3.31.96 \(\int \frac {e^{\frac {3-e^x-5 x+x^2}{x}} (6-5 x-x^2-x^3+e^x (-2+x+x^2)+(-12+e^x (4-4 x)+4 x^2) \log (x^2))}{4 x^2+4 x^3+x^4+(-16 x^2-8 x^3) \log (x^2)+16 x^2 \log ^2(x^2)} \, dx\)

Optimal. Leaf size=30 \[ \frac {e^{-5+\frac {3-e^x}{x}+x}}{-2-x+4 \log \left (x^2\right )} \]

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Rubi [F]  time = 11.19, 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 {e^{\frac {3-e^x-5 x+x^2}{x}} \left (6-5 x-x^2-x^3+e^x \left (-2+x+x^2\right )+\left (-12+e^x (4-4 x)+4 x^2\right ) \log \left (x^2\right )\right )}{4 x^2+4 x^3+x^4+\left (-16 x^2-8 x^3\right ) \log \left (x^2\right )+16 x^2 \log ^2\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((3 - E^x - 5*x + x^2)/x)*(6 - 5*x - x^2 - x^3 + E^x*(-2 + x + x^2) + (-12 + E^x*(4 - 4*x) + 4*x^2)*Log
[x^2]))/(4*x^2 + 4*x^3 + x^4 + (-16*x^2 - 8*x^3)*Log[x^2] + 16*x^2*Log[x^2]^2),x]

[Out]

Defer[Int][E^(-5 + 3/x - E^x/x + x)/(2 + x - 4*Log[x^2])^2, x] - 8*Defer[Int][E^(-5 + 3/x - E^x/x + x)/(x*(2 +
x - 4*Log[x^2])^2), x] - Defer[Int][E^(-5 + 3/x - E^x/x + x)/(2 + x - 4*Log[x^2]), x] + 3*Defer[Int][E^(-5 +
3/x - E^x/x + x)/(x^2*(2 + x - 4*Log[x^2])), x] - Defer[Int][E^(-5 + 3/x - E^x/x + 2*x)/(x^2*(2 + x - 4*Log[x^
2])), x] + Defer[Int][E^(-5 + 3/x - E^x/x + 2*x)/(x*(2 + x - 4*Log[x^2])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} \left (6-5 x-x^2-x^3+e^x \left (-2+x+x^2\right )-4 \left (3+e^x (-1+x)-x^2\right ) \log \left (x^2\right )\right )}{x^2 \left (2+x-4 \log \left (x^2\right )\right )^2} \, dx\\ &=\int \left (-\frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{\left (2+x-4 \log \left (x^2\right )\right )^2}+\frac {6 e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x^2 \left (2+x-4 \log \left (x^2\right )\right )^2}-\frac {5 e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x \left (2+x-4 \log \left (x^2\right )\right )^2}-\frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} x}{\left (2+x-4 \log \left (x^2\right )\right )^2}+\frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+2 x} (-1+x)}{x^2 \left (2+x-4 \log \left (x^2\right )\right )}+\frac {4 e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} \log \left (x^2\right )}{\left (2+x-4 \log \left (x^2\right )\right )^2}-\frac {12 e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} \log \left (x^2\right )}{x^2 \left (2+x-4 \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=4 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} \log \left (x^2\right )}{\left (2+x-4 \log \left (x^2\right )\right )^2} \, dx-5 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x \left (2+x-4 \log \left (x^2\right )\right )^2} \, dx+6 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x^2 \left (2+x-4 \log \left (x^2\right )\right )^2} \, dx-12 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} \log \left (x^2\right )}{x^2 \left (2+x-4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{\left (2+x-4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} x}{\left (2+x-4 \log \left (x^2\right )\right )^2} \, dx+\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+2 x} (-1+x)}{x^2 \left (2+x-4 \log \left (x^2\right )\right )} \, dx\\ &=4 \int \left (\frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} (2+x)}{4 \left (2+x-4 \log \left (x^2\right )\right )^2}-\frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{4 \left (2+x-4 \log \left (x^2\right )\right )}\right ) \, dx-5 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x \left (2+x-4 \log \left (x^2\right )\right )^2} \, dx+6 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x^2 \left (2+x-4 \log \left (x^2\right )\right )^2} \, dx-12 \int \left (\frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} (2+x)}{4 x^2 \left (2+x-4 \log \left (x^2\right )\right )^2}-\frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{4 x^2 \left (2+x-4 \log \left (x^2\right )\right )}\right ) \, dx+\int \left (-\frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+2 x}}{x^2 \left (2+x-4 \log \left (x^2\right )\right )}+\frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+2 x}}{x \left (2+x-4 \log \left (x^2\right )\right )}\right ) \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{\left (2+x-4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} x}{\left (2+x-4 \log \left (x^2\right )\right )^2} \, dx\\ &=-\left (3 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} (2+x)}{x^2 \left (2+x-4 \log \left (x^2\right )\right )^2} \, dx\right )+3 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x^2 \left (2+x-4 \log \left (x^2\right )\right )} \, dx-5 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x \left (2+x-4 \log \left (x^2\right )\right )^2} \, dx+6 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x^2 \left (2+x-4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{\left (2+x-4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} x}{\left (2+x-4 \log \left (x^2\right )\right )^2} \, dx+\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} (2+x)}{\left (2+x-4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{2+x-4 \log \left (x^2\right )} \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+2 x}}{x^2 \left (2+x-4 \log \left (x^2\right )\right )} \, dx+\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+2 x}}{x \left (2+x-4 \log \left (x^2\right )\right )} \, dx\\ &=-\left (3 \int \left (\frac {2 e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x^2 \left (2+x-4 \log \left (x^2\right )\right )^2}+\frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x \left (2+x-4 \log \left (x^2\right )\right )^2}\right ) \, dx\right )+3 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x^2 \left (2+x-4 \log \left (x^2\right )\right )} \, dx-5 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x \left (2+x-4 \log \left (x^2\right )\right )^2} \, dx+6 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x^2 \left (2+x-4 \log \left (x^2\right )\right )^2} \, dx+\int \left (\frac {2 e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{\left (2+x-4 \log \left (x^2\right )\right )^2}+\frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} x}{\left (2+x-4 \log \left (x^2\right )\right )^2}\right ) \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{\left (2+x-4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x} x}{\left (2+x-4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{2+x-4 \log \left (x^2\right )} \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+2 x}}{x^2 \left (2+x-4 \log \left (x^2\right )\right )} \, dx+\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+2 x}}{x \left (2+x-4 \log \left (x^2\right )\right )} \, dx\\ &=2 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{\left (2+x-4 \log \left (x^2\right )\right )^2} \, dx-3 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x \left (2+x-4 \log \left (x^2\right )\right )^2} \, dx+3 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x^2 \left (2+x-4 \log \left (x^2\right )\right )} \, dx-5 \int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{x \left (2+x-4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{\left (2+x-4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{2+x-4 \log \left (x^2\right )} \, dx-\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+2 x}}{x^2 \left (2+x-4 \log \left (x^2\right )\right )} \, dx+\int \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+2 x}}{x \left (2+x-4 \log \left (x^2\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.11, size = 32, normalized size = 1.07 \begin {gather*} \frac {e^{-5+\frac {3}{x}-\frac {e^x}{x}+x}}{-2-x+4 \log \left (x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((3 - E^x - 5*x + x^2)/x)*(6 - 5*x - x^2 - x^3 + E^x*(-2 + x + x^2) + (-12 + E^x*(4 - 4*x) + 4*x^
2)*Log[x^2]))/(4*x^2 + 4*x^3 + x^4 + (-16*x^2 - 8*x^3)*Log[x^2] + 16*x^2*Log[x^2]^2),x]

[Out]

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

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fricas [A]  time = 0.82, size = 30, normalized size = 1.00 \begin {gather*} -\frac {e^{\left (\frac {x^{2} - 5 \, x - e^{x} + 3}{x}\right )}}{x - 4 \, \log \left (x^{2}\right ) + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+4)*exp(x)+4*x^2-12)*log(x^2)+(x^2+x-2)*exp(x)-x^3-x^2-5*x+6)*exp((-exp(x)+x^2-5*x+3)/x)/(16*
x^2*log(x^2)^2+(-8*x^3-16*x^2)*log(x^2)+x^4+4*x^3+4*x^2),x, algorithm="fricas")

[Out]

-e^((x^2 - 5*x - e^x + 3)/x)/(x - 4*log(x^2) + 2)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+4)*exp(x)+4*x^2-12)*log(x^2)+(x^2+x-2)*exp(x)-x^3-x^2-5*x+6)*exp((-exp(x)+x^2-5*x+3)/x)/(16*
x^2*log(x^2)^2+(-8*x^3-16*x^2)*log(x^2)+x^4+4*x^3+4*x^2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{-4,[0,1,9]%%%}+%%%{68,[0,1,8]%%%}+%%%{-320,[0,1,7]%%%}+%
%%{256,[0,1

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (-4 x +4\right ) {\mathrm e}^{x}+4 x^{2}-12\right ) \ln \left (x^{2}\right )+\left (x^{2}+x -2\right ) {\mathrm e}^{x}-x^{3}-x^{2}-5 x +6\right ) {\mathrm e}^{\frac {-{\mathrm e}^{x}+x^{2}-5 x +3}{x}}}{16 x^{2} \ln \left (x^{2}\right )^{2}+\left (-8 x^{3}-16 x^{2}\right ) \ln \left (x^{2}\right )+x^{4}+4 x^{3}+4 x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x+4)*exp(x)+4*x^2-12)*ln(x^2)+(x^2+x-2)*exp(x)-x^3-x^2-5*x+6)*exp((-exp(x)+x^2-5*x+3)/x)/(16*x^2*ln(
x^2)^2+(-8*x^3-16*x^2)*ln(x^2)+x^4+4*x^3+4*x^2),x)

[Out]

int((((-4*x+4)*exp(x)+4*x^2-12)*ln(x^2)+(x^2+x-2)*exp(x)-x^3-x^2-5*x+6)*exp((-exp(x)+x^2-5*x+3)/x)/(16*x^2*ln(
x^2)^2+(-8*x^3-16*x^2)*ln(x^2)+x^4+4*x^3+4*x^2),x)

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maxima [A]  time = 0.64, size = 34, normalized size = 1.13 \begin {gather*} -\frac {e^{\left (x - \frac {e^{x}}{x} + \frac {3}{x}\right )}}{x e^{5} - 8 \, e^{5} \log \relax (x) + 2 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+4)*exp(x)+4*x^2-12)*log(x^2)+(x^2+x-2)*exp(x)-x^3-x^2-5*x+6)*exp((-exp(x)+x^2-5*x+3)/x)/(16*
x^2*log(x^2)^2+(-8*x^3-16*x^2)*log(x^2)+x^4+4*x^3+4*x^2),x, algorithm="maxima")

[Out]

-e^(x - e^x/x + 3/x)/(x*e^5 - 8*e^5*log(x) + 2*e^5)

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mupad [B]  time = 1.95, size = 31, normalized size = 1.03 \begin {gather*} -\frac {{\mathrm {e}}^{-5}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^x}{x}}\,{\mathrm {e}}^{3/x}\,{\mathrm {e}}^x}{x-4\,\ln \left (x^2\right )+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(5*x + exp(x) - x^2 - 3)/x)*(5*x + log(x^2)*(exp(x)*(4*x - 4) - 4*x^2 + 12) + x^2 + x^3 - exp(x)*(x
+ x^2 - 2) - 6))/(4*x^2 - log(x^2)*(16*x^2 + 8*x^3) + 4*x^3 + x^4 + 16*x^2*log(x^2)^2),x)

[Out]

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

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sympy [A]  time = 0.47, size = 26, normalized size = 0.87 \begin {gather*} - \frac {e^{\frac {x^{2} - 5 x - e^{x} + 3}{x}}}{x - 4 \log {\left (x^{2} \right )} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+4)*exp(x)+4*x**2-12)*ln(x**2)+(x**2+x-2)*exp(x)-x**3-x**2-5*x+6)*exp((-exp(x)+x**2-5*x+3)/x)
/(16*x**2*ln(x**2)**2+(-8*x**3-16*x**2)*ln(x**2)+x**4+4*x**3+4*x**2),x)

[Out]

-exp((x**2 - 5*x - exp(x) + 3)/x)/(x - 4*log(x**2) + 2)

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