\(\int \frac {1}{8} e^{\frac {1}{8} (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4))} (8 e^x-24 x^2+e^{x^2} (3 x^2+2 x^4)) \, dx\) [1257]

Optimal result

Integrand size = 62, antiderivative size = 27 \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=\frac {1}{4} e^{e^x-x^3+\frac {1}{8} e^{x^2} x^3} \]

[Out]

exp(1/8*x^3*exp(x^2)+exp(x)-2*ln(2)-x^3)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {12, 2306, 6838} \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=\frac {1}{4} e^{\frac {1}{8} \left (-8 x^3+e^{x^2} x^3+8 e^x\right )} \]

[In]

Int[(E^((8*E^x - 8*x^3 + E^x^2*x^3 - 8*Log[4])/8)*(8*E^x - 24*x^2 + E^x^2*(3*x^2 + 2*x^4)))/8,x]

[Out]

E^((8*E^x - 8*x^3 + E^x^2*x^3)/8)/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2306

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
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{align*} \text {integral}& = \frac {1}{8} \int e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx \\ & = \frac {1}{8} \int \frac {1}{4} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx \\ & = \frac {1}{32} \int e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx \\ & = \frac {1}{4} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=\frac {1}{4} e^{e^x-x^3+\frac {1}{8} e^{x^2} x^3} \]

[In]

Integrate[(E^((8*E^x - 8*x^3 + E^x^2*x^3 - 8*Log[4])/8)*(8*E^x - 24*x^2 + E^x^2*(3*x^2 + 2*x^4)))/8,x]

[Out]

E^(E^x - x^3 + (E^x^2*x^3)/8)/4

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {{\mathrm e}^{\frac {x^{3} {\mathrm e}^{x^{2}}}{8}+{\mathrm e}^{x}-x^{3}}}{4}\)	\(21\)
parallelrisch	\({\mathrm e}^{\frac {x^{3} {\mathrm e}^{x^{2}}}{8}+{\mathrm e}^{x}-2 \ln \left (2\right )-x^{3}}\)	\(23\)

[In]

int(1/8*((2*x^4+3*x^2)*exp(x^2)+8*exp(x)-24*x^2)*exp(1/8*x^3*exp(x^2)+exp(x)-2*ln(2)-x^3),x,method=_RETURNVERB
OSE)

[Out]

1/4*exp(1/8*x^3*exp(x^2)+exp(x)-x^3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=e^{\left (\frac {1}{8} \, x^{3} e^{\left (x^{2}\right )} - x^{3} + e^{x} - 2 \, \log \left (2\right )\right )} \]

[In]

integrate(1/8*((2*x^4+3*x^2)*exp(x^2)+8*exp(x)-24*x^2)*exp(1/8*x^3*exp(x^2)+exp(x)-2*log(2)-x^3),x, algorithm=
"fricas")

[Out]

e^(1/8*x^3*e^(x^2) - x^3 + e^x - 2*log(2))

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=\frac {e^{\frac {x^{3} e^{x^{2}}}{8} - x^{3} + e^{x}}}{4} \]

[In]

integrate(1/8*((2*x**4+3*x**2)*exp(x**2)+8*exp(x)-24*x**2)*exp(1/8*x**3*exp(x**2)+exp(x)-2*ln(2)-x**3),x)

[Out]

exp(x**3*exp(x**2)/8 - x**3 + exp(x))/4

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=\frac {1}{4} \, e^{\left (\frac {1}{8} \, x^{3} e^{\left (x^{2}\right )} - x^{3} + e^{x}\right )} \]

[In]

integrate(1/8*((2*x^4+3*x^2)*exp(x^2)+8*exp(x)-24*x^2)*exp(1/8*x^3*exp(x^2)+exp(x)-2*log(2)-x^3),x, algorithm=
"maxima")

[Out]

1/4*e^(1/8*x^3*e^(x^2) - x^3 + e^x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=e^{\left (\frac {1}{8} \, x^{3} e^{\left (x^{2}\right )} - x^{3} + e^{x} - 2 \, \log \left (2\right )\right )} \]

[In]

integrate(1/8*((2*x^4+3*x^2)*exp(x^2)+8*exp(x)-24*x^2)*exp(1/8*x^3*exp(x^2)+exp(x)-2*log(2)-x^3),x, algorithm=
"giac")

[Out]

e^(1/8*x^3*e^(x^2) - x^3 + e^x - 2*log(2))

Mupad [B] (verification not implemented)

Time = 8.13 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{8} e^{\frac {1}{8} \left (8 e^x-8 x^3+e^{x^2} x^3-8 \log (4)\right )} \left (8 e^x-24 x^2+e^{x^2} \left (3 x^2+2 x^4\right )\right ) \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{-x^3}\,{\mathrm {e}}^{\frac {x^3\,{\mathrm {e}}^{x^2}}{8}}}{4} \]

[In]

int((exp(exp(x) - 2*log(2) + (x^3*exp(x^2))/8 - x^3)*(8*exp(x) + exp(x^2)*(3*x^2 + 2*x^4) - 24*x^2))/8,x)

[Out]

(exp(exp(x))*exp(-x^3)*exp((x^3*exp(x^2))/8))/4