Integrand size = 136, antiderivative size = 27 \[ \int \frac {2+4 x+e^x \left (-2-6 x-4 x^2\right )+\left (2048+1280 x+136 x^2+4 x^3+e^x \left (2048+1280 x+136 x^2+4 x^3\right )\right ) \log (1+2 x)+\left (320+708 x+139 x^2+6 x^3+e^x \left (-192-636 x-535 x^2-63 x^3-2 x^4\right )\right ) \log ^2(1+2 x)}{1+2 x+e^{2 x} (1+2 x)+e^x (2+4 x)} \, dx=\frac {(2+x) \left (2+(16+x)^2 \log ^2(1+2 x)\right )}{1+e^x} \]
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {2+4 x+e^x \left (-2-6 x-4 x^2\right )+\left (2048+1280 x+136 x^2+4 x^3+e^x \left (2048+1280 x+136 x^2+4 x^3\right )\right ) \log (1+2 x)+\left (320+708 x+139 x^2+6 x^3+e^x \left (-192-636 x-535 x^2-63 x^3-2 x^4\right )\right ) \log ^2(1+2 x)}{1+2 x+e^{2 x} (1+2 x)+e^x (2+4 x)} \, dx=\frac {(2+x) \left (2+(16+x)^2 \log ^2(1+2 x)\right )}{1+e^x} \]
Integrate[(2 + 4*x + E^x*(-2 - 6*x - 4*x^2) + (2048 + 1280*x + 136*x^2 + 4 *x^3 + E^x*(2048 + 1280*x + 136*x^2 + 4*x^3))*Log[1 + 2*x] + (320 + 708*x + 139*x^2 + 6*x^3 + E^x*(-192 - 636*x - 535*x^2 - 63*x^3 - 2*x^4))*Log[1 + 2*x]^2)/(1 + 2*x + E^(2*x)*(1 + 2*x) + E^x*(2 + 4*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.
\(\displaystyle \int \frac {e^x \left (-4 x^2-6 x-2\right )+\left (4 x^3+136 x^2+e^x \left (4 x^3+136 x^2+1280 x+2048\right )+1280 x+2048\right ) \log (2 x+1)+\left (6 x^3+139 x^2+e^x \left (-2 x^4-63 x^3-535 x^2-636 x-192\right )+708 x+320\right ) \log ^2(2 x+1)+4 x+2}{2 x+e^{2 x} (2 x+1)+e^x (4 x+2)+1} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^x \left (-4 x^2-6 x-2\right )+\left (4 x^3+136 x^2+e^x \left (4 x^3+136 x^2+1280 x+2048\right )+1280 x+2048\right ) \log (2 x+1)+\left (6 x^3+139 x^2+e^x \left (-2 x^4-63 x^3-535 x^2-636 x-192\right )+708 x+320\right ) \log ^2(2 x+1)+4 x+2}{\left (e^x+1\right )^2 (2 x+1)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {(x+2) \left (x^2 \log ^2(2 x+1)+32 x \log ^2(2 x+1)+256 \log ^2(2 x+1)+2\right )}{\left (e^x+1\right )^2}-\frac {2 x^4 \log ^2(2 x+1)+63 x^3 \log ^2(2 x+1)-4 x^3 \log (2 x+1)+4 x^2+535 x^2 \log ^2(2 x+1)-136 x^2 \log (2 x+1)+6 x+636 x \log ^2(2 x+1)+192 \log ^2(2 x+1)-1280 x \log (2 x+1)-2048 \log (2 x+1)+2}{\left (e^x+1\right ) (2 x+1)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {x^3 \log ^2(2 x+1)}{\left (1+e^x\right )^2}dx-\int \frac {x^3 \log ^2(2 x+1)}{1+e^x}dx+34 \int \frac {x^2 \log ^2(2 x+1)}{\left (1+e^x\right )^2}dx-31 \int \frac {x^2 \log ^2(2 x+1)}{1+e^x}dx+2 \int \frac {x^2 \log (2 x+1)}{1+e^x}dx-2883 \int \frac {\int \frac {1}{\left (1+e^x\right ) (2 x+1)}dx}{2 x+1}dx+512 \int \frac {\log ^2(2 x+1)}{\left (1+e^x\right )^2}dx-192 \int \frac {\log ^2(2 x+1)}{1+e^x}dx+320 \int \frac {x \log ^2(2 x+1)}{\left (1+e^x\right )^2}dx-252 \int \frac {x \log ^2(2 x+1)}{1+e^x}dx+\frac {2883}{2} \log (2 x+1) \int \frac {1}{\left (1+e^x\right ) (2 x+1)}dx+\frac {1213}{2} \int \frac {\log (2 x+1)}{1+e^x}dx+67 \int \frac {x \log (2 x+1)}{1+e^x}dx+\frac {2 x}{e^x+1}+\frac {4}{e^x+1}\) |
Int[(2 + 4*x + E^x*(-2 - 6*x - 4*x^2) + (2048 + 1280*x + 136*x^2 + 4*x^3 + E^x*(2048 + 1280*x + 136*x^2 + 4*x^3))*Log[1 + 2*x] + (320 + 708*x + 139* x^2 + 6*x^3 + E^x*(-192 - 636*x - 535*x^2 - 63*x^3 - 2*x^4))*Log[1 + 2*x]^ 2)/(1 + 2*x + E^(2*x)*(1 + 2*x) + E^x*(2 + 4*x)),x]
3.14.69.3.1 Defintions of rubi rules used
Time = 0.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52
method | result | size |
risch | \(\frac {\left (x^{3}+34 x^{2}+320 x +512\right ) \ln \left (1+2 x \right )^{2}}{{\mathrm e}^{x}+1}+\frac {4+2 x}{{\mathrm e}^{x}+1}\) | \(41\) |
parallelrisch | \(-\frac {-4 \ln \left (1+2 x \right )^{2} x^{3}-136 \ln \left (1+2 x \right )^{2} x^{2}-16-1280 \ln \left (1+2 x \right )^{2} x -2048 \ln \left (1+2 x \right )^{2}-8 x}{4 \left ({\mathrm e}^{x}+1\right )}\) | \(61\) |
int((((-2*x^4-63*x^3-535*x^2-636*x-192)*exp(x)+6*x^3+139*x^2+708*x+320)*ln (1+2*x)^2+((4*x^3+136*x^2+1280*x+2048)*exp(x)+4*x^3+136*x^2+1280*x+2048)*l n(1+2*x)+(-4*x^2-6*x-2)*exp(x)+4*x+2)/((1+2*x)*exp(x)^2+(4*x+2)*exp(x)+2*x +1),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {2+4 x+e^x \left (-2-6 x-4 x^2\right )+\left (2048+1280 x+136 x^2+4 x^3+e^x \left (2048+1280 x+136 x^2+4 x^3\right )\right ) \log (1+2 x)+\left (320+708 x+139 x^2+6 x^3+e^x \left (-192-636 x-535 x^2-63 x^3-2 x^4\right )\right ) \log ^2(1+2 x)}{1+2 x+e^{2 x} (1+2 x)+e^x (2+4 x)} \, dx=\frac {{\left (x^{3} + 34 \, x^{2} + 320 \, x + 512\right )} \log \left (2 \, x + 1\right )^{2} + 2 \, x + 4}{e^{x} + 1} \]
integrate((((-2*x^4-63*x^3-535*x^2-636*x-192)*exp(x)+6*x^3+139*x^2+708*x+3 20)*log(1+2*x)^2+((4*x^3+136*x^2+1280*x+2048)*exp(x)+4*x^3+136*x^2+1280*x+ 2048)*log(1+2*x)+(-4*x^2-6*x-2)*exp(x)+4*x+2)/((1+2*x)*exp(x)^2+(4*x+2)*ex p(x)+2*x+1),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (22) = 44\).
Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07 \[ \int \frac {2+4 x+e^x \left (-2-6 x-4 x^2\right )+\left (2048+1280 x+136 x^2+4 x^3+e^x \left (2048+1280 x+136 x^2+4 x^3\right )\right ) \log (1+2 x)+\left (320+708 x+139 x^2+6 x^3+e^x \left (-192-636 x-535 x^2-63 x^3-2 x^4\right )\right ) \log ^2(1+2 x)}{1+2 x+e^{2 x} (1+2 x)+e^x (2+4 x)} \, dx=\frac {x^{3} \log {\left (2 x + 1 \right )}^{2} + 34 x^{2} \log {\left (2 x + 1 \right )}^{2} + 320 x \log {\left (2 x + 1 \right )}^{2} + 2 x + 512 \log {\left (2 x + 1 \right )}^{2} + 4}{e^{x} + 1} \]
integrate((((-2*x**4-63*x**3-535*x**2-636*x-192)*exp(x)+6*x**3+139*x**2+70 8*x+320)*ln(1+2*x)**2+((4*x**3+136*x**2+1280*x+2048)*exp(x)+4*x**3+136*x** 2+1280*x+2048)*ln(1+2*x)+(-4*x**2-6*x-2)*exp(x)+4*x+2)/((1+2*x)*exp(x)**2+ (4*x+2)*exp(x)+2*x+1),x)
(x**3*log(2*x + 1)**2 + 34*x**2*log(2*x + 1)**2 + 320*x*log(2*x + 1)**2 + 2*x + 512*log(2*x + 1)**2 + 4)/(exp(x) + 1)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {2+4 x+e^x \left (-2-6 x-4 x^2\right )+\left (2048+1280 x+136 x^2+4 x^3+e^x \left (2048+1280 x+136 x^2+4 x^3\right )\right ) \log (1+2 x)+\left (320+708 x+139 x^2+6 x^3+e^x \left (-192-636 x-535 x^2-63 x^3-2 x^4\right )\right ) \log ^2(1+2 x)}{1+2 x+e^{2 x} (1+2 x)+e^x (2+4 x)} \, dx=\frac {{\left (x^{3} + 34 \, x^{2} + 320 \, x + 512\right )} \log \left (2 \, x + 1\right )^{2} + 2 \, x + 4}{e^{x} + 1} \]
integrate((((-2*x^4-63*x^3-535*x^2-636*x-192)*exp(x)+6*x^3+139*x^2+708*x+3 20)*log(1+2*x)^2+((4*x^3+136*x^2+1280*x+2048)*exp(x)+4*x^3+136*x^2+1280*x+ 2048)*log(1+2*x)+(-4*x^2-6*x-2)*exp(x)+4*x+2)/((1+2*x)*exp(x)^2+(4*x+2)*ex p(x)+2*x+1),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (26) = 52\).
Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.15 \[ \int \frac {2+4 x+e^x \left (-2-6 x-4 x^2\right )+\left (2048+1280 x+136 x^2+4 x^3+e^x \left (2048+1280 x+136 x^2+4 x^3\right )\right ) \log (1+2 x)+\left (320+708 x+139 x^2+6 x^3+e^x \left (-192-636 x-535 x^2-63 x^3-2 x^4\right )\right ) \log ^2(1+2 x)}{1+2 x+e^{2 x} (1+2 x)+e^x (2+4 x)} \, dx=\frac {x^{3} \log \left (2 \, x + 1\right )^{2} + 34 \, x^{2} \log \left (2 \, x + 1\right )^{2} + 320 \, x \log \left (2 \, x + 1\right )^{2} + 512 \, \log \left (2 \, x + 1\right )^{2} + 2 \, x + 4}{e^{x} + 1} \]
integrate((((-2*x^4-63*x^3-535*x^2-636*x-192)*exp(x)+6*x^3+139*x^2+708*x+3 20)*log(1+2*x)^2+((4*x^3+136*x^2+1280*x+2048)*exp(x)+4*x^3+136*x^2+1280*x+ 2048)*log(1+2*x)+(-4*x^2-6*x-2)*exp(x)+4*x+2)/((1+2*x)*exp(x)^2+(4*x+2)*ex p(x)+2*x+1),x, algorithm=\
(x^3*log(2*x + 1)^2 + 34*x^2*log(2*x + 1)^2 + 320*x*log(2*x + 1)^2 + 512*l og(2*x + 1)^2 + 2*x + 4)/(e^x + 1)
Time = 13.48 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {2+4 x+e^x \left (-2-6 x-4 x^2\right )+\left (2048+1280 x+136 x^2+4 x^3+e^x \left (2048+1280 x+136 x^2+4 x^3\right )\right ) \log (1+2 x)+\left (320+708 x+139 x^2+6 x^3+e^x \left (-192-636 x-535 x^2-63 x^3-2 x^4\right )\right ) \log ^2(1+2 x)}{1+2 x+e^{2 x} (1+2 x)+e^x (2+4 x)} \, dx=\frac {2\,x+4}{{\mathrm {e}}^x+1}+\frac {{\ln \left (2\,x+1\right )}^2\,\left (x^3+34\,x^2+320\,x+512\right )}{{\mathrm {e}}^x+1} \]