Integrand size = 68, antiderivative size = 21 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx=e^{-\frac {2}{\left (-2+e^3\right ) \left (-\frac {x}{2}+\log (5)\right )}} \] Output:
exp(-2/(-2+exp(3))/(ln(5)-1/2*x))
Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx=e^{\frac {4}{\left (-2+e^3\right ) (x-2 \log (5))}} \] Input:
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]
Output:
E^(4/((-2 + E^3)*(x - 2*Log[5])))
Time = 0.40 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6, 27, 25, 7239, 27, 2638}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int -\frac {4 \exp \left (-\frac {4}{-e^3 x+2 x+\left (2 e^3-4\right ) \log (5)}\right )}{e^3 x^2-2 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (4 e^3-8\right ) \log ^2(5)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int -\frac {4 \exp \left (-\frac {4}{-e^3 x+2 x+\left (2 e^3-4\right ) \log (5)}\right )}{\left (e^3-2\right ) x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (4 e^3-8\right ) \log ^2(5)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -4 \int -\frac {e^{-\frac {4}{-e^3 x+2 x-2 \left (2-e^3\right ) \log (5)}}}{\left (2-e^3\right ) x^2-4 \left (2-e^3\right ) \log (5) x+4 \left (2-e^3\right ) \log ^2(5)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle 4 \int \frac {e^{-\frac {4}{-e^3 x+2 x-2 \left (2-e^3\right ) \log (5)}}}{\left (2-e^3\right ) x^2-4 \left (2-e^3\right ) \log (5) x+4 \left (2-e^3\right ) \log ^2(5)}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle 4 \int \frac {e^{\frac {4}{\left (-2+e^3\right ) (x-2 \log (5))}}}{\left (2-e^3\right ) (x-2 \log (5))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 \int \frac {e^{-\frac {4}{\left (2-e^3\right ) (x-2 \log (5))}}}{(x-2 \log (5))^2}dx}{2-e^3}\) |
\(\Big \downarrow \) 2638 |
\(\displaystyle e^{-\frac {4}{\left (2-e^3\right ) (x-2 \log (5))}}\) |
Input:
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)*Log[5]^2)),x]
Output:
E^(-4/((2 - E^3)*(x - 2*Log[5])))
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.30 (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 | \({\mathrm e}^{-\frac {4}{2 \ln \left (5\right ) {\mathrm e}^{3}-x \,{\mathrm e}^{3}-4 \ln \left (5\right )+2 x}}\) | \(25\) |
parallelrisch | \(\frac {{\mathrm e}^{-\frac {4}{2 \ln \left (5\right ) {\mathrm e}^{3}-x \,{\mathrm e}^{3}-4 \ln \left (5\right )+2 x}} {\mathrm e}^{3}-2 \,{\mathrm e}^{-\frac {4}{2 \ln \left (5\right ) {\mathrm e}^{3}-x \,{\mathrm e}^{3}-4 \ln \left (5\right )+2 x}}}{{\mathrm e}^{3}-2}\) | \(62\) |
norman | \(\frac {-x \,{\mathrm e}^{-\frac {4}{\left (2 \,{\mathrm e}^{3}-4\right ) \ln \left (5\right )-x \,{\mathrm e}^{3}+2 x}}+2 \ln \left (5\right ) {\mathrm e}^{-\frac {4}{\left (2 \,{\mathrm e}^{3}-4\right ) \ln \left (5\right )-x \,{\mathrm e}^{3}+2 x}}}{2 \ln \left (5\right )-x}\) | \(66\) |
orering | \(\frac {\left (2 \ln \left (5\right )-x \right )^{2} \left ({\mathrm e}^{3}-2\right ) {\mathrm e}^{-\frac {4}{\left (2 \,{\mathrm e}^{3}-4\right ) \ln \left (5\right )-x \,{\mathrm e}^{3}+2 x}}}{\left (4 \,{\mathrm e}^{3}-8\right ) \ln \left (5\right )^{2}+\left (-4 x \,{\mathrm e}^{3}+8 x \right ) \ln \left (5\right )+x^{2} {\mathrm e}^{3}-2 x^{2}}\) | \(76\) |
Input:
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)
Output:
exp(4/(x-2*ln(5))/(exp(3)-2))
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx=e^{\left (\frac {4}{x e^{3} - 2 \, {\left (e^{3} - 2\right )} \log \left (5\right ) - 2 \, x}\right )} \] Input:
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*exp(3)-2*x^2),x, algorithm="fricas")
Output:
e^(4/(x*e^3 - 2*(e^3 - 2)*log(5) - 2*x))
Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx=e^{- \frac {4}{- x e^{3} + 2 x + \left (-4 + 2 e^{3}\right ) \log {\left (5 \right )}}} \] Input:
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)
Output:
exp(-4/(-x*exp(3) + 2*x + (-4 + 2*exp(3))*log(5)))
Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx=e^{\left (\frac {4}{x {\left (e^{3} - 2\right )} - 2 \, {\left (e^{3} - 2\right )} \log \left (5\right )}\right )} \] Input:
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*exp(3)-2*x^2),x, algorithm="maxima")
Output:
e^(4/(x*(e^3 - 2) - 2*(e^3 - 2)*log(5)))
Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx=e^{\left (\frac {4}{x e^{3} - 2 \, e^{3} \log \left (5\right ) - 2 \, x + 4 \, \log \left (5\right )}\right )} \] Input:
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*exp(3)-2*x^2),x, algorithm="giac")
Output:
e^(4/(x*e^3 - 2*e^3*log(5) - 2*x + 4*log(5)))
Time = 3.49 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx={\mathrm {e}}^{\frac {4}{\left (x-2\,\ln \left (5\right )\right )\,\left ({\mathrm {e}}^3-2\right )}} \] Input:
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)
Output:
exp(4/((x - 2*log(5))*(exp(3) - 2)))
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int -\frac {4 e^{-\frac {4}{2 x-e^3 x+\left (-4+2 e^3\right ) \log (5)}}}{-2 x^2+e^3 x^2+\left (8 x-4 e^3 x\right ) \log (5)+\left (-8+4 e^3\right ) \log ^2(5)} \, dx=\frac {1}{e^{\frac {4}{2 \,\mathrm {log}\left (5\right ) e^{3}-4 \,\mathrm {log}\left (5\right )-e^{3} x +2 x}}} \] Input:
int(-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*exp(3)-2*x^2),x)
Output:
1/e**(4/(2*log(5)*e**3 - 4*log(5) - e**3*x + 2*x))