Integrand size = 84, antiderivative size = 31 \[ \int \frac {4 x^3+e^{\frac {2 x+3 x^3-x^4+\left (2+3 x^2-x^3\right ) \log (x)}{2 x}} \left (-2-3 x^2-5 x^3+3 x^4+\left (2-3 x^2+2 x^3\right ) \log (x)\right )}{2 x^2} \, dx=-e^{\frac {1}{2} \left (\left (3+\frac {2}{x^2}\right ) x-x^2\right ) (x+\log (x))}+x^2 \]
Time = 6.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {4 x^3+e^{\frac {2 x+3 x^3-x^4+\left (2+3 x^2-x^3\right ) \log (x)}{2 x}} \left (-2-3 x^2-5 x^3+3 x^4+\left (2-3 x^2+2 x^3\right ) \log (x)\right )}{2 x^2} \, dx=x^2-e^{1+\frac {3 x^2}{2}-\frac {x^3}{2}} x^{\frac {1}{x}+\frac {3 x}{2}-\frac {x^2}{2}} \]
Integrate[(4*x^3 + E^((2*x + 3*x^3 - x^4 + (2 + 3*x^2 - x^3)*Log[x])/(2*x) )*(-2 - 3*x^2 - 5*x^3 + 3*x^4 + (2 - 3*x^2 + 2*x^3)*Log[x]))/(2*x^2),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.
\(\displaystyle \int \frac {\left (3 x^4-5 x^3-3 x^2+\left (2 x^3-3 x^2+2\right ) \log (x)-2\right ) \exp \left (\frac {-x^4+3 x^3+\left (-x^3+3 x^2+2\right ) \log (x)+2 x}{2 x}\right )+4 x^3}{2 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {4 x^3-e^{\frac {-x^4+3 x^3+2 x}{2 x}} x^{\frac {-x^3+3 x^2+2}{2 x}} \left (-3 x^4+5 x^3+3 x^2-\left (2 x^3-3 x^2+2\right ) \log (x)+2\right )}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{2} \int \left (e^{-\frac {x^3}{2}+\frac {3 x^2}{2}+1} \left (3 x^4+2 \log (x) x^3-5 x^3-3 \log (x) x^2-3 x^2+2 \log (x)-2\right ) x^{-\frac {x^2}{2}+\frac {3 x}{2}-2+\frac {1}{x}}+4 x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-2 \int e^{-\frac {x^3}{2}+\frac {3 x^2}{2}+1} x^{-\frac {x^2}{2}+\frac {3 x}{2}-2+\frac {1}{x}}dx-5 \int e^{-\frac {x^3}{2}+\frac {3 x^2}{2}+1} x^{-\frac {x^2}{2}+\frac {3 x}{2}+1+\frac {1}{x}}dx+3 \int e^{-\frac {x^3}{2}+\frac {3 x^2}{2}+1} x^{-\frac {x^2}{2}+\frac {3 x}{2}+2+\frac {1}{x}}dx-2 \int \frac {\int e^{-\frac {x^3}{2}+\frac {3 x^2}{2}+1} x^{-\frac {x^2}{2}+\frac {3 x}{2}-2+\frac {1}{x}}dx}{x}dx-2 \int \frac {\int e^{-\frac {x^3}{2}+\frac {3 x^2}{2}+1} x^{-\frac {x^2}{2}+\frac {3 x}{2}+1+\frac {1}{x}}dx}{x}dx+2 \log (x) \int e^{-\frac {x^3}{2}+\frac {3 x^2}{2}+1} x^{-\frac {x^2}{2}+\frac {3 x}{2}-2+\frac {1}{x}}dx+2 \log (x) \int e^{-\frac {x^3}{2}+\frac {3 x^2}{2}+1} x^{-\frac {x^2}{2}+\frac {3 x}{2}+1+\frac {1}{x}}dx-3 \int e^{-\frac {x^3}{2}+\frac {3 x^2}{2}+1} x^{\frac {-\frac {x^3}{2}+\frac {3 x^2}{2}+1}{x}}dx+3 \int \frac {\int e^{-\frac {x^3}{2}+\frac {3 x^2}{2}+1} x^{\frac {-\frac {x^3}{2}+\frac {3 x^2}{2}+1}{x}}dx}{x}dx-3 \log (x) \int e^{-\frac {x^3}{2}+\frac {3 x^2}{2}+1} x^{\frac {-\frac {x^3}{2}+\frac {3 x^2}{2}+1}{x}}dx+2 x^2\right )\) |
Int[(4*x^3 + E^((2*x + 3*x^3 - x^4 + (2 + 3*x^2 - x^3)*Log[x])/(2*x))*(-2 - 3*x^2 - 5*x^3 + 3*x^4 + (2 - 3*x^2 + 2*x^3)*Log[x]))/(2*x^2),x]
3.2.34.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87
method | result | size |
risch | \(-{\mathrm e}^{-\frac {\left (x^{3}-3 x^{2}-2\right ) \left (x +\ln \left (x \right )\right )}{2 x}}+x^{2}\) | \(27\) |
default | \(-{\mathrm e}^{\frac {\left (-x^{3}+3 x^{2}+2\right ) \ln \left (x \right )-x^{4}+3 x^{3}+2 x}{2 x}}+x^{2}\) | \(42\) |
parallelrisch | \(-{\mathrm e}^{\frac {\left (-x^{3}+3 x^{2}+2\right ) \ln \left (x \right )-x^{4}+3 x^{3}+2 x}{2 x}}+x^{2}\) | \(42\) |
parts | \(-{\mathrm e}^{\frac {\left (-x^{3}+3 x^{2}+2\right ) \ln \left (x \right )-x^{4}+3 x^{3}+2 x}{2 x}}+x^{2}\) | \(42\) |
norman | \(\frac {x^{3}-x \,{\mathrm e}^{\frac {\left (-x^{3}+3 x^{2}+2\right ) \ln \left (x \right )-x^{4}+3 x^{3}+2 x}{2 x}}}{x}\) | \(47\) |
int(1/2*(((2*x^3-3*x^2+2)*ln(x)+3*x^4-5*x^3-3*x^2-2)*exp(1/2*((-x^3+3*x^2+ 2)*ln(x)-x^4+3*x^3+2*x)/x)+4*x^3)/x^2,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {4 x^3+e^{\frac {2 x+3 x^3-x^4+\left (2+3 x^2-x^3\right ) \log (x)}{2 x}} \left (-2-3 x^2-5 x^3+3 x^4+\left (2-3 x^2+2 x^3\right ) \log (x)\right )}{2 x^2} \, dx=x^{2} - e^{\left (-\frac {x^{4} - 3 \, x^{3} + {\left (x^{3} - 3 \, x^{2} - 2\right )} \log \left (x\right ) - 2 \, x}{2 \, x}\right )} \]
integrate(1/2*(((2*x^3-3*x^2+2)*log(x)+3*x^4-5*x^3-3*x^2-2)*exp(1/2*((-x^3 +3*x^2+2)*log(x)-x^4+3*x^3+2*x)/x)+4*x^3)/x^2,x, algorithm=\
Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {4 x^3+e^{\frac {2 x+3 x^3-x^4+\left (2+3 x^2-x^3\right ) \log (x)}{2 x}} \left (-2-3 x^2-5 x^3+3 x^4+\left (2-3 x^2+2 x^3\right ) \log (x)\right )}{2 x^2} \, dx=x^{2} - e^{\frac {- \frac {x^{4}}{2} + \frac {3 x^{3}}{2} + x + \frac {\left (- x^{3} + 3 x^{2} + 2\right ) \log {\left (x \right )}}{2}}{x}} \]
integrate(1/2*(((2*x**3-3*x**2+2)*ln(x)+3*x**4-5*x**3-3*x**2-2)*exp(1/2*(( -x**3+3*x**2+2)*ln(x)-x**4+3*x**3+2*x)/x)+4*x**3)/x**2,x)
Time = 0.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {4 x^3+e^{\frac {2 x+3 x^3-x^4+\left (2+3 x^2-x^3\right ) \log (x)}{2 x}} \left (-2-3 x^2-5 x^3+3 x^4+\left (2-3 x^2+2 x^3\right ) \log (x)\right )}{2 x^2} \, dx=x^{2} - e^{\left (-\frac {1}{2} \, x^{3} - \frac {1}{2} \, x^{2} \log \left (x\right ) + \frac {3}{2} \, x^{2} + \frac {3}{2} \, x \log \left (x\right ) + \frac {\log \left (x\right )}{x} + 1\right )} \]
integrate(1/2*(((2*x^3-3*x^2+2)*log(x)+3*x^4-5*x^3-3*x^2-2)*exp(1/2*((-x^3 +3*x^2+2)*log(x)-x^4+3*x^3+2*x)/x)+4*x^3)/x^2,x, algorithm=\
Time = 0.40 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {4 x^3+e^{\frac {2 x+3 x^3-x^4+\left (2+3 x^2-x^3\right ) \log (x)}{2 x}} \left (-2-3 x^2-5 x^3+3 x^4+\left (2-3 x^2+2 x^3\right ) \log (x)\right )}{2 x^2} \, dx=x^{2} - e^{\left (-\frac {1}{2} \, x^{3} - \frac {1}{2} \, x^{2} \log \left (x\right ) + \frac {3}{2} \, x^{2} + \frac {3}{2} \, x \log \left (x\right ) + \frac {\log \left (x\right )}{x} + 1\right )} \]
integrate(1/2*(((2*x^3-3*x^2+2)*log(x)+3*x^4-5*x^3-3*x^2-2)*exp(1/2*((-x^3 +3*x^2+2)*log(x)-x^4+3*x^3+2*x)/x)+4*x^3)/x^2,x, algorithm=\
Time = 11.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {4 x^3+e^{\frac {2 x+3 x^3-x^4+\left (2+3 x^2-x^3\right ) \log (x)}{2 x}} \left (-2-3 x^2-5 x^3+3 x^4+\left (2-3 x^2+2 x^3\right ) \log (x)\right )}{2 x^2} \, dx=x^2-\frac {x^{1/x}\,{\left ({\mathrm {e}}^{x^2}\right )}^{3/2}\,{\mathrm {e}}^{\frac {3\,x\,\ln \left (x\right )}{2}}\,\mathrm {e}\,{\mathrm {e}}^{-\frac {x^2\,\ln \left (x\right )}{2}}}{\sqrt {{\mathrm {e}}^{x^3}}} \]
int(-((exp((x + (log(x)*(3*x^2 - x^3 + 2))/2 + (3*x^3)/2 - x^4/2)/x)*(3*x^ 2 - log(x)*(2*x^3 - 3*x^2 + 2) + 5*x^3 - 3*x^4 + 2))/2 - 2*x^3)/x^2,x)