3.100.19 \(\int \frac {1}{6} e^{-x} (6 e^x+e^{5 e^{\frac {1}{2} (e^{x^2}+x)}} (2+e^{\frac {1}{2} (e^{x^2}+x)} (-5-15 e^x+e^{x^2} (-10 x-30 e^x x)))) \, dx\)

Optimal. Leaf size=36 \[ x-\frac {e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (x+\frac {e^{-x} x}{3}\right )}{x} \]

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Rubi [F]  time = 2.84, 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}{6} e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(6*E^x + E^(5*E^((E^x^2 + x)/2))*(2 + E^((E^x^2 + x)/2)*(-5 - 15*E^x + E^x^2*(-10*x - 30*E^x*x))))/(6*E^x)
,x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{6} \int e^{-x} \left (6 e^x+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}} \left (2+e^{\frac {1}{2} \left (e^{x^2}+x\right )} \left (-5-15 e^x+e^{x^2} \left (-10 x-30 e^x x\right )\right )\right )\right ) \, dx\\ &=\frac {1}{6} \int \left (6+e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x} \left (2-5 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}-15 e^{\frac {e^{x^2}}{2}+\frac {3 x}{2}}-10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}+x^2} x-30 e^{\frac {e^{x^2}}{2}+\frac {3 x}{2}+x^2} x\right )\right ) \, dx\\ &=x+\frac {1}{6} \int e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x} \left (2-5 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}-15 e^{\frac {e^{x^2}}{2}+\frac {3 x}{2}}-10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}+x^2} x-30 e^{\frac {e^{x^2}}{2}+\frac {3 x}{2}+x^2} x\right ) \, dx\\ &=x+\frac {1}{6} \int \left (2 e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x}-5 e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+x\right )}-15 \exp \left (5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+3 x\right )\right )-10 \exp \left (5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+x+2 x^2\right )\right ) x-30 \exp \left (5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+3 x+2 x^2\right )\right ) x\right ) \, dx\\ &=x+\frac {1}{3} \int e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x} \, dx-\frac {5}{6} \int e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+x\right )} \, dx-\frac {5}{3} \int \exp \left (5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+x+2 x^2\right )\right ) x \, dx-\frac {5}{2} \int \exp \left (5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+3 x\right )\right ) \, dx-5 \int \exp \left (5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x+\frac {1}{2} \left (e^{x^2}+3 x+2 x^2\right )\right ) x \, dx\\ &=x+\frac {1}{3} \int e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x} \, dx-\frac {5}{6} \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}-x\right )} \, dx-\frac {5}{3} \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}-x+2 x^2\right )} x \, dx-\frac {5}{2} \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}+x\right )} \, dx-5 \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}+x+2 x^2\right )} x \, dx\\ &=x+\frac {1}{3} \int e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x} \, dx-\frac {5}{3} \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}-x+2 x^2\right )} x \, dx-\frac {5}{3} \operatorname {Subst}\left (\int e^{\frac {e^{4 x^2}}{2}+5 e^{\frac {e^{4 x^2}}{2}+x}-x} \, dx,x,\frac {x}{2}\right )-5 \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}+x+2 x^2\right )} x \, dx-5 \operatorname {Subst}\left (\int e^{\frac {e^{4 x^2}}{2}+5 e^{\frac {e^{4 x^2}}{2}+x}+x} \, dx,x,\frac {x}{2}\right )\\ &=x+\frac {1}{3} \int e^{5 e^{\frac {1}{2} \left (e^{x^2}+x\right )}-x} \, dx-\frac {5}{3} \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}-x+2 x^2\right )} x \, dx-\frac {5}{3} \operatorname {Subst}\left (\int e^{\frac {1}{2} \left (e^{4 x^2}+10 e^{\frac {e^{4 x^2}}{2}+x}-2 x\right )} \, dx,x,\frac {x}{2}\right )-5 \int e^{\frac {1}{2} \left (10 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}+e^{x^2}+x+2 x^2\right )} x \, dx-5 \operatorname {Subst}\left (\int e^{\frac {1}{2} \left (e^{4 x^2}+10 e^{\frac {e^{4 x^2}}{2}+x}+2 x\right )} \, dx,x,\frac {x}{2}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.13, size = 36, normalized size = 1.00 \begin {gather*} \frac {1}{6} e^{5 e^{\frac {e^{x^2}}{2}+\frac {x}{2}}} \left (-6-2 e^{-x}\right )+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6*E^x + E^(5*E^((E^x^2 + x)/2))*(2 + E^((E^x^2 + x)/2)*(-5 - 15*E^x + E^x^2*(-10*x - 30*E^x*x))))/(
6*E^x),x]

[Out]

(E^(5*E^(E^x^2/2 + x/2))*(-6 - 2/E^x))/6 + x

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fricas [A]  time = 0.80, size = 34, normalized size = 0.94 \begin {gather*} \frac {1}{3} \, {\left (3 \, x e^{x} - {\left (3 \, e^{x} + 1\right )} e^{\left (5 \, e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, e^{\left (x^{2}\right )}\right )}\right )}\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((((-30*exp(x)*x-10*x)*exp(x^2)-15*exp(x)-5)*exp(1/2*exp(x^2)+1/2*x)+2)*exp(5*exp(1/2*exp(x^2)+1
/2*x))+6*exp(x))/exp(x),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {1}{6} \, {\left ({\left (5 \, {\left (2 \, {\left (3 \, x e^{x} + x\right )} e^{\left (x^{2}\right )} + 3 \, e^{x} + 1\right )} e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, e^{\left (x^{2}\right )}\right )} - 2\right )} e^{\left (5 \, e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, e^{\left (x^{2}\right )}\right )}\right )} - 6 \, e^{x}\right )} e^{\left (-x\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((((-30*exp(x)*x-10*x)*exp(x^2)-15*exp(x)-5)*exp(1/2*exp(x^2)+1/2*x)+2)*exp(5*exp(1/2*exp(x^2)+1
/2*x))+6*exp(x))/exp(x),x, algorithm="giac")

[Out]

integrate(-1/6*((5*(2*(3*x*e^x + x)*e^(x^2) + 3*e^x + 1)*e^(1/2*x + 1/2*e^(x^2)) - 2)*e^(5*e^(1/2*x + 1/2*e^(x
^2))) - 6*e^x)*e^(-x), x)

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maple [A]  time = 0.08, size = 29, normalized size = 0.81

method	result	size
risch	\(x -\frac {\left (3 \,{\mathrm e}^{x}+1\right ) {\mathrm e}^{-x +5 \,{\mathrm e}^{\frac {{\mathrm e}^{x^{2}}}{2}+\frac {x}{2}}}}{3}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6*((((-30*exp(x)*x-10*x)*exp(x^2)-15*exp(x)-5)*exp(1/2*exp(x^2)+1/2*x)+2)*exp(5*exp(1/2*exp(x^2)+1/2*x))
+6*exp(x))/exp(x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.46, size = 28, normalized size = 0.78 \begin {gather*} -\frac {1}{3} \, {\left (3 \, e^{x} + 1\right )} e^{\left (-x + 5 \, e^{\left (\frac {1}{2} \, x + \frac {1}{2} \, e^{\left (x^{2}\right )}\right )}\right )} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((((-30*exp(x)*x-10*x)*exp(x^2)-15*exp(x)-5)*exp(1/2*exp(x^2)+1/2*x)+2)*exp(5*exp(1/2*exp(x^2)+1
/2*x))+6*exp(x))/exp(x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.16, size = 26, normalized size = 0.72 \begin {gather*} x-{\mathrm {e}}^{5\,\sqrt {{\mathrm {e}}^{{\mathrm {e}}^{x^2}}}\,\sqrt {{\mathrm {e}}^x}-x}\,\left ({\mathrm {e}}^x+\frac {1}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-x)*(exp(x) - (exp(5*exp(x/2 + exp(x^2)/2))*(exp(x/2 + exp(x^2)/2)*(15*exp(x) + exp(x^2)*(10*x + 30*x*
exp(x)) + 5) - 2))/6),x)

[Out]

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

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sympy [A]  time = 5.65, size = 26, normalized size = 0.72 \begin {gather*} x + \frac {\left (-3 - e^{- x}\right ) e^{5 e^{\frac {x}{2} + \frac {e^{x^{2}}}{2}}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*((((-30*exp(x)*x-10*x)*exp(x**2)-15*exp(x)-5)*exp(1/2*exp(x**2)+1/2*x)+2)*exp(5*exp(1/2*exp(x**2
)+1/2*x))+6*exp(x))/exp(x),x)

[Out]

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

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