3.71.56 \(\int \frac {1}{3} e^{-2 x} (-2 e^{-5+e^{\frac {e^{-2 x}}{3}}+\frac {e^{-2 x}}{3}}+3 e^{25-8 x+x^2} (10-2 x)) \, dx\)

Optimal. Leaf size=25 \[ e^{-5+e^{\frac {e^{-2 x}}{3}}}-e^{(-5+x)^2} \]

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Rubi [A]  time = 0.32, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.113, Rules used = {12, 6742, 2282, 2194, 2227, 2209} \begin {gather*} e^{e^{\frac {e^{-2 x}}{3}}-5}-e^{(x-5)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*E^(-5 + E^(1/(3*E^(2*x))) + 1/(3*E^(2*x))) + 3*E^(25 - 8*x + x^2)*(10 - 2*x))/(3*E^(2*x)),x]

[Out]

E^(-5 + E^(1/(3*E^(2*x)))) - E^(-5 + x)^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 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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2227

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, 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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int e^{-2 x} \left (-2 e^{-5+e^{\frac {e^{-2 x}}{3}}+\frac {e^{-2 x}}{3}}+3 e^{25-8 x+x^2} (10-2 x)\right ) \, dx\\ &=\frac {1}{3} \int \left (-2 e^{-5+e^{\frac {e^{-2 x}}{3}}+\frac {e^{-2 x}}{3}-2 x}-6 e^{25-10 x+x^2} (-5+x)\right ) \, dx\\ &=-\left (\frac {2}{3} \int e^{-5+e^{\frac {e^{-2 x}}{3}}+\frac {e^{-2 x}}{3}-2 x} \, dx\right )-2 \int e^{25-10 x+x^2} (-5+x) \, dx\\ &=\frac {1}{3} \operatorname {Subst}\left (\int e^{\frac {1}{3} \left (-15+3 e^{x/3}+x\right )} \, dx,x,e^{-2 x}\right )-2 \int e^{(-5+x)^2} (-5+x) \, dx\\ &=-e^{(-5+x)^2}+\operatorname {Subst}\left (\int e^{-5+e^x+x} \, dx,x,\frac {e^{-2 x}}{3}\right )\\ &=-e^{(-5+x)^2}+\operatorname {Subst}\left (\int e^{-5+x} \, dx,x,e^{\frac {e^{-2 x}}{3}}\right )\\ &=e^{-5+e^{\frac {e^{-2 x}}{3}}}-e^{(-5+x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 25, normalized size = 1.00 \begin {gather*} e^{-5+e^{\frac {e^{-2 x}}{3}}}-e^{(-5+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*E^(-5 + E^(1/(3*E^(2*x))) + 1/(3*E^(2*x))) + 3*E^(25 - 8*x + x^2)*(10 - 2*x))/(3*E^(2*x)),x]

[Out]

E^(-5 + E^(1/(3*E^(2*x)))) - E^(-5 + x)^2

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fricas [B]  time = 0.56, size = 57, normalized size = 2.28 \begin {gather*} -{\left (e^{\left (x^{2} - 10 \, x + e^{\left (-2 \, x - \log \relax (3)\right )} + 25\right )} - e^{\left (e^{\left (-2 \, x - \log \relax (3)\right )} + e^{\left (e^{\left (-2 \, x - \log \relax (3)\right )}\right )} - 5\right )}\right )} e^{\left (-e^{\left (-2 \, x - \log \relax (3)\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(1/exp(2*x+log(3)))*exp(exp(1/exp(2*x+log(3)))-5)+(-2*x+10)*exp(x^2-10*x+25)*exp(2*x+log(3)))
/exp(2*x+log(3)),x, algorithm="fricas")

[Out]

-(e^(x^2 - 10*x + e^(-2*x - log(3)) + 25) - e^(e^(-2*x - log(3)) + e^(e^(-2*x - log(3))) - 5))*e^(-e^(-2*x - l
og(3)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -2 \, {\left ({\left (x - 5\right )} e^{\left (x^{2} - 8 \, x + \log \relax (3) + 25\right )} + e^{\left (e^{\left (-2 \, x - \log \relax (3)\right )} + e^{\left (e^{\left (-2 \, x - \log \relax (3)\right )}\right )} - 5\right )}\right )} e^{\left (-2 \, x - \log \relax (3)\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(1/exp(2*x+log(3)))*exp(exp(1/exp(2*x+log(3)))-5)+(-2*x+10)*exp(x^2-10*x+25)*exp(2*x+log(3)))
/exp(2*x+log(3)),x, algorithm="giac")

[Out]

integrate(-2*((x - 5)*e^(x^2 - 8*x + log(3) + 25) + e^(e^(-2*x - log(3)) + e^(e^(-2*x - log(3))) - 5))*e^(-2*x
- log(3)), x)

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maple [A]  time = 0.04, size = 20, normalized size = 0.80

method	result	size
risch	\(-{\mathrm e}^{\left (x -5\right )^{2}}+{\mathrm e}^{{\mathrm e}^{\frac {{\mathrm e}^{-2 x}}{3}}-5}\)	\(20\)
default	\(-{\mathrm e}^{x^{2}-10 x +25}+{\mathrm e}^{{\mathrm e}^{\frac {{\mathrm e}^{-2 x}}{3}}} {\mathrm e}^{-5}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*exp(1/exp(2*x+ln(3)))*exp(exp(1/exp(2*x+ln(3)))-5)+(-2*x+10)*exp(x^2-10*x+25)*exp(2*x+ln(3)))/exp(2*x+
ln(3)),x,method=_RETURNVERBOSE)

[Out]

-exp((x-5)^2)+exp(exp(1/3*exp(-2*x))-5)

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maxima [C]  time = 0.44, size = 59, normalized size = 2.36 \begin {gather*} -\frac {5 \, \sqrt {\pi } {\left (x - 5\right )} {\left (\operatorname {erf}\left (\sqrt {-{\left (x - 5\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (x - 5\right )}^{2}}} - 5 i \, \sqrt {\pi } \operatorname {erf}\left (i \, x - 5 i\right ) - e^{\left ({\left (x - 5\right )}^{2}\right )} + e^{\left (e^{\left (\frac {1}{3} \, e^{\left (-2 \, x\right )}\right )} - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(1/exp(2*x+log(3)))*exp(exp(1/exp(2*x+log(3)))-5)+(-2*x+10)*exp(x^2-10*x+25)*exp(2*x+log(3)))
/exp(2*x+log(3)),x, algorithm="maxima")

[Out]

-5*sqrt(pi)*(x - 5)*(erf(sqrt(-(x - 5)^2)) - 1)/sqrt(-(x - 5)^2) - 5*I*sqrt(pi)*erf(I*x - 5*I) - e^((x - 5)^2)
+ e^(e^(1/3*e^(-2*x)) - 5)

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mupad [B]  time = 4.27, size = 24, normalized size = 0.96 \begin {gather*} {\mathrm {e}}^{-5}\,{\mathrm {e}}^{{\left ({\mathrm {e}}^{{\mathrm {e}}^{-2\,x}}\right )}^{1/3}}-{\mathrm {e}}^{-10\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(- 2*x - log(3))*(2*exp(exp(exp(- 2*x - log(3))) - 5)*exp(exp(- 2*x - log(3))) + exp(2*x + log(3))*exp
(x^2 - 10*x + 25)*(2*x - 10)),x)

[Out]

exp(-5)*exp(exp(exp(-2*x))^(1/3)) - exp(-10*x)*exp(x^2)*exp(25)

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sympy [A]  time = 0.45, size = 22, normalized size = 0.88 \begin {gather*} e^{e^{\frac {e^{- 2 x}}{3}} - 5} - e^{x^{2} - 10 x + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(1/exp(2*x+ln(3)))*exp(exp(1/exp(2*x+ln(3)))-5)+(-2*x+10)*exp(x**2-10*x+25)*exp(2*x+ln(3)))/e
xp(2*x+ln(3)),x)

[Out]

exp(exp(exp(-2*x)/3) - 5) - exp(x**2 - 10*x + 25)

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