Integrand size = 109, antiderivative size = 31 \[ \int \frac {-8 e^{2 x}-8 x^2-32 x^3+x^4+e^x \left (16 x+44 x^2-13 x^3\right )+\left (3 x^4+e^x \left (-4 x^3+x^4\right )\right ) \log (x)+\left (-8 x^3+e^x \left (12 x^2-4 x^3\right )\right ) \log \left (x^2\right )}{2 e^{2 x}-4 e^x x+2 x^2} \, dx=2 x \left (-2+\frac {x^2 \left (3-\frac {1}{4} x \log (x)+\log \left (x^2\right )\right )}{e^x-x}\right ) \]
Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {-8 e^{2 x}-8 x^2-32 x^3+x^4+e^x \left (16 x+44 x^2-13 x^3\right )+\left (3 x^4+e^x \left (-4 x^3+x^4\right )\right ) \log (x)+\left (-8 x^3+e^x \left (12 x^2-4 x^3\right )\right ) \log \left (x^2\right )}{2 e^{2 x}-4 e^x x+2 x^2} \, dx=\frac {x \left (-8 e^x+8 x+12 x^2-x^3 \log (x)+4 x^2 \log \left (x^2\right )\right )}{2 \left (e^x-x\right )} \]
Integrate[(-8*E^(2*x) - 8*x^2 - 32*x^3 + x^4 + E^x*(16*x + 44*x^2 - 13*x^3 ) + (3*x^4 + E^x*(-4*x^3 + x^4))*Log[x] + (-8*x^3 + E^x*(12*x^2 - 4*x^3))* Log[x^2])/(2*E^(2*x) - 4*E^x*x + 2*x^2),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.
\(\displaystyle \int \frac {x^4-32 x^3-8 x^2+\left (3 x^4+e^x \left (x^4-4 x^3\right )\right ) \log (x)+e^x \left (-13 x^3+44 x^2+16 x\right )+\left (e^x \left (12 x^2-4 x^3\right )-8 x^3\right ) \log \left (x^2\right )-8 e^{2 x}}{2 x^2-4 e^x x+2 e^{2 x}} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {x^4-32 x^3-8 x^2+\left (3 x^4+e^x \left (x^4-4 x^3\right )\right ) \log (x)+e^x \left (-13 x^3+44 x^2+16 x\right )+\left (e^x \left (12 x^2-4 x^3\right )-8 x^3\right ) \log \left (x^2\right )-8 e^{2 x}}{2 \left (e^x-x\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {-x^4+32 x^3+8 x^2+8 e^{2 x}-e^x \left (-13 x^3+44 x^2+16 x\right )-\left (3 x^4-e^x \left (4 x^3-x^4\right )\right ) \log (x)+4 \left (2 x^3-e^x \left (3 x^2-x^3\right )\right ) \log \left (x^2\right )}{\left (e^x-x\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {-x^4+32 x^3+8 x^2+8 e^{2 x}-e^x \left (-13 x^3+44 x^2+16 x\right )-\left (3 x^4-e^x \left (4 x^3-x^4\right )\right ) \log (x)+4 \left (2 x^3-e^x \left (3 x^2-x^3\right )\right ) \log \left (x^2\right )}{\left (e^x-x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{2} \int \left (-\frac {(x-1) \left (x \log (x)-4 \log \left (x^2\right )-12\right ) x^3}{\left (e^x-x\right )^2}-\frac {\left (\log (x) x^2-4 \log (x) x-4 \log \left (x^2\right ) x-13 x+12 \log \left (x^2\right )+44\right ) x^2}{e^x-x}+8\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {\int \frac {x^5}{\left (e^x-x\right )^2}dx}{x}dx+\log (x) \int \frac {x^5}{\left (e^x-x\right )^2}dx-12 \int \frac {x^4}{\left (e^x-x\right )^2}dx+9 \int \frac {\int \frac {x^4}{\left (e^x-x\right )^2}dx}{x}dx-\int \frac {\int \frac {x^4}{e^x-x}dx}{x}dx-\log (x) \int \frac {x^4}{\left (e^x-x\right )^2}dx+\log (x) \int \frac {x^4}{e^x-x}dx+12 \int \frac {x^3}{\left (e^x-x\right )^2}dx-13 \int \frac {x^3}{e^x-x}dx-8 \int \frac {\int \frac {x^3}{\left (e^x-x\right )^2}dx}{x}dx+12 \int \frac {\int \frac {x^3}{e^x-x}dx}{x}dx-4 \log (x) \int \frac {x^3}{e^x-x}dx+44 \int \frac {x^2}{e^x-x}dx-24 \int \frac {\int \frac {x^2}{e^x-x}dx}{x}dx+12 \log \left (x^2\right ) \int \frac {x^2}{e^x-x}dx-4 \log \left (x^2\right ) \int \frac {x^4}{\left (e^x-x\right )^2}dx+4 \log \left (x^2\right ) \int \frac {x^3}{\left (e^x-x\right )^2}dx-4 \log \left (x^2\right ) \int \frac {x^3}{e^x-x}dx-8 x\right )\) |
Int[(-8*E^(2*x) - 8*x^2 - 32*x^3 + x^4 + E^x*(16*x + 44*x^2 - 13*x^3) + (3 *x^4 + E^x*(-4*x^3 + x^4))*Log[x] + (-8*x^3 + E^x*(12*x^2 - 4*x^3))*Log[x^ 2])/(2*E^(2*x) - 4*E^x*x + 2*x^2),x]
3.8.90.3.1 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 = 1.90 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39
method | result | size |
parallelrisch | \(\frac {-480 x^{3} \ln \left (x^{2}\right )+120 x^{4} \ln \left (x \right )+960 \,{\mathrm e}^{x} x -1440 x^{3}-960 x^{2}}{240 x -240 \,{\mathrm e}^{x}}\) | \(43\) |
risch | \(\frac {\left (-8+x \right ) x^{3} \ln \left (x \right )}{-2 \,{\mathrm e}^{x}+2 x}-\frac {x \left (i \pi \,x^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 i \pi \,x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+i \pi \,x^{2} \operatorname {csgn}\left (i x^{2}\right )^{3}-6 x^{2}-4 x +4 \,{\mathrm e}^{x}\right )}{{\mathrm e}^{x}-x}\) | \(102\) |
int((((-4*x^3+12*x^2)*exp(x)-8*x^3)*ln(x^2)+((x^4-4*x^3)*exp(x)+3*x^4)*ln( x)-8*exp(x)^2+(-13*x^3+44*x^2+16*x)*exp(x)+x^4-32*x^3-8*x^2)/(2*exp(x)^2-4 *exp(x)*x+2*x^2),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {-8 e^{2 x}-8 x^2-32 x^3+x^4+e^x \left (16 x+44 x^2-13 x^3\right )+\left (3 x^4+e^x \left (-4 x^3+x^4\right )\right ) \log (x)+\left (-8 x^3+e^x \left (12 x^2-4 x^3\right )\right ) \log \left (x^2\right )}{2 e^{2 x}-4 e^x x+2 x^2} \, dx=-\frac {12 \, x^{3} + 8 \, x^{2} - 8 \, x e^{x} - {\left (x^{4} - 8 \, x^{3}\right )} \log \left (x\right )}{2 \, {\left (x - e^{x}\right )}} \]
integrate((((-4*x^3+12*x^2)*exp(x)-8*x^3)*log(x^2)+((x^4-4*x^3)*exp(x)+3*x ^4)*log(x)-8*exp(x)^2+(-13*x^3+44*x^2+16*x)*exp(x)+x^4-32*x^3-8*x^2)/(2*ex p(x)^2-4*exp(x)*x+2*x^2),x, algorithm=\
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-8 e^{2 x}-8 x^2-32 x^3+x^4+e^x \left (16 x+44 x^2-13 x^3\right )+\left (3 x^4+e^x \left (-4 x^3+x^4\right )\right ) \log (x)+\left (-8 x^3+e^x \left (12 x^2-4 x^3\right )\right ) \log \left (x^2\right )}{2 e^{2 x}-4 e^x x+2 x^2} \, dx=- 4 x + \frac {- x^{4} \log {\left (x \right )} + 8 x^{3} \log {\left (x \right )} + 12 x^{3}}{- 2 x + 2 e^{x}} \]
integrate((((-4*x**3+12*x**2)*exp(x)-8*x**3)*ln(x**2)+((x**4-4*x**3)*exp(x )+3*x**4)*ln(x)-8*exp(x)**2+(-13*x**3+44*x**2+16*x)*exp(x)+x**4-32*x**3-8* x**2)/(2*exp(x)**2-4*exp(x)*x+2*x**2),x)
Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {-8 e^{2 x}-8 x^2-32 x^3+x^4+e^x \left (16 x+44 x^2-13 x^3\right )+\left (3 x^4+e^x \left (-4 x^3+x^4\right )\right ) \log (x)+\left (-8 x^3+e^x \left (12 x^2-4 x^3\right )\right ) \log \left (x^2\right )}{2 e^{2 x}-4 e^x x+2 x^2} \, dx=-\frac {12 \, x^{3} + 8 \, x^{2} - 8 \, x e^{x} - {\left (x^{4} - 8 \, x^{3}\right )} \log \left (x\right )}{2 \, {\left (x - e^{x}\right )}} \]
integrate((((-4*x^3+12*x^2)*exp(x)-8*x^3)*log(x^2)+((x^4-4*x^3)*exp(x)+3*x ^4)*log(x)-8*exp(x)^2+(-13*x^3+44*x^2+16*x)*exp(x)+x^4-32*x^3-8*x^2)/(2*ex p(x)^2-4*exp(x)*x+2*x^2),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {-8 e^{2 x}-8 x^2-32 x^3+x^4+e^x \left (16 x+44 x^2-13 x^3\right )+\left (3 x^4+e^x \left (-4 x^3+x^4\right )\right ) \log (x)+\left (-8 x^3+e^x \left (12 x^2-4 x^3\right )\right ) \log \left (x^2\right )}{2 e^{2 x}-4 e^x x+2 x^2} \, dx=\frac {x^{4} \log \left (x\right ) - 8 \, x^{3} \log \left (x\right ) - 12 \, x^{3} - 8 \, x^{2} + 8 \, x e^{x}}{2 \, {\left (x - e^{x}\right )}} \]
integrate((((-4*x^3+12*x^2)*exp(x)-8*x^3)*log(x^2)+((x^4-4*x^3)*exp(x)+3*x ^4)*log(x)-8*exp(x)^2+(-13*x^3+44*x^2+16*x)*exp(x)+x^4-32*x^3-8*x^2)/(2*ex p(x)^2-4*exp(x)*x+2*x^2),x, algorithm=\
Time = 9.61 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {-8 e^{2 x}-8 x^2-32 x^3+x^4+e^x \left (16 x+44 x^2-13 x^3\right )+\left (3 x^4+e^x \left (-4 x^3+x^4\right )\right ) \log (x)+\left (-8 x^3+e^x \left (12 x^2-4 x^3\right )\right ) \log \left (x^2\right )}{2 e^{2 x}-4 e^x x+2 x^2} \, dx=-\frac {x\,\left (8\,x-8\,{\mathrm {e}}^x-x^3\,\ln \left (x\right )+4\,x^2\,\ln \left (x^2\right )+12\,x^2\right )}{2\,\left (x-{\mathrm {e}}^x\right )} \]