3.85.2 \(\int \frac {1}{5} (10 x+e^{e^{\frac {1}{5} (10 x-2 e^{e^x} x-2 e^x x+2 x^2)}} (-5+e^{\frac {1}{5} (10 x-2 e^{e^x} x-2 e^x x+2 x^2)} (-10 x-4 x^2+e^x (2 x+2 x^2)+e^{e^x} (2 x+2 e^x x^2)))) \, dx\)

Optimal. Leaf size=30 \[ x \left (-e^{e^{\frac {2}{5} x \left (5-e^{e^x}-e^x+x\right )}}+x\right ) \]

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Rubi [B]  time = 0.30, antiderivative size = 105, normalized size of antiderivative = 3.50, number of steps used = 3, number of rules used = 2, integrand size = 112, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {12, 2288} \begin {gather*} x^2-\frac {e^{e^{\frac {2}{5} \left (x^2-e^{e^x} x-e^x x+5 x\right )}} \left (2 x^2-e^x \left (x^2+x\right )-e^{e^x} \left (e^x x^2+x\right )+5 x\right )}{-e^x x-e^{x+e^x} x+2 x-e^{e^x}-e^x+5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 0.47, size = 30, normalized size = 1.00 \begin {gather*} x \left (-e^{e^{\frac {2}{5} x \left (5-e^{e^x}-e^x+x\right )}}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.67, size = 29, normalized size = 0.97 \begin {gather*} x^{2} - x e^{\left (e^{\left (\frac {2}{5} \, x^{2} - \frac {2}{5} \, x e^{x} - \frac {2}{5} \, x e^{\left (e^{x}\right )} + 2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((2*exp(x)*x^2+2*x)*exp(exp(x))+(2*x^2+2*x)*exp(x)-4*x^2-10*x)*exp(-2/5*x*exp(exp(x))-2/5*exp(x
)*x+2/5*x^2+2*x)-5)*exp(exp(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x))+2*x,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((2*exp(x)*x^2+2*x)*exp(exp(x))+(2*x^2+2*x)*exp(x)-4*x^2-10*x)*exp(-2/5*x*exp(exp(x))-2/5*exp(x
)*x+2/5*x^2+2*x)-5)*exp(exp(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x))+2*x,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.13, size = 23, normalized size = 0.77

method	result	size
risch	\(-x \,{\mathrm e}^{{\mathrm e}^{-\frac {2 x \left ({\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}^{x}-x -5\right )}{5}}}+x^{2}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(((2*exp(x)*x^2+2*x)*exp(exp(x))+(2*x^2+2*x)*exp(x)-4*x^2-10*x)*exp(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/
5*x^2+2*x)-5)*exp(exp(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x))+2*x,x,method=_RETURNVERBOSE)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((2*exp(x)*x^2+2*x)*exp(exp(x))+(2*x^2+2*x)*exp(x)-4*x^2-10*x)*exp(-2/5*x*exp(exp(x))-2/5*exp(x
)*x+2/5*x^2+2*x)-5)*exp(exp(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x^2+2*x))+2*x,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.67, size = 31, normalized size = 1.03 \begin {gather*} x\,\left (x-{\mathrm {e}}^{{\mathrm {e}}^{-\frac {2\,x\,{\mathrm {e}}^x}{5}}\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-\frac {2\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}}{5}}\,{\mathrm {e}}^{\frac {2\,x^2}{5}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 11.52, size = 36, normalized size = 1.20 \begin {gather*} x^{2} - x e^{e^{\frac {2 x^{2}}{5} - \frac {2 x e^{x}}{5} - \frac {2 x e^{e^{x}}}{5} + 2 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(((2*exp(x)*x**2+2*x)*exp(exp(x))+(2*x**2+2*x)*exp(x)-4*x**2-10*x)*exp(-2/5*x*exp(exp(x))-2/5*ex
p(x)*x+2/5*x**2+2*x)-5)*exp(exp(-2/5*x*exp(exp(x))-2/5*exp(x)*x+2/5*x**2+2*x))+2*x,x)

[Out]

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

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