3.99.26 \(\int \frac {-2-2 e^{2 e^{\frac {1}{2} (2 x+\log (e^2 x))}}+6 x^2+e^{e^{\frac {1}{2} (2 x+\log (e^2 x))}} (-4+6 x^2+e^{\frac {1}{2} (2 x+\log (e^2 x))} (-x^2-2 x^3))}{2+4 e^{e^{\frac {1}{2} (2 x+\log (e^2 x))}}+2 e^{2 e^{\frac {1}{2} (2 x+\log (e^2 x))}}} \, dx\)

Optimal. Leaf size=25 \[ -x+\frac {x^3}{1+e^{e^{1+x} \sqrt {x}}} \]

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Rubi [F]  time = 3.99, 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 {-2-2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+6 x^2+e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}} \left (-4+6 x^2+e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )} \left (-x^2-2 x^3\right )\right )}{2+4 e^{e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}+2 e^{2 e^{\frac {1}{2} \left (2 x+\log \left (e^2 x\right )\right )}}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2 - 2*E^(2*E^((2*x + Log[E^2*x])/2)) + 6*x^2 + E^E^((2*x + Log[E^2*x])/2)*(-4 + 6*x^2 + E^((2*x + Log[E^
2*x])/2)*(-x^2 - 2*x^3)))/(2 + 4*E^E^((2*x + Log[E^2*x])/2) + 2*E^(2*E^((2*x + Log[E^2*x])/2))),x]

[Out]

-2*Defer[Subst][Defer[Int][x/(1 + E^(E^(1 + x^2)*x))^2, x], x, Sqrt[x]] - 4*Defer[Subst][Defer[Int][(E^(E^(1 +
x^2)*x)*x)/(1 + E^(E^(1 + x^2)*x))^2, x], x, Sqrt[x]] - 2*Defer[Subst][Defer[Int][(E^(2*E^(1 + x^2)*x)*x)/(1
+ E^(E^(1 + x^2)*x))^2, x], x, Sqrt[x]] + 6*Defer[Subst][Defer[Int][x^5/(1 + E^(E^(1 + x^2)*x))^2, x], x, Sqrt
[x]] + 6*Defer[Subst][Defer[Int][(E^(E^(1 + x^2)*x)*x^5)/(1 + E^(E^(1 + x^2)*x))^2, x], x, Sqrt[x]] - Defer[Su
bst][Defer[Int][(E^(1 + E^(1 + x^2)*x + x^2)*x^6)/(1 + E^(E^(1 + x^2)*x))^2, x], x, Sqrt[x]] - 2*Defer[Subst][
Defer[Int][(E^(1 + E^(1 + x^2)*x + x^2)*x^8)/(1 + E^(E^(1 + x^2)*x))^2, x], x, Sqrt[x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2-2 e^{2 e^{1+x} \sqrt {x}}+6 x^2-e^{1+e^{1+x} \sqrt {x}+x} x^{5/2} (1+2 x)+e^{e^{1+x} \sqrt {x}} \left (-4+6 x^2\right )}{2 \left (1+e^{e^{1+x} \sqrt {x}}\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {-2-2 e^{2 e^{1+x} \sqrt {x}}+6 x^2-e^{1+e^{1+x} \sqrt {x}+x} x^{5/2} (1+2 x)+e^{e^{1+x} \sqrt {x}} \left (-4+6 x^2\right )}{\left (1+e^{e^{1+x} \sqrt {x}}\right )^2} \, dx\\ &=-\operatorname {Subst}\left (\int \frac {x \left (2+2 e^{2 e^{1+x^2} x}-6 x^4+e^{e^{1+x^2} x} \left (4-6 x^4\right )+e^{1+e^{1+x^2} x+x^2} \left (x^5+2 x^7\right )\right )}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )\\ &=-\operatorname {Subst}\left (\int \left (\frac {2 x}{\left (1+e^{e^{1+x^2} x}\right )^2}+\frac {2 e^{2 e^{1+x^2} x} x}{\left (1+e^{e^{1+x^2} x}\right )^2}-\frac {6 x^5}{\left (1+e^{e^{1+x^2} x}\right )^2}+\frac {e^{1+e^{1+x^2} x+x^2} x^6 \left (1+2 x^2\right )}{\left (1+e^{e^{1+x^2} x}\right )^2}-\frac {2 e^{e^{1+x^2} x} x \left (-2+3 x^4\right )}{\left (1+e^{e^{1+x^2} x}\right )^2}\right ) \, dx,x,\sqrt {x}\right )\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {x}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )\right )-2 \operatorname {Subst}\left (\int \frac {e^{2 e^{1+x^2} x} x}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )+2 \operatorname {Subst}\left (\int \frac {e^{e^{1+x^2} x} x \left (-2+3 x^4\right )}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )+6 \operatorname {Subst}\left (\int \frac {x^5}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )-\operatorname {Subst}\left (\int \frac {e^{1+e^{1+x^2} x+x^2} x^6 \left (1+2 x^2\right )}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {x}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )\right )-2 \operatorname {Subst}\left (\int \frac {e^{2 e^{1+x^2} x} x}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )+2 \operatorname {Subst}\left (\int \left (-\frac {2 e^{e^{1+x^2} x} x}{\left (1+e^{e^{1+x^2} x}\right )^2}+\frac {3 e^{e^{1+x^2} x} x^5}{\left (1+e^{e^{1+x^2} x}\right )^2}\right ) \, dx,x,\sqrt {x}\right )+6 \operatorname {Subst}\left (\int \frac {x^5}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )-\operatorname {Subst}\left (\int \left (\frac {e^{1+e^{1+x^2} x+x^2} x^6}{\left (1+e^{e^{1+x^2} x}\right )^2}+\frac {2 e^{1+e^{1+x^2} x+x^2} x^8}{\left (1+e^{e^{1+x^2} x}\right )^2}\right ) \, dx,x,\sqrt {x}\right )\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {x}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )\right )-2 \operatorname {Subst}\left (\int \frac {e^{2 e^{1+x^2} x} x}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )-2 \operatorname {Subst}\left (\int \frac {e^{1+e^{1+x^2} x+x^2} x^8}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )-4 \operatorname {Subst}\left (\int \frac {e^{e^{1+x^2} x} x}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )+6 \operatorname {Subst}\left (\int \frac {x^5}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )+6 \operatorname {Subst}\left (\int \frac {e^{e^{1+x^2} x} x^5}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )-\operatorname {Subst}\left (\int \frac {e^{1+e^{1+x^2} x+x^2} x^6}{\left (1+e^{e^{1+x^2} x}\right )^2} \, dx,x,\sqrt {x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 30, normalized size = 1.20 \begin {gather*} \frac {1}{2} \left (-2 x+\frac {2 x^3}{1+e^{e^{1+x} \sqrt {x}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 2*E^(2*E^((2*x + Log[E^2*x])/2)) + 6*x^2 + E^E^((2*x + Log[E^2*x])/2)*(-4 + 6*x^2 + E^((2*x +
Log[E^2*x])/2)*(-x^2 - 2*x^3)))/(2 + 4*E^E^((2*x + Log[E^2*x])/2) + 2*E^(2*E^((2*x + Log[E^2*x])/2))),x]

[Out]

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

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fricas [A]  time = 1.02, size = 37, normalized size = 1.48 \begin {gather*} \frac {x^{3} - x e^{\left (e^{\left (x + \frac {1}{2} \, \log \left (x e^{2}\right )\right )}\right )} - x}{e^{\left (e^{\left (x + \frac {1}{2} \, \log \left (x e^{2}\right )\right )}\right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(exp(1/2*log(exp(2)*x)+x))^2+((-2*x^3-x^2)*exp(1/2*log(exp(2)*x)+x)+6*x^2-4)*exp(exp(1/2*log(
exp(2)*x)+x))+6*x^2-2)/(2*exp(exp(1/2*log(exp(2)*x)+x))^2+4*exp(exp(1/2*log(exp(2)*x)+x))+2),x, algorithm="fri
cas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {6 \, x^{2} + {\left (6 \, x^{2} - {\left (2 \, x^{3} + x^{2}\right )} e^{\left (x + \frac {1}{2} \, \log \left (x e^{2}\right )\right )} - 4\right )} e^{\left (e^{\left (x + \frac {1}{2} \, \log \left (x e^{2}\right )\right )}\right )} - 2 \, e^{\left (2 \, e^{\left (x + \frac {1}{2} \, \log \left (x e^{2}\right )\right )}\right )} - 2}{2 \, {\left (e^{\left (2 \, e^{\left (x + \frac {1}{2} \, \log \left (x e^{2}\right )\right )}\right )} + 2 \, e^{\left (e^{\left (x + \frac {1}{2} \, \log \left (x e^{2}\right )\right )}\right )} + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(exp(1/2*log(exp(2)*x)+x))^2+((-2*x^3-x^2)*exp(1/2*log(exp(2)*x)+x)+6*x^2-4)*exp(exp(1/2*log(
exp(2)*x)+x))+6*x^2-2)/(2*exp(exp(1/2*log(exp(2)*x)+x))^2+4*exp(exp(1/2*log(exp(2)*x)+x))+2),x, algorithm="gia
c")

[Out]

integrate(1/2*(6*x^2 + (6*x^2 - (2*x^3 + x^2)*e^(x + 1/2*log(x*e^2)) - 4)*e^(e^(x + 1/2*log(x*e^2))) - 2*e^(2*
e^(x + 1/2*log(x*e^2))) - 2)/(e^(2*e^(x + 1/2*log(x*e^2))) + 2*e^(e^(x + 1/2*log(x*e^2))) + 1), x)

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maple [A]  time = 0.25, size = 23, normalized size = 0.92

method	result	size
risch	\(\frac {x^{3}}{{\mathrm e}^{\sqrt {{\mathrm e}^{2} x}\, {\mathrm e}^{x}}+1}-x\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.51, size = 45, normalized size = 1.80 \begin {gather*} \frac {{\left (x^{3} e^{6} - x e^{6} - x e^{\left (\sqrt {x} e^{\left (x + 1\right )} + 6\right )}\right )} e^{\left (-2\right )}}{e^{4} + e^{\left (\sqrt {x} e^{\left (x + 1\right )} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(exp(1/2*log(exp(2)*x)+x))^2+((-2*x^3-x^2)*exp(1/2*log(exp(2)*x)+x)+6*x^2-4)*exp(exp(1/2*log(
exp(2)*x)+x))+6*x^2-2)/(2*exp(exp(1/2*log(exp(2)*x)+x))^2+4*exp(exp(1/2*log(exp(2)*x)+x))+2),x, algorithm="max
ima")

[Out]

(x^3*e^6 - x*e^6 - x*e^(sqrt(x)*e^(x + 1) + 6))*e^(-2)/(e^4 + e^(sqrt(x)*e^(x + 1) + 4))

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mupad [B]  time = 5.79, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^3}{{\mathrm {e}}^{\sqrt {x}\,\mathrm {e}\,{\mathrm {e}}^x}+1}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*exp(2*exp(x + log(x*exp(2))/2)) + exp(exp(x + log(x*exp(2))/2))*(exp(x + log(x*exp(2))/2)*(x^2 + 2*x^3
) - 6*x^2 + 4) - 6*x^2 + 2)/(2*exp(2*exp(x + log(x*exp(2))/2)) + 4*exp(exp(x + log(x*exp(2))/2)) + 2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(exp(1/2*ln(exp(2)*x)+x))**2+((-2*x**3-x**2)*exp(1/2*ln(exp(2)*x)+x)+6*x**2-4)*exp(exp(1/2*ln
(exp(2)*x)+x))+6*x**2-2)/(2*exp(exp(1/2*ln(exp(2)*x)+x))**2+4*exp(exp(1/2*ln(exp(2)*x)+x))+2),x)

[Out]

Timed out

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