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

Optimal. Leaf size=30 \[ \left (e^{e^{e^4+e^x-x}}+e^{\frac {2}{x (2+x)}}+x\right )^2 \]

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Rubi [A]  time = 21.16, antiderivative size = 31, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, integrand size = 302, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {1594, 27, 6688, 12, 6686} \begin {gather*} \left (e^{\frac {2}{x^2+2 x}}+x+e^{e^{-x+e^x+e^4}}\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

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Mathematica [A]  time = 0.50, size = 47, normalized size = 1.57 \begin {gather*} e^{-\frac {2}{2+x}} \left (e^{\frac {1}{x}}+e^{e^{e^4+e^x-x}+\frac {1}{2+x}}+e^{\frac {1}{2+x}} x\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.66, size = 69, normalized size = 2.30 \begin {gather*} x^{2} + 2 \, x e^{\left (\frac {2}{x^{2} + 2 \, x}\right )} + 2 \, {\left (x + e^{\left (\frac {2}{x^{2} + 2 \, x}\right )}\right )} e^{\left (e^{\left (-x + e^{4} + e^{x}\right )}\right )} + e^{\left (2 \, e^{\left (-x + e^{4} + e^{x}\right )}\right )} + e^{\left (\frac {4}{x^{2} + 2 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2 + 2*x*e^(2/(x^2 + 2*x)) + 2*(x + e^(2/(x^2 + 2*x)))*e^(e^(-x + e^4 + e^x)) + e^(2*e^(-x + e^4 + e^x)) + e^
(4/(x^2 + 2*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.15, size = 70, normalized size = 2.33

method	result	size
risch	\(x^{2}+2 x \,{\mathrm e}^{\frac {2}{x \left (2+x \right )}}+{\mathrm e}^{\frac {4}{x \left (2+x \right )}}+{\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}^{x}+{\mathrm e}^{4}-x}}+\left (2 x +2 \,{\mathrm e}^{\frac {2}{x \left (2+x \right )}}\right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}+{\mathrm e}^{4}-x}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.68, size = 64, normalized size = 2.13 \begin {gather*} x^{2} + {\left (2 \, x e^{\frac {1}{x}} + 2 \, {\left (x e^{\left (\frac {1}{x + 2}\right )} + e^{\frac {1}{x}}\right )} e^{\left (e^{\left (-x + e^{4} + e^{x}\right )}\right )} + e^{\left (\frac {1}{x + 2} + 2 \, e^{\left (-x + e^{4} + e^{x}\right )}\right )}\right )} e^{\left (-\frac {1}{x + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.55, size = 75, normalized size = 2.50 \begin {gather*} {\mathrm {e}}^{2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{{\mathrm {e}}^4}}+{\mathrm {e}}^{\frac {4}{x^2+2\,x}}+2\,x\,{\mathrm {e}}^{\frac {2}{x^2+2\,x}}+{\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{{\mathrm {e}}^4}}\,\left (2\,x+2\,{\mathrm {e}}^{\frac {2}{x^2+2\,x}}\right )+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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