Integrand size = 131, antiderivative size = 28 \[ \int \frac {-e^{-2 x} x+\left (-3 x^3-2 x^4+e^{e^9} \left (3 x^2+2 x^3\right )\right ) \log \left (4 e^{e^9}-4 x\right )+\left (e^{e^9-2 x}-e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right ) \log \left (\log \left (4 e^{e^9}-4 x\right )\right )}{\left (2 e^{e^9-2 x}-2 e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right )} \, dx=\frac {1}{2} x \left (e^{2 x} x^2+\log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\right ) \]
Time = 0.35 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-e^{-2 x} x+\left (-3 x^3-2 x^4+e^{e^9} \left (3 x^2+2 x^3\right )\right ) \log \left (4 e^{e^9}-4 x\right )+\left (e^{e^9-2 x}-e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right ) \log \left (\log \left (4 e^{e^9}-4 x\right )\right )}{\left (2 e^{e^9-2 x}-2 e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right )} \, dx=\frac {1}{2} x \left (e^{2 x} x^2+\log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\right ) \]
Integrate[(-(x/E^(2*x)) + (-3*x^3 - 2*x^4 + E^E^9*(3*x^2 + 2*x^3))*Log[4*E ^E^9 - 4*x] + (E^(E^9 - 2*x) - x/E^(2*x))*Log[4*E^E^9 - 4*x]*Log[Log[4*E^E ^9 - 4*x]])/((2*E^(E^9 - 2*x) - (2*x)/E^(2*x))*Log[4*E^E^9 - 4*x]),x]
Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(28)=56\).
Time = 0.87 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {7239, 27, 2009}
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.
\(\displaystyle \int \frac {\left (-2 x^4-3 x^3+e^{e^9} \left (2 x^3+3 x^2\right )\right ) \log \left (4 e^{e^9}-4 x\right )-e^{-2 x} x+\left (e^{e^9-2 x}-e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right ) \log \left (\log \left (4 e^{e^9}-4 x\right )\right )}{\left (2 e^{e^9-2 x}-2 e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {1}{2} \left (e^{2 x} (2 x+3) x^2-\frac {x}{\left (e^{e^9}-x\right ) \log \left (4 \left (e^{e^9}-x\right )\right )}+\log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \left (e^{2 x} (2 x+3) x^2-\frac {x}{\left (e^{e^9}-x\right ) \log \left (4 \left (e^{e^9}-x\right )\right )}+\log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (e^{2 x} x^3-\left (e^{e^9}-x\right ) \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )+e^{e^9} \log \left (\log \left (4 \left (e^{e^9}-x\right )\right )\right )\right )\) |
Int[(-(x/E^(2*x)) + (-3*x^3 - 2*x^4 + E^E^9*(3*x^2 + 2*x^3))*Log[4*E^E^9 - 4*x] + (E^(E^9 - 2*x) - x/E^(2*x))*Log[4*E^E^9 - 4*x]*Log[Log[4*E^E^9 - 4 *x]])/((2*E^(E^9 - 2*x) - (2*x)/E^(2*x))*Log[4*E^E^9 - 4*x]),x]
3.13.61.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.78 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89
method | result | size |
risch | \(\frac {x \ln \left (\ln \left (4 \,{\mathrm e}^{{\mathrm e}^{9}}-4 x \right )\right )}{2}+\frac {{\mathrm e}^{2 x} x^{3}}{2}\) | \(25\) |
parallelrisch | \(\frac {\left (\ln \left (\ln \left (4 \,{\mathrm e}^{{\mathrm e}^{9}}-4 x \right )\right ) {\mathrm e}^{-2 x} x +x^{3}\right ) {\mathrm e}^{2 x}}{2}\) | \(32\) |
default | \(\frac {{\mathrm e}^{2 x} x^{3}}{2}+\frac {\operatorname {Ei}_{1}\left (-\ln \left (4 \,{\mathrm e}^{{\mathrm e}^{9}}-4 x \right )\right )}{8}+\frac {{\mathrm e}^{{\mathrm e}^{9}} \ln \left (\ln \left (4 \,{\mathrm e}^{{\mathrm e}^{9}}-4 x \right )\right )}{2}+\frac {\ln \left (2 \ln \left (2\right )+\ln \left ({\mathrm e}^{{\mathrm e}^{9}}-x \right )\right ) x}{2}-\frac {\operatorname {Ei}_{1}\left (-2 \ln \left (2\right )-\ln \left ({\mathrm e}^{{\mathrm e}^{9}}-x \right )\right )}{8}-\frac {{\mathrm e}^{{\mathrm e}^{9}} \ln \left (2 \ln \left (2\right )+\ln \left ({\mathrm e}^{{\mathrm e}^{9}}-x \right )\right )}{2}\) | \(100\) |
parts | \(\frac {{\mathrm e}^{2 x} x^{3}}{2}+\frac {\operatorname {Ei}_{1}\left (-\ln \left (4 \,{\mathrm e}^{{\mathrm e}^{9}}-4 x \right )\right )}{8}+\frac {{\mathrm e}^{{\mathrm e}^{9}} \ln \left (\ln \left (4 \,{\mathrm e}^{{\mathrm e}^{9}}-4 x \right )\right )}{2}+\frac {\ln \left (2 \ln \left (2\right )+\ln \left ({\mathrm e}^{{\mathrm e}^{9}}-x \right )\right ) x}{2}-\frac {\operatorname {Ei}_{1}\left (-2 \ln \left (2\right )-\ln \left ({\mathrm e}^{{\mathrm e}^{9}}-x \right )\right )}{8}-\frac {{\mathrm e}^{{\mathrm e}^{9}} \ln \left (2 \ln \left (2\right )+\ln \left ({\mathrm e}^{{\mathrm e}^{9}}-x \right )\right )}{2}\) | \(100\) |
int(((exp(-x)^2*exp(exp(9))-x*exp(-x)^2)*ln(4*exp(exp(9))-4*x)*ln(ln(4*exp (exp(9))-4*x))+((2*x^3+3*x^2)*exp(exp(9))-2*x^4-3*x^3)*ln(4*exp(exp(9))-4* x)-x*exp(-x)^2)/(2*exp(-x)^2*exp(exp(9))-2*x*exp(-x)^2)/ln(4*exp(exp(9))-4 *x),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {-e^{-2 x} x+\left (-3 x^3-2 x^4+e^{e^9} \left (3 x^2+2 x^3\right )\right ) \log \left (4 e^{e^9}-4 x\right )+\left (e^{e^9-2 x}-e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right ) \log \left (\log \left (4 e^{e^9}-4 x\right )\right )}{\left (2 e^{e^9-2 x}-2 e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right )} \, dx=\frac {1}{2} \, {\left (x^{3} e^{\left (e^{9}\right )} + x e^{\left (-2 \, x + e^{9}\right )} \log \left (\log \left (-4 \, x + 4 \, e^{\left (e^{9}\right )}\right )\right )\right )} e^{\left (2 \, x - e^{9}\right )} \]
integrate(((exp(-x)^2*exp(exp(9))-x*exp(-x)^2)*log(4*exp(exp(9))-4*x)*log( log(4*exp(exp(9))-4*x))+((2*x^3+3*x^2)*exp(exp(9))-2*x^4-3*x^3)*log(4*exp( exp(9))-4*x)-x*exp(-x)^2)/(2*exp(-x)^2*exp(exp(9))-2*x*exp(-x)^2)/log(4*ex p(exp(9))-4*x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (26) = 52\).
Time = 0.39 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89 \[ \int \frac {-e^{-2 x} x+\left (-3 x^3-2 x^4+e^{e^9} \left (3 x^2+2 x^3\right )\right ) \log \left (4 e^{e^9}-4 x\right )+\left (e^{e^9-2 x}-e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right ) \log \left (\log \left (4 e^{e^9}-4 x\right )\right )}{\left (2 e^{e^9-2 x}-2 e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right )} \, dx=\frac {x^{3} e^{2 x}}{2} + \left (\frac {x}{2} - \frac {e^{e^{9}}}{4}\right ) \log {\left (\log {\left (- 4 x + 4 e^{e^{9}} \right )} \right )} + \frac {e^{e^{9}} \log {\left (\log {\left (- 4 x + 4 e^{e^{9}} \right )} \right )}}{4} \]
integrate(((exp(-x)**2*exp(exp(9))-x*exp(-x)**2)*ln(4*exp(exp(9))-4*x)*ln( ln(4*exp(exp(9))-4*x))+((2*x**3+3*x**2)*exp(exp(9))-2*x**4-3*x**3)*ln(4*ex p(exp(9))-4*x)-x*exp(-x)**2)/(2*exp(-x)**2*exp(exp(9))-2*x*exp(-x)**2)/ln( 4*exp(exp(9))-4*x),x)
x**3*exp(2*x)/2 + (x/2 - exp(exp(9))/4)*log(log(-4*x + 4*exp(exp(9)))) + e xp(exp(9))*log(log(-4*x + 4*exp(exp(9))))/4
Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {-e^{-2 x} x+\left (-3 x^3-2 x^4+e^{e^9} \left (3 x^2+2 x^3\right )\right ) \log \left (4 e^{e^9}-4 x\right )+\left (e^{e^9-2 x}-e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right ) \log \left (\log \left (4 e^{e^9}-4 x\right )\right )}{\left (2 e^{e^9-2 x}-2 e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right )} \, dx=\frac {1}{2} \, x^{3} e^{\left (2 \, x\right )} + \frac {1}{2} \, x \log \left (i \, \pi + 2 \, \log \left (2\right ) + \log \left (x - e^{\left (e^{9}\right )}\right )\right ) \]
integrate(((exp(-x)^2*exp(exp(9))-x*exp(-x)^2)*log(4*exp(exp(9))-4*x)*log( log(4*exp(exp(9))-4*x))+((2*x^3+3*x^2)*exp(exp(9))-2*x^4-3*x^3)*log(4*exp( exp(9))-4*x)-x*exp(-x)^2)/(2*exp(-x)^2*exp(exp(9))-2*x*exp(-x)^2)/log(4*ex p(exp(9))-4*x),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {-e^{-2 x} x+\left (-3 x^3-2 x^4+e^{e^9} \left (3 x^2+2 x^3\right )\right ) \log \left (4 e^{e^9}-4 x\right )+\left (e^{e^9-2 x}-e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right ) \log \left (\log \left (4 e^{e^9}-4 x\right )\right )}{\left (2 e^{e^9-2 x}-2 e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right )} \, dx=\frac {1}{2} \, x^{3} e^{\left (2 \, x\right )} + \frac {1}{2} \, x \log \left ({\left | \log \left (-4 \, x + 4 \, e^{\left (e^{9}\right )}\right ) \right |}\right ) \]
integrate(((exp(-x)^2*exp(exp(9))-x*exp(-x)^2)*log(4*exp(exp(9))-4*x)*log( log(4*exp(exp(9))-4*x))+((2*x^3+3*x^2)*exp(exp(9))-2*x^4-3*x^3)*log(4*exp( exp(9))-4*x)-x*exp(-x)^2)/(2*exp(-x)^2*exp(exp(9))-2*x*exp(-x)^2)/log(4*ex p(exp(9))-4*x),x, algorithm=\
Time = 9.52 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {-e^{-2 x} x+\left (-3 x^3-2 x^4+e^{e^9} \left (3 x^2+2 x^3\right )\right ) \log \left (4 e^{e^9}-4 x\right )+\left (e^{e^9-2 x}-e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right ) \log \left (\log \left (4 e^{e^9}-4 x\right )\right )}{\left (2 e^{e^9-2 x}-2 e^{-2 x} x\right ) \log \left (4 e^{e^9}-4 x\right )} \, dx=\frac {x\,\ln \left (\ln \left (4\,{\mathrm {e}}^{{\mathrm {e}}^9}-4\,x\right )\right )}{2}+\frac {x^3\,{\mathrm {e}}^{2\,x}}{2} \]