Integrand size = 191, antiderivative size = 26 \begin {dmath*} \int \frac {6 x^3-4 x^2 \log (2)+e^x \left (-16 x^5+\left (-16 x^3+32 x^4\right ) \log (2)+\left (20 x^2-24 x^3\right ) \log ^2(2)+\left (-8 x+8 x^2\right ) \log ^3(2)+(1-x) \log ^4(2)\right )+e^{2 x} \left (-32 x^5+64 x^4 \log (2)-48 x^3 \log ^2(2)+16 x^2 \log ^3(2)-2 x \log ^4(2)\right )}{x^2+e^x \left (8 x^3-8 x^2 \log (2)+2 x \log ^2(2)\right )+e^{2 x} \left (16 x^4-32 x^3 \log (2)+24 x^2 \log ^2(2)-8 x \log ^3(2)+\log ^4(2)\right )} \, dx=-x^2+\frac {x}{e^x+\frac {x}{(2 x-\log (2))^2}} \end {dmath*}
\begin {dmath*} \int \frac {6 x^3-4 x^2 \log (2)+e^x \left (-16 x^5+\left (-16 x^3+32 x^4\right ) \log (2)+\left (20 x^2-24 x^3\right ) \log ^2(2)+\left (-8 x+8 x^2\right ) \log ^3(2)+(1-x) \log ^4(2)\right )+e^{2 x} \left (-32 x^5+64 x^4 \log (2)-48 x^3 \log ^2(2)+16 x^2 \log ^3(2)-2 x \log ^4(2)\right )}{x^2+e^x \left (8 x^3-8 x^2 \log (2)+2 x \log ^2(2)\right )+e^{2 x} \left (16 x^4-32 x^3 \log (2)+24 x^2 \log ^2(2)-8 x \log ^3(2)+\log ^4(2)\right )} \, dx=\int \frac {6 x^3-4 x^2 \log (2)+e^x \left (-16 x^5+\left (-16 x^3+32 x^4\right ) \log (2)+\left (20 x^2-24 x^3\right ) \log ^2(2)+\left (-8 x+8 x^2\right ) \log ^3(2)+(1-x) \log ^4(2)\right )+e^{2 x} \left (-32 x^5+64 x^4 \log (2)-48 x^3 \log ^2(2)+16 x^2 \log ^3(2)-2 x \log ^4(2)\right )}{x^2+e^x \left (8 x^3-8 x^2 \log (2)+2 x \log ^2(2)\right )+e^{2 x} \left (16 x^4-32 x^3 \log (2)+24 x^2 \log ^2(2)-8 x \log ^3(2)+\log ^4(2)\right )} \, dx \end {dmath*}
Integrate[(6*x^3 - 4*x^2*Log[2] + E^x*(-16*x^5 + (-16*x^3 + 32*x^4)*Log[2] + (20*x^2 - 24*x^3)*Log[2]^2 + (-8*x + 8*x^2)*Log[2]^3 + (1 - x)*Log[2]^4 ) + E^(2*x)*(-32*x^5 + 64*x^4*Log[2] - 48*x^3*Log[2]^2 + 16*x^2*Log[2]^3 - 2*x*Log[2]^4))/(x^2 + E^x*(8*x^3 - 8*x^2*Log[2] + 2*x*Log[2]^2) + E^(2*x) *(16*x^4 - 32*x^3*Log[2] + 24*x^2*Log[2]^2 - 8*x*Log[2]^3 + Log[2]^4)),x]
Integrate[(6*x^3 - 4*x^2*Log[2] + E^x*(-16*x^5 + (-16*x^3 + 32*x^4)*Log[2] + (20*x^2 - 24*x^3)*Log[2]^2 + (-8*x + 8*x^2)*Log[2]^3 + (1 - x)*Log[2]^4 ) + E^(2*x)*(-32*x^5 + 64*x^4*Log[2] - 48*x^3*Log[2]^2 + 16*x^2*Log[2]^3 - 2*x*Log[2]^4))/(x^2 + E^x*(8*x^3 - 8*x^2*Log[2] + 2*x*Log[2]^2) + E^(2*x) *(16*x^4 - 32*x^3*Log[2] + 24*x^2*Log[2]^2 - 8*x*Log[2]^3 + Log[2]^4)), 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.
| \(\displaystyle \int \frac {6 x^3-4 x^2 \log (2)+e^x \left (-16 x^5+\left (8 x^2-8 x\right ) \log ^3(2)+\left (32 x^4-16 x^3\right ) \log (2)+\left (20 x^2-24 x^3\right ) \log ^2(2)+(1-x) \log ^4(2)\right )+e^{2 x} \left (-32 x^5+64 x^4 \log (2)-48 x^3 \log ^2(2)+16 x^2 \log ^3(2)-2 x \log ^4(2)\right )}{x^2+e^x \left (8 x^3-8 x^2 \log (2)+2 x \log ^2(2)\right )+e^{2 x} \left (16 x^4-32 x^3 \log (2)+24 x^2 \log ^2(2)-8 x \log ^3(2)+\log ^4(2)\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {6 x^3-4 x^2 \log (2)+e^x \left (-16 x^5+\left (8 x^2-8 x\right ) \log ^3(2)+\left (32 x^4-16 x^3\right ) \log (2)+\left (20 x^2-24 x^3\right ) \log ^2(2)+(1-x) \log ^4(2)\right )+e^{2 x} \left (-32 x^5+64 x^4 \log (2)-48 x^3 \log ^2(2)+16 x^2 \log ^3(2)-2 x \log ^4(2)\right )}{\left (4 e^x x^2+x+e^x \log ^2(2)-4 e^x x \log (2)\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {(x-1) (2 x-\log (2))^2}{4 e^x x^2+x+e^x \log ^2(2)-4 e^x x \log (2)}+\frac {x \left (4 x^3+4 x^2 (1-\log (2))+x \log ^2(2)-\log ^2(2)\right )}{\left (4 e^x x^2+x+e^x \log ^2(2)-4 e^x x \log (2)\right )^2}-2 x\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\log ^2(2) \int \frac {x}{\left (4 e^x x^2-4 e^x \log (2) x+x+e^x \log ^2(2)\right )^2}dx+\log ^2(2) \int \frac {x^2}{\left (4 e^x x^2-4 e^x \log (2) x+x+e^x \log ^2(2)\right )^2}dx+\log ^2(2) \int \frac {1}{4 e^x x^2-4 e^x \log (2) x+x+e^x \log ^2(2)}dx-\log (2) (4+\log (2)) \int \frac {x}{4 e^x x^2-4 e^x \log (2) x+x+e^x \log ^2(2)}dx+4 (1+\log (2)) \int \frac {x^2}{4 e^x x^2-4 e^x \log (2) x+x+e^x \log ^2(2)}dx+4 \int \frac {x^4}{\left (4 e^x x^2-4 e^x \log (2) x+x+e^x \log ^2(2)\right )^2}dx+4 (1-\log (2)) \int \frac {x^3}{\left (4 e^x x^2-4 e^x \log (2) x+x+e^x \log ^2(2)\right )^2}dx-4 \int \frac {x^3}{4 e^x x^2-4 e^x \log (2) x+x+e^x \log ^2(2)}dx-x^2\) |
Int[(6*x^3 - 4*x^2*Log[2] + E^x*(-16*x^5 + (-16*x^3 + 32*x^4)*Log[2] + (20 *x^2 - 24*x^3)*Log[2]^2 + (-8*x + 8*x^2)*Log[2]^3 + (1 - x)*Log[2]^4) + E^ (2*x)*(-32*x^5 + 64*x^4*Log[2] - 48*x^3*Log[2]^2 + 16*x^2*Log[2]^3 - 2*x*L og[2]^4))/(x^2 + E^x*(8*x^3 - 8*x^2*Log[2] + 2*x*Log[2]^2) + E^(2*x)*(16*x ^4 - 32*x^3*Log[2] + 24*x^2*Log[2]^2 - 8*x*Log[2]^3 + Log[2]^4)),x]
3.13.72.3.1 Defintions of rubi rules used
Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62
| method | result | size |
| risch | \(-x^{2}+\frac {x \left (\ln \left (2\right )-2 x \right )^{2}}{\ln \left (2\right )^{2} {\mathrm e}^{x}-4 x \ln \left (2\right ) {\mathrm e}^{x}+4 \,{\mathrm e}^{x} x^{2}+x}\) | \(42\) |
| norman | \(\frac {\frac {5 x \ln \left (2\right )^{2}}{4}+\frac {\ln \left (2\right )^{4} {\mathrm e}^{x}}{4}-\ln \left (2\right )^{3} {\mathrm e}^{x} x +3 x^{3}-4 x^{2} \ln \left (2\right )-4 \,{\mathrm e}^{x} x^{4}+4 x^{3} \ln \left (2\right ) {\mathrm e}^{x}}{\ln \left (2\right )^{2} {\mathrm e}^{x}-4 x \ln \left (2\right ) {\mathrm e}^{x}+4 \,{\mathrm e}^{x} x^{2}+x}\) | \(80\) |
| parallelrisch | \(\frac {\ln \left (2\right )^{4} {\mathrm e}^{x}-4 \ln \left (2\right )^{3} {\mathrm e}^{x} x +16 x^{3} \ln \left (2\right ) {\mathrm e}^{x}-16 \,{\mathrm e}^{x} x^{4}+5 x \ln \left (2\right )^{2}-16 x^{2} \ln \left (2\right )+12 x^{3}}{4 \ln \left (2\right )^{2} {\mathrm e}^{x}-16 x \ln \left (2\right ) {\mathrm e}^{x}+16 \,{\mathrm e}^{x} x^{2}+4 x}\) | \(80\) |
int(((-2*x*ln(2)^4+16*x^2*ln(2)^3-48*x^3*ln(2)^2+64*x^4*ln(2)-32*x^5)*exp( x)^2+((1-x)*ln(2)^4+(8*x^2-8*x)*ln(2)^3+(-24*x^3+20*x^2)*ln(2)^2+(32*x^4-1 6*x^3)*ln(2)-16*x^5)*exp(x)-4*x^2*ln(2)+6*x^3)/((ln(2)^4-8*x*ln(2)^3+24*x^ 2*ln(2)^2-32*x^3*ln(2)+16*x^4)*exp(x)^2+(2*x*ln(2)^2-8*x^2*ln(2)+8*x^3)*ex p(x)+x^2),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.58 \begin {dmath*} \int \frac {6 x^3-4 x^2 \log (2)+e^x \left (-16 x^5+\left (-16 x^3+32 x^4\right ) \log (2)+\left (20 x^2-24 x^3\right ) \log ^2(2)+\left (-8 x+8 x^2\right ) \log ^3(2)+(1-x) \log ^4(2)\right )+e^{2 x} \left (-32 x^5+64 x^4 \log (2)-48 x^3 \log ^2(2)+16 x^2 \log ^3(2)-2 x \log ^4(2)\right )}{x^2+e^x \left (8 x^3-8 x^2 \log (2)+2 x \log ^2(2)\right )+e^{2 x} \left (16 x^4-32 x^3 \log (2)+24 x^2 \log ^2(2)-8 x \log ^3(2)+\log ^4(2)\right )} \, dx=\frac {3 \, x^{3} - 4 \, x^{2} \log \left (2\right ) + x \log \left (2\right )^{2} - {\left (4 \, x^{4} - 4 \, x^{3} \log \left (2\right ) + x^{2} \log \left (2\right )^{2}\right )} e^{x}}{{\left (4 \, x^{2} - 4 \, x \log \left (2\right ) + \log \left (2\right )^{2}\right )} e^{x} + x} \end {dmath*}
integrate(((-2*x*log(2)^4+16*x^2*log(2)^3-48*x^3*log(2)^2+64*x^4*log(2)-32 *x^5)*exp(x)^2+((1-x)*log(2)^4+(8*x^2-8*x)*log(2)^3+(-24*x^3+20*x^2)*log(2 )^2+(32*x^4-16*x^3)*log(2)-16*x^5)*exp(x)-4*x^2*log(2)+6*x^3)/((log(2)^4-8 *x*log(2)^3+24*x^2*log(2)^2-32*x^3*log(2)+16*x^4)*exp(x)^2+(2*x*log(2)^2-8 *x^2*log(2)+8*x^3)*exp(x)+x^2),x, algorithm=\
(3*x^3 - 4*x^2*log(2) + x*log(2)^2 - (4*x^4 - 4*x^3*log(2) + x^2*log(2)^2) *e^x)/((4*x^2 - 4*x*log(2) + log(2)^2)*e^x + x)
Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \begin {dmath*} \int \frac {6 x^3-4 x^2 \log (2)+e^x \left (-16 x^5+\left (-16 x^3+32 x^4\right ) \log (2)+\left (20 x^2-24 x^3\right ) \log ^2(2)+\left (-8 x+8 x^2\right ) \log ^3(2)+(1-x) \log ^4(2)\right )+e^{2 x} \left (-32 x^5+64 x^4 \log (2)-48 x^3 \log ^2(2)+16 x^2 \log ^3(2)-2 x \log ^4(2)\right )}{x^2+e^x \left (8 x^3-8 x^2 \log (2)+2 x \log ^2(2)\right )+e^{2 x} \left (16 x^4-32 x^3 \log (2)+24 x^2 \log ^2(2)-8 x \log ^3(2)+\log ^4(2)\right )} \, dx=- x^{2} + \frac {4 x^{3} - 4 x^{2} \log {\left (2 \right )} + x \log {\left (2 \right )}^{2}}{x + \left (4 x^{2} - 4 x \log {\left (2 \right )} + \log {\left (2 \right )}^{2}\right ) e^{x}} \end {dmath*}
integrate(((-2*x*ln(2)**4+16*x**2*ln(2)**3-48*x**3*ln(2)**2+64*x**4*ln(2)- 32*x**5)*exp(x)**2+((1-x)*ln(2)**4+(8*x**2-8*x)*ln(2)**3+(-24*x**3+20*x**2 )*ln(2)**2+(32*x**4-16*x**3)*ln(2)-16*x**5)*exp(x)-4*x**2*ln(2)+6*x**3)/(( ln(2)**4-8*x*ln(2)**3+24*x**2*ln(2)**2-32*x**3*ln(2)+16*x**4)*exp(x)**2+(2 *x*ln(2)**2-8*x**2*ln(2)+8*x**3)*exp(x)+x**2),x)
-x**2 + (4*x**3 - 4*x**2*log(2) + x*log(2)**2)/(x + (4*x**2 - 4*x*log(2) + log(2)**2)*exp(x))
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (25) = 50\).
Time = 0.35 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.58 \begin {dmath*} \int \frac {6 x^3-4 x^2 \log (2)+e^x \left (-16 x^5+\left (-16 x^3+32 x^4\right ) \log (2)+\left (20 x^2-24 x^3\right ) \log ^2(2)+\left (-8 x+8 x^2\right ) \log ^3(2)+(1-x) \log ^4(2)\right )+e^{2 x} \left (-32 x^5+64 x^4 \log (2)-48 x^3 \log ^2(2)+16 x^2 \log ^3(2)-2 x \log ^4(2)\right )}{x^2+e^x \left (8 x^3-8 x^2 \log (2)+2 x \log ^2(2)\right )+e^{2 x} \left (16 x^4-32 x^3 \log (2)+24 x^2 \log ^2(2)-8 x \log ^3(2)+\log ^4(2)\right )} \, dx=\frac {3 \, x^{3} - 4 \, x^{2} \log \left (2\right ) + x \log \left (2\right )^{2} - {\left (4 \, x^{4} - 4 \, x^{3} \log \left (2\right ) + x^{2} \log \left (2\right )^{2}\right )} e^{x}}{{\left (4 \, x^{2} - 4 \, x \log \left (2\right ) + \log \left (2\right )^{2}\right )} e^{x} + x} \end {dmath*}
integrate(((-2*x*log(2)^4+16*x^2*log(2)^3-48*x^3*log(2)^2+64*x^4*log(2)-32 *x^5)*exp(x)^2+((1-x)*log(2)^4+(8*x^2-8*x)*log(2)^3+(-24*x^3+20*x^2)*log(2 )^2+(32*x^4-16*x^3)*log(2)-16*x^5)*exp(x)-4*x^2*log(2)+6*x^3)/((log(2)^4-8 *x*log(2)^3+24*x^2*log(2)^2-32*x^3*log(2)+16*x^4)*exp(x)^2+(2*x*log(2)^2-8 *x^2*log(2)+8*x^3)*exp(x)+x^2),x, algorithm=\
(3*x^3 - 4*x^2*log(2) + x*log(2)^2 - (4*x^4 - 4*x^3*log(2) + x^2*log(2)^2) *e^x)/((4*x^2 - 4*x*log(2) + log(2)^2)*e^x + x)
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (25) = 50\).
Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.81 \begin {dmath*} \int \frac {6 x^3-4 x^2 \log (2)+e^x \left (-16 x^5+\left (-16 x^3+32 x^4\right ) \log (2)+\left (20 x^2-24 x^3\right ) \log ^2(2)+\left (-8 x+8 x^2\right ) \log ^3(2)+(1-x) \log ^4(2)\right )+e^{2 x} \left (-32 x^5+64 x^4 \log (2)-48 x^3 \log ^2(2)+16 x^2 \log ^3(2)-2 x \log ^4(2)\right )}{x^2+e^x \left (8 x^3-8 x^2 \log (2)+2 x \log ^2(2)\right )+e^{2 x} \left (16 x^4-32 x^3 \log (2)+24 x^2 \log ^2(2)-8 x \log ^3(2)+\log ^4(2)\right )} \, dx=-\frac {4 \, x^{4} e^{x} - 4 \, x^{3} e^{x} \log \left (2\right ) + x^{2} e^{x} \log \left (2\right )^{2} - 3 \, x^{3} + 4 \, x^{2} \log \left (2\right ) - x \log \left (2\right )^{2}}{4 \, x^{2} e^{x} - 4 \, x e^{x} \log \left (2\right ) + e^{x} \log \left (2\right )^{2} + x} \end {dmath*}
integrate(((-2*x*log(2)^4+16*x^2*log(2)^3-48*x^3*log(2)^2+64*x^4*log(2)-32 *x^5)*exp(x)^2+((1-x)*log(2)^4+(8*x^2-8*x)*log(2)^3+(-24*x^3+20*x^2)*log(2 )^2+(32*x^4-16*x^3)*log(2)-16*x^5)*exp(x)-4*x^2*log(2)+6*x^3)/((log(2)^4-8 *x*log(2)^3+24*x^2*log(2)^2-32*x^3*log(2)+16*x^4)*exp(x)^2+(2*x*log(2)^2-8 *x^2*log(2)+8*x^3)*exp(x)+x^2),x, algorithm=\
-(4*x^4*e^x - 4*x^3*e^x*log(2) + x^2*e^x*log(2)^2 - 3*x^3 + 4*x^2*log(2) - x*log(2)^2)/(4*x^2*e^x - 4*x*e^x*log(2) + e^x*log(2)^2 + x)
Timed out. \begin {dmath*} \int \frac {6 x^3-4 x^2 \log (2)+e^x \left (-16 x^5+\left (-16 x^3+32 x^4\right ) \log (2)+\left (20 x^2-24 x^3\right ) \log ^2(2)+\left (-8 x+8 x^2\right ) \log ^3(2)+(1-x) \log ^4(2)\right )+e^{2 x} \left (-32 x^5+64 x^4 \log (2)-48 x^3 \log ^2(2)+16 x^2 \log ^3(2)-2 x \log ^4(2)\right )}{x^2+e^x \left (8 x^3-8 x^2 \log (2)+2 x \log ^2(2)\right )+e^{2 x} \left (16 x^4-32 x^3 \log (2)+24 x^2 \log ^2(2)-8 x \log ^3(2)+\log ^4(2)\right )} \, dx=\int -\frac {{\mathrm {e}}^x\,\left ({\ln \left (2\right )}^4\,\left (x-1\right )+{\ln \left (2\right )}^3\,\left (8\,x-8\,x^2\right )+\ln \left (2\right )\,\left (16\,x^3-32\,x^4\right )+16\,x^5-{\ln \left (2\right )}^2\,\left (20\,x^2-24\,x^3\right )\right )+{\mathrm {e}}^{2\,x}\,\left (32\,x^5-64\,\ln \left (2\right )\,x^4+48\,{\ln \left (2\right )}^2\,x^3-16\,{\ln \left (2\right )}^3\,x^2+2\,{\ln \left (2\right )}^4\,x\right )+4\,x^2\,\ln \left (2\right )-6\,x^3}{{\mathrm {e}}^x\,\left (8\,x^3-8\,\ln \left (2\right )\,x^2+2\,{\ln \left (2\right )}^2\,x\right )+{\mathrm {e}}^{2\,x}\,\left (16\,x^4-32\,\ln \left (2\right )\,x^3+24\,{\ln \left (2\right )}^2\,x^2-8\,{\ln \left (2\right )}^3\,x+{\ln \left (2\right )}^4\right )+x^2} \,d x \end {dmath*}
int(-(exp(x)*(log(2)^4*(x - 1) + log(2)^3*(8*x - 8*x^2) + log(2)*(16*x^3 - 32*x^4) + 16*x^5 - log(2)^2*(20*x^2 - 24*x^3)) + exp(2*x)*(48*x^3*log(2)^ 2 - 16*x^2*log(2)^3 + 2*x*log(2)^4 - 64*x^4*log(2) + 32*x^5) + 4*x^2*log(2 ) - 6*x^3)/(exp(x)*(2*x*log(2)^2 - 8*x^2*log(2) + 8*x^3) + exp(2*x)*(24*x^ 2*log(2)^2 - 8*x*log(2)^3 - 32*x^3*log(2) + log(2)^4 + 16*x^4) + x^2),x)
int(-(exp(x)*(log(2)^4*(x - 1) + log(2)^3*(8*x - 8*x^2) + log(2)*(16*x^3 - 32*x^4) + 16*x^5 - log(2)^2*(20*x^2 - 24*x^3)) + exp(2*x)*(48*x^3*log(2)^ 2 - 16*x^2*log(2)^3 + 2*x*log(2)^4 - 64*x^4*log(2) + 32*x^5) + 4*x^2*log(2 ) - 6*x^3)/(exp(x)*(2*x*log(2)^2 - 8*x^2*log(2) + 8*x^3) + exp(2*x)*(24*x^ 2*log(2)^2 - 8*x*log(2)^3 - 32*x^3*log(2) + log(2)^4 + 16*x^4) + x^2), x)