Integrand size = 84, antiderivative size = 31 \[ \int \frac {-13 x-4 x^2+e^{x+x^2} \left (14+4 x-3 x^2-2 x^3\right )+\left (e^{x+x^2} (-1-2 x)+x+2 x^2\right ) \log \left (-e^{x+x^2} x+x^2\right )}{3 e^{x+x^2}-3 x} \, dx=x (1+x) \left (\frac {4+x}{x}-\frac {1}{3} \log \left (x \left (-e^{x+x^2}+x\right )\right )\right ) \] Output:
(1+x)*x*((4+x)/x-1/3*ln((x-exp(x^2+x))*x))
Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {-13 x-4 x^2+e^{x+x^2} \left (14+4 x-3 x^2-2 x^3\right )+\left (e^{x+x^2} (-1-2 x)+x+2 x^2\right ) \log \left (-e^{x+x^2} x+x^2\right )}{3 e^{x+x^2}-3 x} \, dx=\frac {1}{3} \left (15 x+3 x^2-x (1+x) \log \left (x \left (-e^{x+x^2}+x\right )\right )\right ) \] Input:
Integrate[(-13*x - 4*x^2 + E^(x + x^2)*(14 + 4*x - 3*x^2 - 2*x^3) + (E^(x + x^2)*(-1 - 2*x) + x + 2*x^2)*Log[-(E^(x + x^2)*x) + x^2])/(3*E^(x + x^2) - 3*x),x]
Output:
(15*x + 3*x^2 - x*(1 + x)*Log[x*(-E^(x + x^2) + x)])/3
Time = 1.89 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {7292, 27, 25, 7293, 2009}
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.
\(\displaystyle \int \frac {-4 x^2+\left (2 x^2+e^{x^2+x} (-2 x-1)+x\right ) \log \left (x^2-e^{x^2+x} x\right )+e^{x^2+x} \left (-2 x^3-3 x^2+4 x+14\right )-13 x}{3 e^{x^2+x}-3 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-4 x^2+\left (2 x^2+e^{x^2+x} (-2 x-1)+x\right ) \log \left (x^2-e^{x^2+x} x\right )+e^{x^2+x} \left (-2 x^3-3 x^2+4 x+14\right )-13 x}{3 \left (e^{x^2+x}-x\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int -\frac {4 x^2+13 x-e^{x^2+x} \left (-2 x^3-3 x^2+4 x+14\right )-\left (2 x^2+x-e^{x^2+x} (2 x+1)\right ) \log \left (x^2-e^{x^2+x} x\right )}{e^{x^2+x}-x}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{3} \int \frac {4 x^2+13 x-e^{x^2+x} \left (-2 x^3-3 x^2+4 x+14\right )-\left (2 x^2+x-e^{x^2+x} (2 x+1)\right ) \log \left (x^2-e^{x^2+x} x\right )}{e^{x^2+x}-x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{3} \int \left (2 x^3+3 x^2+\frac {(x+1)^2 (2 x-1) x}{e^{x^2+x}-x}+2 \log \left (x^2-e^{x^2+x} x\right ) x-4 x+\log \left (x^2-e^{x^2+x} x\right )-14\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (3 x^2+x^2 \left (-\log \left (x^2-e^{x^2+x} x\right )\right )-x \log \left (x^2-e^{x^2+x} x\right )+15 x\right )\) |
Input:
Int[(-13*x - 4*x^2 + E^(x + x^2)*(14 + 4*x - 3*x^2 - 2*x^3) + (E^(x + x^2) *(-1 - 2*x) + x + 2*x^2)*Log[-(E^(x + x^2)*x) + x^2])/(3*E^(x + x^2) - 3*x ),x]
Output:
(15*x + 3*x^2 - x*Log[-(E^(x + x^2)*x) + x^2] - x^2*Log[-(E^(x + x^2)*x) + x^2])/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42
method | result | size |
norman | \(x^{2}+5 x -\frac {\ln \left (-x \,{\mathrm e}^{x^{2}+x}+x^{2}\right ) x}{3}-\frac {\ln \left (-x \,{\mathrm e}^{x^{2}+x}+x^{2}\right ) x^{2}}{3}\) | \(44\) |
parallelrisch | \(-\frac {\ln \left (-x \left ({\mathrm e}^{x^{2}+x}-x \right )\right ) x^{2}}{3}+x^{2}-\frac {\ln \left (-x \left ({\mathrm e}^{x^{2}+x}-x \right )\right ) x}{3}+5 x\) | \(44\) |
risch | \(\left (-\frac {1}{3} x^{2}-\frac {1}{3} x \right ) \ln \left (-{\mathrm e}^{\left (1+x \right ) x}+x \right )-\frac {x^{2} \ln \left (x \right )}{3}-\frac {x \ln \left (x \right )}{3}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (-{\mathrm e}^{\left (1+x \right ) x}+x \right )\right ) \operatorname {csgn}\left (i x \left (-{\mathrm e}^{\left (1+x \right ) x}+x \right )\right )}{6}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (i x \left (-{\mathrm e}^{\left (1+x \right ) x}+x \right )\right )}^{2}}{6}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \left (-{\mathrm e}^{\left (1+x \right ) x}+x \right )\right ) {\operatorname {csgn}\left (i x \left (-{\mathrm e}^{\left (1+x \right ) x}+x \right )\right )}^{2}}{6}+\frac {i \pi \,x^{2} {\operatorname {csgn}\left (i x \left (-{\mathrm e}^{\left (1+x \right ) x}+x \right )\right )}^{3}}{6}+\frac {i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (-{\mathrm e}^{\left (1+x \right ) x}+x \right )\right ) \operatorname {csgn}\left (i x \left (-{\mathrm e}^{\left (1+x \right ) x}+x \right )\right )}{6}-\frac {i \pi x \,\operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (i x \left (-{\mathrm e}^{\left (1+x \right ) x}+x \right )\right )}^{2}}{6}-\frac {i \pi x \,\operatorname {csgn}\left (i \left (-{\mathrm e}^{\left (1+x \right ) x}+x \right )\right ) {\operatorname {csgn}\left (i x \left (-{\mathrm e}^{\left (1+x \right ) x}+x \right )\right )}^{2}}{6}+\frac {i \pi x {\operatorname {csgn}\left (i x \left (-{\mathrm e}^{\left (1+x \right ) x}+x \right )\right )}^{3}}{6}+x^{2}+5 x\) | \(297\) |
Input:
int((((-1-2*x)*exp(x^2+x)+2*x^2+x)*ln(-x*exp(x^2+x)+x^2)+(-2*x^3-3*x^2+4*x +14)*exp(x^2+x)-4*x^2-13*x)/(3*exp(x^2+x)-3*x),x,method=_RETURNVERBOSE)
Output:
x^2+5*x-1/3*ln(-x*exp(x^2+x)+x^2)*x-1/3*ln(-x*exp(x^2+x)+x^2)*x^2
Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-13 x-4 x^2+e^{x+x^2} \left (14+4 x-3 x^2-2 x^3\right )+\left (e^{x+x^2} (-1-2 x)+x+2 x^2\right ) \log \left (-e^{x+x^2} x+x^2\right )}{3 e^{x+x^2}-3 x} \, dx=x^{2} - \frac {1}{3} \, {\left (x^{2} + x\right )} \log \left (x^{2} - x e^{\left (x^{2} + x\right )}\right ) + 5 \, x \] Input:
integrate((((-1-2*x)*exp(x^2+x)+2*x^2+x)*log(-x*exp(x^2+x)+x^2)+(-2*x^3-3* x^2+4*x+14)*exp(x^2+x)-4*x^2-13*x)/(3*exp(x^2+x)-3*x),x, algorithm="fricas ")
Output:
x^2 - 1/3*(x^2 + x)*log(x^2 - x*e^(x^2 + x)) + 5*x
Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {-13 x-4 x^2+e^{x+x^2} \left (14+4 x-3 x^2-2 x^3\right )+\left (e^{x+x^2} (-1-2 x)+x+2 x^2\right ) \log \left (-e^{x+x^2} x+x^2\right )}{3 e^{x+x^2}-3 x} \, dx=x^{2} + 5 x + \left (- \frac {x^{2}}{3} - \frac {x}{3}\right ) \log {\left (x^{2} - x e^{x^{2} + x} \right )} \] Input:
integrate((((-2*x-1)*exp(x**2+x)+2*x**2+x)*ln(-x*exp(x**2+x)+x**2)+(-2*x** 3-3*x**2+4*x+14)*exp(x**2+x)-4*x**2-13*x)/(3*exp(x**2+x)-3*x),x)
Output:
x**2 + 5*x + (-x**2/3 - x/3)*log(x**2 - x*exp(x**2 + x))
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {-13 x-4 x^2+e^{x+x^2} \left (14+4 x-3 x^2-2 x^3\right )+\left (e^{x+x^2} (-1-2 x)+x+2 x^2\right ) \log \left (-e^{x+x^2} x+x^2\right )}{3 e^{x+x^2}-3 x} \, dx=x^{2} - \frac {1}{3} \, {\left (x^{2} + x\right )} \log \left (x - e^{\left (x^{2} + x\right )}\right ) - \frac {1}{3} \, {\left (x^{2} + x\right )} \log \left (x\right ) + 5 \, x \] Input:
integrate((((-1-2*x)*exp(x^2+x)+2*x^2+x)*log(-x*exp(x^2+x)+x^2)+(-2*x^3-3* x^2+4*x+14)*exp(x^2+x)-4*x^2-13*x)/(3*exp(x^2+x)-3*x),x, algorithm="maxima ")
Output:
x^2 - 1/3*(x^2 + x)*log(x - e^(x^2 + x)) - 1/3*(x^2 + x)*log(x) + 5*x
Time = 0.17 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39 \[ \int \frac {-13 x-4 x^2+e^{x+x^2} \left (14+4 x-3 x^2-2 x^3\right )+\left (e^{x+x^2} (-1-2 x)+x+2 x^2\right ) \log \left (-e^{x+x^2} x+x^2\right )}{3 e^{x+x^2}-3 x} \, dx=-\frac {1}{3} \, x^{2} \log \left (x^{2} - x e^{\left (x^{2} + x\right )}\right ) + x^{2} - \frac {1}{3} \, x \log \left (x^{2} - x e^{\left (x^{2} + x\right )}\right ) + 5 \, x \] Input:
integrate((((-1-2*x)*exp(x^2+x)+2*x^2+x)*log(-x*exp(x^2+x)+x^2)+(-2*x^3-3* x^2+4*x+14)*exp(x^2+x)-4*x^2-13*x)/(3*exp(x^2+x)-3*x),x, algorithm="giac")
Output:
-1/3*x^2*log(x^2 - x*e^(x^2 + x)) + x^2 - 1/3*x*log(x^2 - x*e^(x^2 + x)) + 5*x
Time = 3.71 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {-13 x-4 x^2+e^{x+x^2} \left (14+4 x-3 x^2-2 x^3\right )+\left (e^{x+x^2} (-1-2 x)+x+2 x^2\right ) \log \left (-e^{x+x^2} x+x^2\right )}{3 e^{x+x^2}-3 x} \, dx=5\,x-\ln \left (x^2-x\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^x\right )\,\left (\frac {x^2}{3}+\frac {x}{3}\right )+x^2 \] Input:
int((13*x - exp(x + x^2)*(4*x - 3*x^2 - 2*x^3 + 14) - log(x^2 - x*exp(x + x^2))*(x - exp(x + x^2)*(2*x + 1) + 2*x^2) + 4*x^2)/(3*x - 3*exp(x + x^2)) ,x)
Output:
5*x - log(x^2 - x*exp(x^2)*exp(x))*(x/3 + x^2/3) + x^2
Time = 7.82 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.58 \[ \int \frac {-13 x-4 x^2+e^{x+x^2} \left (14+4 x-3 x^2-2 x^3\right )+\left (e^{x+x^2} (-1-2 x)+x+2 x^2\right ) \log \left (-e^{x+x^2} x+x^2\right )}{3 e^{x+x^2}-3 x} \, dx=-\frac {2 \,\mathrm {log}\left (e^{x^{2}+x}-x \right )}{3}-\frac {\mathrm {log}\left (-e^{x^{2}+x} x +x^{2}\right ) x^{2}}{3}-\frac {\mathrm {log}\left (-e^{x^{2}+x} x +x^{2}\right ) x}{3}+\frac {2 \,\mathrm {log}\left (-e^{x^{2}+x} x +x^{2}\right )}{3}-\frac {2 \,\mathrm {log}\left (x \right )}{3}+x^{2}+5 x \] Input:
int((((-2*x-1)*exp(x^2+x)+2*x^2+x)*log(-x*exp(x^2+x)+x^2)+(-2*x^3-3*x^2+4* x+14)*exp(x^2+x)-4*x^2-13*x)/(3*exp(x^2+x)-3*x),x)
Output:
( - 2*log(e**(x**2 + x) - x) - log( - e**(x**2 + x)*x + x**2)*x**2 - log( - e**(x**2 + x)*x + x**2)*x + 2*log( - e**(x**2 + x)*x + x**2) - 2*log(x) + 3*x**2 + 15*x)/3