Integrand size = 62, antiderivative size = 25 \[ \int \frac {-2 x+2 x^2+\left (-2 x+4 x^2\right ) \log (x)+e^{\frac {1}{2} (-8+\log (x))} \left (2 e^{x^2}+e^{x^2} \left (1+4 x^2\right ) \log (x)\right )}{2 x} \, dx=-1+\left (e^{-4+x^2} \sqrt {x}-x+x^2\right ) \log (x) \] Output:
(exp(1/4*ln(x)-2)^2*exp(x^2)+x^2-x)*ln(x)-1
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-2 x+2 x^2+\left (-2 x+4 x^2\right ) \log (x)+e^{\frac {1}{2} (-8+\log (x))} \left (2 e^{x^2}+e^{x^2} \left (1+4 x^2\right ) \log (x)\right )}{2 x} \, dx=e^{-4+x^2} \sqrt {x} \log (x)-x \log (x)+x^2 \log (x) \] Input:
Integrate[(-2*x + 2*x^2 + (-2*x + 4*x^2)*Log[x] + E^((-8 + Log[x])/2)*(2*E ^x^2 + E^x^2*(1 + 4*x^2)*Log[x]))/(2*x),x]
Output:
E^(-4 + x^2)*Sqrt[x]*Log[x] - x*Log[x] + x^2*Log[x]
Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {27, 25, 2010, 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.
\(\displaystyle \int \frac {2 x^2+\left (4 x^2-2 x\right ) \log (x)+e^{\frac {1}{2} (\log (x)-8)} \left (2 e^{x^2}+e^{x^2} \left (4 x^2+1\right ) \log (x)\right )-2 x}{2 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {-2 x^2+2 x-\frac {\left (e^{x^2} \left (4 x^2+1\right ) \log (x)+2 e^{x^2}\right ) \sqrt {x}}{e^4}+2 \left (x-2 x^2\right ) \log (x)}{x}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {-2 x^2+2 x-\frac {\left (e^{x^2} \left (4 x^2+1\right ) \log (x)+2 e^{x^2}\right ) \sqrt {x}}{e^4}+2 \left (x-2 x^2\right ) \log (x)}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{2} \int \left (-2 (2 \log (x) x+x-\log (x)-1)-\frac {e^{x^2-4} \left (4 \log (x) x^2+\log (x)+2\right )}{\sqrt {x}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (2 x^2 \log (x)+2 e^{x^2-4} \sqrt {x} \log (x)-2 x \log (x)\right )\) |
Input:
Int[(-2*x + 2*x^2 + (-2*x + 4*x^2)*Log[x] + E^((-8 + Log[x])/2)*(2*E^x^2 + E^x^2*(1 + 4*x^2)*Log[x]))/(2*x),x]
Output:
(2*E^(-4 + x^2)*Sqrt[x]*Log[x] - 2*x*Log[x] + 2*x^2*Log[x])/2
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 = 1.88 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08
method | result | size |
default | \(x^{2} \ln \left (x \right )-x \ln \left (x \right )+\sqrt {x}\, {\mathrm e}^{\left (-2+x \right ) \left (2+x \right )} \ln \left (x \right )\) | \(27\) |
parts | \(x^{2} \ln \left (x \right )-x \ln \left (x \right )+\sqrt {x}\, {\mathrm e}^{\left (-2+x \right ) \left (2+x \right )} \ln \left (x \right )\) | \(27\) |
parallelrisch | \(-\frac {-2 \ln \left (x \right ) {\mathrm e}^{x^{2}} {\mathrm e}^{\ln \left (\sqrt {x}\right )-4} x^{2}-2 x^{4} \ln \left (x \right )+2 x^{3} \ln \left (x \right )}{2 x^{2}}\) | \(41\) |
orering | \(\text {Expression too large to display}\) | \(1987\) |
Input:
int(1/2*(((4*x^2+1)*exp(x^2)*ln(x)+2*exp(x^2))*exp(1/4*ln(x)-2)^2+(4*x^2-2 *x)*ln(x)+2*x^2-2*x)/x,x,method=_RETURNVERBOSE)
Output:
x^2*ln(x)-x*ln(x)+x^(1/2)*exp((-2+x)*(2+x))*ln(x)
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-2 x+2 x^2+\left (-2 x+4 x^2\right ) \log (x)+e^{\frac {1}{2} (-8+\log (x))} \left (2 e^{x^2}+e^{x^2} \left (1+4 x^2\right ) \log (x)\right )}{2 x} \, dx={\left ({\left (x^{2} - x\right )} e^{4} \log \left (x\right ) + \sqrt {x} e^{\left (x^{2}\right )} \log \left (x\right )\right )} e^{\left (-4\right )} \] Input:
integrate(1/2*(((4*x^2+1)*exp(x^2)*log(x)+2*exp(x^2))*exp(1/4*log(x)-2)^2+ (4*x^2-2*x)*log(x)+2*x^2-2*x)/x,x, algorithm="fricas")
Output:
((x^2 - x)*e^4*log(x) + sqrt(x)*e^(x^2)*log(x))*e^(-4)
Timed out. \[ \int \frac {-2 x+2 x^2+\left (-2 x+4 x^2\right ) \log (x)+e^{\frac {1}{2} (-8+\log (x))} \left (2 e^{x^2}+e^{x^2} \left (1+4 x^2\right ) \log (x)\right )}{2 x} \, dx=\text {Timed out} \] Input:
integrate(1/2*(((4*x**2+1)*exp(x**2)*ln(x)+2*exp(x**2))*exp(1/4*ln(x)-2)** 2+(4*x**2-2*x)*ln(x)+2*x**2-2*x)/x,x)
Output:
Timed out
\[ \int \frac {-2 x+2 x^2+\left (-2 x+4 x^2\right ) \log (x)+e^{\frac {1}{2} (-8+\log (x))} \left (2 e^{x^2}+e^{x^2} \left (1+4 x^2\right ) \log (x)\right )}{2 x} \, dx=\int { \frac {2 \, x^{2} + {\left ({\left (4 \, x^{2} + 1\right )} e^{\left (x^{2}\right )} \log \left (x\right ) + 2 \, e^{\left (x^{2}\right )}\right )} e^{\left (\frac {1}{2} \, \log \left (x\right ) - 4\right )} + 2 \, {\left (2 \, x^{2} - x\right )} \log \left (x\right ) - 2 \, x}{2 \, x} \,d x } \] Input:
integrate(1/2*(((4*x^2+1)*exp(x^2)*log(x)+2*exp(x^2))*exp(1/4*log(x)-2)^2+ (4*x^2-2*x)*log(x)+2*x^2-2*x)/x,x, algorithm="maxima")
Output:
2*e^(-4)*integrate(x^(3/2)*e^(x^2)*log(x), x) - x*log(x) + 1/2*integrate(2 *x*(2*log(x) + 1), x) + 1/2*integrate((log(x) + 2)*e^(x^2 - 4)/sqrt(x), x)
\[ \int \frac {-2 x+2 x^2+\left (-2 x+4 x^2\right ) \log (x)+e^{\frac {1}{2} (-8+\log (x))} \left (2 e^{x^2}+e^{x^2} \left (1+4 x^2\right ) \log (x)\right )}{2 x} \, dx=\int { \frac {2 \, x^{2} + {\left ({\left (4 \, x^{2} + 1\right )} e^{\left (x^{2}\right )} \log \left (x\right ) + 2 \, e^{\left (x^{2}\right )}\right )} e^{\left (\frac {1}{2} \, \log \left (x\right ) - 4\right )} + 2 \, {\left (2 \, x^{2} - x\right )} \log \left (x\right ) - 2 \, x}{2 \, x} \,d x } \] Input:
integrate(1/2*(((4*x^2+1)*exp(x^2)*log(x)+2*exp(x^2))*exp(1/4*log(x)-2)^2+ (4*x^2-2*x)*log(x)+2*x^2-2*x)/x,x, algorithm="giac")
Output:
integrate(1/2*(2*x^2 + ((4*x^2 + 1)*e^(x^2)*log(x) + 2*e^(x^2))*e^(1/2*log (x) - 4) + 2*(2*x^2 - x)*log(x) - 2*x)/x, x)
Timed out. \[ \int \frac {-2 x+2 x^2+\left (-2 x+4 x^2\right ) \log (x)+e^{\frac {1}{2} (-8+\log (x))} \left (2 e^{x^2}+e^{x^2} \left (1+4 x^2\right ) \log (x)\right )}{2 x} \, dx=-\int \frac {x-\frac {{\mathrm {e}}^{\frac {\ln \left (x\right )}{2}-4}\,\left (2\,{\mathrm {e}}^{x^2}+{\mathrm {e}}^{x^2}\,\ln \left (x\right )\,\left (4\,x^2+1\right )\right )}{2}+\frac {\ln \left (x\right )\,\left (2\,x-4\,x^2\right )}{2}-x^2}{x} \,d x \] Input:
int(-(x - (exp(log(x)/2 - 4)*(2*exp(x^2) + exp(x^2)*log(x)*(4*x^2 + 1)))/2 + (log(x)*(2*x - 4*x^2))/2 - x^2)/x,x)
Output:
-int((x - (exp(log(x)/2 - 4)*(2*exp(x^2) + exp(x^2)*log(x)*(4*x^2 + 1)))/2 + (log(x)*(2*x - 4*x^2))/2 - x^2)/x, x)
Time = 0.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {-2 x+2 x^2+\left (-2 x+4 x^2\right ) \log (x)+e^{\frac {1}{2} (-8+\log (x))} \left (2 e^{x^2}+e^{x^2} \left (1+4 x^2\right ) \log (x)\right )}{2 x} \, dx=\frac {\mathrm {log}\left (x \right ) \left (\sqrt {x}\, e^{x^{2}}+e^{4} x^{2}-e^{4} x \right )}{e^{4}} \] Input:
int(1/2*(((4*x^2+1)*exp(x^2)*log(x)+2*exp(x^2))*exp(1/4*log(x)-2)^2+(4*x^2 -2*x)*log(x)+2*x^2-2*x)/x,x)
Output:
(log(x)*(sqrt(x)*e**(x**2) + e**4*x**2 - e**4*x))/e**4