3.64.94 \(\int \frac {1}{8} e^{\frac {5}{4} e^{\frac {1}{2} (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3)}+\frac {1}{2} (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3)} (10 x+e^{2 e^{\frac {e^5+x^2}{x}}} (-15 x^2+e^{\frac {e^5+x^2}{x}} (10 e^5 x-10 x^3))) \, dx\)

Optimal. Leaf size=35 \[ e^{\frac {5}{4} e^{\frac {1}{2} x \left (x-e^{2 e^{\frac {e^5}{x}+x}} x^2\right )}} \]

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Rubi [F]  time = 9.48, 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}{8} \exp \left (\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) \left (10 x+e^{2 e^{\frac {e^5+x^2}{x}}} \left (-15 x^2+e^{\frac {e^5+x^2}{x}} \left (10 e^5 x-10 x^3\right )\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((5*E^((x^2 - E^(2*E^((E^5 + x^2)/x))*x^3)/2))/4 + (x^2 - E^(2*E^((E^5 + x^2)/x))*x^3)/2)*(10*x + E^(2*
E^((E^5 + x^2)/x))*(-15*x^2 + E^((E^5 + x^2)/x)*(10*E^5*x - 10*x^3))))/8,x]

[Out]

(5*Defer[Int][E^((5*E^((x^2 - E^(2*E^((E^5 + x^2)/x))*x^3)/2))/4 + (x^2 - E^(2*E^((E^5 + x^2)/x))*x^3)/2)*x, x
])/4 + (5*Defer[Int][E^(5 + 2*E^(E^5/x + x) + (5*E^((x^2 - E^(2*E^((E^5 + x^2)/x))*x^3)/2))/4 + E^5/x + x + (x
^2 - E^(2*E^((E^5 + x^2)/x))*x^3)/2)*x, x])/4 - (15*Defer[Int][E^(2*E^(E^5/x + x) + (5*E^((x^2 - E^(2*E^((E^5
+ x^2)/x))*x^3)/2))/4 + (x^2 - E^(2*E^((E^5 + x^2)/x))*x^3)/2)*x^2, x])/8 - (5*Defer[Int][E^(2*E^(E^5/x + x) +
(5*E^((x^2 - E^(2*E^((E^5 + x^2)/x))*x^3)/2))/4 + E^5/x + x + (x^2 - E^(2*E^((E^5 + x^2)/x))*x^3)/2)*x^3, x])
/4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} \int \exp \left (\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) \left (10 x+e^{2 e^{\frac {e^5+x^2}{x}}} \left (-15 x^2+e^{\frac {e^5+x^2}{x}} \left (10 e^5 x-10 x^3\right )\right )\right ) \, dx\\ &=\frac {1}{8} \int \left (10 \exp \left (\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x-5 \exp \left (2 e^{\frac {e^5}{x}+x}+\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x \left (-2 e^{5+\frac {e^5}{x}+x}+3 x+2 e^{\frac {e^5}{x}+x} x^2\right )\right ) \, dx\\ &=-\left (\frac {5}{8} \int \exp \left (2 e^{\frac {e^5}{x}+x}+\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x \left (-2 e^{5+\frac {e^5}{x}+x}+3 x+2 e^{\frac {e^5}{x}+x} x^2\right ) \, dx\right )+\frac {5}{4} \int \exp \left (\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x \, dx\\ &=-\left (\frac {5}{8} \int \left (3 \exp \left (2 e^{\frac {e^5}{x}+x}+\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x^2+2 \exp \left (2 e^{\frac {e^5}{x}+x}+\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {e^5}{x}+x+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x \left (-e^5+x^2\right )\right ) \, dx\right )+\frac {5}{4} \int \exp \left (\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x \, dx\\ &=\frac {5}{4} \int \exp \left (\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x \, dx-\frac {5}{4} \int \exp \left (2 e^{\frac {e^5}{x}+x}+\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {e^5}{x}+x+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x \left (-e^5+x^2\right ) \, dx-\frac {15}{8} \int \exp \left (2 e^{\frac {e^5}{x}+x}+\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x^2 \, dx\\ &=\frac {5}{4} \int \exp \left (\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x \, dx-\frac {5}{4} \int \left (-\exp \left (5+2 e^{\frac {e^5}{x}+x}+\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {e^5}{x}+x+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x+\exp \left (2 e^{\frac {e^5}{x}+x}+\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {e^5}{x}+x+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x^3\right ) \, dx-\frac {15}{8} \int \exp \left (2 e^{\frac {e^5}{x}+x}+\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x^2 \, dx\\ &=\frac {5}{4} \int \exp \left (\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x \, dx+\frac {5}{4} \int \exp \left (5+2 e^{\frac {e^5}{x}+x}+\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {e^5}{x}+x+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x \, dx-\frac {5}{4} \int \exp \left (2 e^{\frac {e^5}{x}+x}+\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {e^5}{x}+x+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x^3 \, dx-\frac {15}{8} \int \exp \left (2 e^{\frac {e^5}{x}+x}+\frac {5}{4} e^{\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )}+\frac {1}{2} \left (x^2-e^{2 e^{\frac {e^5+x^2}{x}}} x^3\right )\right ) x^2 \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 34, normalized size = 0.97 \begin {gather*} e^{\frac {5}{4} e^{-\frac {1}{2} x^2 \left (-1+e^{2 e^{\frac {e^5}{x}+x}} x\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((5*E^((x^2 - E^(2*E^((E^5 + x^2)/x))*x^3)/2))/4 + (x^2 - E^(2*E^((E^5 + x^2)/x))*x^3)/2)*(10*x +
E^(2*E^((E^5 + x^2)/x))*(-15*x^2 + E^((E^5 + x^2)/x)*(10*E^5*x - 10*x^3))))/8,x]

[Out]

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

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fricas [A]  time = 0.68, size = 29, normalized size = 0.83 \begin {gather*} e^{\left (\frac {5}{4} \, e^{\left (-\frac {1}{2} \, x^{3} e^{\left (2 \, e^{\left (\frac {x^{2} + e^{5}}{x}\right )}\right )} + \frac {1}{2} \, x^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(((10*x*exp(5)-10*x^3)*exp((x^2+exp(5))/x)-15*x^2)*exp(exp((x^2+exp(5))/x))^2+10*x)*exp(-1/2*x^3
*exp(exp((x^2+exp(5))/x))^2+1/2*x^2)*exp(5/4*exp(-1/2*x^3*exp(exp((x^2+exp(5))/x))^2+1/2*x^2)),x, algorithm="f
ricas")

[Out]

e^(5/4*e^(-1/2*x^3*e^(2*e^((x^2 + e^5)/x)) + 1/2*x^2))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {5}{8} \, {\left ({\left (3 \, x^{2} + 2 \, {\left (x^{3} - x e^{5}\right )} e^{\left (\frac {x^{2} + e^{5}}{x}\right )}\right )} e^{\left (2 \, e^{\left (\frac {x^{2} + e^{5}}{x}\right )}\right )} - 2 \, x\right )} e^{\left (-\frac {1}{2} \, x^{3} e^{\left (2 \, e^{\left (\frac {x^{2} + e^{5}}{x}\right )}\right )} + \frac {1}{2} \, x^{2} + \frac {5}{4} \, e^{\left (-\frac {1}{2} \, x^{3} e^{\left (2 \, e^{\left (\frac {x^{2} + e^{5}}{x}\right )}\right )} + \frac {1}{2} \, x^{2}\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(((10*x*exp(5)-10*x^3)*exp((x^2+exp(5))/x)-15*x^2)*exp(exp((x^2+exp(5))/x))^2+10*x)*exp(-1/2*x^3
*exp(exp((x^2+exp(5))/x))^2+1/2*x^2)*exp(5/4*exp(-1/2*x^3*exp(exp((x^2+exp(5))/x))^2+1/2*x^2)),x, algorithm="g
iac")

[Out]

integrate(-5/8*((3*x^2 + 2*(x^3 - x*e^5)*e^((x^2 + e^5)/x))*e^(2*e^((x^2 + e^5)/x)) - 2*x)*e^(-1/2*x^3*e^(2*e^
((x^2 + e^5)/x)) + 1/2*x^2 + 5/4*e^(-1/2*x^3*e^(2*e^((x^2 + e^5)/x)) + 1/2*x^2)), x)

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maple [A]  time = 0.18, size = 28, normalized size = 0.80

method	result	size
risch	\({\mathrm e}^{\frac {5 \,{\mathrm e}^{-\frac {x^{2} \left (x \,{\mathrm e}^{2 \,{\mathrm e}^{\frac {x^{2}+{\mathrm e}^{5}}{x}}}-1\right )}{2}}}{4}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*(((10*x*exp(5)-10*x^3)*exp((x^2+exp(5))/x)-15*x^2)*exp(exp((x^2+exp(5))/x))^2+10*x)*exp(-1/2*x^3*exp(e
xp((x^2+exp(5))/x))^2+1/2*x^2)*exp(5/4*exp(-1/2*x^3*exp(exp((x^2+exp(5))/x))^2+1/2*x^2)),x,method=_RETURNVERBO
SE)

[Out]

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

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maxima [A]  time = 0.85, size = 27, normalized size = 0.77 \begin {gather*} e^{\left (\frac {5}{4} \, e^{\left (-\frac {1}{2} \, x^{3} e^{\left (2 \, e^{\left (x + \frac {e^{5}}{x}\right )}\right )} + \frac {1}{2} \, x^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(((10*x*exp(5)-10*x^3)*exp((x^2+exp(5))/x)-15*x^2)*exp(exp((x^2+exp(5))/x))^2+10*x)*exp(-1/2*x^3
*exp(exp((x^2+exp(5))/x))^2+1/2*x^2)*exp(5/4*exp(-1/2*x^3*exp(exp((x^2+exp(5))/x))^2+1/2*x^2)),x, algorithm="m
axima")

[Out]

e^(5/4*e^(-1/2*x^3*e^(2*e^(x + e^5/x)) + 1/2*x^2))

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mupad [B]  time = 4.70, size = 27, normalized size = 0.77 \begin {gather*} {\mathrm {e}}^{\frac {5\,{\mathrm {e}}^{-\frac {x^3\,{\mathrm {e}}^{2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^5}{x}}\,{\mathrm {e}}^x}}{2}}\,{\mathrm {e}}^{\frac {x^2}{2}}}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x^2/2 - (x^3*exp(2*exp((exp(5) + x^2)/x)))/2)*exp((5*exp(x^2/2 - (x^3*exp(2*exp((exp(5) + x^2)/x)))/2
))/4)*(10*x + exp(2*exp((exp(5) + x^2)/x))*(exp((exp(5) + x^2)/x)*(10*x*exp(5) - 10*x^3) - 15*x^2)))/8,x)

[Out]

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

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sympy [A]  time = 39.04, size = 29, normalized size = 0.83 \begin {gather*} e^{\frac {5 e^{- \frac {x^{3} e^{2 e^{\frac {x^{2} + e^{5}}{x}}}}{2} + \frac {x^{2}}{2}}}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(((10*x*exp(5)-10*x**3)*exp((x**2+exp(5))/x)-15*x**2)*exp(exp((x**2+exp(5))/x))**2+10*x)*exp(-1/
2*x**3*exp(exp((x**2+exp(5))/x))**2+1/2*x**2)*exp(5/4*exp(-1/2*x**3*exp(exp((x**2+exp(5))/x))**2+1/2*x**2)),x)

[Out]

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

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