3.103.22 \(\int \frac {-2 x+2 e^{2/x} x+20 x^3+12 x^4+2 x^5+e^{\frac {1}{x}} (-2-2 x+12 x^2+4 x^3)+(2 x-4 e^{\frac {1}{x}} x-12 x^2-4 x^3) \log (x)+2 x \log ^2(x)}{e^{2/x} x+9 x^3+6 x^4+x^5+e^{\frac {1}{x}} (6 x^2+2 x^3)+(-2 e^{\frac {1}{x}} x-6 x^2-2 x^3) \log (x)+x \log ^2(x)} \, dx\) [10222]

Optimal. Leaf size=25 \[ 2 \left (-2+x-\frac {x}{e^{\frac {1}{x}}+x (3+x)-\log (x)}\right ) \]

[Out]

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

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Rubi [F]
time = 2.42, 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 x+2 e^{2/x} x+20 x^3+12 x^4+2 x^5+e^{\frac {1}{x}} \left (-2-2 x+12 x^2+4 x^3\right )+\left (2 x-4 e^{\frac {1}{x}} x-12 x^2-4 x^3\right ) \log (x)+2 x \log ^2(x)}{e^{2/x} x+9 x^3+6 x^4+x^5+e^{\frac {1}{x}} \left (6 x^2+2 x^3\right )+\left (-2 e^{\frac {1}{x}} x-6 x^2-2 x^3\right ) \log (x)+x \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*x + 2*E^(2/x)*x + 20*x^3 + 12*x^4 + 2*x^5 + E^x^(-1)*(-2 - 2*x + 12*x^2 + 4*x^3) + (2*x - 4*E^x^(-1)*x
- 12*x^2 - 4*x^3)*Log[x] + 2*x*Log[x]^2)/(E^(2/x)*x + 9*x^3 + 6*x^4 + x^5 + E^x^(-1)*(6*x^2 + 2*x^3) + (-2*E^
x^(-1)*x - 6*x^2 - 2*x^3)*Log[x] + x*Log[x]^2),x]

[Out]

2*x + 4*Defer[Int][(E^x^(-1) + 3*x + x^2 - Log[x])^(-2), x] + 8*Defer[Int][x/(E^x^(-1) + 3*x + x^2 - Log[x])^2
, x] + 4*Defer[Int][x^2/(E^x^(-1) + 3*x + x^2 - Log[x])^2, x] - 2*Defer[Int][(E^x^(-1) + 3*x + x^2 - Log[x])^(
-1), x] - 2*Defer[Int][1/(x*(E^x^(-1) + 3*x + x^2 - Log[x])), x] - 2*Defer[Int][Log[x]/(x*(E^x^(-1) + 3*x + x^
2 - Log[x])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (e^{2/x} x+e^{\frac {1}{x}} \left (-1-x+6 x^2+2 x^3\right )+x \left (-1+10 x^2+6 x^3+x^4\right )-x \left (-1+2 e^{\frac {1}{x}}+6 x+2 x^2\right ) \log (x)+x \log ^2(x)\right )}{x \left (e^{\frac {1}{x}}+x (3+x)-\log (x)\right )^2} \, dx\\ &=2 \int \frac {e^{2/x} x+e^{\frac {1}{x}} \left (-1-x+6 x^2+2 x^3\right )+x \left (-1+10 x^2+6 x^3+x^4\right )-x \left (-1+2 e^{\frac {1}{x}}+6 x+2 x^2\right ) \log (x)+x \log ^2(x)}{x \left (e^{\frac {1}{x}}+x (3+x)-\log (x)\right )^2} \, dx\\ &=2 \int \left (1-\frac {1+x}{x \left (e^{\frac {1}{x}}+3 x+x^2-\log (x)\right )}+\frac {2 x+4 x^2+2 x^3-\log (x)}{x \left (e^{\frac {1}{x}}+3 x+x^2-\log (x)\right )^2}\right ) \, dx\\ &=2 x-2 \int \frac {1+x}{x \left (e^{\frac {1}{x}}+3 x+x^2-\log (x)\right )} \, dx+2 \int \frac {2 x+4 x^2+2 x^3-\log (x)}{x \left (e^{\frac {1}{x}}+3 x+x^2-\log (x)\right )^2} \, dx\\ &=2 x-2 \int \left (\frac {1}{e^{\frac {1}{x}}+3 x+x^2-\log (x)}+\frac {1}{x \left (e^{\frac {1}{x}}+3 x+x^2-\log (x)\right )}\right ) \, dx+2 \int \left (\frac {2}{\left (e^{\frac {1}{x}}+3 x+x^2-\log (x)\right )^2}+\frac {4 x}{\left (e^{\frac {1}{x}}+3 x+x^2-\log (x)\right )^2}+\frac {2 x^2}{\left (e^{\frac {1}{x}}+3 x+x^2-\log (x)\right )^2}-\frac {\log (x)}{x \left (e^{\frac {1}{x}}+3 x+x^2-\log (x)\right )^2}\right ) \, dx\\ &=2 x-2 \int \frac {1}{e^{\frac {1}{x}}+3 x+x^2-\log (x)} \, dx-2 \int \frac {1}{x \left (e^{\frac {1}{x}}+3 x+x^2-\log (x)\right )} \, dx-2 \int \frac {\log (x)}{x \left (e^{\frac {1}{x}}+3 x+x^2-\log (x)\right )^2} \, dx+4 \int \frac {1}{\left (e^{\frac {1}{x}}+3 x+x^2-\log (x)\right )^2} \, dx+4 \int \frac {x^2}{\left (e^{\frac {1}{x}}+3 x+x^2-\log (x)\right )^2} \, dx+8 \int \frac {x}{\left (e^{\frac {1}{x}}+3 x+x^2-\log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.06, size = 26, normalized size = 1.04 \begin {gather*} 2 \left (x+\frac {x}{-e^{\frac {1}{x}}-3 x-x^2+\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x + 2*E^(2/x)*x + 20*x^3 + 12*x^4 + 2*x^5 + E^x^(-1)*(-2 - 2*x + 12*x^2 + 4*x^3) + (2*x - 4*E^x^
(-1)*x - 12*x^2 - 4*x^3)*Log[x] + 2*x*Log[x]^2)/(E^(2/x)*x + 9*x^3 + 6*x^4 + x^5 + E^x^(-1)*(6*x^2 + 2*x^3) +
(-2*E^x^(-1)*x - 6*x^2 - 2*x^3)*Log[x] + x*Log[x]^2),x]

[Out]

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

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Maple [A]
time = 0.14, size = 25, normalized size = 1.00

method	result	size
risch	\(2 x -\frac {2 x}{x^{2}+3 x +{\mathrm e}^{\frac {1}{x}}-\ln \left (x \right )}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*ln(x)^2+(-4*x*exp(1/x)-4*x^3-12*x^2+2*x)*ln(x)+2*x*exp(1/x)^2+(4*x^3+12*x^2-2*x-2)*exp(1/x)+2*x^5+12*
x^4+20*x^3-2*x)/(x*ln(x)^2+(-2*x*exp(1/x)-2*x^3-6*x^2)*ln(x)+x*exp(1/x)^2+(2*x^3+6*x^2)*exp(1/x)+x^5+6*x^4+9*x
^3),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 42, normalized size = 1.68 \begin {gather*} \frac {2 \, {\left (x^{3} + 3 \, x^{2} + x e^{\frac {1}{x}} - x \log \left (x\right ) - x\right )}}{x^{2} + 3 \, x + e^{\frac {1}{x}} - \log \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(x)^2+(-4*x*exp(1/x)-4*x^3-12*x^2+2*x)*log(x)+2*x*exp(1/x)^2+(4*x^3+12*x^2-2*x-2)*exp(1/x)+2
*x^5+12*x^4+20*x^3-2*x)/(x*log(x)^2+(-2*x*exp(1/x)-2*x^3-6*x^2)*log(x)+x*exp(1/x)^2+(2*x^3+6*x^2)*exp(1/x)+x^5
+6*x^4+9*x^3),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.42, size = 42, normalized size = 1.68 \begin {gather*} \frac {2 \, {\left (x^{3} + 3 \, x^{2} + x e^{\frac {1}{x}} - x \log \left (x\right ) - x\right )}}{x^{2} + 3 \, x + e^{\frac {1}{x}} - \log \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(x)^2+(-4*x*exp(1/x)-4*x^3-12*x^2+2*x)*log(x)+2*x*exp(1/x)^2+(4*x^3+12*x^2-2*x-2)*exp(1/x)+2
*x^5+12*x^4+20*x^3-2*x)/(x*log(x)^2+(-2*x*exp(1/x)-2*x^3-6*x^2)*log(x)+x*exp(1/x)^2+(2*x^3+6*x^2)*exp(1/x)+x^5
+6*x^4+9*x^3),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.08, size = 20, normalized size = 0.80 \begin {gather*} 2 x - \frac {2 x}{x^{2} + 3 x + e^{\frac {1}{x}} - \log {\left (x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*ln(x)**2+(-4*x*exp(1/x)-4*x**3-12*x**2+2*x)*ln(x)+2*x*exp(1/x)**2+(4*x**3+12*x**2-2*x-2)*exp(1/
x)+2*x**5+12*x**4+20*x**3-2*x)/(x*ln(x)**2+(-2*x*exp(1/x)-2*x**3-6*x**2)*ln(x)+x*exp(1/x)**2+(2*x**3+6*x**2)*e
xp(1/x)+x**5+6*x**4+9*x**3),x)

[Out]

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

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Giac [A]
time = 0.43, size = 42, normalized size = 1.68 \begin {gather*} \frac {2 \, {\left (x^{3} + 3 \, x^{2} + x e^{\frac {1}{x}} - x \log \left (x\right ) - x\right )}}{x^{2} + 3 \, x + e^{\frac {1}{x}} - \log \left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(x)^2+(-4*x*exp(1/x)-4*x^3-12*x^2+2*x)*log(x)+2*x*exp(1/x)^2+(4*x^3+12*x^2-2*x-2)*exp(1/x)+2
*x^5+12*x^4+20*x^3-2*x)/(x*log(x)^2+(-2*x*exp(1/x)-2*x^3-6*x^2)*log(x)+x*exp(1/x)^2+(2*x^3+6*x^2)*exp(1/x)+x^5
+6*x^4+9*x^3),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 6.83, size = 24, normalized size = 0.96 \begin {gather*} 2\,x-\frac {2\,x}{3\,x+{\mathrm {e}}^{1/x}-\ln \left (x\right )+x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*log(x)^2 - 2*x - exp(1/x)*(2*x - 12*x^2 - 4*x^3 + 2) + 2*x*exp(2/x) + 20*x^3 + 12*x^4 + 2*x^5 - log(x
)*(4*x*exp(1/x) - 2*x + 12*x^2 + 4*x^3))/(x*log(x)^2 - log(x)*(2*x*exp(1/x) + 6*x^2 + 2*x^3) + exp(1/x)*(6*x^2
+ 2*x^3) + x*exp(2/x) + 9*x^3 + 6*x^4 + x^5),x)

[Out]

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

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