Integrand size = 90, antiderivative size = 24 \[ \int \frac {-2 e^{1+x} x+e^{5+2 x} \left (-1+x+x^2\right )+\left (-e^{1+x} x^2+e^{5+2 x} \left (-x+x^2\right )\right ) \log \left (x^2+e^{4+x} \left (x-x^2\right )\right )}{-x^2+e^{4+x} \left (-x+x^2\right )} \, dx=e^{1+x} \log \left (x^2+e^{4+x} \left (x-x^2\right )\right ) \] Output:
exp(1+x)*ln((-x^2+x)*exp(4+x)+x^2)
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {-2 e^{1+x} x+e^{5+2 x} \left (-1+x+x^2\right )+\left (-e^{1+x} x^2+e^{5+2 x} \left (-x+x^2\right )\right ) \log \left (x^2+e^{4+x} \left (x-x^2\right )\right )}{-x^2+e^{4+x} \left (-x+x^2\right )} \, dx=e^{1+x} \log \left (x \left (-e^{4+x} (-1+x)+x\right )\right ) \] Input:
Integrate[(-2*E^(1 + x)*x + E^(5 + 2*x)*(-1 + x + x^2) + (-(E^(1 + x)*x^2) + E^(5 + 2*x)*(-x + x^2))*Log[x^2 + E^(4 + x)*(x - x^2)])/(-x^2 + E^(4 + x)*(-x + x^2)),x]
Output:
E^(1 + x)*Log[x*(-(E^(4 + x)*(-1 + x)) + 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.
\(\displaystyle \int \frac {e^{2 x+5} \left (x^2+x-1\right )+\left (e^{2 x+5} \left (x^2-x\right )-e^{x+1} x^2\right ) \log \left (x^2+e^{x+4} \left (x-x^2\right )\right )-2 e^{x+1} x}{e^{x+4} \left (x^2-x\right )-x^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x \left (x^2-x+1\right )}{e^3 (x-1)^2 \left (e^{x+4} x-x-e^{x+4}\right )}+\frac {x^2-x+1}{e^3 (x-1)^2}+\frac {e^{x+1} \left (x^2+x^2 \log \left (x \left (x-e^{x+4} (x-1)\right )\right )+x-x \log \left (x \left (x-e^{x+4} (x-1)\right )\right )-1\right )}{(x-1) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\int \frac {1}{e^{x+4} x-x-e^{x+4}}dx}{e^3}+\frac {\int \frac {1}{(x-1)^2 \left (e^{x+4} x-x-e^{x+4}\right )}dx}{e^3}+\frac {2 \int \frac {1}{(x-1) \left (e^{x+4} x-x-e^{x+4}\right )}dx}{e^3}-\int \frac {e^{x+1}}{(x-1) \left (e^{x+4} x-x-e^{x+4}\right )}dx+\frac {\int \frac {x}{e^{x+4} x-x-e^{x+4}}dx}{e^3}-\int \frac {e^{x+1} x}{e^{x+4} x-x-e^{x+4}}dx+e^{x+1} \log \left (x^2+e^{x+4} (1-x) x\right )+\frac {x}{e^3}+\frac {1}{e^3 (1-x)}+\frac {\log (1-x)}{e^3}\) |
Input:
Int[(-2*E^(1 + x)*x + E^(5 + 2*x)*(-1 + x + x^2) + (-(E^(1 + x)*x^2) + E^( 5 + 2*x)*(-x + x^2))*Log[x^2 + E^(4 + x)*(x - x^2)])/(-x^2 + E^(4 + x)*(-x + x^2)),x]
Output:
$Aborted
Time = 1.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \({\mathrm e}^{1+x} \ln \left (\left (-x^{2}+x \right ) {\mathrm e}^{4+x}+x^{2}\right )\) | \(23\) |
norman | \({\mathrm e}^{1+x} \ln \left (\left (-x^{2}+x \right ) {\mathrm e}^{3} {\mathrm e}^{1+x}+x^{2}\right )\) | \(25\) |
risch | \({\mathrm e}^{1+x} \ln \left (x \,{\mathrm e}^{4+x}-{\mathrm e}^{4+x}-x \right )+{\mathrm e}^{1+x} \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (x \,{\mathrm e}^{4+x}-{\mathrm e}^{4+x}-x \right )\right ) \operatorname {csgn}\left (i x \left (x \,{\mathrm e}^{4+x}-{\mathrm e}^{4+x}-x \right )\right ) {\mathrm e}^{1+x}}{2}+\frac {i \pi \,\operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (i x \left (x \,{\mathrm e}^{4+x}-{\mathrm e}^{4+x}-x \right )\right )}^{2} {\mathrm e}^{1+x}}{2}-i \pi {\operatorname {csgn}\left (i x \left (x \,{\mathrm e}^{4+x}-{\mathrm e}^{4+x}-x \right )\right )}^{2} {\mathrm e}^{1+x}+\frac {i \pi \,\operatorname {csgn}\left (i \left (x \,{\mathrm e}^{4+x}-{\mathrm e}^{4+x}-x \right )\right ) {\operatorname {csgn}\left (i x \left (x \,{\mathrm e}^{4+x}-{\mathrm e}^{4+x}-x \right )\right )}^{2} {\mathrm e}^{1+x}}{2}+\frac {i \pi {\operatorname {csgn}\left (i x \left (x \,{\mathrm e}^{4+x}-{\mathrm e}^{4+x}-x \right )\right )}^{3} {\mathrm e}^{1+x}}{2}+i \pi \,{\mathrm e}^{1+x}\) | \(242\) |
Input:
int((((x^2-x)*exp(1+x)*exp(4+x)-x^2*exp(1+x))*ln((-x^2+x)*exp(4+x)+x^2)+(x ^2+x-1)*exp(1+x)*exp(4+x)-2*x*exp(1+x))/((x^2-x)*exp(4+x)-x^2),x,method=_R ETURNVERBOSE)
Output:
exp(1+x)*ln((-x^2+x)*exp(4+x)+x^2)
Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-2 e^{1+x} x+e^{5+2 x} \left (-1+x+x^2\right )+\left (-e^{1+x} x^2+e^{5+2 x} \left (-x+x^2\right )\right ) \log \left (x^2+e^{4+x} \left (x-x^2\right )\right )}{-x^2+e^{4+x} \left (-x+x^2\right )} \, dx=e^{\left (x + 1\right )} \log \left (x^{2} - {\left (x^{2} - x\right )} e^{\left (x + 4\right )}\right ) \] Input:
integrate((((x^2-x)*exp(1+x)*exp(4+x)-x^2*exp(1+x))*log((-x^2+x)*exp(4+x)+ x^2)+(x^2+x-1)*exp(1+x)*exp(4+x)-2*x*exp(1+x))/((x^2-x)*exp(4+x)-x^2),x, a lgorithm="fricas")
Output:
e^(x + 1)*log(x^2 - (x^2 - x)*e^(x + 4))
Timed out. \[ \int \frac {-2 e^{1+x} x+e^{5+2 x} \left (-1+x+x^2\right )+\left (-e^{1+x} x^2+e^{5+2 x} \left (-x+x^2\right )\right ) \log \left (x^2+e^{4+x} \left (x-x^2\right )\right )}{-x^2+e^{4+x} \left (-x+x^2\right )} \, dx=\text {Timed out} \] Input:
integrate((((x**2-x)*exp(1+x)*exp(4+x)-x**2*exp(1+x))*ln((-x**2+x)*exp(4+x )+x**2)+(x**2+x-1)*exp(1+x)*exp(4+x)-2*x*exp(1+x))/((x**2-x)*exp(4+x)-x**2 ),x)
Output:
Timed out
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-2 e^{1+x} x+e^{5+2 x} \left (-1+x+x^2\right )+\left (-e^{1+x} x^2+e^{5+2 x} \left (-x+x^2\right )\right ) \log \left (x^2+e^{4+x} \left (x-x^2\right )\right )}{-x^2+e^{4+x} \left (-x+x^2\right )} \, dx=e^{\left (x + 1\right )} \log \left (-{\left (x e^{4} - e^{4}\right )} e^{x} + x\right ) + e^{\left (x + 1\right )} \log \left (x\right ) \] Input:
integrate((((x^2-x)*exp(1+x)*exp(4+x)-x^2*exp(1+x))*log((-x^2+x)*exp(4+x)+ x^2)+(x^2+x-1)*exp(1+x)*exp(4+x)-2*x*exp(1+x))/((x^2-x)*exp(4+x)-x^2),x, a lgorithm="maxima")
Output:
e^(x + 1)*log(-(x*e^4 - e^4)*e^x + x) + e^(x + 1)*log(x)
Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {-2 e^{1+x} x+e^{5+2 x} \left (-1+x+x^2\right )+\left (-e^{1+x} x^2+e^{5+2 x} \left (-x+x^2\right )\right ) \log \left (x^2+e^{4+x} \left (x-x^2\right )\right )}{-x^2+e^{4+x} \left (-x+x^2\right )} \, dx={\left (e^{\left (x + 4\right )} \log \left ({\left (x + 4\right )} e^{\left (x + 4\right )} - x - 5 \, e^{\left (x + 4\right )}\right ) + e^{\left (x + 4\right )} \log \left (-x\right )\right )} e^{\left (-3\right )} \] Input:
integrate((((x^2-x)*exp(1+x)*exp(4+x)-x^2*exp(1+x))*log((-x^2+x)*exp(4+x)+ x^2)+(x^2+x-1)*exp(1+x)*exp(4+x)-2*x*exp(1+x))/((x^2-x)*exp(4+x)-x^2),x, a lgorithm="giac")
Output:
(e^(x + 4)*log((x + 4)*e^(x + 4) - x - 5*e^(x + 4)) + e^(x + 4)*log(-x))*e ^(-3)
Time = 4.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-2 e^{1+x} x+e^{5+2 x} \left (-1+x+x^2\right )+\left (-e^{1+x} x^2+e^{5+2 x} \left (-x+x^2\right )\right ) \log \left (x^2+e^{4+x} \left (x-x^2\right )\right )}{-x^2+e^{4+x} \left (-x+x^2\right )} \, dx=\ln \left (x^2+{\mathrm {e}}^4\,{\mathrm {e}}^x\,\left (x-x^2\right )\right )\,{\mathrm {e}}^{x+1} \] Input:
int((2*x*exp(x + 1) + log(exp(x + 4)*(x - x^2) + x^2)*(x^2*exp(x + 1) + ex p(x + 1)*exp(x + 4)*(x - x^2)) - exp(x + 1)*exp(x + 4)*(x + x^2 - 1))/(exp (x + 4)*(x - x^2) + x^2),x)
Output:
log(x^2 + exp(4)*exp(x)*(x - x^2))*exp(x + 1)
Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-2 e^{1+x} x+e^{5+2 x} \left (-1+x+x^2\right )+\left (-e^{1+x} x^2+e^{5+2 x} \left (-x+x^2\right )\right ) \log \left (x^2+e^{4+x} \left (x-x^2\right )\right )}{-x^2+e^{4+x} \left (-x+x^2\right )} \, dx=e^{x} \mathrm {log}\left (-e^{x} e^{4} x^{2}+e^{x} e^{4} x +x^{2}\right ) e \] Input:
int((((x^2-x)*exp(1+x)*exp(4+x)-x^2*exp(1+x))*log((-x^2+x)*exp(4+x)+x^2)+( x^2+x-1)*exp(1+x)*exp(4+x)-2*x*exp(1+x))/((x^2-x)*exp(4+x)-x^2),x)
Output:
e**x*log( - e**x*e**4*x**2 + e**x*e**4*x + x**2)*e