3.79.94 \(\int \frac {1}{5} (10+e^{e^{10+x^2}} (-3-6 e^{10+x^2} x^2)) \, dx\)

Optimal. Leaf size=23 \[ -e^{10}+2 x-\frac {3}{5} e^{e^{10+x^2}} x \]

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 0.78, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 2288} \begin {gather*} 2 x-\frac {3}{5} e^{e^{x^2+10}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 + E^E^(10 + x^2)*(-3 - 6*E^(10 + x^2)*x^2))/5,x]

[Out]

2*x - (3*E^E^(10 + x^2)*x)/5

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+e^{e^{10+x^2}} \left (-3-6 e^{10+x^2} x^2\right )\right ) \, dx\\ &=2 x+\frac {1}{5} \int e^{e^{10+x^2}} \left (-3-6 e^{10+x^2} x^2\right ) \, dx\\ &=2 x-\frac {3}{5} e^{e^{10+x^2}} x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.78 \begin {gather*} 2 x-\frac {3}{5} e^{e^{10+x^2}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 + E^E^(10 + x^2)*(-3 - 6*E^(10 + x^2)*x^2))/5,x]

[Out]

2*x - (3*E^E^(10 + x^2)*x)/5

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fricas [A]  time = 0.62, size = 14, normalized size = 0.61 \begin {gather*} -\frac {3}{5} \, x e^{\left (e^{\left (x^{2} + 10\right )}\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-6*x^2*exp(5)^2*exp(x^2)-3)*exp(exp(5)^2*exp(x^2))+2,x, algorithm="fricas")

[Out]

-3/5*x*e^(e^(x^2 + 10)) + 2*x

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giac [A]  time = 0.25, size = 14, normalized size = 0.61 \begin {gather*} -\frac {3}{5} \, x e^{\left (e^{\left (x^{2} + 10\right )}\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-6*x^2*exp(5)^2*exp(x^2)-3)*exp(exp(5)^2*exp(x^2))+2,x, algorithm="giac")

[Out]

-3/5*x*e^(e^(x^2 + 10)) + 2*x

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maple [A]  time = 0.09, size = 15, normalized size = 0.65

method	result	size
risch	\(2 x -\frac {3 \,{\mathrm e}^{{\mathrm e}^{x^{2}+10}} x}{5}\)	\(15\)
default	\(2 x -\frac {3 \,{\mathrm e}^{{\mathrm e}^{10} {\mathrm e}^{x^{2}}} x}{5}\)	\(18\)
norman	\(2 x -\frac {3 \,{\mathrm e}^{{\mathrm e}^{10} {\mathrm e}^{x^{2}}} x}{5}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-6*x^2*exp(5)^2*exp(x^2)-3)*exp(exp(5)^2*exp(x^2))+2,x,method=_RETURNVERBOSE)

[Out]

2*x-3/5*exp(exp(x^2+10))*x

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maxima [A]  time = 0.41, size = 14, normalized size = 0.61 \begin {gather*} -\frac {3}{5} \, x e^{\left (e^{\left (x^{2} + 10\right )}\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-6*x^2*exp(5)^2*exp(x^2)-3)*exp(exp(5)^2*exp(x^2))+2,x, algorithm="maxima")

[Out]

-3/5*x*e^(e^(x^2 + 10)) + 2*x

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mupad [B]  time = 4.97, size = 15, normalized size = 0.65 \begin {gather*} -\frac {x\,\left (3\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{10}}-10\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2 - (exp(exp(x^2)*exp(10))*(6*x^2*exp(x^2)*exp(10) + 3))/5,x)

[Out]

-(x*(3*exp(exp(x^2)*exp(10)) - 10))/5

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sympy [A]  time = 1.11, size = 17, normalized size = 0.74 \begin {gather*} - \frac {3 x e^{e^{10} e^{x^{2}}}}{5} + 2 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*x*exp(exp(10)*exp(x**2))/5 + 2*x

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