Integrand size = 171, antiderivative size = 35 \[ \int \frac {e^{\frac {5+\left (-2-36 x^2-e^{2 x} x^2-24 x^3-4 x^4+e^x \left (12 x^2+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (4+e^x\right )\right )}{\log \left (\frac {1}{2} \left (4+e^x\right )\right )}} \left (-5 e^x+\left (-288 x-288 x^2-64 x^3+e^{3 x} \left (-2 x-2 x^2\right )+e^x \left (24 x+24 x^2\right )+e^{2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )\right )}{\left (4+e^x\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )} \, dx=e^{-2-x^2 \left (6-e^x+2 x\right )^2+\frac {5}{\log \left (\frac {1}{2} \left (4+e^x\right )\right )}} \]
Time = 0.38 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.29 \[ \int \frac {e^{\frac {5+\left (-2-36 x^2-e^{2 x} x^2-24 x^3-4 x^4+e^x \left (12 x^2+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (4+e^x\right )\right )}{\log \left (\frac {1}{2} \left (4+e^x\right )\right )}} \left (-5 e^x+\left (-288 x-288 x^2-64 x^3+e^{3 x} \left (-2 x-2 x^2\right )+e^x \left (24 x+24 x^2\right )+e^{2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )\right )}{\left (4+e^x\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )} \, dx=e^{-2-\left (-6+e^x\right )^2 x^2+4 \left (-6+e^x\right ) x^3-4 x^4+\frac {5}{\log \left (\frac {1}{2} \left (4+e^x\right )\right )}} \]
Integrate[(E^((5 + (-2 - 36*x^2 - E^(2*x)*x^2 - 24*x^3 - 4*x^4 + E^x*(12*x ^2 + 4*x^3))*Log[(4 + E^x)/2])/Log[(4 + E^x)/2])*(-5*E^x + (-288*x - 288*x ^2 - 64*x^3 + E^(3*x)*(-2*x - 2*x^2) + E^x*(24*x + 24*x^2) + E^(2*x)*(16*x + 16*x^2 + 4*x^3))*Log[(4 + E^x)/2]^2))/((4 + E^x)*Log[(4 + E^x)/2]^2),x]
Time = 16.96 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.40, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {7293, 7239, 7257}
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.
| \(\displaystyle \int \frac {\left (\left (-64 x^3-288 x^2+e^{3 x} \left (-2 x^2-2 x\right )+e^x \left (24 x^2+24 x\right )+e^{2 x} \left (4 x^3+16 x^2+16 x\right )-288 x\right ) \log ^2\left (\frac {1}{2} \left (e^x+4\right )\right )-5 e^x\right ) \exp \left (\frac {\left (-4 x^4-24 x^3-e^{2 x} x^2-36 x^2+e^x \left (4 x^3+12 x^2\right )-2\right ) \log \left (\frac {1}{2} \left (e^x+4\right )\right )+5}{\log \left (\frac {1}{2} \left (e^x+4\right )\right )}\right )}{\left (e^x+4\right ) \log ^2\left (\frac {1}{2} \left (e^x+4\right )\right )} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {\left (-16 x^3 \log ^2\left (\frac {1}{2} \left (e^x+4\right )\right )-72 x^2 \log ^2\left (\frac {1}{2} \left (e^x+4\right )\right )-72 x \log ^2\left (\frac {1}{2} \left (e^x+4\right )\right )-5\right ) \exp \left (\frac {\left (-4 x^4-24 x^3-e^{2 x} x^2-36 x^2+e^x \left (4 x^3+12 x^2\right )-2\right ) \log \left (\frac {1}{2} \left (e^x+4\right )\right )+5}{\log \left (\frac {1}{2} \left (e^x+4\right )\right )}\right )}{\log ^2\left (\frac {1}{2} \left (e^x+4\right )\right )}+\frac {20 \exp \left (\frac {\left (-4 x^4-24 x^3-e^{2 x} x^2-36 x^2+e^x \left (4 x^3+12 x^2\right )-2\right ) \log \left (\frac {1}{2} \left (e^x+4\right )\right )+5}{\log \left (\frac {1}{2} \left (e^x+4\right )\right )}\right )}{\left (e^x+4\right ) \log ^2\left (\frac {1}{2} \left (e^x+4\right )\right )}-2 x (x+1) \exp \left (\frac {\left (-4 x^4-24 x^3-e^{2 x} x^2-36 x^2+e^x \left (4 x^3+12 x^2\right )-2\right ) \log \left (\frac {1}{2} \left (e^x+4\right )\right )+5}{\log \left (\frac {1}{2} \left (e^x+4\right )\right )}+2 x\right )+4 x \left (x^2+6 x+6\right ) \exp \left (\frac {\left (-4 x^4-24 x^3-e^{2 x} x^2-36 x^2+e^x \left (4 x^3+12 x^2\right )-2\right ) \log \left (\frac {1}{2} \left (e^x+4\right )\right )+5}{\log \left (\frac {1}{2} \left (e^x+4\right )\right )}+x\right )\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (-2 \left (e^x+4\right ) x \left (-2 e^x \left (x^2+6 x+6\right )+4 \left (2 x^2+9 x+9\right )+e^{2 x} (x+1)\right ) \log ^2\left (\frac {1}{2} \left (e^x+4\right )\right )-5 e^x\right ) \exp \left (-4 x^4+4 \left (e^x-6\right ) x^3-\left (e^x-6\right )^2 x^2+\frac {5}{\log \left (\frac {1}{2} \left (e^x+4\right )\right )}-2\right )}{\left (e^x+4\right ) \log ^2\left (\frac {1}{2} \left (e^x+4\right )\right )}dx\) |
| \(\Big \downarrow \) 7257 |
| \(\displaystyle \exp \left (-4 x^4-4 \left (6-e^x\right ) x^3-\left (6-e^x\right )^2 x^2+\frac {5}{\log \left (\frac {1}{2} \left (e^x+4\right )\right )}-2\right )\) |
Int[(E^((5 + (-2 - 36*x^2 - E^(2*x)*x^2 - 24*x^3 - 4*x^4 + E^x*(12*x^2 + 4 *x^3))*Log[(4 + E^x)/2])/Log[(4 + E^x)/2])*(-5*E^x + (-288*x - 288*x^2 - 6 4*x^3 + E^(3*x)*(-2*x - 2*x^2) + E^x*(24*x + 24*x^2) + E^(2*x)*(16*x + 16* x^2 + 4*x^3))*Log[(4 + E^x)/2]^2))/((4 + E^x)*Log[(4 + E^x)/2]^2),x]
3.3.86.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Leaf count of result is larger than twice the leaf count of optimal. \(61\) vs. \(2(30)=60\).
Time = 3.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.77
| method | result | size |
| parallelrisch | \({\mathrm e}^{\frac {\left (-{\mathrm e}^{2 x} x^{2}+\left (4 x^{3}+12 x^{2}\right ) {\mathrm e}^{x}-4 x^{4}-24 x^{3}-36 x^{2}-2\right ) \ln \left (\frac {{\mathrm e}^{x}}{2}+2\right )+5}{\ln \left (\frac {{\mathrm e}^{x}}{2}+2\right )}}\) | \(62\) |
| risch | \({\mathrm e}^{\frac {4 \ln \left (\frac {{\mathrm e}^{x}}{2}+2\right ) {\mathrm e}^{x} x^{3}-4 \ln \left (\frac {{\mathrm e}^{x}}{2}+2\right ) x^{4}+12 \ln \left (\frac {{\mathrm e}^{x}}{2}+2\right ) {\mathrm e}^{x} x^{2}-\ln \left (\frac {{\mathrm e}^{x}}{2}+2\right ) {\mathrm e}^{2 x} x^{2}-24 \ln \left (\frac {{\mathrm e}^{x}}{2}+2\right ) x^{3}-36 \ln \left (\frac {{\mathrm e}^{x}}{2}+2\right ) x^{2}-2 \ln \left (\frac {{\mathrm e}^{x}}{2}+2\right )+5}{\ln \left (\frac {{\mathrm e}^{x}}{2}+2\right )}}\) | \(103\) |
int((((-2*x^2-2*x)*exp(x)^3+(4*x^3+16*x^2+16*x)*exp(x)^2+(24*x^2+24*x)*exp (x)-64*x^3-288*x^2-288*x)*ln(1/2*exp(x)+2)^2-5*exp(x))*exp(((-exp(x)^2*x^2 +(4*x^3+12*x^2)*exp(x)-4*x^4-24*x^3-36*x^2-2)*ln(1/2*exp(x)+2)+5)/ln(1/2*e xp(x)+2))/(exp(x)+4)/ln(1/2*exp(x)+2)^2,x,method=_RETURNVERBOSE)
exp(((-exp(x)^2*x^2+(4*x^3+12*x^2)*exp(x)-4*x^4-24*x^3-36*x^2-2)*ln(1/2*ex p(x)+2)+5)/ln(1/2*exp(x)+2))
Time = 0.24 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.71 \[ \int \frac {e^{\frac {5+\left (-2-36 x^2-e^{2 x} x^2-24 x^3-4 x^4+e^x \left (12 x^2+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (4+e^x\right )\right )}{\log \left (\frac {1}{2} \left (4+e^x\right )\right )}} \left (-5 e^x+\left (-288 x-288 x^2-64 x^3+e^{3 x} \left (-2 x-2 x^2\right )+e^x \left (24 x+24 x^2\right )+e^{2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )\right )}{\left (4+e^x\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )} \, dx=e^{\left (-\frac {{\left (4 \, x^{4} + 24 \, x^{3} + x^{2} e^{\left (2 \, x\right )} + 36 \, x^{2} - 4 \, {\left (x^{3} + 3 \, x^{2}\right )} e^{x} + 2\right )} \log \left (\frac {1}{2} \, e^{x} + 2\right ) - 5}{\log \left (\frac {1}{2} \, e^{x} + 2\right )}\right )} \]
integrate((((-2*x^2-2*x)*exp(x)^3+(4*x^3+16*x^2+16*x)*exp(x)^2+(24*x^2+24* x)*exp(x)-64*x^3-288*x^2-288*x)*log(1/2*exp(x)+2)^2-5*exp(x))*exp(((-exp(x )^2*x^2+(4*x^3+12*x^2)*exp(x)-4*x^4-24*x^3-36*x^2-2)*log(1/2*exp(x)+2)+5)/ log(1/2*exp(x)+2))/(exp(x)+4)/log(1/2*exp(x)+2)^2,x, algorithm=\
e^(-((4*x^4 + 24*x^3 + x^2*e^(2*x) + 36*x^2 - 4*(x^3 + 3*x^2)*e^x + 2)*log (1/2*e^x + 2) - 5)/log(1/2*e^x + 2))
Timed out. \[ \int \frac {e^{\frac {5+\left (-2-36 x^2-e^{2 x} x^2-24 x^3-4 x^4+e^x \left (12 x^2+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (4+e^x\right )\right )}{\log \left (\frac {1}{2} \left (4+e^x\right )\right )}} \left (-5 e^x+\left (-288 x-288 x^2-64 x^3+e^{3 x} \left (-2 x-2 x^2\right )+e^x \left (24 x+24 x^2\right )+e^{2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )\right )}{\left (4+e^x\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )} \, dx=\text {Timed out} \]
integrate((((-2*x**2-2*x)*exp(x)**3+(4*x**3+16*x**2+16*x)*exp(x)**2+(24*x* *2+24*x)*exp(x)-64*x**3-288*x**2-288*x)*ln(1/2*exp(x)+2)**2-5*exp(x))*exp( ((-exp(x)**2*x**2+(4*x**3+12*x**2)*exp(x)-4*x**4-24*x**3-36*x**2-2)*ln(1/2 *exp(x)+2)+5)/ln(1/2*exp(x)+2))/(exp(x)+4)/ln(1/2*exp(x)+2)**2,x)
Time = 0.54 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.57 \[ \int \frac {e^{\frac {5+\left (-2-36 x^2-e^{2 x} x^2-24 x^3-4 x^4+e^x \left (12 x^2+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (4+e^x\right )\right )}{\log \left (\frac {1}{2} \left (4+e^x\right )\right )}} \left (-5 e^x+\left (-288 x-288 x^2-64 x^3+e^{3 x} \left (-2 x-2 x^2\right )+e^x \left (24 x+24 x^2\right )+e^{2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )\right )}{\left (4+e^x\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )} \, dx=e^{\left (-4 \, x^{4} + 4 \, x^{3} e^{x} - 24 \, x^{3} - x^{2} e^{\left (2 \, x\right )} + 12 \, x^{2} e^{x} - 36 \, x^{2} - \frac {5}{\log \left (2\right ) - \log \left (e^{x} + 4\right )} - 2\right )} \]
integrate((((-2*x^2-2*x)*exp(x)^3+(4*x^3+16*x^2+16*x)*exp(x)^2+(24*x^2+24* x)*exp(x)-64*x^3-288*x^2-288*x)*log(1/2*exp(x)+2)^2-5*exp(x))*exp(((-exp(x )^2*x^2+(4*x^3+12*x^2)*exp(x)-4*x^4-24*x^3-36*x^2-2)*log(1/2*exp(x)+2)+5)/ log(1/2*exp(x)+2))/(exp(x)+4)/log(1/2*exp(x)+2)^2,x, algorithm=\
e^(-4*x^4 + 4*x^3*e^x - 24*x^3 - x^2*e^(2*x) + 12*x^2*e^x - 36*x^2 - 5/(lo g(2) - log(e^x + 4)) - 2)
Exception generated. \[ \int \frac {e^{\frac {5+\left (-2-36 x^2-e^{2 x} x^2-24 x^3-4 x^4+e^x \left (12 x^2+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (4+e^x\right )\right )}{\log \left (\frac {1}{2} \left (4+e^x\right )\right )}} \left (-5 e^x+\left (-288 x-288 x^2-64 x^3+e^{3 x} \left (-2 x-2 x^2\right )+e^x \left (24 x+24 x^2\right )+e^{2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )\right )}{\left (4+e^x\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )} \, dx=\text {Exception raised: TypeError} \]
integrate((((-2*x^2-2*x)*exp(x)^3+(4*x^3+16*x^2+16*x)*exp(x)^2+(24*x^2+24* x)*exp(x)-64*x^3-288*x^2-288*x)*log(1/2*exp(x)+2)^2-5*exp(x))*exp(((-exp(x )^2*x^2+(4*x^3+12*x^2)*exp(x)-4*x^4-24*x^3-36*x^2-2)*log(1/2*exp(x)+2)+5)/ log(1/2*exp(x)+2))/(exp(x)+4)/log(1/2*exp(x)+2)^2,x, algorithm=\
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{-8000,[0,3,3]%%%}+%%%{-36000,[0,3,2]%%%}+%%%{-36000,[0,3,1 ]%%%} / %
Time = 9.73 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.69 \[ \int \frac {e^{\frac {5+\left (-2-36 x^2-e^{2 x} x^2-24 x^3-4 x^4+e^x \left (12 x^2+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (4+e^x\right )\right )}{\log \left (\frac {1}{2} \left (4+e^x\right )\right )}} \left (-5 e^x+\left (-288 x-288 x^2-64 x^3+e^{3 x} \left (-2 x-2 x^2\right )+e^x \left (24 x+24 x^2\right )+e^{2 x} \left (16 x+16 x^2+4 x^3\right )\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )\right )}{\left (4+e^x\right ) \log ^2\left (\frac {1}{2} \left (4+e^x\right )\right )} \, dx={\mathrm {e}}^{\frac {5}{\ln \left (\frac {{\mathrm {e}}^x}{2}+2\right )}}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{4\,x^3\,{\mathrm {e}}^x}\,{\mathrm {e}}^{12\,x^2\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-4\,x^4}\,{\mathrm {e}}^{-24\,x^3}\,{\mathrm {e}}^{-36\,x^2}\,{\mathrm {e}}^{-x^2\,{\mathrm {e}}^{2\,x}} \]
int(-(exp(-(log(exp(x)/2 + 2)*(x^2*exp(2*x) - exp(x)*(12*x^2 + 4*x^3) + 36 *x^2 + 24*x^3 + 4*x^4 + 2) - 5)/log(exp(x)/2 + 2))*(5*exp(x) + log(exp(x)/ 2 + 2)^2*(288*x + exp(3*x)*(2*x + 2*x^2) - exp(2*x)*(16*x + 16*x^2 + 4*x^3 ) - exp(x)*(24*x + 24*x^2) + 288*x^2 + 64*x^3)))/(log(exp(x)/2 + 2)^2*(exp (x) + 4)),x)