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

Optimal. Leaf size=21 \[ e^{-\frac {2}{\left (-2+e^3\right ) \left (-\frac {x}{2}+\log (5)\right )}} \]

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Rubi [A]  time = 0.13, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {6, 12, 6688, 2209} \begin {gather*} e^{-\frac {4}{\left (2-e^3\right ) (x-2 \log (5))}} \end {gather*}

Antiderivative was successfully verified.

[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 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 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} &=\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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 0.90 \begin {gather*} e^{\frac {4}{\left (-2+e^3\right ) (x-2 \log (5))}} \end {gather*}

Antiderivative was successfully verified.

[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])))

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fricas [A]  time = 0.59, size = 21, normalized size = 1.00 \begin {gather*} e^{\left (\frac {4}{x e^{3} - 2 \, {\left (e^{3} - 2\right )} \log \relax (5) - 2 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.24, size = 23, normalized size = 1.10 \begin {gather*} e^{\left (\frac {4}{x e^{3} - 2 \, e^{3} \log \relax (5) - 2 \, x + 4 \, \log \relax (5)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maple [A]  time = 0.33, size = 18, normalized size = 0.86

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.47, size = 20, normalized size = 0.95 \begin {gather*} e^{\left (\frac {4}{x {\left (e^{3} - 2\right )} - 2 \, {\left (e^{3} - 2\right )} \log \relax (5)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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mupad [B]  time = 5.02, size = 17, normalized size = 0.81 \begin {gather*} {\mathrm {e}}^{\frac {4}{\left (x-2\,\ln \relax (5)\right )\,\left ({\mathrm {e}}^3-2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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sympy [A]  time = 0.36, size = 22, normalized size = 1.05 \begin {gather*} e^{- \frac {4}{- x e^{3} + 2 x + \left (-4 + 2 e^{3}\right ) \log {\relax (5 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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