Integrand size = 55, antiderivative size = 25 \[ \int \frac {3-x^2+\left (-2 x^2+e^x (2+2 x)\right ) \log (x)+\left (e^x (1+x)+e^x \left (2 x+x^2\right ) \log (x)\right ) \log \left (x^2\right )}{x} \, dx=\log (x) \left (3-x \left (x-\frac {e^x (1+x) \log \left (x^2\right )}{x}\right )\right ) \] Output:
ln(x)*(-x*(x-exp(x)*(1+x)*ln(x^2)/x)+3)
Time = 0.15 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {3-x^2+\left (-2 x^2+e^x (2+2 x)\right ) \log (x)+\left (e^x (1+x)+e^x \left (2 x+x^2\right ) \log (x)\right ) \log \left (x^2\right )}{x} \, dx=-\log (x) \left (-3+x^2-e^x (1+x) \log \left (x^2\right )\right ) \] Input:
Integrate[(3 - x^2 + (-2*x^2 + E^x*(2 + 2*x))*Log[x] + (E^x*(1 + x) + E^x* (2*x + x^2)*Log[x])*Log[x^2])/x,x]
Output:
-(Log[x]*(-3 + x^2 - E^x*(1 + x)*Log[x^2]))
Time = 1.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {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.
\(\displaystyle \int \frac {-x^2+\left (e^x (2 x+2)-2 x^2\right ) \log (x)+\left (e^x \left (x^2+2 x\right ) \log (x)+e^x (x+1)\right ) \log \left (x^2\right )+3}{x} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {-x^2-2 x^2 \log (x)+3}{x}+\frac {e^x \left (x^2 \log (x) \log \left (x^2\right )+2 x \log (x) \log \left (x^2\right )+x \log \left (x^2\right )+\log \left (x^2\right )+2 x \log (x)+2 \log (x)\right )}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^2 (-\log (x))+e^x x \log (x) \log \left (x^2\right )+e^x \log (x) \log \left (x^2\right )+3 \log (x)\) |
Input:
Int[(3 - x^2 + (-2*x^2 + E^x*(2 + 2*x))*Log[x] + (E^x*(1 + x) + E^x*(2*x + x^2)*Log[x])*Log[x^2])/x,x]
Output:
3*Log[x] - x^2*Log[x] + E^x*Log[x]*Log[x^2] + E^x*x*Log[x]*Log[x^2]
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 = 2.68 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28
method | result | size |
parallelrisch | \(\ln \left (x \right ) {\mathrm e}^{x} \ln \left (x^{2}\right ) x -x^{2} \ln \left (x \right )+3 \ln \left (x \right )+\ln \left (x \right ) {\mathrm e}^{x} \ln \left (x^{2}\right )\) | \(32\) |
default | \(\left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right ) {\mathrm e}^{x} \ln \left (x \right )+x \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right ) {\mathrm e}^{x} \ln \left (x \right )+2 \,{\mathrm e}^{x} \ln \left (x \right )^{2}+2 x \,{\mathrm e}^{x} \ln \left (x \right )^{2}-x^{2} \ln \left (x \right )+3 \ln \left (x \right )\) | \(59\) |
parts | \(\left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right ) {\mathrm e}^{x} \ln \left (x \right )+x \left (\ln \left (x^{2}\right )-2 \ln \left (x \right )\right ) {\mathrm e}^{x} \ln \left (x \right )+2 \,{\mathrm e}^{x} \ln \left (x \right )^{2}+2 x \,{\mathrm e}^{x} \ln \left (x \right )^{2}-x^{2} \ln \left (x \right )+3 \ln \left (x \right )\) | \(59\) |
risch | \(2 \left (1+x \right ) {\mathrm e}^{x} \ln \left (x \right )^{2}+\left (-x^{2}-\frac {i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}}{2}+i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x}-\frac {i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x}}{2}-\frac {i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}}{2}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x}-\frac {i \pi \operatorname {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x}}{2}\right ) \ln \left (x \right )+3 \ln \left (x \right )\) | \(139\) |
orering | \(\text {Expression too large to display}\) | \(3245\) |
Input:
int((((x^2+2*x)*exp(x)*ln(x)+(1+x)*exp(x))*ln(x^2)+((2+2*x)*exp(x)-2*x^2)* ln(x)-x^2+3)/x,x,method=_RETURNVERBOSE)
Output:
ln(x)*exp(x)*ln(x^2)*x-x^2*ln(x)+3*ln(x)+ln(x)*exp(x)*ln(x^2)
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {3-x^2+\left (-2 x^2+e^x (2+2 x)\right ) \log (x)+\left (e^x (1+x)+e^x \left (2 x+x^2\right ) \log (x)\right ) \log \left (x^2\right )}{x} \, dx=2 \, {\left (x + 1\right )} e^{x} \log \left (x\right )^{2} - {\left (x^{2} - 3\right )} \log \left (x\right ) \] Input:
integrate((((x^2+2*x)*exp(x)*log(x)+(1+x)*exp(x))*log(x^2)+((2+2*x)*exp(x) -2*x^2)*log(x)-x^2+3)/x,x, algorithm="fricas")
Output:
2*(x + 1)*e^x*log(x)^2 - (x^2 - 3)*log(x)
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {3-x^2+\left (-2 x^2+e^x (2+2 x)\right ) \log (x)+\left (e^x (1+x)+e^x \left (2 x+x^2\right ) \log (x)\right ) \log \left (x^2\right )}{x} \, dx=- x^{2} \log {\left (x \right )} + \left (2 x \log {\left (x \right )}^{2} + 2 \log {\left (x \right )}^{2}\right ) e^{x} + 3 \log {\left (x \right )} \] Input:
integrate((((x**2+2*x)*exp(x)*ln(x)+(1+x)*exp(x))*ln(x**2)+((2+2*x)*exp(x) -2*x**2)*ln(x)-x**2+3)/x,x)
Output:
-x**2*log(x) + (2*x*log(x)**2 + 2*log(x)**2)*exp(x) + 3*log(x)
\[ \int \frac {3-x^2+\left (-2 x^2+e^x (2+2 x)\right ) \log (x)+\left (e^x (1+x)+e^x \left (2 x+x^2\right ) \log (x)\right ) \log \left (x^2\right )}{x} \, dx=\int { -\frac {x^{2} - {\left ({\left (x^{2} + 2 \, x\right )} e^{x} \log \left (x\right ) + {\left (x + 1\right )} e^{x}\right )} \log \left (x^{2}\right ) + 2 \, {\left (x^{2} - {\left (x + 1\right )} e^{x}\right )} \log \left (x\right ) - 3}{x} \,d x } \] Input:
integrate((((x^2+2*x)*exp(x)*log(x)+(1+x)*exp(x))*log(x^2)+((2+2*x)*exp(x) -2*x^2)*log(x)-x^2+3)/x,x, algorithm="maxima")
Output:
-x^2*log(x) + 2*((x + 1)*log(x)^2 - log(x))*e^x + 2*e^x*log(x) - 2*Ei(x) + 2*integrate(e^x/x, x) + 3*log(x)
Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {3-x^2+\left (-2 x^2+e^x (2+2 x)\right ) \log (x)+\left (e^x (1+x)+e^x \left (2 x+x^2\right ) \log (x)\right ) \log \left (x^2\right )}{x} \, dx=2 \, x e^{x} \log \left (x\right )^{2} - x^{2} \log \left (x\right ) + 2 \, e^{x} \log \left (x\right )^{2} + 3 \, \log \left (x\right ) \] Input:
integrate((((x^2+2*x)*exp(x)*log(x)+(1+x)*exp(x))*log(x^2)+((2+2*x)*exp(x) -2*x^2)*log(x)-x^2+3)/x,x, algorithm="giac")
Output:
2*x*e^x*log(x)^2 - x^2*log(x) + 2*e^x*log(x)^2 + 3*log(x)
Time = 3.66 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {3-x^2+\left (-2 x^2+e^x (2+2 x)\right ) \log (x)+\left (e^x (1+x)+e^x \left (2 x+x^2\right ) \log (x)\right ) \log \left (x^2\right )}{x} \, dx=\ln \left (x\right )\,\left (\ln \left (x^2\right )\,{\mathrm {e}}^x-x^2+x\,\ln \left (x^2\right )\,{\mathrm {e}}^x+3\right ) \] Input:
int((log(x^2)*(exp(x)*(x + 1) + exp(x)*log(x)*(2*x + x^2)) + log(x)*(exp(x )*(2*x + 2) - 2*x^2) - x^2 + 3)/x,x)
Output:
log(x)*(log(x^2)*exp(x) - x^2 + x*log(x^2)*exp(x) + 3)
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {3-x^2+\left (-2 x^2+e^x (2+2 x)\right ) \log (x)+\left (e^x (1+x)+e^x \left (2 x+x^2\right ) \log (x)\right ) \log \left (x^2\right )}{x} \, dx=\mathrm {log}\left (x \right ) \left (e^{x} \mathrm {log}\left (x^{2}\right ) x +e^{x} \mathrm {log}\left (x^{2}\right )-x^{2}+3\right ) \] Input:
int((((x^2+2*x)*exp(x)*log(x)+(1+x)*exp(x))*log(x^2)+((2+2*x)*exp(x)-2*x^2 )*log(x)-x^2+3)/x,x)
Output:
log(x)*(e**x*log(x**2)*x + e**x*log(x**2) - x**2 + 3)