Integrand size = 101, antiderivative size = 28 \[ \int \frac {-1536-400 x+96 x^2+25 x^3+e^x \left (-768-8 x+146 x^2+25 x^3\right )+\left (-8 x-4 e^x x+4 x^2\right ) \log \left (\frac {4 x^2}{2+e^x-x}\right )}{160 x-30 x^3-5 x^4+e^x \left (80 x+40 x^2+5 x^3\right )} \, dx=\left (-5+\frac {4}{5 (4+x)}\right ) \log \left (\frac {4 x^2}{2+e^x-x}\right ) \]
Time = 0.74 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {-1536-400 x+96 x^2+25 x^3+e^x \left (-768-8 x+146 x^2+25 x^3\right )+\left (-8 x-4 e^x x+4 x^2\right ) \log \left (\frac {4 x^2}{2+e^x-x}\right )}{160 x-30 x^3-5 x^4+e^x \left (80 x+40 x^2+5 x^3\right )} \, dx=\frac {1}{5} \left (25 \log \left (2+e^x-x\right )-50 \log (x)+\frac {4 \log \left (\frac {4 x^2}{2+e^x-x}\right )}{4+x}\right ) \]
Integrate[(-1536 - 400*x + 96*x^2 + 25*x^3 + E^x*(-768 - 8*x + 146*x^2 + 2 5*x^3) + (-8*x - 4*E^x*x + 4*x^2)*Log[(4*x^2)/(2 + E^x - x)])/(160*x - 30* x^3 - 5*x^4 + E^x*(80*x + 40*x^2 + 5*x^3)),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.
\(\displaystyle \int \frac {25 x^3+96 x^2+\left (4 x^2-4 e^x x-8 x\right ) \log \left (\frac {4 x^2}{-x+e^x+2}\right )+e^x \left (25 x^3+146 x^2-8 x-768\right )-400 x-1536}{-5 x^4-30 x^3+e^x \left (5 x^3+40 x^2+80 x\right )+160 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {25 x^3+96 x^2+\left (4 x^2-4 e^x x-8 x\right ) \log \left (\frac {4 x^2}{-x+e^x+2}\right )+e^x \left (25 x^3+146 x^2-8 x-768\right )-400 x-1536}{5 \left (-x+e^x+2\right ) x (x+4)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {-25 x^3-96 x^2+400 x+e^x \left (-25 x^3-146 x^2+8 x+768\right )+4 \left (-x^2+e^x x+2 x\right ) \log \left (\frac {4 x^2}{-x+e^x+2}\right )+1536}{\left (-x+e^x+2\right ) x (x+4)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int \frac {-25 x^3-96 x^2+400 x+e^x \left (-25 x^3-146 x^2+8 x+768\right )+4 \left (-x^2+e^x x+2 x\right ) \log \left (\frac {4 x^2}{-x+e^x+2}\right )+1536}{\left (-x+e^x+2\right ) x (x+4)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{5} \int \left (\frac {-25 x^3-146 x^2+4 \log \left (\frac {4 x^2}{-x+e^x+2}\right ) x+8 x+768}{x (x+4)^2}-\frac {25 x^2+21 x-288}{\left (-x+e^x+2\right ) (x+4)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-75 \int \frac {1}{-x+e^x+2}dx+25 \int \frac {x}{-x+e^x+2}dx+\frac {4 \log \left (\frac {4 x^2}{-x+e^x+2}\right )}{x+4}+25 x-50 \log (x)\right )\) |
Int[(-1536 - 400*x + 96*x^2 + 25*x^3 + E^x*(-768 - 8*x + 146*x^2 + 25*x^3) + (-8*x - 4*E^x*x + 4*x^2)*Log[(4*x^2)/(2 + E^x - x)])/(160*x - 30*x^3 - 5*x^4 + E^x*(80*x + 40*x^2 + 5*x^3)),x]
3.17.65.3.1 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 = 0.42 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54
method | result | size |
norman | \(\frac {-\frac {96 \ln \left (\frac {4 x^{2}}{{\mathrm e}^{x}+2-x}\right )}{5}-5 \ln \left (\frac {4 x^{2}}{{\mathrm e}^{x}+2-x}\right ) x}{4+x}\) | \(43\) |
parallelrisch | \(-\frac {25 \ln \left (\frac {4 x^{2}}{{\mathrm e}^{x}+2-x}\right ) x +96 \ln \left (\frac {4 x^{2}}{{\mathrm e}^{x}+2-x}\right )}{5 \left (4+x \right )}\) | \(44\) |
risch | \(-\frac {4 \ln \left (-{\mathrm e}^{x}+x -2\right )}{5 \left (4+x \right )}+\frac {-2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-2 i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}+2-x}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{{\mathrm e}^{x}+2-x}\right )^{2}-2 i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{x}+2-x}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{{\mathrm e}^{x}+2-x}\right )-4 i \pi \operatorname {csgn}\left (\frac {i x^{2}}{{\mathrm e}^{x}+2-x}\right )^{2}-2 i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+4 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+2 i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{{\mathrm e}^{x}+2-x}\right )^{2}+4 i \pi -2 i \pi \operatorname {csgn}\left (\frac {i x^{2}}{{\mathrm e}^{x}+2-x}\right )^{3}-50 x \ln \left (x \right )+25 \ln \left ({\mathrm e}^{x}+2-x \right ) x -192 \ln \left (x \right )+8 \ln \left (2\right )+100 \ln \left ({\mathrm e}^{x}+2-x \right )}{20+5 x}\) | \(260\) |
int(((-4*exp(x)*x+4*x^2-8*x)*ln(4*x^2/(exp(x)+2-x))+(25*x^3+146*x^2-8*x-76 8)*exp(x)+25*x^3+96*x^2-400*x-1536)/((5*x^3+40*x^2+80*x)*exp(x)-5*x^4-30*x ^3+160*x),x,method=_RETURNVERBOSE)
Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {-1536-400 x+96 x^2+25 x^3+e^x \left (-768-8 x+146 x^2+25 x^3\right )+\left (-8 x-4 e^x x+4 x^2\right ) \log \left (\frac {4 x^2}{2+e^x-x}\right )}{160 x-30 x^3-5 x^4+e^x \left (80 x+40 x^2+5 x^3\right )} \, dx=-\frac {{\left (25 \, x + 96\right )} \log \left (-\frac {4 \, x^{2}}{x - e^{x} - 2}\right )}{5 \, {\left (x + 4\right )}} \]
integrate(((-4*exp(x)*x+4*x^2-8*x)*log(4*x^2/(exp(x)+2-x))+(25*x^3+146*x^2 -8*x-768)*exp(x)+25*x^3+96*x^2-400*x-1536)/((5*x^3+40*x^2+80*x)*exp(x)-5*x ^4-30*x^3+160*x),x, algorithm=\
Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {-1536-400 x+96 x^2+25 x^3+e^x \left (-768-8 x+146 x^2+25 x^3\right )+\left (-8 x-4 e^x x+4 x^2\right ) \log \left (\frac {4 x^2}{2+e^x-x}\right )}{160 x-30 x^3-5 x^4+e^x \left (80 x+40 x^2+5 x^3\right )} \, dx=- 10 \log {\left (x \right )} + 5 \log {\left (- x + e^{x} + 2 \right )} + \frac {4 \log {\left (\frac {4 x^{2}}{- x + e^{x} + 2} \right )}}{5 x + 20} \]
integrate(((-4*exp(x)*x+4*x**2-8*x)*ln(4*x**2/(exp(x)+2-x))+(25*x**3+146*x **2-8*x-768)*exp(x)+25*x**3+96*x**2-400*x-1536)/((5*x**3+40*x**2+80*x)*exp (x)-5*x**4-30*x**3+160*x),x)
Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-1536-400 x+96 x^2+25 x^3+e^x \left (-768-8 x+146 x^2+25 x^3\right )+\left (-8 x-4 e^x x+4 x^2\right ) \log \left (\frac {4 x^2}{2+e^x-x}\right )}{160 x-30 x^3-5 x^4+e^x \left (80 x+40 x^2+5 x^3\right )} \, dx=-\frac {2 \, {\left (25 \, x + 96\right )} \log \left (x\right ) - {\left (25 \, x + 96\right )} \log \left (-x + e^{x} + 2\right ) - 8 \, \log \left (2\right )}{5 \, {\left (x + 4\right )}} \]
integrate(((-4*exp(x)*x+4*x^2-8*x)*log(4*x^2/(exp(x)+2-x))+(25*x^3+146*x^2 -8*x-768)*exp(x)+25*x^3+96*x^2-400*x-1536)/((5*x^3+40*x^2+80*x)*exp(x)-5*x ^4-30*x^3+160*x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (26) = 52\).
Time = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.96 \[ \int \frac {-1536-400 x+96 x^2+25 x^3+e^x \left (-768-8 x+146 x^2+25 x^3\right )+\left (-8 x-4 e^x x+4 x^2\right ) \log \left (\frac {4 x^2}{2+e^x-x}\right )}{160 x-30 x^3-5 x^4+e^x \left (80 x+40 x^2+5 x^3\right )} \, dx=\frac {25 \, x \log \left (x - e^{x} - 2\right ) - 50 \, x \log \left (x\right ) + 100 \, \log \left (x - e^{x} - 2\right ) - 200 \, \log \left (x\right ) + 4 \, \log \left (-\frac {4 \, x^{2}}{x - e^{x} - 2}\right )}{5 \, {\left (x + 4\right )}} \]
integrate(((-4*exp(x)*x+4*x^2-8*x)*log(4*x^2/(exp(x)+2-x))+(25*x^3+146*x^2 -8*x-768)*exp(x)+25*x^3+96*x^2-400*x-1536)/((5*x^3+40*x^2+80*x)*exp(x)-5*x ^4-30*x^3+160*x),x, algorithm=\
1/5*(25*x*log(x - e^x - 2) - 50*x*log(x) + 100*log(x - e^x - 2) - 200*log( x) + 4*log(-4*x^2/(x - e^x - 2)))/(x + 4)
Time = 13.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {-1536-400 x+96 x^2+25 x^3+e^x \left (-768-8 x+146 x^2+25 x^3\right )+\left (-8 x-4 e^x x+4 x^2\right ) \log \left (\frac {4 x^2}{2+e^x-x}\right )}{160 x-30 x^3-5 x^4+e^x \left (80 x+40 x^2+5 x^3\right )} \, dx=5\,\ln \left ({\mathrm {e}}^x-x+2\right )-10\,\ln \left (x\right )+\frac {4\,\ln \left (\frac {4\,x^2}{{\mathrm {e}}^x-x+2}\right )}{5\,\left (x+4\right )} \]