\(\int \frac {1}{8} e^{2 e^{80 e^{4+x+x^2}}} (x+e^{4+80 e^{4+x+x^2}+x+x^2} (80 x^2+160 x^3)) \, dx\) [3146]

Optimal result

Integrand size = 52, antiderivative size = 23 \[ \int \frac {1}{8} e^{2 e^{80 e^{4+x+x^2}}} \left (x+e^{4+80 e^{4+x+x^2}+x+x^2} \left (80 x^2+160 x^3\right )\right ) \, dx=\frac {1}{16} e^{2 e^{80 e^{4+x+x^2}}} x^2 \]

[Out]

1/16*x^2*exp(exp(80*exp(4)*exp(x^2+x)))^2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {12, 2326} \[ \int \frac {1}{8} e^{2 e^{80 e^{4+x+x^2}}} \left (x+e^{4+80 e^{4+x+x^2}+x+x^2} \left (80 x^2+160 x^3\right )\right ) \, dx=\frac {e^{2 e^{80 e^{x^2+x+4}}} \left (2 x^3+x^2\right )}{16 (2 x+1)} \]

[In]

Int[(E^(2*E^(80*E^(4 + x + x^2)))*(x + E^(4 + 80*E^(4 + x + x^2) + x + x^2)*(80*x^2 + 160*x^3)))/8,x]

[Out]

(E^(2*E^(80*E^(4 + x + x^2)))*(x^2 + 2*x^3))/(16*(1 + 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 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}{8} \int e^{2 e^{80 e^{4+x+x^2}}} \left (x+e^{4+80 e^{4+x+x^2}+x+x^2} \left (80 x^2+160 x^3\right )\right ) \, dx \\ & = \frac {e^{2 e^{80 e^{4+x+x^2}}} \left (x^2+2 x^3\right )}{16 (1+2 x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1}{8} e^{2 e^{80 e^{4+x+x^2}}} \left (x+e^{4+80 e^{4+x+x^2}+x+x^2} \left (80 x^2+160 x^3\right )\right ) \, dx=\frac {1}{16} e^{2 e^{80 e^{4+x+x^2}}} x^2 \]

[In]

Integrate[(E^(2*E^(80*E^(4 + x + x^2)))*(x + E^(4 + 80*E^(4 + x + x^2) + x + x^2)*(80*x^2 + 160*x^3)))/8,x]

[Out]

(E^(2*E^(80*E^(4 + x + x^2)))*x^2)/16

Maple [A] (verified)

Time = 3.33 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {x^{2} {\mathrm e}^{2 \,{\mathrm e}^{80 \,{\mathrm e}^{x^{2}+x +4}}}}{16}\)	\(19\)
parallelrisch	\(\frac {x^{2} {\mathrm e}^{2 \,{\mathrm e}^{80 \,{\mathrm e}^{x^{2}+x +4}}}}{16}\)	\(20\)

[In]

int(1/8*((160*x^3+80*x^2)*exp(4)*exp(x^2+x)*exp(80*exp(4)*exp(x^2+x))+x)*exp(exp(80*exp(4)*exp(x^2+x)))^2,x,me
thod=_RETURNVERBOSE)

[Out]

1/16*x^2*exp(2*exp(80*exp(x^2+x+4)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {1}{8} e^{2 e^{80 e^{4+x+x^2}}} \left (x+e^{4+80 e^{4+x+x^2}+x+x^2} \left (80 x^2+160 x^3\right )\right ) \, dx=\frac {1}{16} \, x^{2} e^{\left (2 \, e^{\left (80 \, e^{\left (x^{2} + x + 4\right )}\right )}\right )} \]

[In]

integrate(1/8*((160*x^3+80*x^2)*exp(4)*exp(x^2+x)*exp(80*exp(4)*exp(x^2+x))+x)*exp(exp(80*exp(4)*exp(x^2+x)))^
2,x, algorithm="fricas")

[Out]

1/16*x^2*e^(2*e^(80*e^(x^2 + x + 4)))

Sympy [A] (verification not implemented)

Time = 57.70 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {1}{8} e^{2 e^{80 e^{4+x+x^2}}} \left (x+e^{4+80 e^{4+x+x^2}+x+x^2} \left (80 x^2+160 x^3\right )\right ) \, dx=\frac {x^{2} e^{2 e^{80 e^{4} e^{x^{2} + x}}}}{16} \]

[In]

integrate(1/8*((160*x**3+80*x**2)*exp(4)*exp(x**2+x)*exp(80*exp(4)*exp(x**2+x))+x)*exp(exp(80*exp(4)*exp(x**2+
x)))**2,x)

[Out]

x**2*exp(2*exp(80*exp(4)*exp(x**2 + x)))/16

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {1}{8} e^{2 e^{80 e^{4+x+x^2}}} \left (x+e^{4+80 e^{4+x+x^2}+x+x^2} \left (80 x^2+160 x^3\right )\right ) \, dx=\frac {1}{16} \, x^{2} e^{\left (2 \, e^{\left (80 \, e^{\left (x^{2} + x + 4\right )}\right )}\right )} \]

[In]

integrate(1/8*((160*x^3+80*x^2)*exp(4)*exp(x^2+x)*exp(80*exp(4)*exp(x^2+x))+x)*exp(exp(80*exp(4)*exp(x^2+x)))^
2,x, algorithm="maxima")

[Out]

1/16*x^2*e^(2*e^(80*e^(x^2 + x + 4)))

Giac [F]

\[ \int \frac {1}{8} e^{2 e^{80 e^{4+x+x^2}}} \left (x+e^{4+80 e^{4+x+x^2}+x+x^2} \left (80 x^2+160 x^3\right )\right ) \, dx=\int { \frac {1}{8} \, {\left (80 \, {\left (2 \, x^{3} + x^{2}\right )} e^{\left (x^{2} + x + 80 \, e^{\left (x^{2} + x + 4\right )} + 4\right )} + x\right )} e^{\left (2 \, e^{\left (80 \, e^{\left (x^{2} + x + 4\right )}\right )}\right )} \,d x } \]

[In]

integrate(1/8*((160*x^3+80*x^2)*exp(4)*exp(x^2+x)*exp(80*exp(4)*exp(x^2+x))+x)*exp(exp(80*exp(4)*exp(x^2+x)))^
2,x, algorithm="giac")

[Out]

integrate(1/8*(80*(2*x^3 + x^2)*e^(x^2 + x + 80*e^(x^2 + x + 4) + 4) + x)*e^(2*e^(80*e^(x^2 + x + 4))), x)

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1}{8} e^{2 e^{80 e^{4+x+x^2}}} \left (x+e^{4+80 e^{4+x+x^2}+x+x^2} \left (80 x^2+160 x^3\right )\right ) \, dx=\frac {x^2\,{\mathrm {e}}^{2\,{\mathrm {e}}^{80\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^4\,{\mathrm {e}}^x}}}{16} \]

[In]

int((exp(2*exp(80*exp(x + x^2)*exp(4)))*(x + exp(x + x^2)*exp(80*exp(x + x^2)*exp(4))*exp(4)*(80*x^2 + 160*x^3
)))/8,x)

[Out]

(x^2*exp(2*exp(80*exp(x^2)*exp(4)*exp(x))))/16