3.81.75 \(\int \frac {-8+e^5 (8-2 x)+2 x+x^3+e^{2 e^x} (2-2 e^5+2 e^x x^3)}{-4 x^3+e^{2 e^x} x^3+x^4} \, dx\)

Optimal. Leaf size=26 \[ -4+\frac {-1+e^5}{x^2}+\log \left (4-e^{2 e^x}-x\right ) \]

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Rubi [F]  time = 1.07, 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 {-8+e^5 (8-2 x)+2 x+x^3+e^{2 e^x} \left (2-2 e^5+2 e^x x^3\right )}{-4 x^3+e^{2 e^x} x^3+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

Defer[Int][(-4 + E^(2*E^x) + x)^(-1), x] + 2*Defer[Int][E^(2*E^x + x)/(-4 + E^(2*E^x) + x), x] - 8*Defer[Int][
1/(x^3*(-4 + E^(2*E^x) + x)), x] + 8*E^5*Defer[Int][1/(x^3*(-4 + E^(2*E^x) + x)), x] + 2*(1 - E^5)*Defer[Int][
E^(2*E^x)/(x^3*(-4 + E^(2*E^x) + x)), x] + 2*Defer[Int][1/(x^2*(-4 + E^(2*E^x) + x)), x] - 2*E^5*Defer[Int][1/
(x^2*(-4 + E^(2*E^x) + x)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {1}{-4+e^{2 e^x}+x}+\frac {2 e^{2 e^x+x}}{-4+e^{2 e^x}+x}-\frac {8}{x^3 \left (-4+e^{2 e^x}+x\right )}+\frac {2 e^{2 e^x} \left (1-e^5\right )}{x^3 \left (-4+e^{2 e^x}+x\right )}-\frac {2 e^5 (-4+x)}{x^3 \left (-4+e^{2 e^x}+x\right )}+\frac {2}{x^2 \left (-4+e^{2 e^x}+x\right )}\right ) \, dx\\ &=2 \int \frac {e^{2 e^x+x}}{-4+e^{2 e^x}+x} \, dx+2 \int \frac {1}{x^2 \left (-4+e^{2 e^x}+x\right )} \, dx-8 \int \frac {1}{x^3 \left (-4+e^{2 e^x}+x\right )} \, dx-\left (2 e^5\right ) \int \frac {-4+x}{x^3 \left (-4+e^{2 e^x}+x\right )} \, dx+\left (2 \left (1-e^5\right )\right ) \int \frac {e^{2 e^x}}{x^3 \left (-4+e^{2 e^x}+x\right )} \, dx+\int \frac {1}{-4+e^{2 e^x}+x} \, dx\\ &=2 \int \frac {e^{2 e^x+x}}{-4+e^{2 e^x}+x} \, dx+2 \int \frac {1}{x^2 \left (-4+e^{2 e^x}+x\right )} \, dx-8 \int \frac {1}{x^3 \left (-4+e^{2 e^x}+x\right )} \, dx-\left (2 e^5\right ) \int \left (-\frac {4}{x^3 \left (-4+e^{2 e^x}+x\right )}+\frac {1}{x^2 \left (-4+e^{2 e^x}+x\right )}\right ) \, dx+\left (2 \left (1-e^5\right )\right ) \int \frac {e^{2 e^x}}{x^3 \left (-4+e^{2 e^x}+x\right )} \, dx+\int \frac {1}{-4+e^{2 e^x}+x} \, dx\\ &=2 \int \frac {e^{2 e^x+x}}{-4+e^{2 e^x}+x} \, dx+2 \int \frac {1}{x^2 \left (-4+e^{2 e^x}+x\right )} \, dx-8 \int \frac {1}{x^3 \left (-4+e^{2 e^x}+x\right )} \, dx-\left (2 e^5\right ) \int \frac {1}{x^2 \left (-4+e^{2 e^x}+x\right )} \, dx+\left (8 e^5\right ) \int \frac {1}{x^3 \left (-4+e^{2 e^x}+x\right )} \, dx+\left (2 \left (1-e^5\right )\right ) \int \frac {e^{2 e^x}}{x^3 \left (-4+e^{2 e^x}+x\right )} \, dx+\int \frac {1}{-4+e^{2 e^x}+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 25, normalized size = 0.96 \begin {gather*} \frac {-1+e^5}{x^2}+\log \left (4-e^{2 e^x}-x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.65, size = 21, normalized size = 0.81 \begin {gather*} \frac {x^{2} \log \left (x + e^{\left (2 \, e^{x}\right )} - 4\right ) + e^{5} - 1}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.30, size = 21, normalized size = 0.81 \begin {gather*} \frac {x^{2} \log \left (x + e^{\left (2 \, e^{x}\right )} - 4\right ) + e^{5} - 1}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 19, normalized size = 0.73

method	result	size
norman	\(\frac {{\mathrm e}^{5}-1}{x^{2}}+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}+x -4\right )\)	\(19\)
risch	\(\frac {{\mathrm e}^{5}}{x^{2}}-\frac {1}{x^{2}}+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}+x -4\right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x)*x^3-2*exp(5)+2)*exp(exp(x))^2+(-2*x+8)*exp(5)+x^3+2*x-8)/(x^3*exp(exp(x))^2+x^4-4*x^3),x,method
=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.40, size = 18, normalized size = 0.69 \begin {gather*} \frac {e^{5} - 1}{x^{2}} + \log \left (x + e^{\left (2 \, e^{x}\right )} - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.18, size = 18, normalized size = 0.69 \begin {gather*} \ln \left (x+{\mathrm {e}}^{2\,{\mathrm {e}}^x}-4\right )+\frac {{\mathrm {e}}^5-1}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.18, size = 22, normalized size = 0.85 \begin {gather*} \log {\left (x + e^{2 e^{x}} - 4 \right )} - \frac {2 - 2 e^{5}}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x**3-2*exp(5)+2)*exp(exp(x))**2+(-2*x+8)*exp(5)+x**3+2*x-8)/(x**3*exp(exp(x))**2+x**4-4*x
**3),x)

[Out]

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

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