Integrand size = 123, antiderivative size = 32 \begin {dmath*} \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\frac {x^2}{-x+\frac {1}{4} e^5 x^2 \left (2 x+\log \left (1-\frac {x}{2}\right )\right )} \end {dmath*}
Time = 0.80 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \begin {dmath*} \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\frac {4 x}{-4+2 e^5 x^2+e^5 x \log \left (1-\frac {x}{2}\right )} \end {dmath*}
Integrate[(32 - 16*x + E^5*(12*x^2 - 8*x^3))/(-32 + 16*x + E^5*(32*x^2 - 1 6*x^3) + E^10*(-8*x^4 + 4*x^5) + (E^5*(16*x - 8*x^2) + E^10*(-8*x^3 + 4*x^ 4))*Log[(2 - x)/2] + E^10*(-2*x^2 + x^3)*Log[(2 - x)/2]^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.
\(\displaystyle \int \frac {e^5 \left (12 x^2-8 x^3\right )-16 x+32}{e^{10} \left (4 x^5-8 x^4\right )+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (x^3-2 x^2\right ) \log ^2\left (\frac {2-x}{2}\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (4 x^4-8 x^3\right )\right ) \log \left (\frac {2-x}{2}\right )+16 x-32} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 \left (2 e^5 x^3-3 e^5 x^2+4 x-8\right )}{(2-x) \left (-2 e^5 x^2-e^5 x \log \left (1-\frac {x}{2}\right )+4\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int -\frac {-2 e^5 x^3+3 e^5 x^2-4 x+8}{(2-x) \left (-2 e^5 x^2-e^5 \log \left (1-\frac {x}{2}\right ) x+4\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -4 \int \frac {-2 e^5 x^3+3 e^5 x^2-4 x+8}{(2-x) \left (-2 e^5 x^2-e^5 \log \left (1-\frac {x}{2}\right ) x+4\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -4 \int \left (\frac {2 e^5 x^2}{\left (2 e^5 x^2+e^5 \log \left (1-\frac {x}{2}\right ) x-4\right )^2}+\frac {e^5 x}{\left (2 e^5 x^2+e^5 \log \left (1-\frac {x}{2}\right ) x-4\right )^2}+\frac {4 e^5}{(x-2) \left (2 e^5 x^2+e^5 \log \left (1-\frac {x}{2}\right ) x-4\right )^2}+\frac {2 \left (2+e^5\right )}{\left (2 e^5 x^2+e^5 \log \left (1-\frac {x}{2}\right ) x-4\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \left (2 \left (2+e^5\right ) \int \frac {1}{\left (2 e^5 x^2+e^5 \log \left (1-\frac {x}{2}\right ) x-4\right )^2}dx+4 e^5 \int \frac {1}{(x-2) \left (2 e^5 x^2+e^5 \log \left (1-\frac {x}{2}\right ) x-4\right )^2}dx+e^5 \int \frac {x}{\left (2 e^5 x^2+e^5 \log \left (1-\frac {x}{2}\right ) x-4\right )^2}dx+2 e^5 \int \frac {x^2}{\left (2 e^5 x^2+e^5 \log \left (1-\frac {x}{2}\right ) x-4\right )^2}dx\right )\) |
Int[(32 - 16*x + E^5*(12*x^2 - 8*x^3))/(-32 + 16*x + E^5*(32*x^2 - 16*x^3) + E^10*(-8*x^4 + 4*x^5) + (E^5*(16*x - 8*x^2) + E^10*(-8*x^3 + 4*x^4))*Lo g[(2 - x)/2] + E^10*(-2*x^2 + x^3)*Log[(2 - x)/2]^2),x]
3.6.3.3.1 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]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.96 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78
method | result | size |
norman | \(\frac {4 x}{{\mathrm e}^{5} \ln \left (1-\frac {x}{2}\right ) x +2 x^{2} {\mathrm e}^{5}-4}\) | \(25\) |
risch | \(\frac {4 x}{{\mathrm e}^{5} \ln \left (1-\frac {x}{2}\right ) x +2 x^{2} {\mathrm e}^{5}-4}\) | \(25\) |
derivativedivides | \(-\frac {2 x}{{\mathrm e}^{5} \ln \left (1-\frac {x}{2}\right ) \left (1-\frac {x}{2}\right )-4 \,{\mathrm e}^{5} \left (1-\frac {x}{2}\right )^{2}-{\mathrm e}^{5} \ln \left (1-\frac {x}{2}\right )+8 \,{\mathrm e}^{5} \left (1-\frac {x}{2}\right )-4 \,{\mathrm e}^{5}+2}\) | \(56\) |
default | \(-\frac {2 x}{{\mathrm e}^{5} \ln \left (1-\frac {x}{2}\right ) \left (1-\frac {x}{2}\right )-4 \,{\mathrm e}^{5} \left (1-\frac {x}{2}\right )^{2}-{\mathrm e}^{5} \ln \left (1-\frac {x}{2}\right )+8 \,{\mathrm e}^{5} \left (1-\frac {x}{2}\right )-4 \,{\mathrm e}^{5}+2}\) | \(56\) |
parallelrisch | \(\frac {256+64 \,{\mathrm e}^{5} \ln \left (-2+x \right )+256 x -128 x^{2} {\mathrm e}^{5}-64 \,{\mathrm e}^{5} \ln \left (1-\frac {x}{2}\right ) x +16 x \,{\mathrm e}^{10} \ln \left (1-\frac {x}{2}\right )^{2}+32 x^{2} {\mathrm e}^{10} \ln \left (1-\frac {x}{2}\right )-64 \,{\mathrm e}^{5} \ln \left (1-\frac {x}{2}\right )-16 \,{\mathrm e}^{10} \ln \left (-2+x \right ) x \ln \left (1-\frac {x}{2}\right )-32 \,{\mathrm e}^{10} \ln \left (-2+x \right ) x^{2}}{64 \,{\mathrm e}^{5} \ln \left (1-\frac {x}{2}\right ) x +128 x^{2} {\mathrm e}^{5}-256}\) | \(125\) |
int(((-8*x^3+12*x^2)*exp(5)-16*x+32)/((x^3-2*x^2)*exp(5)^2*ln(1-1/2*x)^2+( (4*x^4-8*x^3)*exp(5)^2+(-8*x^2+16*x)*exp(5))*ln(1-1/2*x)+(4*x^5-8*x^4)*exp (5)^2+(-16*x^3+32*x^2)*exp(5)+16*x-32),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \begin {dmath*} \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\frac {4 \, x}{2 \, x^{2} e^{5} + x e^{5} \log \left (-\frac {1}{2} \, x + 1\right ) - 4} \end {dmath*}
integrate(((-8*x^3+12*x^2)*exp(5)-16*x+32)/((x^3-2*x^2)*exp(5)^2*log(1-1/2 *x)^2+((4*x^4-8*x^3)*exp(5)^2+(-8*x^2+16*x)*exp(5))*log(1-1/2*x)+(4*x^5-8* x^4)*exp(5)^2+(-16*x^3+32*x^2)*exp(5)+16*x-32),x, algorithm=\
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \begin {dmath*} \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\frac {4 x}{2 x^{2} e^{5} + x e^{5} \log {\left (1 - \frac {x}{2} \right )} - 4} \end {dmath*}
integrate(((-8*x**3+12*x**2)*exp(5)-16*x+32)/((x**3-2*x**2)*exp(5)**2*ln(1 -1/2*x)**2+((4*x**4-8*x**3)*exp(5)**2+(-8*x**2+16*x)*exp(5))*ln(1-1/2*x)+( 4*x**5-8*x**4)*exp(5)**2+(-16*x**3+32*x**2)*exp(5)+16*x-32),x)
Time = 0.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \begin {dmath*} \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\frac {4 \, x}{2 \, x^{2} e^{5} - x e^{5} \log \left (2\right ) + x e^{5} \log \left (-x + 2\right ) - 4} \end {dmath*}
integrate(((-8*x^3+12*x^2)*exp(5)-16*x+32)/((x^3-2*x^2)*exp(5)^2*log(1-1/2 *x)^2+((4*x^4-8*x^3)*exp(5)^2+(-8*x^2+16*x)*exp(5))*log(1-1/2*x)+(4*x^5-8* x^4)*exp(5)^2+(-16*x^3+32*x^2)*exp(5)+16*x-32),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \begin {dmath*} \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\frac {4 \, x}{2 \, {\left (x - 2\right )}^{2} e^{5} + {\left (x - 2\right )} e^{5} \log \left (-\frac {1}{2} \, x + 1\right ) + 8 \, {\left (x - 2\right )} e^{5} + 2 \, e^{5} \log \left (-\frac {1}{2} \, x + 1\right ) + 8 \, e^{5} - 4} \end {dmath*}
integrate(((-8*x^3+12*x^2)*exp(5)-16*x+32)/((x^3-2*x^2)*exp(5)^2*log(1-1/2 *x)^2+((4*x^4-8*x^3)*exp(5)^2+(-8*x^2+16*x)*exp(5))*log(1-1/2*x)+(4*x^5-8* x^4)*exp(5)^2+(-16*x^3+32*x^2)*exp(5)+16*x-32),x, algorithm=\
4*x/(2*(x - 2)^2*e^5 + (x - 2)*e^5*log(-1/2*x + 1) + 8*(x - 2)*e^5 + 2*e^5 *log(-1/2*x + 1) + 8*e^5 - 4)
Timed out. \begin {dmath*} \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\int \frac {{\mathrm {e}}^5\,\left (12\,x^2-8\,x^3\right )-16\,x+32}{-{\mathrm {e}}^{10}\,\left (2\,x^2-x^3\right )\,{\ln \left (1-\frac {x}{2}\right )}^2+\left ({\mathrm {e}}^5\,\left (16\,x-8\,x^2\right )-{\mathrm {e}}^{10}\,\left (8\,x^3-4\,x^4\right )\right )\,\ln \left (1-\frac {x}{2}\right )+16\,x-{\mathrm {e}}^{10}\,\left (8\,x^4-4\,x^5\right )+{\mathrm {e}}^5\,\left (32\,x^2-16\,x^3\right )-32} \,d x \end {dmath*}
int((exp(5)*(12*x^2 - 8*x^3) - 16*x + 32)/(16*x + log(1 - x/2)*(exp(5)*(16 *x - 8*x^2) - exp(10)*(8*x^3 - 4*x^4)) - exp(10)*(8*x^4 - 4*x^5) + exp(5)* (32*x^2 - 16*x^3) - exp(10)*log(1 - x/2)^2*(2*x^2 - x^3) - 32),x)