3.49.32 \(\int \frac {2 e^{\frac {-1+x^2}{x^2}} x^2+2 x^5+e^{2 x} (2 e^{\frac {-1+x^2}{x^2}}+2 x^3)+e^x (-4 e^{\frac {-1+x^2}{x^2}} x-4 x^4)+2 (-2 x^3+e^x (x^3+x^4)) \log (\log (5))}{e^{2 x} x^3-2 e^x x^4+x^5} \, dx\)

Optimal. Leaf size=34 \[ e^{\frac {-\frac {1}{x}+x}{x}}+(2+x) \left (2+\frac {2 \log (\log (5))}{-e^x+x}\right ) \]

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Rubi [F]  time = 1.59, 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 {2 e^{\frac {-1+x^2}{x^2}} x^2+2 x^5+e^{2 x} \left (2 e^{\frac {-1+x^2}{x^2}}+2 x^3\right )+e^x \left (-4 e^{\frac {-1+x^2}{x^2}} x-4 x^4\right )+2 \left (-2 x^3+e^x \left (x^3+x^4\right )\right ) \log (\log (5))}{e^{2 x} x^3-2 e^x x^4+x^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 34, normalized size = 1.00 \begin {gather*} 2 \left (\frac {1}{2} e^{1-\frac {1}{x^2}}+x-\frac {(2+x) \log (\log (5))}{e^x-x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.63, size = 53, normalized size = 1.56 \begin {gather*} \frac {2 \, x^{2} - {\left (2 \, x + e^{\left (\frac {x^{2} - 1}{x^{2}}\right )}\right )} e^{x} + x e^{\left (\frac {x^{2} - 1}{x^{2}}\right )} + 2 \, {\left (x + 2\right )} \log \left (\log \relax (5)\right )}{x - e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*((x^4+x^3)*exp(x)-2*x^3)*log(log(5))+(2*exp((x^2-1)/x^2)+2*x^3)*exp(x)^2+(-4*x*exp((x^2-1)/x^2)-4
*x^4)*exp(x)+2*x^2*exp((x^2-1)/x^2)+2*x^5)/(exp(x)^2*x^3-2*exp(x)*x^4+x^5),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 38, normalized size = 1.12 \begin {gather*} \frac {2 \, {\left (x^{2} - x e^{x} + x \log \left (\log \relax (5)\right ) + 2 \, \log \left (\log \relax (5)\right )\right )}}{x - e^{x}} + e^{\left (-\frac {1}{x^{2}} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*((x^4+x^3)*exp(x)-2*x^3)*log(log(5))+(2*exp((x^2-1)/x^2)+2*x^3)*exp(x)^2+(-4*x*exp((x^2-1)/x^2)-4
*x^4)*exp(x)+2*x^2*exp((x^2-1)/x^2)+2*x^5)/(exp(x)^2*x^3-2*exp(x)*x^4+x^5),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 32, normalized size = 0.94

method	result	size
risch	\(2 x +\frac {2 \left (2+x \right ) \ln \left (\ln \relax (5)\right )}{x -{\mathrm e}^{x}}+{\mathrm e}^{\frac {\left (x -1\right ) \left (x +1\right )}{x^{2}}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*((x^4+x^3)*exp(x)-2*x^3)*ln(ln(5))+(2*exp((x^2-1)/x^2)+2*x^3)*exp(x)^2+(-4*x*exp((x^2-1)/x^2)-4*x^4)*ex
p(x)+2*x^2*exp((x^2-1)/x^2)+2*x^5)/(exp(x)^2*x^3-2*exp(x)*x^4+x^5),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 51, normalized size = 1.50 \begin {gather*} \frac {{\left (x e + 2 \, {\left (x^{2} - x e^{x} + x \log \left (\log \relax (5)\right ) + 2 \, \log \left (\log \relax (5)\right )\right )} e^{\left (\frac {1}{x^{2}}\right )} - e^{\left (x + 1\right )}\right )} e^{\left (-\frac {1}{x^{2}}\right )}}{x - e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*((x^4+x^3)*exp(x)-2*x^3)*log(log(5))+(2*exp((x^2-1)/x^2)+2*x^3)*exp(x)^2+(-4*x*exp((x^2-1)/x^2)-4
*x^4)*exp(x)+2*x^2*exp((x^2-1)/x^2)+2*x^5)/(exp(x)^2*x^3-2*exp(x)*x^4+x^5),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.67, size = 45, normalized size = 1.32 \begin {gather*} 2\,x+{\mathrm {e}}^{1-\frac {1}{x^2}}+\frac {2\,\left (\ln \left (\ln \relax (5)\right )\,x^2+\ln \left (\ln \relax (5)\right )\,x-2\,\ln \left (\ln \relax (5)\right )\right )}{\left (x-{\mathrm {e}}^x\right )\,\left (x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2*exp((x^2 - 1)/x^2) + 2*log(log(5))*(exp(x)*(x^3 + x^4) - 2*x^3) + exp(2*x)*(2*exp((x^2 - 1)/x^2) +
2*x^3) - exp(x)*(4*x*exp((x^2 - 1)/x^2) + 4*x^4) + 2*x^5)/(x^3*exp(2*x) - 2*x^4*exp(x) + x^5),x)

[Out]

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

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sympy [A]  time = 0.30, size = 34, normalized size = 1.00 \begin {gather*} 2 x + e^{\frac {x^{2} - 1}{x^{2}}} + \frac {- 2 x \log {\left (\log {\relax (5 )} \right )} - 4 \log {\left (\log {\relax (5 )} \right )}}{- x + e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*((x**4+x**3)*exp(x)-2*x**3)*ln(ln(5))+(2*exp((x**2-1)/x**2)+2*x**3)*exp(x)**2+(-4*x*exp((x**2-1)/
x**2)-4*x**4)*exp(x)+2*x**2*exp((x**2-1)/x**2)+2*x**5)/(exp(x)**2*x**3-2*exp(x)*x**4+x**5),x)

[Out]

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

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