3.88.79 \(\int \frac {1}{5} e^{3+\frac {1}{5} (-10+e^3 (-e^{43/16} x-x^2))} (-e^{43/16}-2 x) \, dx\)

Optimal. Leaf size=19 \[ e^{-2-\frac {1}{5} e^3 x \left (e^{43/16}+x\right )} \]

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Rubi [A]  time = 0.05, antiderivative size = 24, normalized size of antiderivative = 1.26, number of steps used = 3, number of rules used = 3, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 2244, 2236} \begin {gather*} e^{-\frac {1}{5} e^3 x^2-\frac {1}{5} e^{91/16} x-2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(3 + (-10 + E^3*(-(E^(43/16)*x) - x^2))/5)*(-E^(43/16) - 2*x))/5,x]

[Out]

E^(-2 - (E^(91/16)*x)/5 - (E^3*x^2)/5)

Rule 12

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

Rule 2236

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]

Rule 2244

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \exp \left (3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )\right ) \left (-e^{43/16}-2 x\right ) \, dx\\ &=\frac {1}{5} \int e^{1-\frac {1}{5} e^{91/16} x-\frac {e^3 x^2}{5}} \left (-e^{43/16}-2 x\right ) \, dx\\ &=e^{-2-\frac {1}{5} e^{91/16} x-\frac {e^3 x^2}{5}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 24, normalized size = 1.26 \begin {gather*} e^{-2-\frac {1}{5} e^{91/16} x-\frac {e^3 x^2}{5}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(3 + (-10 + E^3*(-(E^(43/16)*x) - x^2))/5)*(-E^(43/16) - 2*x))/5,x]

[Out]

E^(-2 - (E^(91/16)*x)/5 - (E^3*x^2)/5)

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fricas [A]  time = 0.52, size = 15, normalized size = 0.79 \begin {gather*} e^{\left (-\frac {1}{5} \, x^{2} e^{3} - \frac {1}{5} \, x e^{\frac {91}{16}} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(43/16)-2*x)*exp(3)*exp(1/5*(-x*exp(43/16)-x^2)*exp(3)-2),x, algorithm="fricas")

[Out]

e^(-1/5*x^2*e^3 - 1/5*x*e^(91/16) - 2)

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giac [A]  time = 0.22, size = 15, normalized size = 0.79 \begin {gather*} e^{\left (-\frac {1}{5} \, x^{2} e^{3} - \frac {1}{5} \, x e^{\frac {91}{16}} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(43/16)-2*x)*exp(3)*exp(1/5*(-x*exp(43/16)-x^2)*exp(3)-2),x, algorithm="giac")

[Out]

e^(-1/5*x^2*e^3 - 1/5*x*e^(91/16) - 2)

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maple [A]  time = 0.04, size = 16, normalized size = 0.84

method	result	size
risch	\({\mathrm e}^{-\frac {x \,{\mathrm e}^{\frac {91}{16}}}{5}-\frac {x^{2} {\mathrm e}^{3}}{5}-2}\)	\(16\)
gosper	\({\mathrm e}^{-\frac {{\mathrm e}^{3} {\mathrm e}^{\frac {43}{16}} x}{5}-\frac {x^{2} {\mathrm e}^{3}}{5}-2}\)	\(18\)
derivativedivides	\({\mathrm e}^{\frac {\left (-x \,{\mathrm e}^{\frac {43}{16}}-x^{2}\right ) {\mathrm e}^{3}}{5}-2}\)	\(19\)
default	\({\mathrm e}^{\frac {\left (-x \,{\mathrm e}^{\frac {43}{16}}-x^{2}\right ) {\mathrm e}^{3}}{5}-2}\)	\(19\)
norman	\({\mathrm e}^{\frac {\left (-x \,{\mathrm e}^{\frac {43}{16}}-x^{2}\right ) {\mathrm e}^{3}}{5}-2}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-exp(43/16)-2*x)*exp(3)*exp(1/5*(-x*exp(43/16)-x^2)*exp(3)-2),x,method=_RETURNVERBOSE)

[Out]

exp(-1/5*x*exp(91/16)-1/5*x^2*exp(3)-2)

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maxima [A]  time = 0.35, size = 15, normalized size = 0.79 \begin {gather*} e^{\left (-\frac {1}{5} \, {\left (x^{2} + x e^{\frac {43}{16}}\right )} e^{3} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(43/16)-2*x)*exp(3)*exp(1/5*(-x*exp(43/16)-x^2)*exp(3)-2),x, algorithm="maxima")

[Out]

e^(-1/5*(x^2 + x*e^(43/16))*e^3 - 2)

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mupad [B]  time = 0.15, size = 17, normalized size = 0.89 \begin {gather*} {\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^3}{5}}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{91/16}}{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3)*exp(- (exp(3)*(x*exp(43/16) + x^2))/5 - 2)*(2*x + exp(43/16)))/5,x)

[Out]

exp(-(x^2*exp(3))/5)*exp(-2)*exp(-(x*exp(91/16))/5)

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sympy [A]  time = 0.13, size = 20, normalized size = 1.05 \begin {gather*} e^{\left (- \frac {x^{2}}{5} - \frac {x e^{\frac {43}{16}}}{5}\right ) e^{3} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-exp(43/16)-2*x)*exp(3)*exp(1/5*(-x*exp(43/16)-x**2)*exp(3)-2),x)

[Out]

exp((-x**2/5 - x*exp(43/16)/5)*exp(3) - 2)

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