\(\int \frac {e^{\frac {1}{2} (24 x^2-e^3 x^2+x^3)} (48 x-2 e^3 x+3 x^2)}{2+2 e^2} \, dx\) [5413]

Optimal result

Integrand size = 48, antiderivative size = 25 \[ \int \frac {e^{\frac {1}{2} \left (24 x^2-e^3 x^2+x^3\right )} \left (48 x-2 e^3 x+3 x^2\right )}{2+2 e^2} \, dx=\frac {e^{\frac {1}{2} x^2 \left (24-e^3+x\right )}}{1+e^2} \]

[Out]

exp(1/2*(x-exp(3)+24)*x^2)/(exp(2)+1)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {6, 12, 1607, 6820, 6838} \[ \int \frac {e^{\frac {1}{2} \left (24 x^2-e^3 x^2+x^3\right )} \left (48 x-2 e^3 x+3 x^2\right )}{2+2 e^2} \, dx=\frac {e^{\frac {1}{2} x^2 \left (x-e^3+24\right )}}{1+e^2} \]

[In]

Int[(E^((24*x^2 - E^3*x^2 + x^3)/2)*(48*x - 2*E^3*x + 3*x^2))/(2 + 2*E^2),x]

[Out]

E^((x^2*(24 - E^3 + x))/2)/(1 + E^2)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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}& = \int \frac {e^{\frac {1}{2} \left (24 x^2-e^3 x^2+x^3\right )} \left (\left (48-2 e^3\right ) x+3 x^2\right )}{2+2 e^2} \, dx \\ & = \frac {\int e^{\frac {1}{2} \left (24 x^2-e^3 x^2+x^3\right )} \left (\left (48-2 e^3\right ) x+3 x^2\right ) \, dx}{2 \left (1+e^2\right )} \\ & = \frac {\int e^{\frac {1}{2} \left (24 x^2-e^3 x^2+x^3\right )} x \left (48-2 e^3+3 x\right ) \, dx}{2 \left (1+e^2\right )} \\ & = \frac {\int e^{\frac {1}{2} x^2 \left (24-e^3+x\right )} x \left (48-2 e^3+3 x\right ) \, dx}{2 \left (1+e^2\right )} \\ & = \frac {e^{\frac {1}{2} x^2 \left (24-e^3+x\right )}}{1+e^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {1}{2} \left (24 x^2-e^3 x^2+x^3\right )} \left (48 x-2 e^3 x+3 x^2\right )}{2+2 e^2} \, dx=\frac {e^{\frac {1}{2} x^2 \left (24-e^3+x\right )}}{1+e^2} \]

[In]

Integrate[(E^((24*x^2 - E^3*x^2 + x^3)/2)*(48*x - 2*E^3*x + 3*x^2))/(2 + 2*E^2),x]

[Out]

E^((x^2*(24 - E^3 + x))/2)/(1 + E^2)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{-\frac {x^{2} \left ({\mathrm e}^{3}-x -24\right )}{2}}}{2 \,{\mathrm e}^{2}+2}\)	\(24\)
parallelrisch	\(\frac {2 \,{\mathrm e}^{-\frac {x^{2} \left ({\mathrm e}^{3}-x -24\right )}{2}}}{2 \,{\mathrm e}^{2}+2}\)	\(24\)
gosper	\(\frac {{\mathrm e}^{-\frac {x^{2} {\mathrm e}^{3}}{2}+\frac {x^{3}}{2}+12 x^{2}}}{{\mathrm e}^{2}+1}\)	\(27\)
norman	\(\frac {{\mathrm e}^{-\frac {x^{2} {\mathrm e}^{3}}{2}+\frac {x^{3}}{2}+12 x^{2}}}{{\mathrm e}^{2}+1}\)	\(27\)

[In]

int((-2*x*exp(3)+3*x^2+48*x)*exp(-1/2*x^2*exp(3)+1/2*x^3+12*x^2)/(2*exp(2)+2),x,method=_RETURNVERBOSE)

[Out]

2/(2*exp(2)+2)*exp(-1/2*x^2*(exp(3)-x-24))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {1}{2} \left (24 x^2-e^3 x^2+x^3\right )} \left (48 x-2 e^3 x+3 x^2\right )}{2+2 e^2} \, dx=\frac {e^{\left (\frac {1}{2} \, x^{3} - \frac {1}{2} \, x^{2} e^{3} + 12 \, x^{2}\right )}}{e^{2} + 1} \]

[In]

integrate((-2*x*exp(3)+3*x^2+48*x)*exp(-1/2*x^2*exp(3)+1/2*x^3+12*x^2)/(2*exp(2)+2),x, algorithm="fricas")

[Out]

e^(1/2*x^3 - 1/2*x^2*e^3 + 12*x^2)/(e^2 + 1)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {1}{2} \left (24 x^2-e^3 x^2+x^3\right )} \left (48 x-2 e^3 x+3 x^2\right )}{2+2 e^2} \, dx=\frac {e^{\frac {x^{3}}{2} - \frac {x^{2} e^{3}}{2} + 12 x^{2}}}{1 + e^{2}} \]

[In]

integrate((-2*x*exp(3)+3*x**2+48*x)*exp(-1/2*x**2*exp(3)+1/2*x**3+12*x**2)/(2*exp(2)+2),x)

[Out]

exp(x**3/2 - x**2*exp(3)/2 + 12*x**2)/(1 + exp(2))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {1}{2} \left (24 x^2-e^3 x^2+x^3\right )} \left (48 x-2 e^3 x+3 x^2\right )}{2+2 e^2} \, dx=\frac {e^{\left (\frac {1}{2} \, x^{3} - \frac {1}{2} \, x^{2} e^{3} + 12 \, x^{2}\right )}}{e^{2} + 1} \]

[In]

integrate((-2*x*exp(3)+3*x^2+48*x)*exp(-1/2*x^2*exp(3)+1/2*x^3+12*x^2)/(2*exp(2)+2),x, algorithm="maxima")

[Out]

e^(1/2*x^3 - 1/2*x^2*e^3 + 12*x^2)/(e^2 + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {1}{2} \left (24 x^2-e^3 x^2+x^3\right )} \left (48 x-2 e^3 x+3 x^2\right )}{2+2 e^2} \, dx=\frac {e^{\left (\frac {1}{2} \, x^{3} - \frac {1}{2} \, x^{2} e^{3} + 12 \, x^{2}\right )}}{e^{2} + 1} \]

[In]

integrate((-2*x*exp(3)+3*x^2+48*x)*exp(-1/2*x^2*exp(3)+1/2*x^3+12*x^2)/(2*exp(2)+2),x, algorithm="giac")

[Out]

e^(1/2*x^3 - 1/2*x^2*e^3 + 12*x^2)/(e^2 + 1)

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {1}{2} \left (24 x^2-e^3 x^2+x^3\right )} \left (48 x-2 e^3 x+3 x^2\right )}{2+2 e^2} \, dx=\frac {{\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^3}{2}}\,{\mathrm {e}}^{\frac {x^3}{2}}\,{\mathrm {e}}^{12\,x^2}}{{\mathrm {e}}^2+1} \]

[In]

int((exp(12*x^2 - (x^2*exp(3))/2 + x^3/2)*(48*x - 2*x*exp(3) + 3*x^2))/(2*exp(2) + 2),x)

[Out]

(exp(-(x^2*exp(3))/2)*exp(x^3/2)*exp(12*x^2))/(exp(2) + 1)