\(\int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+(2 e^{4+x+x^2}+10 x-5 x^2-10 x^3) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx\) [3229]

Optimal result

Integrand size = 99, antiderivative size = 25 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=x \left (1+\frac {5 e^{-4-x-x^2} x}{1+\log (x)}\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=\int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx \]

[In]

Int[(E^(4 + x + x^2) + 5*x - 5*x^2 - 10*x^3 + (2*E^(4 + x + x^2) + 10*x - 5*x^2 - 10*x^3)*Log[x] + E^(4 + x +
x^2)*Log[x]^2)/(E^(4 + x + x^2) + 2*E^(4 + x + x^2)*Log[x] + E^(4 + x + x^2)*Log[x]^2),x]

[Out]

x - 5*Defer[Int][(E^(-4 - x - x^2)*x)/(1 + Log[x])^2, x] + 10*Defer[Int][(E^(-4 - x - x^2)*x)/(1 + Log[x]), x]
- 5*Defer[Int][(E^(-4 - x - x^2)*x^2)/(1 + Log[x]), x] - 10*Defer[Int][(E^(-4 - x - x^2)*x^3)/(1 + Log[x]), x
]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-4-x-x^2} \left (e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)\right )}{(1+\log (x))^2} \, dx \\ & = \int \left (1+\frac {5 e^{-4-x-x^2} x}{(1+\log (x))^2}-\frac {5 e^{-4-x-x^2} x^2}{(1+\log (x))^2}-\frac {10 e^{-4-x-x^2} x^3}{(1+\log (x))^2}+\frac {10 e^{-4-x-x^2} x \log (x)}{(1+\log (x))^2}-\frac {5 e^{-4-x-x^2} x^2 \log (x)}{(1+\log (x))^2}-\frac {10 e^{-4-x-x^2} x^3 \log (x)}{(1+\log (x))^2}\right ) \, dx \\ & = x+5 \int \frac {e^{-4-x-x^2} x}{(1+\log (x))^2} \, dx-5 \int \frac {e^{-4-x-x^2} x^2}{(1+\log (x))^2} \, dx-5 \int \frac {e^{-4-x-x^2} x^2 \log (x)}{(1+\log (x))^2} \, dx-10 \int \frac {e^{-4-x-x^2} x^3}{(1+\log (x))^2} \, dx+10 \int \frac {e^{-4-x-x^2} x \log (x)}{(1+\log (x))^2} \, dx-10 \int \frac {e^{-4-x-x^2} x^3 \log (x)}{(1+\log (x))^2} \, dx \\ & = x+5 \int \frac {e^{-4-x-x^2} x}{(1+\log (x))^2} \, dx-5 \int \frac {e^{-4-x-x^2} x^2}{(1+\log (x))^2} \, dx-5 \int \left (-\frac {e^{-4-x-x^2} x^2}{(1+\log (x))^2}+\frac {e^{-4-x-x^2} x^2}{1+\log (x)}\right ) \, dx-10 \int \frac {e^{-4-x-x^2} x^3}{(1+\log (x))^2} \, dx+10 \int \left (-\frac {e^{-4-x-x^2} x}{(1+\log (x))^2}+\frac {e^{-4-x-x^2} x}{1+\log (x)}\right ) \, dx-10 \int \left (-\frac {e^{-4-x-x^2} x^3}{(1+\log (x))^2}+\frac {e^{-4-x-x^2} x^3}{1+\log (x)}\right ) \, dx \\ & = x+5 \int \frac {e^{-4-x-x^2} x}{(1+\log (x))^2} \, dx-5 \int \frac {e^{-4-x-x^2} x^2}{1+\log (x)} \, dx-10 \int \frac {e^{-4-x-x^2} x}{(1+\log (x))^2} \, dx+10 \int \frac {e^{-4-x-x^2} x}{1+\log (x)} \, dx-10 \int \frac {e^{-4-x-x^2} x^3}{1+\log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=\frac {x \left (1+5 e^{-4-x-x^2} x+\log (x)\right )}{1+\log (x)} \]

[In]

Integrate[(E^(4 + x + x^2) + 5*x - 5*x^2 - 10*x^3 + (2*E^(4 + x + x^2) + 10*x - 5*x^2 - 10*x^3)*Log[x] + E^(4
+ x + x^2)*Log[x]^2)/(E^(4 + x + x^2) + 2*E^(4 + x + x^2)*Log[x] + E^(4 + x + x^2)*Log[x]^2),x]

[Out]

(x*(1 + 5*E^(-4 - x - x^2)*x + Log[x]))/(1 + Log[x])

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00

method	result	size
risch	\(x +\frac {5 x^{2} {\mathrm e}^{-x^{2}-x -4}}{\ln \left (x \right )+1}\)	\(25\)
parallelrisch	\(\frac {\left (x \ln \left (x \right ) {\mathrm e}^{x^{2}+x +4}+5 x^{2}+x \,{\mathrm e}^{x^{2}+x +4}\right ) {\mathrm e}^{-x^{2}-x -4}}{\ln \left (x \right )+1}\)	\(43\)

[In]

int((exp(x^2+x+4)*ln(x)^2+(2*exp(x^2+x+4)-10*x^3-5*x^2+10*x)*ln(x)+exp(x^2+x+4)-10*x^3-5*x^2+5*x)/(exp(x^2+x+4
)*ln(x)^2+2*exp(x^2+x+4)*ln(x)+exp(x^2+x+4)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=\frac {x e^{\left (x^{2} + x + 4\right )} \log \left (x\right ) + 5 \, x^{2} + x e^{\left (x^{2} + x + 4\right )}}{e^{\left (x^{2} + x + 4\right )} \log \left (x\right ) + e^{\left (x^{2} + x + 4\right )}} \]

[In]

integrate((exp(x^2+x+4)*log(x)^2+(2*exp(x^2+x+4)-10*x^3-5*x^2+10*x)*log(x)+exp(x^2+x+4)-10*x^3-5*x^2+5*x)/(exp
(x^2+x+4)*log(x)^2+2*exp(x^2+x+4)*log(x)+exp(x^2+x+4)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=\frac {5 x^{2} e^{- x^{2} - x - 4}}{\log {\left (x \right )} + 1} + x \]

[In]

integrate((exp(x**2+x+4)*ln(x)**2+(2*exp(x**2+x+4)-10*x**3-5*x**2+10*x)*ln(x)+exp(x**2+x+4)-10*x**3-5*x**2+5*x
)/(exp(x**2+x+4)*ln(x)**2+2*exp(x**2+x+4)*ln(x)+exp(x**2+x+4)),x)

[Out]

5*x**2*exp(-x**2 - x - 4)/(log(x) + 1) + x

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=\frac {{\left (5 \, x^{2} e^{\left (-x^{2}\right )} + {\left (x e^{4} \log \left (x\right ) + x e^{4}\right )} e^{x}\right )} e^{\left (-x\right )}}{e^{4} \log \left (x\right ) + e^{4}} \]

[In]

integrate((exp(x^2+x+4)*log(x)^2+(2*exp(x^2+x+4)-10*x^3-5*x^2+10*x)*log(x)+exp(x^2+x+4)-10*x^3-5*x^2+5*x)/(exp
(x^2+x+4)*log(x)^2+2*exp(x^2+x+4)*log(x)+exp(x^2+x+4)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=\frac {5 \, x^{2} e^{\left (-x^{2} - x\right )} + x e^{4} \log \left (x\right ) + x e^{4}}{e^{4} \log \left (x\right ) + e^{4}} \]

[In]

integrate((exp(x^2+x+4)*log(x)^2+(2*exp(x^2+x+4)-10*x^3-5*x^2+10*x)*log(x)+exp(x^2+x+4)-10*x^3-5*x^2+5*x)/(exp
(x^2+x+4)*log(x)^2+2*exp(x^2+x+4)*log(x)+exp(x^2+x+4)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.98 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=x+\frac {5\,x^2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^{-x^2}}{\ln \left (x\right )+1} \]

[In]

int((5*x + exp(x + x^2 + 4) + log(x)*(10*x + 2*exp(x + x^2 + 4) - 5*x^2 - 10*x^3) + exp(x + x^2 + 4)*log(x)^2
- 5*x^2 - 10*x^3)/(exp(x + x^2 + 4) + 2*exp(x + x^2 + 4)*log(x) + exp(x + x^2 + 4)*log(x)^2),x)

[Out]

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