3.87.25 \(\int \frac {1+x+e^x (1+2 x-x^2)+e^{x^2} (-2 x^2+e^x (x-2 x^2))-e^x x \log (x)}{x+e^{2 x^2} x-2 x^2+x^3+e^{x^2} (2 x-2 x^2)+(-2 x-2 e^{x^2} x+2 x^2) \log (x)+x \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {1+e^x}{1+e^{x^2}-x-\log (x)} \]

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Rubi [F]  time = 3.79, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1+x+e^x \left (1+2 x-x^2\right )+e^{x^2} \left (-2 x^2+e^x \left (x-2 x^2\right )\right )-e^x x \log (x)}{x+e^{2 x^2} x-2 x^2+x^3+e^{x^2} \left (2 x-2 x^2\right )+\left (-2 x-2 e^{x^2} x+2 x^2\right ) \log (x)+x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1+x-2 e^{x^2} x^2+e^{x+x^2} \left (x-2 x^2\right )+e^x \left (1+2 x-x^2\right )-e^x x \log (x)}{x \left (1+e^{x^2}-x-\log (x)\right )^2} \, dx\\ &=\int \left (\frac {-e^x+2 x+2 e^x x}{-1-e^{x^2}+x+\log (x)}-\frac {\left (1+e^x\right ) \left (-1-x-2 x^2+2 x^3+2 x^2 \log (x)\right )}{x \left (-1-e^{x^2}+x+\log (x)\right )^2}\right ) \, dx\\ &=\int \frac {-e^x+2 x+2 e^x x}{-1-e^{x^2}+x+\log (x)} \, dx-\int \frac {\left (1+e^x\right ) \left (-1-x-2 x^2+2 x^3+2 x^2 \log (x)\right )}{x \left (-1-e^{x^2}+x+\log (x)\right )^2} \, dx\\ &=-\int \left (-\frac {1}{\left (1+e^{x^2}-x-\log (x)\right )^2}-\frac {e^x}{\left (1+e^{x^2}-x-\log (x)\right )^2}-\frac {1}{x \left (-1-e^{x^2}+x+\log (x)\right )^2}-\frac {e^x}{x \left (-1-e^{x^2}+x+\log (x)\right )^2}-\frac {2 x}{\left (-1-e^{x^2}+x+\log (x)\right )^2}-\frac {2 e^x x}{\left (-1-e^{x^2}+x+\log (x)\right )^2}+\frac {2 x^2}{\left (-1-e^{x^2}+x+\log (x)\right )^2}+\frac {2 e^x x^2}{\left (-1-e^{x^2}+x+\log (x)\right )^2}+\frac {2 x \log (x)}{\left (-1-e^{x^2}+x+\log (x)\right )^2}+\frac {2 e^x x \log (x)}{\left (-1-e^{x^2}+x+\log (x)\right )^2}\right ) \, dx+\int \left (\frac {e^x}{1+e^{x^2}-x-\log (x)}+\frac {2 x}{-1-e^{x^2}+x+\log (x)}+\frac {2 e^x x}{-1-e^{x^2}+x+\log (x)}\right ) \, dx\\ &=2 \int \frac {x}{\left (-1-e^{x^2}+x+\log (x)\right )^2} \, dx+2 \int \frac {e^x x}{\left (-1-e^{x^2}+x+\log (x)\right )^2} \, dx-2 \int \frac {x^2}{\left (-1-e^{x^2}+x+\log (x)\right )^2} \, dx-2 \int \frac {e^x x^2}{\left (-1-e^{x^2}+x+\log (x)\right )^2} \, dx-2 \int \frac {x \log (x)}{\left (-1-e^{x^2}+x+\log (x)\right )^2} \, dx-2 \int \frac {e^x x \log (x)}{\left (-1-e^{x^2}+x+\log (x)\right )^2} \, dx+2 \int \frac {x}{-1-e^{x^2}+x+\log (x)} \, dx+2 \int \frac {e^x x}{-1-e^{x^2}+x+\log (x)} \, dx+\int \frac {1}{\left (1+e^{x^2}-x-\log (x)\right )^2} \, dx+\int \frac {e^x}{\left (1+e^{x^2}-x-\log (x)\right )^2} \, dx+\int \frac {e^x}{1+e^{x^2}-x-\log (x)} \, dx+\int \frac {1}{x \left (-1-e^{x^2}+x+\log (x)\right )^2} \, dx+\int \frac {e^x}{x \left (-1-e^{x^2}+x+\log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.69, size = 22, normalized size = 1.00 \begin {gather*} \frac {1+e^x}{1+e^{x^2}-x-\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x + E^x*(1 + 2*x - x^2) + E^x^2*(-2*x^2 + E^x*(x - 2*x^2)) - E^x*x*Log[x])/(x + E^(2*x^2)*x - 2
*x^2 + x^3 + E^x^2*(2*x - 2*x^2) + (-2*x - 2*E^x^2*x + 2*x^2)*Log[x] + x*Log[x]^2),x]

[Out]

(1 + E^x)/(1 + E^x^2 - x - Log[x])

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fricas [A]  time = 0.52, size = 19, normalized size = 0.86 \begin {gather*} -\frac {e^{x} + 1}{x - e^{\left (x^{2}\right )} + \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(x)*log(x)+((-2*x^2+x)*exp(x)-2*x^2)*exp(x^2)+(-x^2+2*x+1)*exp(x)+x+1)/(x*log(x)^2+(-2*exp(x^
2)*x+2*x^2-2*x)*log(x)+x*exp(x^2)^2+(-2*x^2+2*x)*exp(x^2)+x^3-2*x^2+x),x, algorithm="fricas")

[Out]

-(e^x + 1)/(x - e^(x^2) + log(x) - 1)

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giac [A]  time = 0.28, size = 19, normalized size = 0.86 \begin {gather*} -\frac {e^{x} + 1}{x - e^{\left (x^{2}\right )} + \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(x)*log(x)+((-2*x^2+x)*exp(x)-2*x^2)*exp(x^2)+(-x^2+2*x+1)*exp(x)+x+1)/(x*log(x)^2+(-2*exp(x^
2)*x+2*x^2-2*x)*log(x)+x*exp(x^2)^2+(-2*x^2+2*x)*exp(x^2)+x^3-2*x^2+x),x, algorithm="giac")

[Out]

-(e^x + 1)/(x - e^(x^2) + log(x) - 1)

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maple [A]  time = 0.06, size = 20, normalized size = 0.91

method	result	size
risch	\(-\frac {{\mathrm e}^{x}+1}{x -1+\ln \relax (x )-{\mathrm e}^{x^{2}}}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*exp(x)*ln(x)+((-2*x^2+x)*exp(x)-2*x^2)*exp(x^2)+(-x^2+2*x+1)*exp(x)+x+1)/(x*ln(x)^2+(-2*exp(x^2)*x+2*x
^2-2*x)*ln(x)+x*exp(x^2)^2+(-2*x^2+2*x)*exp(x^2)+x^3-2*x^2+x),x,method=_RETURNVERBOSE)

[Out]

-(exp(x)+1)/(x-1+ln(x)-exp(x^2))

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maxima [A]  time = 0.41, size = 19, normalized size = 0.86 \begin {gather*} -\frac {e^{x} + 1}{x - e^{\left (x^{2}\right )} + \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(x)*log(x)+((-2*x^2+x)*exp(x)-2*x^2)*exp(x^2)+(-x^2+2*x+1)*exp(x)+x+1)/(x*log(x)^2+(-2*exp(x^
2)*x+2*x^2-2*x)*log(x)+x*exp(x^2)^2+(-2*x^2+2*x)*exp(x^2)+x^3-2*x^2+x),x, algorithm="maxima")

[Out]

-(e^x + 1)/(x - e^(x^2) + log(x) - 1)

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mupad [B]  time = 5.42, size = 85, normalized size = 3.86 \begin {gather*} -\frac {x\,\left ({\mathrm {e}}^x+1\right )+x^2\,\left ({\mathrm {e}}^x+1\right )+x^3\,\left (2\,{\mathrm {e}}^x-\ln \relax (x)\,\left (2\,{\mathrm {e}}^x+2\right )+2\right )-x^4\,\left (2\,{\mathrm {e}}^x+2\right )}{\left (x-{\mathrm {e}}^{x^2}+\ln \relax (x)-1\right )\,\left (x-2\,x^3\,\ln \relax (x)+x^2+2\,x^3-2\,x^4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + exp(x^2)*(exp(x)*(x - 2*x^2) - 2*x^2) + exp(x)*(2*x - x^2 + 1) - x*exp(x)*log(x) + 1)/(x + exp(x^2)*(
2*x - 2*x^2) + x*log(x)^2 + x*exp(2*x^2) - log(x)*(2*x + 2*x*exp(x^2) - 2*x^2) - 2*x^2 + x^3),x)

[Out]

-(x*(exp(x) + 1) + x^2*(exp(x) + 1) + x^3*(2*exp(x) - log(x)*(2*exp(x) + 2) + 2) - x^4*(2*exp(x) + 2))/((x - e
xp(x^2) + log(x) - 1)*(x - 2*x^3*log(x) + x^2 + 2*x^3 - 2*x^4))

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sympy [A]  time = 0.31, size = 15, normalized size = 0.68 \begin {gather*} \frac {e^{x} + 1}{- x + e^{x^{2}} - \log {\relax (x )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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