Integrand size = 88, antiderivative size = 23 \[ \int \frac {4 e^{x^2+\frac {3+x^3}{x}} x^3+e^{\frac {3+x^3}{x}} \left (9-3 x-6 x^3+e^{x^2} \left (-6+2 x+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (-3+2 e^{x^2}\right )\right )}{-3 x+2 e^{x^2} x} \, dx=e^{\frac {3}{x}+x^2} x \log \left (-\frac {3}{2}+e^{x^2}\right ) \] Output:
exp(x^2+3/x)*ln(exp(x^2)-3/2)*x
Time = 1.34 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {4 e^{x^2+\frac {3+x^3}{x}} x^3+e^{\frac {3+x^3}{x}} \left (9-3 x-6 x^3+e^{x^2} \left (-6+2 x+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (-3+2 e^{x^2}\right )\right )}{-3 x+2 e^{x^2} x} \, dx=e^{\frac {3}{x}+x^2} x \log \left (-\frac {3}{2}+e^{x^2}\right ) \] Input:
Integrate[(4*E^(x^2 + (3 + x^3)/x)*x^3 + E^((3 + x^3)/x)*(9 - 3*x - 6*x^3 + E^x^2*(-6 + 2*x + 4*x^3))*Log[(-3 + 2*E^x^2)/2])/(-3*x + 2*E^x^2*x),x]
Output:
E^(3/x + x^2)*x*Log[-3/2 + E^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.
\(\displaystyle \int \frac {4 e^{\frac {x^3+3}{x}+x^2} x^3+e^{\frac {x^3+3}{x}} \left (-6 x^3+e^{x^2} \left (4 x^3+2 x-6\right )-3 x+9\right ) \log \left (\frac {1}{2} \left (2 e^{x^2}-3\right )\right )}{2 e^{x^2} x-3 x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (3 e^{3/x} x^2+\frac {9 e^{3/x} x^2}{2 e^{x^2}-3}+\frac {e^{x^2+\frac {3}{x}} \left (2 x^3+x \log \left (e^{x^2}-\frac {3}{2}\right )-3 \log \left (e^{x^2}-\frac {3}{2}\right )+2 x^3 \log \left (e^{x^2}-\frac {3}{2}\right )\right )}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int e^{x^2+\frac {3}{x}} x^2dx+9 \int \frac {e^{3/x} x^2}{-3+2 e^{x^2}}dx-4 \int \frac {e^{x^2} x \int e^{x^2+\frac {3}{x}}dx}{-3+2 e^{x^2}}dx+12 \int \frac {e^{x^2} x \int \frac {e^{x^2+\frac {3}{x}}}{x}dx}{-3+2 e^{x^2}}dx-8 \int \frac {e^{x^2} x \int e^{x^2+\frac {3}{x}} x^2dx}{-3+2 e^{x^2}}dx+\log \left (e^{x^2}-\frac {3}{2}\right ) \int e^{x^2+\frac {3}{x}}dx-3 \log \left (e^{x^2}-\frac {3}{2}\right ) \int \frac {e^{x^2+\frac {3}{x}}}{x}dx+2 \log \left (e^{x^2}-\frac {3}{2}\right ) \int e^{x^2+\frac {3}{x}} x^2dx-\frac {27 \operatorname {ExpIntegralEi}\left (\frac {3}{x}\right )}{2}+e^{3/x} x^3+\frac {3}{2} e^{3/x} x^2+\frac {9}{2} e^{3/x} x\) |
Input:
Int[(4*E^(x^2 + (3 + x^3)/x)*x^3 + E^((3 + x^3)/x)*(9 - 3*x - 6*x^3 + E^x^ 2*(-6 + 2*x + 4*x^3))*Log[(-3 + 2*E^x^2)/2])/(-3*x + 2*E^x^2*x),x]
Output:
$Aborted
Time = 0.81 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87
method | result | size |
risch | \({\mathrm e}^{\frac {x^{3}+3}{x}} \ln \left ({\mathrm e}^{x^{2}}-\frac {3}{2}\right ) x\) | \(20\) |
parallelrisch | \({\mathrm e}^{\frac {x^{3}+3}{x}} \ln \left ({\mathrm e}^{x^{2}}-\frac {3}{2}\right ) x\) | \(20\) |
Input:
int((((4*x^3+2*x-6)*exp(x^2)-6*x^3-3*x+9)*exp((x^3+3)/x)*ln(exp(x^2)-3/2)+ 4*x^3*exp(x^2)*exp((x^3+3)/x))/(2*exp(x^2)*x-3*x),x,method=_RETURNVERBOSE)
Output:
exp((x^3+3)/x)*ln(exp(x^2)-3/2)*x
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (19) = 38\).
Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.30 \[ \int \frac {4 e^{x^2+\frac {3+x^3}{x}} x^3+e^{\frac {3+x^3}{x}} \left (9-3 x-6 x^3+e^{x^2} \left (-6+2 x+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (-3+2 e^{x^2}\right )\right )}{-3 x+2 e^{x^2} x} \, dx=x e^{\left (\frac {x^{3} + 3}{x}\right )} \log \left (\frac {1}{2} \, {\left (2 \, e^{\left (\frac {2 \, x^{3} + 3}{x}\right )} - 3 \, e^{\left (\frac {x^{3} + 3}{x}\right )}\right )} e^{\left (-\frac {x^{3} + 3}{x}\right )}\right ) \] Input:
integrate((((4*x^3+2*x-6)*exp(x^2)-6*x^3-3*x+9)*exp((x^3+3)/x)*log(exp(x^2 )-3/2)+4*x^3*exp(x^2)*exp((x^3+3)/x))/(2*exp(x^2)*x-3*x),x, algorithm="fri cas")
Output:
x*e^((x^3 + 3)/x)*log(1/2*(2*e^((2*x^3 + 3)/x) - 3*e^((x^3 + 3)/x))*e^(-(x ^3 + 3)/x))
Timed out. \[ \int \frac {4 e^{x^2+\frac {3+x^3}{x}} x^3+e^{\frac {3+x^3}{x}} \left (9-3 x-6 x^3+e^{x^2} \left (-6+2 x+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (-3+2 e^{x^2}\right )\right )}{-3 x+2 e^{x^2} x} \, dx=\text {Timed out} \] Input:
integrate((((4*x**3+2*x-6)*exp(x**2)-6*x**3-3*x+9)*exp((x**3+3)/x)*ln(exp( x**2)-3/2)+4*x**3*exp(x**2)*exp((x**3+3)/x))/(2*exp(x**2)*x-3*x),x)
Output:
Timed out
Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {4 e^{x^2+\frac {3+x^3}{x}} x^3+e^{\frac {3+x^3}{x}} \left (9-3 x-6 x^3+e^{x^2} \left (-6+2 x+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (-3+2 e^{x^2}\right )\right )}{-3 x+2 e^{x^2} x} \, dx=-x e^{\left (x^{2} + \frac {3}{x}\right )} \log \left (2\right ) + x e^{\left (x^{2} + \frac {3}{x}\right )} \log \left (2 \, e^{\left (x^{2}\right )} - 3\right ) \] Input:
integrate((((4*x^3+2*x-6)*exp(x^2)-6*x^3-3*x+9)*exp((x^3+3)/x)*log(exp(x^2 )-3/2)+4*x^3*exp(x^2)*exp((x^3+3)/x))/(2*exp(x^2)*x-3*x),x, algorithm="max ima")
Output:
-x*e^(x^2 + 3/x)*log(2) + x*e^(x^2 + 3/x)*log(2*e^(x^2) - 3)
Time = 0.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {4 e^{x^2+\frac {3+x^3}{x}} x^3+e^{\frac {3+x^3}{x}} \left (9-3 x-6 x^3+e^{x^2} \left (-6+2 x+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (-3+2 e^{x^2}\right )\right )}{-3 x+2 e^{x^2} x} \, dx=-x e^{\left (\frac {x^{3} + 3}{x}\right )} \log \left (2\right ) + x e^{\left (\frac {x^{3} + 3}{x}\right )} \log \left (2 \, e^{\left (x^{2}\right )} - 3\right ) \] Input:
integrate((((4*x^3+2*x-6)*exp(x^2)-6*x^3-3*x+9)*exp((x^3+3)/x)*log(exp(x^2 )-3/2)+4*x^3*exp(x^2)*exp((x^3+3)/x))/(2*exp(x^2)*x-3*x),x, algorithm="gia c")
Output:
-x*e^((x^3 + 3)/x)*log(2) + x*e^((x^3 + 3)/x)*log(2*e^(x^2) - 3)
Timed out. \[ \int \frac {4 e^{x^2+\frac {3+x^3}{x}} x^3+e^{\frac {3+x^3}{x}} \left (9-3 x-6 x^3+e^{x^2} \left (-6+2 x+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (-3+2 e^{x^2}\right )\right )}{-3 x+2 e^{x^2} x} \, dx=\int -\frac {4\,x^3\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{\frac {x^3+3}{x}}-\ln \left ({\mathrm {e}}^{x^2}-\frac {3}{2}\right )\,{\mathrm {e}}^{\frac {x^3+3}{x}}\,\left (3\,x-{\mathrm {e}}^{x^2}\,\left (4\,x^3+2\,x-6\right )+6\,x^3-9\right )}{3\,x-2\,x\,{\mathrm {e}}^{x^2}} \,d x \] Input:
int(-(4*x^3*exp(x^2)*exp((x^3 + 3)/x) - log(exp(x^2) - 3/2)*exp((x^3 + 3)/ x)*(3*x - exp(x^2)*(2*x + 4*x^3 - 6) + 6*x^3 - 9))/(3*x - 2*x*exp(x^2)),x)
Output:
int(-(4*x^3*exp(x^2)*exp((x^3 + 3)/x) - log(exp(x^2) - 3/2)*exp((x^3 + 3)/ x)*(3*x - exp(x^2)*(2*x + 4*x^3 - 6) + 6*x^3 - 9))/(3*x - 2*x*exp(x^2)), x )
Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {4 e^{x^2+\frac {3+x^3}{x}} x^3+e^{\frac {3+x^3}{x}} \left (9-3 x-6 x^3+e^{x^2} \left (-6+2 x+4 x^3\right )\right ) \log \left (\frac {1}{2} \left (-3+2 e^{x^2}\right )\right )}{-3 x+2 e^{x^2} x} \, dx=e^{\frac {x^{3}+3}{x}} \mathrm {log}\left (e^{x^{2}}-\frac {3}{2}\right ) x \] Input:
int((((4*x^3+2*x-6)*exp(x^2)-6*x^3-3*x+9)*exp((x^3+3)/x)*log(exp(x^2)-3/2) +4*x^3*exp(x^2)*exp((x^3+3)/x))/(2*exp(x^2)*x-3*x),x)
Output:
e**((x**3 + 3)/x)*log((2*e**(x**2) - 3)/2)*x