Integrand size = 95, antiderivative size = 30 \[ \int \frac {-10+2 x^2+e^{-4 x^2+x^3} \left (-68 x+30 x^2-26 x^3+9 x^4+8 x^5-3 x^6\right )+e^{-4 x^2+x^3} \left (40 x-15 x^2-8 x^3+3 x^4\right ) \log \left (5-x^2\right )}{-5+x^2} \, dx=2 x-e^{(-4+x) x^2} \left (2+x^2-\log \left (5-x^2\right )\right ) \] Output:
2*x-exp((-4+x)*x^2)*(2-ln(-x^2+5)+x^2)
Time = 0.06 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {-10+2 x^2+e^{-4 x^2+x^3} \left (-68 x+30 x^2-26 x^3+9 x^4+8 x^5-3 x^6\right )+e^{-4 x^2+x^3} \left (40 x-15 x^2-8 x^3+3 x^4\right ) \log \left (5-x^2\right )}{-5+x^2} \, dx=2 x-e^{(-4+x) x^2} \left (2+x^2\right )+e^{(-4+x) x^2} \log \left (5-x^2\right ) \] Input:
Integrate[(-10 + 2*x^2 + E^(-4*x^2 + x^3)*(-68*x + 30*x^2 - 26*x^3 + 9*x^4 + 8*x^5 - 3*x^6) + E^(-4*x^2 + x^3)*(40*x - 15*x^2 - 8*x^3 + 3*x^4)*Log[5 - x^2])/(-5 + x^2),x]
Output:
2*x - E^((-4 + x)*x^2)*(2 + x^2) + E^((-4 + x)*x^2)*Log[5 - x^2]
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.
\(\displaystyle \int \frac {2 x^2+e^{x^3-4 x^2} \left (3 x^4-8 x^3-15 x^2+40 x\right ) \log \left (5-x^2\right )+e^{x^3-4 x^2} \left (-3 x^6+8 x^5+9 x^4-26 x^3+30 x^2-68 x\right )-10}{x^2-5} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {e^{x^3-4 x^2} x \left (3 x^5-8 x^4-9 x^3+26 x^2+8 x^2 \log \left (5-x^2\right )+15 x \log \left (5-x^2\right )-40 \log \left (5-x^2\right )-3 x^3 \log \left (5-x^2\right )-30 x+68\right )}{5-x^2}+2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 14 \int e^{x^3-4 x^2} xdx-6 \int e^{x^3-4 x^2} x^2dx+8 \int e^{x^3-4 x^2} x^3dx-3 \int e^{x^3-4 x^2} x^4dx+e^{x^3-4 x^2} \log \left (5-x^2\right )+2 x\) |
Input:
Int[(-10 + 2*x^2 + E^(-4*x^2 + x^3)*(-68*x + 30*x^2 - 26*x^3 + 9*x^4 + 8*x ^5 - 3*x^6) + E^(-4*x^2 + x^3)*(40*x - 15*x^2 - 8*x^3 + 3*x^4)*Log[5 - x^2 ])/(-5 + x^2),x]
Output:
$Aborted
Time = 0.84 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50
method | result | size |
risch | \(-{\mathrm e}^{\left (x -4\right ) x^{2}} x^{2}+{\mathrm e}^{\left (x -4\right ) x^{2}} \ln \left (-x^{2}+5\right )+2 x -2 \,{\mathrm e}^{\left (x -4\right ) x^{2}}\) | \(45\) |
parallelrisch | \(-{\mathrm e}^{\left (x -4\right ) x^{2}} x^{2}+{\mathrm e}^{\left (x -4\right ) x^{2}} \ln \left (-x^{2}+5\right )+2 x -2 \,{\mathrm e}^{\left (x -4\right ) x^{2}}\) | \(45\) |
orering | \(\text {Expression too large to display}\) | \(1615\) |
Input:
int(((3*x^4-8*x^3-15*x^2+40*x)*exp(x^3-4*x^2)*ln(-x^2+5)+(-3*x^6+8*x^5+9*x ^4-26*x^3+30*x^2-68*x)*exp(x^3-4*x^2)+2*x^2-10)/(x^2-5),x,method=_RETURNVE RBOSE)
Output:
-exp((x-4)*x^2)*x^2+exp((x-4)*x^2)*ln(-x^2+5)+2*x-2*exp((x-4)*x^2)
Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {-10+2 x^2+e^{-4 x^2+x^3} \left (-68 x+30 x^2-26 x^3+9 x^4+8 x^5-3 x^6\right )+e^{-4 x^2+x^3} \left (40 x-15 x^2-8 x^3+3 x^4\right ) \log \left (5-x^2\right )}{-5+x^2} \, dx=-{\left (x^{2} + 2\right )} e^{\left (x^{3} - 4 \, x^{2}\right )} + e^{\left (x^{3} - 4 \, x^{2}\right )} \log \left (-x^{2} + 5\right ) + 2 \, x \] Input:
integrate(((3*x^4-8*x^3-15*x^2+40*x)*exp(x^3-4*x^2)*log(-x^2+5)+(-3*x^6+8* x^5+9*x^4-26*x^3+30*x^2-68*x)*exp(x^3-4*x^2)+2*x^2-10)/(x^2-5),x, algorith m="fricas")
Output:
-(x^2 + 2)*e^(x^3 - 4*x^2) + e^(x^3 - 4*x^2)*log(-x^2 + 5) + 2*x
Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-10+2 x^2+e^{-4 x^2+x^3} \left (-68 x+30 x^2-26 x^3+9 x^4+8 x^5-3 x^6\right )+e^{-4 x^2+x^3} \left (40 x-15 x^2-8 x^3+3 x^4\right ) \log \left (5-x^2\right )}{-5+x^2} \, dx=2 x + \left (- x^{2} + \log {\left (5 - x^{2} \right )} - 2\right ) e^{x^{3} - 4 x^{2}} \] Input:
integrate(((3*x**4-8*x**3-15*x**2+40*x)*exp(x**3-4*x**2)*ln(-x**2+5)+(-3*x **6+8*x**5+9*x**4-26*x**3+30*x**2-68*x)*exp(x**3-4*x**2)+2*x**2-10)/(x**2- 5),x)
Output:
2*x + (-x**2 + log(5 - x**2) - 2)*exp(x**3 - 4*x**2)
Time = 0.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {-10+2 x^2+e^{-4 x^2+x^3} \left (-68 x+30 x^2-26 x^3+9 x^4+8 x^5-3 x^6\right )+e^{-4 x^2+x^3} \left (40 x-15 x^2-8 x^3+3 x^4\right ) \log \left (5-x^2\right )}{-5+x^2} \, dx=-{\left ({\left (x^{2} + 2\right )} e^{\left (x^{3}\right )} - e^{\left (x^{3}\right )} \log \left (-x^{2} + 5\right )\right )} e^{\left (-4 \, x^{2}\right )} + 2 \, x \] Input:
integrate(((3*x^4-8*x^3-15*x^2+40*x)*exp(x^3-4*x^2)*log(-x^2+5)+(-3*x^6+8* x^5+9*x^4-26*x^3+30*x^2-68*x)*exp(x^3-4*x^2)+2*x^2-10)/(x^2-5),x, algorith m="maxima")
Output:
-((x^2 + 2)*e^(x^3) - e^(x^3)*log(-x^2 + 5))*e^(-4*x^2) + 2*x
Time = 0.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67 \[ \int \frac {-10+2 x^2+e^{-4 x^2+x^3} \left (-68 x+30 x^2-26 x^3+9 x^4+8 x^5-3 x^6\right )+e^{-4 x^2+x^3} \left (40 x-15 x^2-8 x^3+3 x^4\right ) \log \left (5-x^2\right )}{-5+x^2} \, dx=-x^{2} e^{\left (x^{3} - 4 \, x^{2}\right )} + e^{\left (x^{3} - 4 \, x^{2}\right )} \log \left (-x^{2} + 5\right ) + 2 \, x - 2 \, e^{\left (x^{3} - 4 \, x^{2}\right )} \] Input:
integrate(((3*x^4-8*x^3-15*x^2+40*x)*exp(x^3-4*x^2)*log(-x^2+5)+(-3*x^6+8* x^5+9*x^4-26*x^3+30*x^2-68*x)*exp(x^3-4*x^2)+2*x^2-10)/(x^2-5),x, algorith m="giac")
Output:
-x^2*e^(x^3 - 4*x^2) + e^(x^3 - 4*x^2)*log(-x^2 + 5) + 2*x - 2*e^(x^3 - 4* x^2)
Timed out. \[ \int \frac {-10+2 x^2+e^{-4 x^2+x^3} \left (-68 x+30 x^2-26 x^3+9 x^4+8 x^5-3 x^6\right )+e^{-4 x^2+x^3} \left (40 x-15 x^2-8 x^3+3 x^4\right ) \log \left (5-x^2\right )}{-5+x^2} \, dx=\int -\frac {{\mathrm {e}}^{x^3-4\,x^2}\,\left (3\,x^6-8\,x^5-9\,x^4+26\,x^3-30\,x^2+68\,x\right )-2\,x^2-\ln \left (5-x^2\right )\,{\mathrm {e}}^{x^3-4\,x^2}\,\left (3\,x^4-8\,x^3-15\,x^2+40\,x\right )+10}{x^2-5} \,d x \] Input:
int(-(exp(x^3 - 4*x^2)*(68*x - 30*x^2 + 26*x^3 - 9*x^4 - 8*x^5 + 3*x^6) - 2*x^2 - log(5 - x^2)*exp(x^3 - 4*x^2)*(40*x - 15*x^2 - 8*x^3 + 3*x^4) + 10 )/(x^2 - 5),x)
Output:
int(-(exp(x^3 - 4*x^2)*(68*x - 30*x^2 + 26*x^3 - 9*x^4 - 8*x^5 + 3*x^6) - 2*x^2 - log(5 - x^2)*exp(x^3 - 4*x^2)*(40*x - 15*x^2 - 8*x^3 + 3*x^4) + 10 )/(x^2 - 5), x)
Time = 0.17 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.73 \[ \int \frac {-10+2 x^2+e^{-4 x^2+x^3} \left (-68 x+30 x^2-26 x^3+9 x^4+8 x^5-3 x^6\right )+e^{-4 x^2+x^3} \left (40 x-15 x^2-8 x^3+3 x^4\right ) \log \left (5-x^2\right )}{-5+x^2} \, dx=\frac {e^{x^{3}} \mathrm {log}\left (-x^{2}+5\right )-e^{x^{3}} x^{2}-2 e^{x^{3}}+2 e^{4 x^{2}} x}{e^{4 x^{2}}} \] Input:
int(((3*x^4-8*x^3-15*x^2+40*x)*exp(x^3-4*x^2)*log(-x^2+5)+(-3*x^6+8*x^5+9* x^4-26*x^3+30*x^2-68*x)*exp(x^3-4*x^2)+2*x^2-10)/(x^2-5),x)
Output:
(e**(x**3)*log( - x**2 + 5) - e**(x**3)*x**2 - 2*e**(x**3) + 2*e**(4*x**2) *x)/e**(4*x**2)