\(\int e^{e^{1+x^2}} (e^x (-18-22 x-2 x^2)+e^{1+x+x^2} (-36 x^2-4 x^3)) \, dx\) [5721]

Optimal result

Integrand size = 45, antiderivative size = 17 \[ \int e^{e^{1+x^2}} \left (e^x \left (-18-22 x-2 x^2\right )+e^{1+x+x^2} \left (-36 x^2-4 x^3\right )\right ) \, dx=-2 e^{e^{1+x^2}+x} x (9+x) \]

[Out]

-2*x*exp(x)*(x+9)*exp(exp(x^2+1))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {2326} \[ \int e^{e^{1+x^2}} \left (e^x \left (-18-22 x-2 x^2\right )+e^{1+x+x^2} \left (-36 x^2-4 x^3\right )\right ) \, dx=-\frac {2 e^{e^{x^2+1}+x} \left (x^3+9 x^2\right )}{x} \]

[In]

Int[E^E^(1 + x^2)*(E^x*(-18 - 22*x - 2*x^2) + E^(1 + x + x^2)*(-36*x^2 - 4*x^3)),x]

[Out]

(-2*E^(E^(1 + x^2) + x)*(9*x^2 + x^3))/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 {2 e^{e^{1+x^2}+x} \left (9 x^2+x^3\right )}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int e^{e^{1+x^2}} \left (e^x \left (-18-22 x-2 x^2\right )+e^{1+x+x^2} \left (-36 x^2-4 x^3\right )\right ) \, dx=-2 e^{e^{1+x^2}+x} x (9+x) \]

[In]

Integrate[E^E^(1 + x^2)*(E^x*(-18 - 22*x - 2*x^2) + E^(1 + x + x^2)*(-36*x^2 - 4*x^3)),x]

[Out]

-2*E^(E^(1 + x^2) + x)*x*(9 + x)

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
risch	\(-2 \left (x +9\right ) x \,{\mathrm e}^{x +{\mathrm e}^{x^{2}+1}}\)	\(16\)
parallelrisch	\(-2 \,{\mathrm e}^{{\mathrm e}^{x^{2}+1}} {\mathrm e}^{x} x^{2}-18 x \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x^{2}+1}}\)	\(28\)

[In]

int(((-4*x^3-36*x^2)*exp(x)*exp(x^2+1)+(-2*x^2-22*x-18)*exp(x))*exp(exp(x^2+1)),x,method=_RETURNVERBOSE)

[Out]

-2*(x+9)*x*exp(x+exp(x^2+1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int e^{e^{1+x^2}} \left (e^x \left (-18-22 x-2 x^2\right )+e^{1+x+x^2} \left (-36 x^2-4 x^3\right )\right ) \, dx=-2 \, {\left (x^{2} + 9 \, x\right )} e^{\left (x + e^{\left (x^{2} + 1\right )}\right )} \]

[In]

integrate(((-4*x^3-36*x^2)*exp(x)*exp(x^2+1)+(-2*x^2-22*x-18)*exp(x))*exp(exp(x^2+1)),x, algorithm="fricas")

[Out]

-2*(x^2 + 9*x)*e^(x + e^(x^2 + 1))

Sympy [A] (verification not implemented)

Time = 5.74 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.41 \[ \int e^{e^{1+x^2}} \left (e^x \left (-18-22 x-2 x^2\right )+e^{1+x+x^2} \left (-36 x^2-4 x^3\right )\right ) \, dx=\left (- 2 x^{2} e^{x} - 18 x e^{x}\right ) e^{e^{x^{2} + 1}} \]

[In]

integrate(((-4*x**3-36*x**2)*exp(x)*exp(x**2+1)+(-2*x**2-22*x-18)*exp(x))*exp(exp(x**2+1)),x)

[Out]

(-2*x**2*exp(x) - 18*x*exp(x))*exp(exp(x**2 + 1))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int e^{e^{1+x^2}} \left (e^x \left (-18-22 x-2 x^2\right )+e^{1+x+x^2} \left (-36 x^2-4 x^3\right )\right ) \, dx=-2 \, {\left (x^{2} + 9 \, x\right )} e^{\left (x + e^{\left (x^{2} + 1\right )}\right )} \]

[In]

integrate(((-4*x^3-36*x^2)*exp(x)*exp(x^2+1)+(-2*x^2-22*x-18)*exp(x))*exp(exp(x^2+1)),x, algorithm="maxima")

[Out]

-2*(x^2 + 9*x)*e^(x + e^(x^2 + 1))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.59 \[ \int e^{e^{1+x^2}} \left (e^x \left (-18-22 x-2 x^2\right )+e^{1+x+x^2} \left (-36 x^2-4 x^3\right )\right ) \, dx=-2 \, x^{2} e^{\left (x + e^{\left (x^{2} + 1\right )}\right )} - 18 \, x e^{\left (x + e^{\left (x^{2} + 1\right )}\right )} \]

[In]

integrate(((-4*x^3-36*x^2)*exp(x)*exp(x^2+1)+(-2*x^2-22*x-18)*exp(x))*exp(exp(x^2+1)),x, algorithm="giac")

[Out]

-2*x^2*e^(x + e^(x^2 + 1)) - 18*x*e^(x + e^(x^2 + 1))

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35 \[ \int e^{e^{1+x^2}} \left (e^x \left (-18-22 x-2 x^2\right )+e^{1+x+x^2} \left (-36 x^2-4 x^3\right )\right ) \, dx=-{\mathrm {e}}^{{\mathrm {e}}^{x^2}\,\mathrm {e}}\,\left (2\,x^2\,{\mathrm {e}}^x+18\,x\,{\mathrm {e}}^x\right ) \]

[In]

int(-exp(exp(x^2 + 1))*(exp(x)*(22*x + 2*x^2 + 18) + exp(x^2 + 1)*exp(x)*(36*x^2 + 4*x^3)),x)

[Out]

-exp(exp(x^2)*exp(1))*(2*x^2*exp(x) + 18*x*exp(x))