 x^2+e^{2 x} (5+x)+e^x \left (10 x+x^2\right )} \, dx=2358774 \, \log \left (x e^{x} + 5 \, x + 5 \, e^{x}\right ) - 2358774 \, \log \left (x + e^{x}\right ) \]

[In]

integrate((2358774*exp(x)^2+2358774*exp(x)*x^2)/((5+x)*exp(x)^2+(x^2+10*x)*exp(x)+5*x^2),x, algorithm="giac")

[Out]

2358774*log(x*e^x + 5*x + 5*e^x) - 2358774*log(x + e^x)

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {2358774 e^{2 x}+2358774 e^x x^2}{5 x^2+e^{2 x} (5+x)+e^x \left (10 x+x^2\right )} \, dx=2358774\,\ln \left (5\,x+5\,{\mathrm {e}}^x+x\,{\mathrm {e}}^x\right )-2358774\,\ln \left (x+{\mathrm {e}}^x\right ) \]

[In]

int((2358774*exp(2*x) + 2358774*x^2*exp(x))/(exp(x)*(10*x + x^2) + exp(2*x)*(x + 5) + 5*x^2),x)

[Out]

2358774*log(5*x + 5*exp(x) + x*exp(x)) - 2358774*log(x + exp(x))