\(\int \frac {1}{5} (10+e^{e^{10+x^2}} (-3-6 e^{10+x^2} x^2)) \, dx\) [7894]

Optimal result

Integrand size = 30, antiderivative size = 23 \[ \int \frac {1}{5} \left (10+e^{e^{10+x^2}} \left (-3-6 e^{10+x^2} x^2\right )\right ) \, dx=-e^{10}+2 x-\frac {3}{5} e^{e^{10+x^2}} x \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 2326} \[ \int \frac {1}{5} \left (10+e^{e^{10+x^2}} \left (-3-6 e^{10+x^2} x^2\right )\right ) \, dx=2 x-\frac {3}{5} e^{e^{x^2+10}} x \]

[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 2326

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {1}{5} \left (10+e^{e^{10+x^2}} \left (-3-6 e^{10+x^2} x^2\right )\right ) \, dx=2 x-\frac {3}{5} e^{e^{10+x^2}} x \]

[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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 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	\(2 x -\frac {3 \,{\mathrm e}^{{\mathrm e}^{10} {\mathrm e}^{x^{2}}} x}{5}\)	\(18\)
parallelrisch	\(2 x -\frac {3 \,{\mathrm e}^{{\mathrm e}^{10} {\mathrm e}^{x^{2}}} x}{5}\)	\(18\)

[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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61 \[ \int \frac {1}{5} \left (10+e^{e^{10+x^2}} \left (-3-6 e^{10+x^2} x^2\right )\right ) \, dx=-\frac {3}{5} \, x e^{\left (e^{\left (x^{2} + 10\right )}\right )} + 2 \, x \]

[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

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {1}{5} \left (10+e^{e^{10+x^2}} \left (-3-6 e^{10+x^2} x^2\right )\right ) \, dx=- \frac {3 x e^{e^{10} e^{x^{2}}}}{5} + 2 x \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61 \[ \int \frac {1}{5} \left (10+e^{e^{10+x^2}} \left (-3-6 e^{10+x^2} x^2\right )\right ) \, dx=-\frac {3}{5} \, x e^{\left (e^{\left (x^{2} + 10\right )}\right )} + 2 \, x \]

[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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61 \[ \int \frac {1}{5} \left (10+e^{e^{10+x^2}} \left (-3-6 e^{10+x^2} x^2\right )\right ) \, dx=-\frac {3}{5} \, x e^{\left (e^{\left (x^{2} + 10\right )}\right )} + 2 \, x \]

[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

Mupad [B] (verification not implemented)

Time = 11.63 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {1}{5} \left (10+e^{e^{10+x^2}} \left (-3-6 e^{10+x^2} x^2\right )\right ) \, dx=-\frac {x\,\left (3\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{10}}-10\right )}{5} \]

[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