Integrand size = 89, antiderivative size = 28 \[ \int \frac {e^{e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )} \left (-8+e^x (8+24 x)\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{-4 x-3 x^2+e^x (16+12 x)} \, dx=e^{e^{\log ^2\left (\frac {-4 e^x+x}{2 (3+x+5 (1+x))}\right )}} \]
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^{e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )} \left (-8+e^x (8+24 x)\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{-4 x-3 x^2+e^x (16+12 x)} \, dx=e^{e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}} \]
Integrate[(E^(E^Log[(-4*E^x + x)/(16 + 12*x)]^2 + Log[(-4*E^x + x)/(16 + 1 2*x)]^2)*(-8 + E^x*(8 + 24*x))*Log[(-4*E^x + x)/(16 + 12*x)])/(-4*x - 3*x^ 2 + E^x*(16 + 12*x)),x]
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.
\(\displaystyle \int \frac {\left (e^x (24 x+8)-8\right ) \log \left (\frac {x-4 e^x}{12 x+16}\right ) \exp \left (\log ^2\left (\frac {x-4 e^x}{12 x+16}\right )+e^{\log ^2\left (\frac {x-4 e^x}{12 x+16}\right )}\right )}{-3 x^2-4 x+e^x (12 x+16)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {8 \left (3 e^x x+e^x-1\right ) \log \left (\frac {x-4 e^x}{12 x+16}\right ) \exp \left (\log ^2\left (\frac {x-4 e^x}{12 x+16}\right )+e^{\log ^2\left (\frac {x-4 e^x}{12 x+16}\right )}\right )}{\left (4 e^x-x\right ) (3 x+4)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 8 \int -\frac {\exp \left (\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )+e^{\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )}\right ) \left (-3 e^x x-e^x+1\right ) \log \left (-\frac {4 e^x-x}{4 (3 x+4)}\right )}{\left (4 e^x-x\right ) (3 x+4)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -8 \int \frac {\exp \left (\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )+e^{\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )}\right ) \left (-3 e^x x-e^x+1\right ) \log \left (-\frac {4 e^x-x}{4 (3 x+4)}\right )}{\left (4 e^x-x\right ) (3 x+4)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -8 \int \left (-\frac {\exp \left (\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )+e^{\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )}\right ) (x-1) \log \left (\frac {x-4 e^x}{4 (3 x+4)}\right )}{4 \left (4 e^x-x\right )}-\frac {\exp \left (\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )+e^{\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )}\right ) (3 x+1) \log \left (\frac {x-4 e^x}{4 (3 x+4)}\right )}{4 (3 x+4)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -8 \left (-\frac {1}{4} \int \exp \left (\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )+e^{\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )}\right ) \log \left (\frac {x-4 e^x}{4 (3 x+4)}\right )dx+\frac {1}{4} \int \frac {\exp \left (\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )+e^{\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )}\right ) \log \left (\frac {x-4 e^x}{4 (3 x+4)}\right )}{4 e^x-x}dx-\frac {1}{4} \int \frac {\exp \left (\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )+e^{\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )}\right ) x \log \left (\frac {x-4 e^x}{4 (3 x+4)}\right )}{4 e^x-x}dx+\frac {3}{4} \int \frac {\exp \left (\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )+e^{\log ^2\left (-\frac {4 e^x-x}{4 (3 x+4)}\right )}\right ) \log \left (\frac {x-4 e^x}{4 (3 x+4)}\right )}{3 x+4}dx\right )\) |
Int[(E^(E^Log[(-4*E^x + x)/(16 + 12*x)]^2 + Log[(-4*E^x + x)/(16 + 12*x)]^ 2)*(-8 + E^x*(8 + 24*x))*Log[(-4*E^x + x)/(16 + 12*x)])/(-4*x - 3*x^2 + E^ x*(16 + 12*x)),x]
3.17.88.3.1 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]]
Time = 8.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82
method | result | size |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{\ln \left (-\frac {4 \,{\mathrm e}^{x}-x}{4 \left (4+3 x \right )}\right )^{2}}}\) | \(23\) |
risch | \({\mathrm e}^{{\mathrm e}^{\frac {{\left (i \operatorname {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right )\right ) \pi {\operatorname {csgn}\left (\frac {i \left (4 \,{\mathrm e}^{x}-x \right )}{x +\frac {4}{3}}\right )}^{2}+i \operatorname {csgn}\left (i \left (4 \,{\mathrm e}^{x}-x \right )\right ) \pi \,\operatorname {csgn}\left (\frac {i \left (4 \,{\mathrm e}^{x}-x \right )}{x +\frac {4}{3}}\right ) \operatorname {csgn}\left (\frac {i}{x +\frac {4}{3}}\right )-i \pi {\operatorname {csgn}\left (\frac {i \left (4 \,{\mathrm e}^{x}-x \right )}{x +\frac {4}{3}}\right )}^{3}-i \pi {\operatorname {csgn}\left (\frac {i \left (4 \,{\mathrm e}^{x}-x \right )}{x +\frac {4}{3}}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x +\frac {4}{3}}\right )+4 \ln \left (2\right )+2 \ln \left (x +\frac {4}{3}\right )-2 \ln \left (-4 \,{\mathrm e}^{x}+x \right )+2 \ln \left (3\right )\right )}^{2}}{4}}}\) | \(163\) |
int(((24*x+8)*exp(x)-8)*ln((-4*exp(x)+x)/(12*x+16))*exp(ln((-4*exp(x)+x)/( 12*x+16))^2)*exp(exp(ln((-4*exp(x)+x)/(12*x+16))^2))/((12*x+16)*exp(x)-3*x ^2-4*x),x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {e^{e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )} \left (-8+e^x (8+24 x)\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{-4 x-3 x^2+e^x (16+12 x)} \, dx=e^{\left (e^{\left (\log \left (\frac {x - 4 \, e^{x}}{4 \, {\left (3 \, x + 4\right )}}\right )^{2}\right )}\right )} \]
integrate(((24*x+8)*exp(x)-8)*log((-4*exp(x)+x)/(12*x+16))*exp(log((-4*exp (x)+x)/(12*x+16))^2)*exp(exp(log((-4*exp(x)+x)/(12*x+16))^2))/((12*x+16)*e xp(x)-3*x^2-4*x),x, algorithm=\
Timed out. \[ \int \frac {e^{e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )} \left (-8+e^x (8+24 x)\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{-4 x-3 x^2+e^x (16+12 x)} \, dx=\text {Timed out} \]
integrate(((24*x+8)*exp(x)-8)*ln((-4*exp(x)+x)/(12*x+16))*exp(ln((-4*exp(x )+x)/(12*x+16))**2)*exp(exp(ln((-4*exp(x)+x)/(12*x+16))**2))/((12*x+16)*ex p(x)-3*x**2-4*x),x)
Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (20) = 40\).
Time = 0.65 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.21 \[ \int \frac {e^{e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )} \left (-8+e^x (8+24 x)\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{-4 x-3 x^2+e^x (16+12 x)} \, dx=e^{\left (e^{\left (4 \, \log \left (2\right )^{2} + 4 \, \log \left (2\right ) \log \left (3 \, x + 4\right ) + \log \left (3 \, x + 4\right )^{2} - 4 \, \log \left (2\right ) \log \left (x - 4 \, e^{x}\right ) - 2 \, \log \left (3 \, x + 4\right ) \log \left (x - 4 \, e^{x}\right ) + \log \left (x - 4 \, e^{x}\right )^{2}\right )}\right )} \]
integrate(((24*x+8)*exp(x)-8)*log((-4*exp(x)+x)/(12*x+16))*exp(log((-4*exp (x)+x)/(12*x+16))^2)*exp(exp(log((-4*exp(x)+x)/(12*x+16))^2))/((12*x+16)*e xp(x)-3*x^2-4*x),x, algorithm=\
e^(e^(4*log(2)^2 + 4*log(2)*log(3*x + 4) + log(3*x + 4)^2 - 4*log(2)*log(x - 4*e^x) - 2*log(3*x + 4)*log(x - 4*e^x) + log(x - 4*e^x)^2))
\[ \int \frac {e^{e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )} \left (-8+e^x (8+24 x)\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{-4 x-3 x^2+e^x (16+12 x)} \, dx=\int { -\frac {8 \, {\left ({\left (3 \, x + 1\right )} e^{x} - 1\right )} e^{\left (\log \left (\frac {x - 4 \, e^{x}}{4 \, {\left (3 \, x + 4\right )}}\right )^{2} + e^{\left (\log \left (\frac {x - 4 \, e^{x}}{4 \, {\left (3 \, x + 4\right )}}\right )^{2}\right )}\right )} \log \left (\frac {x - 4 \, e^{x}}{4 \, {\left (3 \, x + 4\right )}}\right )}{3 \, x^{2} - 4 \, {\left (3 \, x + 4\right )} e^{x} + 4 \, x} \,d x } \]
integrate(((24*x+8)*exp(x)-8)*log((-4*exp(x)+x)/(12*x+16))*exp(log((-4*exp (x)+x)/(12*x+16))^2)*exp(exp(log((-4*exp(x)+x)/(12*x+16))^2))/((12*x+16)*e xp(x)-3*x^2-4*x),x, algorithm=\
integrate(-8*((3*x + 1)*e^x - 1)*e^(log(1/4*(x - 4*e^x)/(3*x + 4))^2 + e^( log(1/4*(x - 4*e^x)/(3*x + 4))^2))*log(1/4*(x - 4*e^x)/(3*x + 4))/(3*x^2 - 4*(3*x + 4)*e^x + 4*x), x)
Time = 11.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {e^{e^{\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )}+\log ^2\left (\frac {-4 e^x+x}{16+12 x}\right )} \left (-8+e^x (8+24 x)\right ) \log \left (\frac {-4 e^x+x}{16+12 x}\right )}{-4 x-3 x^2+e^x (16+12 x)} \, dx={\mathrm {e}}^{{\mathrm {e}}^{{\ln \left (\frac {x-4\,{\mathrm {e}}^x}{12\,x+16}\right )}^2}} \]