\(\int \frac {1}{5} e^{3+\frac {1}{5} (-10+e^3 (-e^{43/16} x-x^2))} (-e^{43/16}-2 x) \, dx\) [8779]

Optimal result

Integrand size = 43, antiderivative size = 19 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx=e^{-2-\frac {1}{5} e^3 x \left (e^{43/16}+x\right )} \]

[Out]

exp(-2-1/15*exp(3)*(3*x+3*exp(43/16))*x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 2276, 2268} \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx=e^{-\frac {1}{5} e^3 x^2-\frac {1}{5} e^{91/16} x-2} \]

[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 2268

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 2276

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx=e^{-2-\frac {1}{5} e^{91/16} x-\frac {e^3 x^2}{5}} \]

[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)

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 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	\({\mathrm e}^{\frac {\left (-x \,{\mathrm e}^{\frac {43}{16}}-x^{2}\right ) {\mathrm e}^{3}}{5}-2}\)	\(19\)
parallelrisch	\({\mathrm e}^{\frac {\left (-x \,{\mathrm e}^{\frac {43}{16}}-x^{2}\right ) {\mathrm e}^{3}}{5}-2}\)	\(19\)
parts	\(-\frac {{\mathrm e}^{3} \sqrt {\pi }\, {\mathrm e}^{-2+\frac {{\mathrm e}^{3} {\mathrm e}^{\frac {43}{8}}}{20}} \sqrt {5}\, {\mathrm e}^{-\frac {3}{2}} \operatorname {erf}\left (\frac {\sqrt {5}\, {\mathrm e}^{\frac {3}{2}} x}{5}+\frac {{\mathrm e}^{3} {\mathrm e}^{\frac {43}{16}} \sqrt {5}\, {\mathrm e}^{-\frac {3}{2}}}{10}\right ) {\mathrm e}^{\frac {43}{16}}}{10}-\frac {{\mathrm e}^{3} \sqrt {\pi }\, {\mathrm e}^{-2+\frac {{\mathrm e}^{3} {\mathrm e}^{\frac {43}{8}}}{20}} \sqrt {5}\, {\mathrm e}^{-\frac {3}{2}} \operatorname {erf}\left (\frac {\sqrt {5}\, {\mathrm e}^{\frac {3}{2}} x}{5}+\frac {{\mathrm e}^{3} {\mathrm e}^{\frac {43}{16}} \sqrt {5}\, {\mathrm e}^{-\frac {3}{2}}}{10}\right ) x}{5}+{\mathrm e}^{-2+\frac {{\mathrm e}^{3} {\mathrm e}^{\frac {43}{8}}}{20}} {\mathrm e}^{-3} {\mathrm e}^{3} \left (\operatorname {erf}\left (\frac {\sqrt {5}\, {\mathrm e}^{\frac {3}{2}} x}{5}+\frac {{\mathrm e}^{3} {\mathrm e}^{\frac {43}{16}} \sqrt {5}\, {\mathrm e}^{-\frac {3}{2}}}{10}\right ) \left (\frac {\sqrt {5}\, {\mathrm e}^{\frac {3}{2}} x}{5}+\frac {{\mathrm e}^{3} {\mathrm e}^{\frac {43}{16}} \sqrt {5}\, {\mathrm e}^{-\frac {3}{2}}}{10}\right ) \sqrt {\pi }+{\mathrm e}^{-\left (\frac {\sqrt {5}\, {\mathrm e}^{\frac {3}{2}} x}{5}+\frac {{\mathrm e}^{3} {\mathrm e}^{\frac {43}{16}} \sqrt {5}\, {\mathrm e}^{-\frac {3}{2}}}{10}\right )^{2}}\right )\)	\(196\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx=e^{\left (-\frac {1}{5} \, x^{2} e^{3} - \frac {1}{5} \, x e^{\frac {91}{16}} - 2\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx=e^{\left (- \frac {x^{2}}{5} - \frac {x e^{\frac {43}{16}}}{5}\right ) e^{3} - 2} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx=e^{\left (-\frac {1}{5} \, {\left (x^{2} + x e^{\frac {43}{16}}\right )} e^{3} - 2\right )} \]

[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)

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.26 (sec) , antiderivative size = 96, normalized size of antiderivative = 5.05 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx=-\frac {1}{10} \, \sqrt {5} \sqrt {\pi } \operatorname {erf}\left (\frac {1}{10} \, \sqrt {5} {\left (2 \, x + e^{\frac {43}{16}}\right )} e^{\frac {3}{2}}\right ) e^{\left (\frac {1}{80} \, {\left (4 \, e^{\frac {91}{8}} + 295 \, e^{3}\right )} e^{\left (-3\right )} - \frac {3}{2}\right )} + \frac {1}{10} \, {\left (\sqrt {5} \sqrt {\pi } \operatorname {erf}\left (\frac {1}{10} \, \sqrt {5} {\left (2 \, x + e^{\frac {43}{16}}\right )} e^{\frac {3}{2}}\right ) e^{\left (\frac {1}{20} \, {\left (e^{\frac {91}{8}} + 20 \, e^{3}\right )} e^{\left (-3\right )} + \frac {67}{16}\right )} + 10 \, e^{\left (-\frac {1}{5} \, x^{2} e^{3} - \frac {1}{5} \, x e^{\frac {91}{16}} + 1\right )}\right )} e^{\left (-3\right )} \]

[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]

-1/10*sqrt(5)*sqrt(pi)*erf(1/10*sqrt(5)*(2*x + e^(43/16))*e^(3/2))*e^(1/80*(4*e^(91/8) + 295*e^3)*e^(-3) - 3/2
) + 1/10*(sqrt(5)*sqrt(pi)*erf(1/10*sqrt(5)*(2*x + e^(43/16))*e^(3/2))*e^(1/20*(e^(91/8) + 20*e^3)*e^(-3) + 67
/16) + 10*e^(-1/5*x^2*e^3 - 1/5*x*e^(91/16) + 1))*e^(-3)

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx={\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^3}{5}}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{91/16}}{5}} \]

[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)