Integrand size = 101, antiderivative size = 30 \[ \int \frac {e^{1-e^x-e^{\log ^2(-2+x)}+2 x+\log ^2\left (16-8 x+x^2\right )} \left (16-12 x+2 x^2+e^x \left (-8+6 x-x^2\right )+e^{\log ^2(-2+x)} (8-2 x) \log (-2+x)+(-8+4 x) \log \left (16-8 x+x^2\right )\right )}{8-6 x+x^2} \, dx=e^{1-e^x-e^{\log ^2(-2+x)}+2 x+\log ^2\left ((-4+x)^2\right )} \] Output:
exp(2*x+1-exp(x)+ln((-4+x)^2)^2-exp(ln(-2+x)^2))
Time = 0.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1-e^x-e^{\log ^2(-2+x)}+2 x+\log ^2\left (16-8 x+x^2\right )} \left (16-12 x+2 x^2+e^x \left (-8+6 x-x^2\right )+e^{\log ^2(-2+x)} (8-2 x) \log (-2+x)+(-8+4 x) \log \left (16-8 x+x^2\right )\right )}{8-6 x+x^2} \, dx=e^{1-e^x-e^{\log ^2(-2+x)}+2 x+\log ^2\left ((-4+x)^2\right )} \] Input:
Integrate[(E^(1 - E^x - E^Log[-2 + x]^2 + 2*x + Log[16 - 8*x + x^2]^2)*(16 - 12*x + 2*x^2 + E^x*(-8 + 6*x - x^2) + E^Log[-2 + x]^2*(8 - 2*x)*Log[-2 + x] + (-8 + 4*x)*Log[16 - 8*x + x^2]))/(8 - 6*x + x^2),x]
Output:
E^(1 - E^x - E^Log[-2 + x]^2 + 2*x + Log[(-4 + x)^2]^2)
Time = 3.14 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {7257}
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.
\(\displaystyle \int \frac {e^{\log ^2\left (x^2-8 x+16\right )-e^x+2 x-e^{\log ^2(x-2)}+1} \left (2 x^2+e^x \left (-x^2+6 x-8\right )+(4 x-8) \log \left (x^2-8 x+16\right )-12 x+(8-2 x) e^{\log ^2(x-2)} \log (x-2)+16\right )}{x^2-6 x+8} \, dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle e^{\log ^2\left (x^2-8 x+16\right )-e^x+2 x-e^{\log ^2(x-2)}+1}\) |
Input:
Int[(E^(1 - E^x - E^Log[-2 + x]^2 + 2*x + Log[16 - 8*x + x^2]^2)*(16 - 12* x + 2*x^2 + E^x*(-8 + 6*x - x^2) + E^Log[-2 + x]^2*(8 - 2*x)*Log[-2 + x] + (-8 + 4*x)*Log[16 - 8*x + x^2]))/(8 - 6*x + x^2),x]
Output:
E^(1 - E^x - E^Log[-2 + x]^2 + 2*x + Log[16 - 8*x + x^2]^2)
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 18.90 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03
method | result | size |
parallelrisch | \({\mathrm e}^{-{\mathrm e}^{\ln \left (-2+x \right )^{2}}+\ln \left (x^{2}-8 x +16\right )^{2}-{\mathrm e}^{x}+2 x +1}\) | \(31\) |
risch | \(\left (x -4\right )^{-4 i \operatorname {csgn}\left (i \left (x -4\right )^{2}\right ) \pi } \left (x -4\right )^{4 i \operatorname {csgn}\left (i \left (x -4\right )\right ) \pi } {\mathrm e}^{-{\mathrm e}^{\ln \left (-2+x \right )^{2}}+4 \ln \left (x -4\right )^{2}+1-\frac {\operatorname {csgn}\left (i \left (x -4\right )^{2}\right )^{6} \pi ^{2}}{4}+\operatorname {csgn}\left (i \left (x -4\right )^{2}\right )^{5} \pi ^{2} \operatorname {csgn}\left (i \left (x -4\right )\right )-\frac {3 \operatorname {csgn}\left (i \left (x -4\right )^{2}\right )^{4} \pi ^{2} \operatorname {csgn}\left (i \left (x -4\right )\right )^{2}}{2}+\operatorname {csgn}\left (i \left (x -4\right )^{2}\right )^{3} \operatorname {csgn}\left (i \left (x -4\right )\right )^{3} \pi ^{2}-\frac {\operatorname {csgn}\left (i \left (x -4\right )^{2}\right )^{2} \operatorname {csgn}\left (i \left (x -4\right )\right )^{4} \pi ^{2}}{4}-{\mathrm e}^{x}+2 x}\) | \(175\) |
Input:
int(((-2*x+8)*ln(-2+x)*exp(ln(-2+x)^2)+(4*x-8)*ln(x^2-8*x+16)+(-x^2+6*x-8) *exp(x)+2*x^2-12*x+16)*exp(-exp(ln(-2+x)^2)+ln(x^2-8*x+16)^2-exp(x)+2*x+1) /(x^2-6*x+8),x,method=_RETURNVERBOSE)
Output:
exp(-exp(ln(-2+x)^2)+ln(x^2-8*x+16)^2-exp(x)+2*x+1)
Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1-e^x-e^{\log ^2(-2+x)}+2 x+\log ^2\left (16-8 x+x^2\right )} \left (16-12 x+2 x^2+e^x \left (-8+6 x-x^2\right )+e^{\log ^2(-2+x)} (8-2 x) \log (-2+x)+(-8+4 x) \log \left (16-8 x+x^2\right )\right )}{8-6 x+x^2} \, dx=e^{\left (\log \left (x^{2} - 8 \, x + 16\right )^{2} + 2 \, x - e^{\left (\log \left (x - 2\right )^{2}\right )} - e^{x} + 1\right )} \] Input:
integrate(((-2*x+8)*log(-2+x)*exp(log(-2+x)^2)+(4*x-8)*log(x^2-8*x+16)+(-x ^2+6*x-8)*exp(x)+2*x^2-12*x+16)*exp(-exp(log(-2+x)^2)+log(x^2-8*x+16)^2-ex p(x)+2*x+1)/(x^2-6*x+8),x, algorithm="fricas")
Output:
e^(log(x^2 - 8*x + 16)^2 + 2*x - e^(log(x - 2)^2) - e^x + 1)
Time = 2.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {e^{1-e^x-e^{\log ^2(-2+x)}+2 x+\log ^2\left (16-8 x+x^2\right )} \left (16-12 x+2 x^2+e^x \left (-8+6 x-x^2\right )+e^{\log ^2(-2+x)} (8-2 x) \log (-2+x)+(-8+4 x) \log \left (16-8 x+x^2\right )\right )}{8-6 x+x^2} \, dx=e^{2 x - e^{x} - e^{\log {\left (x - 2 \right )}^{2}} + \log {\left (x^{2} - 8 x + 16 \right )}^{2} + 1} \] Input:
integrate(((-2*x+8)*ln(-2+x)*exp(ln(-2+x)**2)+(4*x-8)*ln(x**2-8*x+16)+(-x* *2+6*x-8)*exp(x)+2*x**2-12*x+16)*exp(-exp(ln(-2+x)**2)+ln(x**2-8*x+16)**2- exp(x)+2*x+1)/(x**2-6*x+8),x)
Output:
exp(2*x - exp(x) - exp(log(x - 2)**2) + log(x**2 - 8*x + 16)**2 + 1)
Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^{1-e^x-e^{\log ^2(-2+x)}+2 x+\log ^2\left (16-8 x+x^2\right )} \left (16-12 x+2 x^2+e^x \left (-8+6 x-x^2\right )+e^{\log ^2(-2+x)} (8-2 x) \log (-2+x)+(-8+4 x) \log \left (16-8 x+x^2\right )\right )}{8-6 x+x^2} \, dx=e^{\left (4 \, \log \left (x - 4\right )^{2} + 2 \, x - e^{\left (\log \left (x - 2\right )^{2}\right )} - e^{x} + 1\right )} \] Input:
integrate(((-2*x+8)*log(-2+x)*exp(log(-2+x)^2)+(4*x-8)*log(x^2-8*x+16)+(-x ^2+6*x-8)*exp(x)+2*x^2-12*x+16)*exp(-exp(log(-2+x)^2)+log(x^2-8*x+16)^2-ex p(x)+2*x+1)/(x^2-6*x+8),x, algorithm="maxima")
Output:
e^(4*log(x - 4)^2 + 2*x - e^(log(x - 2)^2) - e^x + 1)
Time = 0.14 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1-e^x-e^{\log ^2(-2+x)}+2 x+\log ^2\left (16-8 x+x^2\right )} \left (16-12 x+2 x^2+e^x \left (-8+6 x-x^2\right )+e^{\log ^2(-2+x)} (8-2 x) \log (-2+x)+(-8+4 x) \log \left (16-8 x+x^2\right )\right )}{8-6 x+x^2} \, dx=e^{\left (\log \left (x^{2} - 8 \, x + 16\right )^{2} + 2 \, x - e^{\left (\log \left (x - 2\right )^{2}\right )} - e^{x} + 1\right )} \] Input:
integrate(((-2*x+8)*log(-2+x)*exp(log(-2+x)^2)+(4*x-8)*log(x^2-8*x+16)+(-x ^2+6*x-8)*exp(x)+2*x^2-12*x+16)*exp(-exp(log(-2+x)^2)+log(x^2-8*x+16)^2-ex p(x)+2*x+1)/(x^2-6*x+8),x, algorithm="giac")
Output:
e^(log(x^2 - 8*x + 16)^2 + 2*x - e^(log(x - 2)^2) - e^x + 1)
Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^{1-e^x-e^{\log ^2(-2+x)}+2 x+\log ^2\left (16-8 x+x^2\right )} \left (16-12 x+2 x^2+e^x \left (-8+6 x-x^2\right )+e^{\log ^2(-2+x)} (8-2 x) \log (-2+x)+(-8+4 x) \log \left (16-8 x+x^2\right )\right )}{8-6 x+x^2} \, dx={\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{{\ln \left (x^2-8\,x+16\right )}^2}\,\mathrm {e}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,{\mathrm {e}}^{-{\mathrm {e}}^{{\ln \left (x-2\right )}^2}} \] Input:
int(-(exp(2*x - exp(log(x - 2)^2) - exp(x) + log(x^2 - 8*x + 16)^2 + 1)*(1 2*x - log(x^2 - 8*x + 16)*(4*x - 8) + exp(x)*(x^2 - 6*x + 8) - 2*x^2 + log (x - 2)*exp(log(x - 2)^2)*(2*x - 8) - 16))/(x^2 - 6*x + 8),x)
Output:
exp(2*x)*exp(log(x^2 - 8*x + 16)^2)*exp(1)*exp(-exp(x))*exp(-exp(log(x - 2 )^2))
Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {e^{1-e^x-e^{\log ^2(-2+x)}+2 x+\log ^2\left (16-8 x+x^2\right )} \left (16-12 x+2 x^2+e^x \left (-8+6 x-x^2\right )+e^{\log ^2(-2+x)} (8-2 x) \log (-2+x)+(-8+4 x) \log \left (16-8 x+x^2\right )\right )}{8-6 x+x^2} \, dx=\frac {e^{\mathrm {log}\left (x^{2}-8 x +16\right )^{2}+2 x} e}{e^{e^{\mathrm {log}\left (x -2\right )^{2}}+e^{x}}} \] Input:
int(((-2*x+8)*log(-2+x)*exp(log(-2+x)^2)+(4*x-8)*log(x^2-8*x+16)+(-x^2+6*x -8)*exp(x)+2*x^2-12*x+16)*exp(-exp(log(-2+x)^2)+log(x^2-8*x+16)^2-exp(x)+2 *x+1)/(x^2-6*x+8),x)
Output:
(e**(log(x**2 - 8*x + 16)**2 + 2*x)*e)/e**(e**(log(x - 2)**2) + e**x)