Integrand size = 52, antiderivative size = 25 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {e^x}{x+\frac {2 x}{4-\log (\log (5) (4+\log (25)))}} \]
Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {e^x (-4+\log (\log (5) (4+\log (25))))}{x (-6+\log (\log (5) (4+\log (25))))} \]
Integrate[(E^x*(4 - 4*x) + E^x*(-1 + x)*Log[4*Log[5] + Log[5]*Log[25]])/(- 6*x^2 + x^2*Log[4*Log[5] + Log[5]*Log[25]]),x]
Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {6, 27, 6, 27, 2627}
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.
\(\displaystyle \int \frac {e^x (4-4 x)+e^x (x-1) \log (4 \log (5)+\log (5) \log (25))}{x^2 \log (4 \log (5)+\log (5) \log (25))-6 x^2} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^x (4-4 x)+e^x (x-1) \log (4 \log (5)+\log (5) \log (25))}{x^2 (\log (4 \log (5)+\log (5) \log (25))-6)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {4 e^x (1-x)-e^x (1-x) \log (\log (5) (4+\log (25)))}{x^2}dx}{6-\log (\log (5) (4+\log (25)))}\) |
\(\Big \downarrow \) 6 |
\(\displaystyle -\frac {\int \frac {e^x (1-x) (4-\log (\log (5) (4+\log (25))))}{x^2}dx}{6-\log (\log (5) (4+\log (25)))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {(4-\log (\log (5) (4+\log (25)))) \int \frac {e^x (1-x)}{x^2}dx}{6-\log (\log (5) (4+\log (25)))}\) |
\(\Big \downarrow \) 2627 |
\(\displaystyle \frac {e^x (4-\log (\log (5) (4+\log (25))))}{x (6-\log (\log (5) (4+\log (25))))}\) |
Int[(E^x*(4 - 4*x) + E^x*(-1 + x)*Log[4*Log[5] + Log[5]*Log[25]])/(-6*x^2 + x^2*Log[4*Log[5] + Log[5]*Log[25]]),x]
3.10.67.3.1 Defintions of rubi rules used
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_)^(v_)*((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[g*(d + e*x)^(m + 1)*(F^v/(D[v, x]*e*Log[F])), x] /; FreeQ[{F, d, e, f , g, m}, x] && LinearQ[v, x] && EqQ[e*g*(m + 1) - D[v, x]*(e*f - d*g)*Log[F ], 0]
Time = 2.76 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32
method | result | size |
risch | \(\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+\ln \left (2+\ln \left (5\right )\right )-4\right ) {\mathrm e}^{x}}{x \left (\ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+\ln \left (2+\ln \left (5\right )\right )-6\right )}\) | \(33\) |
gosper | \(\frac {{\mathrm e}^{x} \left (\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-4\right )}{x \left (\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-6\right )}\) | \(37\) |
norman | \(\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-4\right ) {\mathrm e}^{x}}{\left (\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6\right ) x}\) | \(37\) |
parallelrisch | \(\frac {{\mathrm e}^{x} \ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-4 \,{\mathrm e}^{x}}{\left (\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-6\right ) x}\) | \(41\) |
meijerg | \(\frac {\left (\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-4\right ) \left (\ln \left (x \right )+i \pi -\ln \left (-x \right )-\operatorname {Ei}_{1}\left (-x \right )\right )}{\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-6}-\frac {\left (-\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )+4\right ) \left (\frac {1}{x}+1-\ln \left (x \right )-i \pi -\frac {2+2 x}{2 x}+\frac {{\mathrm e}^{x}}{x}+\ln \left (-x \right )+\operatorname {Ei}_{1}\left (-x \right )\right )}{\ln \left (2 \ln \left (5\right )^{2}+4 \ln \left (5\right )\right )-6}\) | \(125\) |
default | \(-\frac {\ln \left (2\right ) \operatorname {Ei}_{1}\left (-x \right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}-\frac {\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right ) \operatorname {Ei}_{1}\left (-x \right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}+\frac {-\frac {4 \,{\mathrm e}^{x}}{x}-4 \,\operatorname {Ei}_{1}\left (-x \right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}+\frac {4 \,\operatorname {Ei}_{1}\left (-x \right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}-\frac {\ln \left (2\right ) \left (-\frac {{\mathrm e}^{x}}{x}-\operatorname {Ei}_{1}\left (-x \right )\right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}-\frac {\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right ) \left (-\frac {{\mathrm e}^{x}}{x}-\operatorname {Ei}_{1}\left (-x \right )\right )}{\ln \left (2\right )+\ln \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )\right )-6}\) | \(194\) |
int(((-1+x)*exp(x)*ln(2*ln(5)^2+4*ln(5))+(4-4*x)*exp(x))/(x^2*ln(2*ln(5)^2 +4*ln(5))-6*x^2),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {e^{x} \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right ) - 4 \, e^{x}}{x \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right ) - 6 \, x} \]
integrate(((-1+x)*exp(x)*log(2*log(5)^2+4*log(5))+(4-4*x)*exp(x))/(x^2*log (2*log(5)^2+4*log(5))-6*x^2),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.68 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {\left (-4 + \log {\left (\log {\left (5 \right )} \right )} + \log {\left (2 \right )} + \log {\left (\log {\left (5 \right )} + 2 \right )}\right ) e^{x}}{- 6 x + x \log {\left (\log {\left (5 \right )} \right )} + x \log {\left (2 \right )} + x \log {\left (\log {\left (5 \right )} + 2 \right )}} \]
integrate(((-1+x)*exp(x)*ln(2*ln(5)**2+4*ln(5))+(4-4*x)*exp(x))/(x**2*ln(2 *ln(5)**2+4*ln(5))-6*x**2),x)
(-4 + log(log(5)) + log(2) + log(log(5) + 2))*exp(x)/(-6*x + x*log(log(5)) + x*log(2) + x*log(log(5) + 2))
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.31 (sec) , antiderivative size = 98, normalized size of antiderivative = 3.92 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {{\rm Ei}\left (x\right ) \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right )}{\log \left (2 \, {\left (\log \left (5\right ) + 2\right )} \log \left (5\right )\right ) - 6} - \frac {\Gamma \left (-1, -x\right ) \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right )}{\log \left (2 \, {\left (\log \left (5\right ) + 2\right )} \log \left (5\right )\right ) - 6} - \frac {4 \, {\rm Ei}\left (x\right )}{\log \left (2 \, {\left (\log \left (5\right ) + 2\right )} \log \left (5\right )\right ) - 6} + \frac {4 \, \Gamma \left (-1, -x\right )}{\log \left (2 \, {\left (\log \left (5\right ) + 2\right )} \log \left (5\right )\right ) - 6} \]
integrate(((-1+x)*exp(x)*log(2*log(5)^2+4*log(5))+(4-4*x)*exp(x))/(x^2*log (2*log(5)^2+4*log(5))-6*x^2),x, algorithm=\
Ei(x)*log(2*log(5)^2 + 4*log(5))/(log(2*(log(5) + 2)*log(5)) - 6) - gamma( -1, -x)*log(2*log(5)^2 + 4*log(5))/(log(2*(log(5) + 2)*log(5)) - 6) - 4*Ei (x)/(log(2*(log(5) + 2)*log(5)) - 6) + 4*gamma(-1, -x)/(log(2*(log(5) + 2) *log(5)) - 6)
Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {e^{x} \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right ) - 4 \, e^{x}}{x \log \left (2 \, \log \left (5\right )^{2} + 4 \, \log \left (5\right )\right ) - 6 \, x} \]
integrate(((-1+x)*exp(x)*log(2*log(5)^2+4*log(5))+(4-4*x)*exp(x))/(x^2*log (2*log(5)^2+4*log(5))-6*x^2),x, algorithm=\
Time = 9.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {e^x (4-4 x)+e^x (-1+x) \log (4 \log (5)+\log (5) \log (25))}{-6 x^2+x^2 \log (4 \log (5)+\log (5) \log (25))} \, dx=\frac {{\mathrm {e}}^x\,\left (\ln \left (\ln \left (625\right )+2\,{\ln \left (5\right )}^2\right )-4\right )}{x\,\left (\ln \left (\ln \left (625\right )+2\,{\ln \left (5\right )}^2\right )-6\right )} \]