Integrand size = 148, antiderivative size = 30 \[ \int \frac {2-2 x+e^{16 x^2+8 x^3+x^4} \left (-2 x+193 x^2-16 x^3-64 x^4+4 x^5+4 x^6\right )}{e^{32 x^2+16 x^3+2 x^4} \left (-18 x+21 x^2-8 x^3+x^4\right )+e^{16 x^2+8 x^3+x^4} \left (-12 x+10 x^2-2 x^3\right ) \log \left (-2 x+x^2\right )+\left (-2 x+x^2\right ) \log ^2\left (-2 x+x^2\right )} \, dx=-1+\frac {1}{e^{x^2 (4+x)^2} (3-x)+\log \left (-2 x+x^2\right )} \] Output:
1/(ln(x^2-2*x)+(3-x)*exp((4+x)^2*x^2))-1
Time = 0.12 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {2-2 x+e^{16 x^2+8 x^3+x^4} \left (-2 x+193 x^2-16 x^3-64 x^4+4 x^5+4 x^6\right )}{e^{32 x^2+16 x^3+2 x^4} \left (-18 x+21 x^2-8 x^3+x^4\right )+e^{16 x^2+8 x^3+x^4} \left (-12 x+10 x^2-2 x^3\right ) \log \left (-2 x+x^2\right )+\left (-2 x+x^2\right ) \log ^2\left (-2 x+x^2\right )} \, dx=-\frac {1}{e^{x^2 (4+x)^2} (-3+x)-\log ((-2+x) x)} \] Input:
Integrate[(2 - 2*x + E^(16*x^2 + 8*x^3 + x^4)*(-2*x + 193*x^2 - 16*x^3 - 6 4*x^4 + 4*x^5 + 4*x^6))/(E^(32*x^2 + 16*x^3 + 2*x^4)*(-18*x + 21*x^2 - 8*x ^3 + x^4) + E^(16*x^2 + 8*x^3 + x^4)*(-12*x + 10*x^2 - 2*x^3)*Log[-2*x + x ^2] + (-2*x + x^2)*Log[-2*x + x^2]^2),x]
Output:
-(E^(x^2*(4 + x)^2)*(-3 + x) - Log[(-2 + x)*x])^(-1)
Time = 1.51 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.30, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {7292, 7237}
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.
\(\displaystyle \int \frac {e^{x^4+8 x^3+16 x^2} \left (4 x^6+4 x^5-64 x^4-16 x^3+193 x^2-2 x\right )-2 x+2}{\left (x^2-2 x\right ) \log ^2\left (x^2-2 x\right )+e^{2 x^4+16 x^3+32 x^2} \left (x^4-8 x^3+21 x^2-18 x\right )+e^{x^4+8 x^3+16 x^2} \left (-2 x^3+10 x^2-12 x\right ) \log \left (x^2-2 x\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-e^{x^4+8 x^3+16 x^2} \left (4 x^6+4 x^5-64 x^4-16 x^3+193 x^2-2 x\right )+2 x-2}{(2-x) x \left (-e^{x^2 (x+4)^2} x+3 e^{x^2 (x+4)^2}+\log ((x-2) x)\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {1}{-e^{x^2 (x+4)^2} x+3 e^{x^2 (x+4)^2}+\log (-((2-x) x))}\) |
Input:
Int[(2 - 2*x + E^(16*x^2 + 8*x^3 + x^4)*(-2*x + 193*x^2 - 16*x^3 - 64*x^4 + 4*x^5 + 4*x^6))/(E^(32*x^2 + 16*x^3 + 2*x^4)*(-18*x + 21*x^2 - 8*x^3 + x ^4) + E^(16*x^2 + 8*x^3 + x^4)*(-12*x + 10*x^2 - 2*x^3)*Log[-2*x + x^2] + (-2*x + x^2)*Log[-2*x + x^2]^2),x]
Output:
(3*E^(x^2*(4 + x)^2) - E^(x^2*(4 + x)^2)*x + Log[-((2 - x)*x)])^(-1)
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Time = 1.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53
method | result | size |
parallelrisch | \(-\frac {1}{{\mathrm e}^{x^{2} \left (x^{2}+8 x +16\right )} x -\ln \left (x^{2}-2 x \right )-3 \,{\mathrm e}^{x^{2} \left (x^{2}+8 x +16\right )}}\) | \(46\) |
risch | \(\frac {2 i}{\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (-2+x \right )\right ) \operatorname {csgn}\left (i x \left (-2+x \right )\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \left (-2+x \right )\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (-2+x \right )\right ) \operatorname {csgn}\left (i x \left (-2+x \right )\right )^{2}+\pi \operatorname {csgn}\left (i x \left (-2+x \right )\right )^{3}-2 i x \,{\mathrm e}^{\left (4+x \right )^{2} x^{2}}+6 i {\mathrm e}^{\left (4+x \right )^{2} x^{2}}+2 i \ln \left (x \right )+2 i \ln \left (-2+x \right )}\) | \(118\) |
Input:
int(((4*x^6+4*x^5-64*x^4-16*x^3+193*x^2-2*x)*exp(x^4+8*x^3+16*x^2)-2*x+2)/ ((x^2-2*x)*ln(x^2-2*x)^2+(-2*x^3+10*x^2-12*x)*exp(x^4+8*x^3+16*x^2)*ln(x^2 -2*x)+(x^4-8*x^3+21*x^2-18*x)*exp(x^4+8*x^3+16*x^2)^2),x,method=_RETURNVER BOSE)
Output:
-1/(exp(x^2*(x^2+8*x+16))*x-ln(x^2-2*x)-3*exp(x^2*(x^2+8*x+16)))
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {2-2 x+e^{16 x^2+8 x^3+x^4} \left (-2 x+193 x^2-16 x^3-64 x^4+4 x^5+4 x^6\right )}{e^{32 x^2+16 x^3+2 x^4} \left (-18 x+21 x^2-8 x^3+x^4\right )+e^{16 x^2+8 x^3+x^4} \left (-12 x+10 x^2-2 x^3\right ) \log \left (-2 x+x^2\right )+\left (-2 x+x^2\right ) \log ^2\left (-2 x+x^2\right )} \, dx=-\frac {1}{{\left (x - 3\right )} e^{\left (x^{4} + 8 \, x^{3} + 16 \, x^{2}\right )} - \log \left (x^{2} - 2 \, x\right )} \] Input:
integrate(((4*x^6+4*x^5-64*x^4-16*x^3+193*x^2-2*x)*exp(x^4+8*x^3+16*x^2)-2 *x+2)/((x^2-2*x)*log(x^2-2*x)^2+(-2*x^3+10*x^2-12*x)*exp(x^4+8*x^3+16*x^2) *log(x^2-2*x)+(x^4-8*x^3+21*x^2-18*x)*exp(x^4+8*x^3+16*x^2)^2),x, algorith m="fricas")
Output:
-1/((x - 3)*e^(x^4 + 8*x^3 + 16*x^2) - log(x^2 - 2*x))
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {2-2 x+e^{16 x^2+8 x^3+x^4} \left (-2 x+193 x^2-16 x^3-64 x^4+4 x^5+4 x^6\right )}{e^{32 x^2+16 x^3+2 x^4} \left (-18 x+21 x^2-8 x^3+x^4\right )+e^{16 x^2+8 x^3+x^4} \left (-12 x+10 x^2-2 x^3\right ) \log \left (-2 x+x^2\right )+\left (-2 x+x^2\right ) \log ^2\left (-2 x+x^2\right )} \, dx=- \frac {1}{\left (x - 3\right ) e^{x^{4} + 8 x^{3} + 16 x^{2}} - \log {\left (x^{2} - 2 x \right )}} \] Input:
integrate(((4*x**6+4*x**5-64*x**4-16*x**3+193*x**2-2*x)*exp(x**4+8*x**3+16 *x**2)-2*x+2)/((x**2-2*x)*ln(x**2-2*x)**2+(-2*x**3+10*x**2-12*x)*exp(x**4+ 8*x**3+16*x**2)*ln(x**2-2*x)+(x**4-8*x**3+21*x**2-18*x)*exp(x**4+8*x**3+16 *x**2)**2),x)
Output:
-1/((x - 3)*exp(x**4 + 8*x**3 + 16*x**2) - log(x**2 - 2*x))
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {2-2 x+e^{16 x^2+8 x^3+x^4} \left (-2 x+193 x^2-16 x^3-64 x^4+4 x^5+4 x^6\right )}{e^{32 x^2+16 x^3+2 x^4} \left (-18 x+21 x^2-8 x^3+x^4\right )+e^{16 x^2+8 x^3+x^4} \left (-12 x+10 x^2-2 x^3\right ) \log \left (-2 x+x^2\right )+\left (-2 x+x^2\right ) \log ^2\left (-2 x+x^2\right )} \, dx=-\frac {1}{{\left (x - 3\right )} e^{\left (x^{4} + 8 \, x^{3} + 16 \, x^{2}\right )} - \log \left (x - 2\right ) - \log \left (x\right )} \] Input:
integrate(((4*x^6+4*x^5-64*x^4-16*x^3+193*x^2-2*x)*exp(x^4+8*x^3+16*x^2)-2 *x+2)/((x^2-2*x)*log(x^2-2*x)^2+(-2*x^3+10*x^2-12*x)*exp(x^4+8*x^3+16*x^2) *log(x^2-2*x)+(x^4-8*x^3+21*x^2-18*x)*exp(x^4+8*x^3+16*x^2)^2),x, algorith m="maxima")
Output:
-1/((x - 3)*e^(x^4 + 8*x^3 + 16*x^2) - log(x - 2) - log(x))
Time = 0.23 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63 \[ \int \frac {2-2 x+e^{16 x^2+8 x^3+x^4} \left (-2 x+193 x^2-16 x^3-64 x^4+4 x^5+4 x^6\right )}{e^{32 x^2+16 x^3+2 x^4} \left (-18 x+21 x^2-8 x^3+x^4\right )+e^{16 x^2+8 x^3+x^4} \left (-12 x+10 x^2-2 x^3\right ) \log \left (-2 x+x^2\right )+\left (-2 x+x^2\right ) \log ^2\left (-2 x+x^2\right )} \, dx=-\frac {1}{x e^{\left (x^{4} + 8 \, x^{3} + 16 \, x^{2}\right )} - 3 \, e^{\left (x^{4} + 8 \, x^{3} + 16 \, x^{2}\right )} - \log \left (x^{2} - 2 \, x\right )} \] Input:
integrate(((4*x^6+4*x^5-64*x^4-16*x^3+193*x^2-2*x)*exp(x^4+8*x^3+16*x^2)-2 *x+2)/((x^2-2*x)*log(x^2-2*x)^2+(-2*x^3+10*x^2-12*x)*exp(x^4+8*x^3+16*x^2) *log(x^2-2*x)+(x^4-8*x^3+21*x^2-18*x)*exp(x^4+8*x^3+16*x^2)^2),x, algorith m="giac")
Output:
-1/(x*e^(x^4 + 8*x^3 + 16*x^2) - 3*e^(x^4 + 8*x^3 + 16*x^2) - log(x^2 - 2* x))
Time = 2.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {2-2 x+e^{16 x^2+8 x^3+x^4} \left (-2 x+193 x^2-16 x^3-64 x^4+4 x^5+4 x^6\right )}{e^{32 x^2+16 x^3+2 x^4} \left (-18 x+21 x^2-8 x^3+x^4\right )+e^{16 x^2+8 x^3+x^4} \left (-12 x+10 x^2-2 x^3\right ) \log \left (-2 x+x^2\right )+\left (-2 x+x^2\right ) \log ^2\left (-2 x+x^2\right )} \, dx=\frac {1}{\ln \left (x^2-2\,x\right )-{\mathrm {e}}^{x^4+8\,x^3+16\,x^2}\,\left (x-3\right )} \] Input:
int((2*x + exp(16*x^2 + 8*x^3 + x^4)*(2*x - 193*x^2 + 16*x^3 + 64*x^4 - 4* x^5 - 4*x^6) - 2)/(exp(32*x^2 + 16*x^3 + 2*x^4)*(18*x - 21*x^2 + 8*x^3 - x ^4) + log(x^2 - 2*x)^2*(2*x - x^2) + exp(16*x^2 + 8*x^3 + x^4)*log(x^2 - 2 *x)*(12*x - 10*x^2 + 2*x^3)),x)
Output:
1/(log(x^2 - 2*x) - exp(16*x^2 + 8*x^3 + x^4)*(x - 3))
\[ \int \frac {2-2 x+e^{16 x^2+8 x^3+x^4} \left (-2 x+193 x^2-16 x^3-64 x^4+4 x^5+4 x^6\right )}{e^{32 x^2+16 x^3+2 x^4} \left (-18 x+21 x^2-8 x^3+x^4\right )+e^{16 x^2+8 x^3+x^4} \left (-12 x+10 x^2-2 x^3\right ) \log \left (-2 x+x^2\right )+\left (-2 x+x^2\right ) \log ^2\left (-2 x+x^2\right )} \, dx=\int \frac {\left (4 x^{6}+4 x^{5}-64 x^{4}-16 x^{3}+193 x^{2}-2 x \right ) {\mathrm e}^{x^{4}+8 x^{3}+16 x^{2}}-2 x +2}{\left (x^{2}-2 x \right ) \mathrm {log}\left (x^{2}-2 x \right )^{2}+\left (-2 x^{3}+10 x^{2}-12 x \right ) {\mathrm e}^{x^{4}+8 x^{3}+16 x^{2}} \mathrm {log}\left (x^{2}-2 x \right )+\left (x^{4}-8 x^{3}+21 x^{2}-18 x \right ) \left ({\mathrm e}^{x^{4}+8 x^{3}+16 x^{2}}\right )^{2}}d x \] Input:
int(((4*x^6+4*x^5-64*x^4-16*x^3+193*x^2-2*x)*exp(x^4+8*x^3+16*x^2)-2*x+2)/ ((x^2-2*x)*log(x^2-2*x)^2+(-2*x^3+10*x^2-12*x)*exp(x^4+8*x^3+16*x^2)*log(x ^2-2*x)+(x^4-8*x^3+21*x^2-18*x)*exp(x^4+8*x^3+16*x^2)^2),x)
Output:
int(((4*x^6+4*x^5-64*x^4-16*x^3+193*x^2-2*x)*exp(x^4+8*x^3+16*x^2)-2*x+2)/ ((x^2-2*x)*log(x^2-2*x)^2+(-2*x^3+10*x^2-12*x)*exp(x^4+8*x^3+16*x^2)*log(x ^2-2*x)+(x^4-8*x^3+21*x^2-18*x)*exp(x^4+8*x^3+16*x^2)^2),x)