3.15.18 \(\int \frac {e^{-x-e^{2 x} (1-x-x^2)^2} (e^x (2-2 x-2 x^2)+e^{3 x} (1-x-x^2)^2 (-6 x^2-2 x^3))}{-3 x^3+3 x^4+3 x^5} \, dx\)

Optimal. Leaf size=28 \[ \frac {e^{-e^{2 x} \left (1-x-x^2\right )^2}}{3 x^2} \]

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Rubi [B]  time = 1.01, antiderivative size = 90, normalized size of antiderivative = 3.21, number of steps used = 4, number of rules used = 4, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {1594, 6688, 12, 2288} \begin {gather*} \frac {e^{2 x-e^{2 x} \left (-x^2-x+1\right )^2} \left (-x^3-4 x^2-2 x+3\right )}{3 x \left (e^{2 x} (2 x+1) \left (-x^2-x+1\right )-e^{2 x} \left (-x^2-x+1\right )^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-x - E^(2*x)*(1 - x - x^2)^2)*(E^x*(2 - 2*x - 2*x^2) + E^(3*x)*(1 - x - x^2)^2*(-6*x^2 - 2*x^3)))/(-3*
x^3 + 3*x^4 + 3*x^5),x]

[Out]

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

Rule 12

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

Rule 1594

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

Rule 2288

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]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x-e^{2 x} \left (1-x-x^2\right )^2} \left (e^x \left (2-2 x-2 x^2\right )+e^{3 x} \left (1-x-x^2\right )^2 \left (-6 x^2-2 x^3\right )\right )}{x^3 \left (-3+3 x+3 x^2\right )} \, dx\\ &=\int \frac {2 e^{-e^{2 x} \left (-1+x+x^2\right )^2} \left (-1-e^{2 x} x^2 \left (-3+2 x+4 x^2+x^3\right )\right )}{3 x^3} \, dx\\ &=\frac {2}{3} \int \frac {e^{-e^{2 x} \left (-1+x+x^2\right )^2} \left (-1-e^{2 x} x^2 \left (-3+2 x+4 x^2+x^3\right )\right )}{x^3} \, dx\\ &=\frac {e^{2 x-e^{2 x} \left (1-x-x^2\right )^2} \left (3-2 x-4 x^2-x^3\right )}{3 x \left (e^{2 x} (1+2 x) \left (1-x-x^2\right )-e^{2 x} \left (1-x-x^2\right )^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 24, normalized size = 0.86 \begin {gather*} \frac {e^{-e^{2 x} \left (-1+x+x^2\right )^2}}{3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-x - E^(2*x)*(1 - x - x^2)^2)*(E^x*(2 - 2*x - 2*x^2) + E^(3*x)*(1 - x - x^2)^2*(-6*x^2 - 2*x^3))
)/(-3*x^3 + 3*x^4 + 3*x^5),x]

[Out]

1/(3*E^(E^(2*x)*(-1 + x + x^2)^2)*x^2)

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fricas [A]  time = 0.78, size = 30, normalized size = 1.07 \begin {gather*} \frac {e^{\left (-{\left (x^{4} + 2 \, x^{3} - x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )}\right )}}{3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-6*x^2)*exp(x)*exp(log(-x^2-x+1)+x)^2+(-2*x^2-2*x+2)*exp(x))/(3*x^5+3*x^4-3*x^3)/exp(exp(log
(-x^2-x+1)+x)^2+x),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left ({\left (x^{3} + 3 \, x^{2}\right )} e^{\left (3 \, x + 2 \, \log \left (-x^{2} - x + 1\right )\right )} + {\left (x^{2} + x - 1\right )} e^{x}\right )} e^{\left (-x - e^{\left (2 \, x + 2 \, \log \left (-x^{2} - x + 1\right )\right )}\right )}}{3 \, {\left (x^{5} + x^{4} - x^{3}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-6*x^2)*exp(x)*exp(log(-x^2-x+1)+x)^2+(-2*x^2-2*x+2)*exp(x))/(3*x^5+3*x^4-3*x^3)/exp(exp(log
(-x^2-x+1)+x)^2+x),x, algorithm="giac")

[Out]

integrate(-2/3*((x^3 + 3*x^2)*e^(3*x + 2*log(-x^2 - x + 1)) + (x^2 + x - 1)*e^x)*e^(-x - e^(2*x + 2*log(-x^2 -
x + 1)))/(x^5 + x^4 - x^3), x)

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maple [A]  time = 0.45, size = 21, normalized size = 0.75

method	result	size
risch	\(\frac {{\mathrm e}^{-{\mathrm e}^{2 x} \left (x^{2}+x -1\right )^{2}}}{3 x^{2}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^3-6*x^2)*exp(x)*exp(ln(-x^2-x+1)+x)^2+(-2*x^2-2*x+2)*exp(x))/(3*x^5+3*x^4-3*x^3)/exp(exp(ln(-x^2-x+
1)+x)^2+x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.77, size = 46, normalized size = 1.64 \begin {gather*} \frac {e^{\left (-x^{4} e^{\left (2 \, x\right )} - 2 \, x^{3} e^{\left (2 \, x\right )} + x^{2} e^{\left (2 \, x\right )} + 2 \, x e^{\left (2 \, x\right )} - e^{\left (2 \, x\right )}\right )}}{3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^3-6*x^2)*exp(x)*exp(log(-x^2-x+1)+x)^2+(-2*x^2-2*x+2)*exp(x))/(3*x^5+3*x^4-3*x^3)/exp(exp(log
(-x^2-x+1)+x)^2+x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.15, size = 49, normalized size = 1.75 \begin {gather*} \frac {{\mathrm {e}}^{-{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{-2\,x^3\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{-x^4\,{\mathrm {e}}^{2\,x}}}{3\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- x - exp(2*x + 2*log(1 - x^2 - x)))*(exp(x)*(2*x + 2*x^2 - 2) + exp(2*x + 2*log(1 - x^2 - x))*exp(x
)*(6*x^2 + 2*x^3)))/(3*x^4 - 3*x^3 + 3*x^5),x)

[Out]

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

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sympy [A]  time = 0.31, size = 26, normalized size = 0.93 \begin {gather*} \frac {e^{x} e^{- x - \left (- x^{2} - x + 1\right )^{2} e^{2 x}}}{3 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**3-6*x**2)*exp(x)*exp(ln(-x**2-x+1)+x)**2+(-2*x**2-2*x+2)*exp(x))/(3*x**5+3*x**4-3*x**3)/exp(
exp(ln(-x**2-x+1)+x)**2+x),x)

[Out]

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

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