Integrand size = 226, antiderivative size = 34 \begin {dmath*} \int \frac {4 x^4+24 x^5-24 x^6+\left (-8 x^3-48 x^4+48 x^5\right ) \log (4)+\left (4 x^2+24 x^3-24 x^4\right ) \log ^2(4)+e^x \left (-12 x^2-20 x^3+\left (8 x+20 x^2\right ) \log (4)\right )+\left (-8 x^4+8 x^5+\left (16 x^3-16 x^4\right ) \log (4)+\left (-8 x^2+8 x^3\right ) \log ^2(4)+e^x \left (8 x^2-8 x \log (4)\right )\right ) \log \left (\frac {e^x-x^2+x^3+\left (x-x^2\right ) \log (4)}{-x+\log (4)}\right )}{5 x^3-5 x^4+\left (-10 x^2+10 x^3\right ) \log (4)+\left (5 x-5 x^2\right ) \log ^2(4)+e^x (-5 x+5 \log (4))} \, dx=\frac {4}{5} x^2 \left (1+2 x-\log \left (x-x^2+\frac {e^x}{-x+\log (4)}\right )\right ) \end {dmath*}
Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \begin {dmath*} \int \frac {4 x^4+24 x^5-24 x^6+\left (-8 x^3-48 x^4+48 x^5\right ) \log (4)+\left (4 x^2+24 x^3-24 x^4\right ) \log ^2(4)+e^x \left (-12 x^2-20 x^3+\left (8 x+20 x^2\right ) \log (4)\right )+\left (-8 x^4+8 x^5+\left (16 x^3-16 x^4\right ) \log (4)+\left (-8 x^2+8 x^3\right ) \log ^2(4)+e^x \left (8 x^2-8 x \log (4)\right )\right ) \log \left (\frac {e^x-x^2+x^3+\left (x-x^2\right ) \log (4)}{-x+\log (4)}\right )}{5 x^3-5 x^4+\left (-10 x^2+10 x^3\right ) \log (4)+\left (5 x-5 x^2\right ) \log ^2(4)+e^x (-5 x+5 \log (4))} \, dx=\frac {4}{5} \left (x^2+2 x^3-x^2 \log \left (x-x^2-\frac {e^x}{x-\log (4)}\right )\right ) \end {dmath*}
Integrate[(4*x^4 + 24*x^5 - 24*x^6 + (-8*x^3 - 48*x^4 + 48*x^5)*Log[4] + ( 4*x^2 + 24*x^3 - 24*x^4)*Log[4]^2 + E^x*(-12*x^2 - 20*x^3 + (8*x + 20*x^2) *Log[4]) + (-8*x^4 + 8*x^5 + (16*x^3 - 16*x^4)*Log[4] + (-8*x^2 + 8*x^3)*L og[4]^2 + E^x*(8*x^2 - 8*x*Log[4]))*Log[(E^x - x^2 + x^3 + (x - x^2)*Log[4 ])/(-x + Log[4])])/(5*x^3 - 5*x^4 + (-10*x^2 + 10*x^3)*Log[4] + (5*x - 5*x ^2)*Log[4]^2 + E^x*(-5*x + 5*Log[4])),x]
Time = 4.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {7292, 27, 27, 7293, 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.
| \(\displaystyle \int \frac {-24 x^6+24 x^5+4 x^4+e^x \left (-20 x^3-12 x^2+\left (20 x^2+8 x\right ) \log (4)\right )+\left (48 x^5-48 x^4-8 x^3\right ) \log (4)+\left (-24 x^4+24 x^3+4 x^2\right ) \log ^2(4)+\left (8 x^5-8 x^4+e^x \left (8 x^2-8 x \log (4)\right )+\left (16 x^3-16 x^4\right ) \log (4)+\left (8 x^3-8 x^2\right ) \log ^2(4)\right ) \log \left (\frac {x^3-x^2+\left (x-x^2\right ) \log (4)+e^x}{\log (4)-x}\right )}{-5 x^4+5 x^3+\left (5 x-5 x^2\right ) \log ^2(4)+\left (10 x^3-10 x^2\right ) \log (4)+e^x (5 \log (4)-5 x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {24 x^6-24 x^5-4 x^4-e^x \left (-20 x^3-12 x^2+\left (20 x^2+8 x\right ) \log (4)\right )-\left (48 x^5-48 x^4-8 x^3\right ) \log (4)-\left (-24 x^4+24 x^3+4 x^2\right ) \log ^2(4)-\left (8 x^5-8 x^4+e^x \left (8 x^2-8 x \log (4)\right )+\left (16 x^3-16 x^4\right ) \log (4)+\left (8 x^3-8 x^2\right ) \log ^2(4)\right ) \log \left (\frac {x^3-x^2+\left (x-x^2\right ) \log (4)+e^x}{\log (4)-x}\right )}{5 (x-\log (4)) \left (x^3-x^2 (1+\log (4))+e^x+x \log (4)\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int -\frac {4 \left (-6 x^6+6 x^5+x^4-e^x \left (5 x^3+3 x^2-\left (5 x^2+2 x\right ) \log (4)\right )-2 \left (-x^5+x^4-e^x \left (x^2-x \log (4)\right )+\left (x^2-x^3\right ) \log ^2(4)-2 \left (x^3-x^4\right ) \log (4)\right ) \log \left (-\frac {x^3-x^2+e^x+\left (x-x^2\right ) \log (4)}{x-\log (4)}\right )+\left (-6 x^4+6 x^3+x^2\right ) \log ^2(4)-2 \left (-6 x^5+6 x^4+x^3\right ) \log (4)\right )}{(x-\log (4)) \left (x^3-(1+\log (4)) x^2+\log (4) x+e^x\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {4}{5} \int \frac {-6 x^6+6 x^5+x^4-e^x \left (5 x^3+3 x^2-\left (5 x^2+2 x\right ) \log (4)\right )-2 \left (-x^5+x^4-e^x \left (x^2-x \log (4)\right )+\left (x^2-x^3\right ) \log ^2(4)-2 \left (x^3-x^4\right ) \log (4)\right ) \log \left (-\frac {x^3-x^2+e^x+\left (x-x^2\right ) \log (4)}{x-\log (4)}\right )+\left (-6 x^4+6 x^3+x^2\right ) \log ^2(4)-2 \left (-6 x^5+6 x^4+x^3\right ) \log (4)}{(x-\log (4)) \left (x^3-(1+\log (4)) x^2+\log (4) x+e^x\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {4}{5} \int \left (\frac {\left (-x^3+(4+\log (4)) x^2-(2+\log (64)) x+\log (4)\right ) x^2}{x^3-(1+\log (4)) x^2+\log (4) x+e^x}+\frac {\left (-5 x^2+2 \log \left (-x^2+x-\frac {e^x}{x-\log (4)}\right ) x-3 \left (1-\frac {5 \log (4)}{3}\right ) x-2 \log (4) \log \left (-x^2+x-\frac {e^x}{x-\log (4)}\right )+2 \log (4)\right ) x}{x-\log (4)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4}{5} \left (-2 x^3-x^2+x^2 \log \left (-x^2+x-\frac {e^x}{x-\log (4)}\right )\right )\) |
Int[(4*x^4 + 24*x^5 - 24*x^6 + (-8*x^3 - 48*x^4 + 48*x^5)*Log[4] + (4*x^2 + 24*x^3 - 24*x^4)*Log[4]^2 + E^x*(-12*x^2 - 20*x^3 + (8*x + 20*x^2)*Log[4 ]) + (-8*x^4 + 8*x^5 + (16*x^3 - 16*x^4)*Log[4] + (-8*x^2 + 8*x^3)*Log[4]^ 2 + E^x*(8*x^2 - 8*x*Log[4]))*Log[(E^x - x^2 + x^3 + (x - x^2)*Log[4])/(-x + Log[4])])/(5*x^3 - 5*x^4 + (-10*x^2 + 10*x^3)*Log[4] + (5*x - 5*x^2)*Lo g[4]^2 + E^x*(-5*x + 5*Log[4])),x]
3.7.89.3.1 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 = 5.58 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.68
| method | result | size |
| parallelrisch | \(\frac {8 x^{3}}{5}-\frac {4 \ln \left (\frac {{\mathrm e}^{x}+2 \left (-x^{2}+x \right ) \ln \left (2\right )+x^{3}-x^{2}}{2 \ln \left (2\right )-x}\right ) x^{2}}{5}-\frac {16 \ln \left (2\right )^{2}}{5}+\frac {4 x^{2}}{5}\) | \(57\) |
| risch | \(-\frac {4 x^{2} \ln \left (\left (x^{2}-x \right ) \ln \left (2\right )-\frac {x^{3}}{2}+\frac {x^{2}}{2}-\frac {{\mathrm e}^{x}}{2}\right )}{5}+\frac {4 x^{2} \ln \left (\ln \left (2\right )-\frac {x}{2}\right )}{5}+\frac {8 x^{3}}{5}+\frac {4 i \pi \,x^{2} {\operatorname {csgn}\left (\frac {i \left (-\left (x^{2}-x \right ) \ln \left (2\right )+\frac {x^{3}}{2}-\frac {x^{2}}{2}+\frac {{\mathrm e}^{x}}{2}\right )}{\ln \left (2\right )-\frac {x}{2}}\right )}^{2}}{5}+\frac {2 i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (2\right )-\frac {x}{2}}\right ) \operatorname {csgn}\left (i \left (-\left (x^{2}-x \right ) \ln \left (2\right )+\frac {x^{3}}{2}-\frac {x^{2}}{2}+\frac {{\mathrm e}^{x}}{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left (-\left (x^{2}-x \right ) \ln \left (2\right )+\frac {x^{3}}{2}-\frac {x^{2}}{2}+\frac {{\mathrm e}^{x}}{2}\right )}{\ln \left (2\right )-\frac {x}{2}}\right )}{5}-\frac {2 i \pi \,x^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (2\right )-\frac {x}{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (-\left (x^{2}-x \right ) \ln \left (2\right )+\frac {x^{3}}{2}-\frac {x^{2}}{2}+\frac {{\mathrm e}^{x}}{2}\right )}{\ln \left (2\right )-\frac {x}{2}}\right )}^{2}}{5}+\frac {2 i \pi \,x^{2} \operatorname {csgn}\left (i \left (-\left (x^{2}-x \right ) \ln \left (2\right )+\frac {x^{3}}{2}-\frac {x^{2}}{2}+\frac {{\mathrm e}^{x}}{2}\right )\right ) {\operatorname {csgn}\left (\frac {i \left (-\left (x^{2}-x \right ) \ln \left (2\right )+\frac {x^{3}}{2}-\frac {x^{2}}{2}+\frac {{\mathrm e}^{x}}{2}\right )}{\ln \left (2\right )-\frac {x}{2}}\right )}^{2}}{5}+\frac {2 i \pi \,x^{2} {\operatorname {csgn}\left (\frac {i \left (-\left (x^{2}-x \right ) \ln \left (2\right )+\frac {x^{3}}{2}-\frac {x^{2}}{2}+\frac {{\mathrm e}^{x}}{2}\right )}{\ln \left (2\right )-\frac {x}{2}}\right )}^{3}}{5}-\frac {4 i \pi \,x^{2}}{5}+\frac {4 x^{2}}{5}\) | \(379\) |
int((((-16*x*ln(2)+8*x^2)*exp(x)+4*(8*x^3-8*x^2)*ln(2)^2+2*(-16*x^4+16*x^3 )*ln(2)+8*x^5-8*x^4)*ln((exp(x)+2*(-x^2+x)*ln(2)+x^3-x^2)/(2*ln(2)-x))+(2* (20*x^2+8*x)*ln(2)-20*x^3-12*x^2)*exp(x)+4*(-24*x^4+24*x^3+4*x^2)*ln(2)^2+ 2*(48*x^5-48*x^4-8*x^3)*ln(2)-24*x^6+24*x^5+4*x^4)/((10*ln(2)-5*x)*exp(x)+ 4*(-5*x^2+5*x)*ln(2)^2+2*(10*x^3-10*x^2)*ln(2)-5*x^4+5*x^3),x,method=_RETU RNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.44 \begin {dmath*} \int \frac {4 x^4+24 x^5-24 x^6+\left (-8 x^3-48 x^4+48 x^5\right ) \log (4)+\left (4 x^2+24 x^3-24 x^4\right ) \log ^2(4)+e^x \left (-12 x^2-20 x^3+\left (8 x+20 x^2\right ) \log (4)\right )+\left (-8 x^4+8 x^5+\left (16 x^3-16 x^4\right ) \log (4)+\left (-8 x^2+8 x^3\right ) \log ^2(4)+e^x \left (8 x^2-8 x \log (4)\right )\right ) \log \left (\frac {e^x-x^2+x^3+\left (x-x^2\right ) \log (4)}{-x+\log (4)}\right )}{5 x^3-5 x^4+\left (-10 x^2+10 x^3\right ) \log (4)+\left (5 x-5 x^2\right ) \log ^2(4)+e^x (-5 x+5 \log (4))} \, dx=\frac {8}{5} \, x^{3} - \frac {4}{5} \, x^{2} \log \left (-\frac {x^{3} - x^{2} - 2 \, {\left (x^{2} - x\right )} \log \left (2\right ) + e^{x}}{x - 2 \, \log \left (2\right )}\right ) + \frac {4}{5} \, x^{2} \end {dmath*}
integrate((((-16*x*log(2)+8*x^2)*exp(x)+4*(8*x^3-8*x^2)*log(2)^2+2*(-16*x^ 4+16*x^3)*log(2)+8*x^5-8*x^4)*log((exp(x)+2*(-x^2+x)*log(2)+x^3-x^2)/(2*lo g(2)-x))+(2*(20*x^2+8*x)*log(2)-20*x^3-12*x^2)*exp(x)+4*(-24*x^4+24*x^3+4* x^2)*log(2)^2+2*(48*x^5-48*x^4-8*x^3)*log(2)-24*x^6+24*x^5+4*x^4)/((10*log (2)-5*x)*exp(x)+4*(-5*x^2+5*x)*log(2)^2+2*(10*x^3-10*x^2)*log(2)-5*x^4+5*x ^3),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (31) = 62\).
Time = 0.60 (sec) , antiderivative size = 110, normalized size of antiderivative = 3.24 \begin {dmath*} \int \frac {4 x^4+24 x^5-24 x^6+\left (-8 x^3-48 x^4+48 x^5\right ) \log (4)+\left (4 x^2+24 x^3-24 x^4\right ) \log ^2(4)+e^x \left (-12 x^2-20 x^3+\left (8 x+20 x^2\right ) \log (4)\right )+\left (-8 x^4+8 x^5+\left (16 x^3-16 x^4\right ) \log (4)+\left (-8 x^2+8 x^3\right ) \log ^2(4)+e^x \left (8 x^2-8 x \log (4)\right )\right ) \log \left (\frac {e^x-x^2+x^3+\left (x-x^2\right ) \log (4)}{-x+\log (4)}\right )}{5 x^3-5 x^4+\left (-10 x^2+10 x^3\right ) \log (4)+\left (5 x-5 x^2\right ) \log ^2(4)+e^x (-5 x+5 \log (4))} \, dx=\frac {8 x^{3}}{5} + \frac {4 x^{2}}{5} + \left (- \frac {4 x^{2}}{5} + \frac {16 \log {\left (2 \right )}^{2}}{15}\right ) \log {\left (\frac {x^{3} - x^{2} + \left (- 2 x^{2} + 2 x\right ) \log {\left (2 \right )} + e^{x}}{- x + 2 \log {\left (2 \right )}} \right )} + \frac {16 \log {\left (2 \right )}^{2} \log {\left (x - 2 \log {\left (2 \right )} \right )}}{15} - \frac {16 \log {\left (2 \right )}^{2} \log {\left (x^{3} - 2 x^{2} \log {\left (2 \right )} - x^{2} + 2 x \log {\left (2 \right )} + e^{x} \right )}}{15} \end {dmath*}
integrate((((-16*x*ln(2)+8*x**2)*exp(x)+4*(8*x**3-8*x**2)*ln(2)**2+2*(-16* x**4+16*x**3)*ln(2)+8*x**5-8*x**4)*ln((exp(x)+2*(-x**2+x)*ln(2)+x**3-x**2) /(2*ln(2)-x))+(2*(20*x**2+8*x)*ln(2)-20*x**3-12*x**2)*exp(x)+4*(-24*x**4+2 4*x**3+4*x**2)*ln(2)**2+2*(48*x**5-48*x**4-8*x**3)*ln(2)-24*x**6+24*x**5+4 *x**4)/((10*ln(2)-5*x)*exp(x)+4*(-5*x**2+5*x)*ln(2)**2+2*(10*x**3-10*x**2) *ln(2)-5*x**4+5*x**3),x)
8*x**3/5 + 4*x**2/5 + (-4*x**2/5 + 16*log(2)**2/15)*log((x**3 - x**2 + (-2 *x**2 + 2*x)*log(2) + exp(x))/(-x + 2*log(2))) + 16*log(2)**2*log(x - 2*lo g(2))/15 - 16*log(2)**2*log(x**3 - 2*x**2*log(2) - x**2 + 2*x*log(2) + exp (x))/15
Time = 0.33 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.59 \begin {dmath*} \int \frac {4 x^4+24 x^5-24 x^6+\left (-8 x^3-48 x^4+48 x^5\right ) \log (4)+\left (4 x^2+24 x^3-24 x^4\right ) \log ^2(4)+e^x \left (-12 x^2-20 x^3+\left (8 x+20 x^2\right ) \log (4)\right )+\left (-8 x^4+8 x^5+\left (16 x^3-16 x^4\right ) \log (4)+\left (-8 x^2+8 x^3\right ) \log ^2(4)+e^x \left (8 x^2-8 x \log (4)\right )\right ) \log \left (\frac {e^x-x^2+x^3+\left (x-x^2\right ) \log (4)}{-x+\log (4)}\right )}{5 x^3-5 x^4+\left (-10 x^2+10 x^3\right ) \log (4)+\left (5 x-5 x^2\right ) \log ^2(4)+e^x (-5 x+5 \log (4))} \, dx=\frac {8}{5} \, x^{3} - \frac {4}{5} \, x^{2} \log \left (-x^{3} + x^{2} {\left (2 \, \log \left (2\right ) + 1\right )} - 2 \, x \log \left (2\right ) - e^{x}\right ) + \frac {4}{5} \, x^{2} \log \left (x - 2 \, \log \left (2\right )\right ) + \frac {4}{5} \, x^{2} \end {dmath*}
integrate((((-16*x*log(2)+8*x^2)*exp(x)+4*(8*x^3-8*x^2)*log(2)^2+2*(-16*x^ 4+16*x^3)*log(2)+8*x^5-8*x^4)*log((exp(x)+2*(-x^2+x)*log(2)+x^3-x^2)/(2*lo g(2)-x))+(2*(20*x^2+8*x)*log(2)-20*x^3-12*x^2)*exp(x)+4*(-24*x^4+24*x^3+4* x^2)*log(2)^2+2*(48*x^5-48*x^4-8*x^3)*log(2)-24*x^6+24*x^5+4*x^4)/((10*log (2)-5*x)*exp(x)+4*(-5*x^2+5*x)*log(2)^2+2*(10*x^3-10*x^2)*log(2)-5*x^4+5*x ^3),x, algorithm=\
8/5*x^3 - 4/5*x^2*log(-x^3 + x^2*(2*log(2) + 1) - 2*x*log(2) - e^x) + 4/5* x^2*log(x - 2*log(2)) + 4/5*x^2
Time = 0.38 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.59 \begin {dmath*} \int \frac {4 x^4+24 x^5-24 x^6+\left (-8 x^3-48 x^4+48 x^5\right ) \log (4)+\left (4 x^2+24 x^3-24 x^4\right ) \log ^2(4)+e^x \left (-12 x^2-20 x^3+\left (8 x+20 x^2\right ) \log (4)\right )+\left (-8 x^4+8 x^5+\left (16 x^3-16 x^4\right ) \log (4)+\left (-8 x^2+8 x^3\right ) \log ^2(4)+e^x \left (8 x^2-8 x \log (4)\right )\right ) \log \left (\frac {e^x-x^2+x^3+\left (x-x^2\right ) \log (4)}{-x+\log (4)}\right )}{5 x^3-5 x^4+\left (-10 x^2+10 x^3\right ) \log (4)+\left (5 x-5 x^2\right ) \log ^2(4)+e^x (-5 x+5 \log (4))} \, dx=\frac {8}{5} \, x^{3} - \frac {4}{5} \, x^{2} \log \left (-x^{3} + 2 \, x^{2} \log \left (2\right ) + x^{2} - 2 \, x \log \left (2\right ) - e^{x}\right ) + \frac {4}{5} \, x^{2} \log \left (x - 2 \, \log \left (2\right )\right ) + \frac {4}{5} \, x^{2} \end {dmath*}
integrate((((-16*x*log(2)+8*x^2)*exp(x)+4*(8*x^3-8*x^2)*log(2)^2+2*(-16*x^ 4+16*x^3)*log(2)+8*x^5-8*x^4)*log((exp(x)+2*(-x^2+x)*log(2)+x^3-x^2)/(2*lo g(2)-x))+(2*(20*x^2+8*x)*log(2)-20*x^3-12*x^2)*exp(x)+4*(-24*x^4+24*x^3+4* x^2)*log(2)^2+2*(48*x^5-48*x^4-8*x^3)*log(2)-24*x^6+24*x^5+4*x^4)/((10*log (2)-5*x)*exp(x)+4*(-5*x^2+5*x)*log(2)^2+2*(10*x^3-10*x^2)*log(2)-5*x^4+5*x ^3),x, algorithm=\
8/5*x^3 - 4/5*x^2*log(-x^3 + 2*x^2*log(2) + x^2 - 2*x*log(2) - e^x) + 4/5* x^2*log(x - 2*log(2)) + 4/5*x^2
Time = 14.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.32 \begin {dmath*} \int \frac {4 x^4+24 x^5-24 x^6+\left (-8 x^3-48 x^4+48 x^5\right ) \log (4)+\left (4 x^2+24 x^3-24 x^4\right ) \log ^2(4)+e^x \left (-12 x^2-20 x^3+\left (8 x+20 x^2\right ) \log (4)\right )+\left (-8 x^4+8 x^5+\left (16 x^3-16 x^4\right ) \log (4)+\left (-8 x^2+8 x^3\right ) \log ^2(4)+e^x \left (8 x^2-8 x \log (4)\right )\right ) \log \left (\frac {e^x-x^2+x^3+\left (x-x^2\right ) \log (4)}{-x+\log (4)}\right )}{5 x^3-5 x^4+\left (-10 x^2+10 x^3\right ) \log (4)+\left (5 x-5 x^2\right ) \log ^2(4)+e^x (-5 x+5 \log (4))} \, dx=\frac {4\,x^2\,\left (2\,x-\ln \left (-\frac {{\mathrm {e}}^x-x^2+x^3+2\,\ln \left (2\right )\,\left (x-x^2\right )}{x-2\,\ln \left (2\right )}\right )+1\right )}{5} \end {dmath*}
int((exp(x)*(12*x^2 - 2*log(2)*(8*x + 20*x^2) + 20*x^3) - 4*log(2)^2*(4*x^ 2 + 24*x^3 - 24*x^4) + log(-(exp(x) - x^2 + x^3 + 2*log(2)*(x - x^2))/(x - 2*log(2)))*(exp(x)*(16*x*log(2) - 8*x^2) - 2*log(2)*(16*x^3 - 16*x^4) + 8 *x^4 - 8*x^5 + 4*log(2)^2*(8*x^2 - 8*x^3)) + 2*log(2)*(8*x^3 + 48*x^4 - 48 *x^5) - 4*x^4 - 24*x^5 + 24*x^6)/(exp(x)*(5*x - 10*log(2)) - 4*log(2)^2*(5 *x - 5*x^2) + 2*log(2)*(10*x^2 - 10*x^3) - 5*x^3 + 5*x^4),x)