3.4.81 \(\int \frac {1}{2} e^{-2 x} (-e^{e^{-2 x}}+e^{2 x} (36 x-6 x^2)) \, dx\)

Optimal. Leaf size=22 \[ -2+\frac {1}{4} e^{e^{-2 x}}+(9-x) x^2 \]

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Rubi [A]  time = 0.09, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {12, 6688, 2282, 2194, 43} \begin {gather*} -x^3+9 x^2+\frac {1}{4} e^{e^{-2 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^E^(-2*x) + E^(2*x)*(36*x - 6*x^2))/(2*E^(2*x)),x]

[Out]

E^E^(-2*x)/4 + 9*x^2 - x^3

Rule 12

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[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} &=\frac {1}{2} \int e^{-2 x} \left (-e^{e^{-2 x}}+e^{2 x} \left (36 x-6 x^2\right )\right ) \, dx\\ &=\frac {1}{2} \int \left (-e^{e^{-2 x}-2 x}-6 (-6+x) x\right ) \, dx\\ &=-\left (\frac {1}{2} \int e^{e^{-2 x}-2 x} \, dx\right )-3 \int (-6+x) x \, dx\\ &=\frac {1}{4} \operatorname {Subst}\left (\int e^x \, dx,x,e^{-2 x}\right )-3 \int \left (-6 x+x^2\right ) \, dx\\ &=\frac {1}{4} e^{e^{-2 x}}+9 x^2-x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 22, normalized size = 1.00 \begin {gather*} \frac {1}{4} e^{e^{-2 x}}+9 x^2-x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^E^(-2*x) + E^(2*x)*(36*x - 6*x^2))/(2*E^(2*x)),x]

[Out]

E^E^(-2*x)/4 + 9*x^2 - x^3

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fricas [A]  time = 0.56, size = 18, normalized size = 0.82 \begin {gather*} -x^{3} + 9 \, x^{2} + \frac {1}{4} \, e^{\left (e^{\left (-2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-exp(1/exp(x)^2)+(-6*x^2+36*x)*exp(x)^2)/exp(x)^2,x, algorithm="fricas")

[Out]

-x^3 + 9*x^2 + 1/4*e^(e^(-2*x))

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giac [B]  time = 0.36, size = 36, normalized size = 1.64 \begin {gather*} -\frac {1}{4} \, {\left (4 \, x^{3} e^{\left (-2 \, x\right )} - 36 \, x^{2} e^{\left (-2 \, x\right )} - e^{\left (-2 \, x + e^{\left (-2 \, x\right )}\right )}\right )} e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-exp(1/exp(x)^2)+(-6*x^2+36*x)*exp(x)^2)/exp(x)^2,x, algorithm="giac")

[Out]

-1/4*(4*x^3*e^(-2*x) - 36*x^2*e^(-2*x) - e^(-2*x + e^(-2*x)))*e^(2*x)

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maple [A]  time = 0.02, size = 19, normalized size = 0.86

method	result	size
default	\(-x^{3}+9 x^{2}+\frac {{\mathrm e}^{{\mathrm e}^{-2 x}}}{4}\)	\(19\)
risch	\(-x^{3}+9 x^{2}+\frac {{\mathrm e}^{{\mathrm e}^{-2 x}}}{4}\)	\(19\)
norman	\(\left (9 \,{\mathrm e}^{2 x} x^{2}-{\mathrm e}^{2 x} x^{3}+\frac {{\mathrm e}^{2 x} {\mathrm e}^{{\mathrm e}^{-2 x}}}{4}\right ) {\mathrm e}^{-2 x}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-exp(1/exp(x)^2)+(-6*x^2+36*x)*exp(x)^2)/exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

-x^3+9*x^2+1/4*exp(1/exp(x)^2)

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maxima [A]  time = 0.78, size = 18, normalized size = 0.82 \begin {gather*} -x^{3} + 9 \, x^{2} + \frac {1}{4} \, e^{\left (e^{\left (-2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-exp(1/exp(x)^2)+(-6*x^2+36*x)*exp(x)^2)/exp(x)^2,x, algorithm="maxima")

[Out]

-x^3 + 9*x^2 + 1/4*e^(e^(-2*x))

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mupad [B]  time = 0.43, size = 18, normalized size = 0.82 \begin {gather*} \frac {{\mathrm {e}}^{{\mathrm {e}}^{-2\,x}}}{4}+9\,x^2-x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-2*x)*(exp(exp(-2*x))/2 - (exp(2*x)*(36*x - 6*x^2))/2),x)

[Out]

exp(exp(-2*x))/4 + 9*x^2 - x^3

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sympy [A]  time = 0.15, size = 17, normalized size = 0.77 \begin {gather*} - x^{3} + 9 x^{2} + \frac {e^{e^{- 2 x}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-exp(1/exp(x)**2)+(-6*x**2+36*x)*exp(x)**2)/exp(x)**2,x)

[Out]

-x**3 + 9*x**2 + exp(exp(-2*x))/4

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