3.15.74 \(\int \frac {1}{5} (10+e^{e^{1+e^{\frac {e x^2}{5}}}-x} (5-2 e^{2+e^{\frac {e x^2}{5}}+\frac {e x^2}{5}} x)) \, dx\) [1474]

Optimal. Leaf size=26 \[ -e^{e^{1+e^{\frac {e x^2}{5}}}-x}+2 x \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {12, 6838} \begin {gather*} 2 x-e^{e^{e^{\frac {e x^2}{5}}+1}-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 + E^(E^(1 + E^((E*x^2)/5)) - x)*(5 - 2*E^(2 + E^((E*x^2)/5) + (E*x^2)/5)*x))/5,x]

[Out]

-E^(E^(1 + E^((E*x^2)/5)) - x) + 2*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 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 26, normalized size = 1.00 \begin {gather*} -e^{e^{1+e^{\frac {e x^2}{5}}}-x}+2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 + E^(E^(1 + E^((E*x^2)/5)) - x)*(5 - 2*E^(2 + E^((E*x^2)/5) + (E*x^2)/5)*x))/5,x]

[Out]

-E^(E^(1 + E^((E*x^2)/5)) - x) + 2*x

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Maple [A]
time = 0.22, size = 23, normalized size = 0.88

method	result	size
default	\(2 x -{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{\frac {x^{2} {\mathrm e}}{5}}+1}-x}\)	\(23\)
norman	\(2 x -{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{\frac {x^{2} {\mathrm e}}{5}}+1}-x}\)	\(23\)
risch	\(2 x -{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{\frac {x^{2} {\mathrm e}}{5}}+1}-x}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-2*x*exp(1)*exp(1/5*x^2*exp(1))*exp(exp(1/5*x^2*exp(1))+1)+5)*exp(exp(exp(1/5*x^2*exp(1))+1)-x)+2,x,m
ethod=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 22, normalized size = 0.85 \begin {gather*} 2 \, x - e^{\left (-x + e^{\left (e^{\left (\frac {1}{5} \, x^{2} e\right )} + 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - e^(-x + e^(e^(1/5*x^2*e) + 1))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (22) = 44\).
time = 0.39, size = 52, normalized size = 2.00 \begin {gather*} 2 \, x - e^{\left (-{\left (x e^{\left (\frac {1}{5} \, x^{2} e + 1\right )} - e^{\left (\frac {1}{5} \, x^{2} e + e^{\left (\frac {1}{5} \, x^{2} e\right )} + 2\right )}\right )} e^{\left (-\frac {1}{5} \, x^{2} e - 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.19, size = 19, normalized size = 0.73 \begin {gather*} 2 x - e^{- x + e^{e^{\frac {e x^{2}}{5}} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - exp(-x + exp(exp(E*x**2/5) + 1))

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Giac [A]
time = 0.41, size = 22, normalized size = 0.85 \begin {gather*} 2 \, x - e^{\left (-x + e^{\left (e^{\left (\frac {1}{5} \, x^{2} e\right )} + 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x - e^(-x + e^(e^(1/5*x^2*e) + 1))

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Mupad [B]
time = 1.12, size = 24, normalized size = 0.92 \begin {gather*} 2\,x-{\mathrm {e}}^{-x}\,{\mathrm {e}}^{\mathrm {e}\,{\mathrm {e}}^{{\left ({\mathrm {e}}^{x^2\,\mathrm {e}}\right )}^{1/5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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