\(\int -\frac {4 e^{-\frac {4}{2 x-e^3 x+(-4+2 e^3) \log (5)}}}{-2 x^2+e^3 x^2+(8 x-4 e^3 x) \log (5)+(-8+4 e^3) \log ^2(5)} \, dx\) [7523]

Optimal result

Integrand size = 68, antiderivative size = 21 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx=e^{-\frac {2}{\left (-2+e^3\right ) \left (-\frac {x}{2}+\log (5)\right )}} \]

[Out]

exp(-2/(exp(3)-2)/(ln(5)-1/2*x))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {6, 12, 6820, 2240} \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx=e^{-\frac {4}{\left (2-e^3\right ) (x-2 \log (5))}} \]

[In]

Int[-4/(E^(4/(2*x - E^3*x + (-4 + 2*E^3)*Log[5]))*(-2*x^2 + E^3*x^2 + (8*x - 4*E^3*x)*Log[5] + (-8 + 4*E^3)*Lo
g[5]^2)),x]

[Out]

E^(-4/((2 - E^3)*(x - 2*Log[5])))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

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

Rule 2240

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 6820

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

Rubi steps \begin{align*} \text {integral}& = \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{\left (-2+e^3\right ) x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx \\ & = -\left (4 \int \frac {e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{\left (-2+e^3\right ) x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx\right ) \\ & = -\left (4 \int \frac {e^{\frac {4}{\left (-2+e^3\right ) (x-2 \log (5))}}}{\left (-2+e^3\right ) (x-2 \log (5))^2} \, dx\right ) \\ & = \frac {4 \int \frac {e^{\frac {4}{\left (-2+e^3\right ) (x-2 \log (5))}}}{(x-2 \log (5))^2} \, dx}{2-e^3} \\ & = e^{-\frac {4}{\left (2-e^3\right ) (x-2 \log (5))}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx=e^{\frac {4}{\left (-2+e^3\right ) (x-2 \log (5))}} \]

[In]

Integrate[-4/(E^(4/(2*x - E^3*x + (-4 + 2*E^3)*Log[5]))*(-2*x^2 + E^3*x^2 + (8*x - 4*E^3*x)*Log[5] + (-8 + 4*E
^3)*Log[5]^2)),x]

[Out]

E^(4/((-2 + E^3)*(x - 2*Log[5])))

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\({\mathrm e}^{\frac {4}{\left (x -2 \ln \left (5\right )\right ) \left ({\mathrm e}^{3}-2\right )}}\)	\(18\)
default	\({\mathrm e}^{\frac {4}{\left (x -2 \ln \left (5\right )\right ) \left ({\mathrm e}^{3}-2\right )}}\)	\(18\)
risch	\({\mathrm e}^{-\frac {4}{\left (2 \ln \left (5\right )-x \right ) \left ({\mathrm e}^{3}-2\right )}}\)	\(20\)
gosper	\({\mathrm e}^{-\frac {4}{2 \,{\mathrm e}^{3} \ln \left (5\right )-x \,{\mathrm e}^{3}-4 \ln \left (5\right )+2 x}}\)	\(25\)
parallelrisch	\(\frac {{\mathrm e}^{3} {\mathrm e}^{-\frac {4}{2 \,{\mathrm e}^{3} \ln \left (5\right )-x \,{\mathrm e}^{3}-4 \ln \left (5\right )+2 x}}-2 \,{\mathrm e}^{-\frac {4}{2 \,{\mathrm e}^{3} \ln \left (5\right )-x \,{\mathrm e}^{3}-4 \ln \left (5\right )+2 x}}}{{\mathrm e}^{3}-2}\)	\(62\)
norman	\(\frac {-x \,{\mathrm e}^{-\frac {4}{\left (2 \,{\mathrm e}^{3}-4\right ) \ln \left (5\right )-x \,{\mathrm e}^{3}+2 x}}+2 \ln \left (5\right ) {\mathrm e}^{-\frac {4}{\left (2 \,{\mathrm e}^{3}-4\right ) \ln \left (5\right )-x \,{\mathrm e}^{3}+2 x}}}{2 \ln \left (5\right )-x}\)	\(66\)

[In]

int(-4*exp(-4/((2*exp(3)-4)*ln(5)-x*exp(3)+2*x))/((4*exp(3)-8)*ln(5)^2+(-4*x*exp(3)+8*x)*ln(5)+x^2*exp(3)-2*x^
2),x,method=_RETURNVERBOSE)

[Out]

exp(4/(x-2*ln(5))/(exp(3)-2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx=e^{\left (\frac {4}{x e^{3} - 2 \, {\left (e^{3} - 2\right )} \log \left (5\right ) - 2 \, x}\right )} \]

[In]

integrate(-4*exp(-4/((2*exp(3)-4)*log(5)-x*exp(3)+2*x))/((4*exp(3)-8)*log(5)^2+(-4*x*exp(3)+8*x)*log(5)+x^2*ex
p(3)-2*x^2),x, algorithm="fricas")

[Out]

e^(4/(x*e^3 - 2*(e^3 - 2)*log(5) - 2*x))

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx=e^{- \frac {4}{- x e^{3} + 2 x + \left (-4 + 2 e^{3}\right ) \log {\left (5 \right )}}} \]

[In]

integrate(-4*exp(-4/((2*exp(3)-4)*ln(5)-x*exp(3)+2*x))/((4*exp(3)-8)*ln(5)**2+(-4*x*exp(3)+8*x)*ln(5)+x**2*exp
(3)-2*x**2),x)

[Out]

exp(-4/(-x*exp(3) + 2*x + (-4 + 2*exp(3))*log(5)))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx=e^{\left (\frac {4}{x {\left (e^{3} - 2\right )} - 2 \, {\left (e^{3} - 2\right )} \log \left (5\right )}\right )} \]

[In]

integrate(-4*exp(-4/((2*exp(3)-4)*log(5)-x*exp(3)+2*x))/((4*exp(3)-8)*log(5)^2+(-4*x*exp(3)+8*x)*log(5)+x^2*ex
p(3)-2*x^2),x, algorithm="maxima")

[Out]

e^(4/(x*(e^3 - 2) - 2*(e^3 - 2)*log(5)))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx=e^{\left (\frac {4}{x e^{3} - 2 \, e^{3} \log \left (5\right ) - 2 \, x + 4 \, \log \left (5\right )}\right )} \]

[In]

integrate(-4*exp(-4/((2*exp(3)-4)*log(5)-x*exp(3)+2*x))/((4*exp(3)-8)*log(5)^2+(-4*x*exp(3)+8*x)*log(5)+x^2*ex
p(3)-2*x^2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx={\mathrm {e}}^{\frac {4}{\left (x-2\,\ln \left (5\right )\right )\,\left ({\mathrm {e}}^3-2\right )}} \]

[In]

int(-(4*exp(-4/(2*x - x*exp(3) + log(5)*(2*exp(3) - 4))))/(log(5)*(8*x - 4*x*exp(3)) + log(5)^2*(4*exp(3) - 8)
+ x^2*exp(3) - 2*x^2),x)

[Out]

exp(4/((x - 2*log(5))*(exp(3) - 2)))