Integrand size = 106, antiderivative size = 26 \[ \int \frac {-4 x^4-2 x^5+\left (4 x^2+4 x^3+2 x^4+2 x^5\right ) \log \left (2+x^2\right )+\left (4 x^2+4 x^3+2 x^4+2 x^5+e^x \left (-4+4 x+2 x^3+x^4\right )\right ) \log ^2\left (2+x^2\right )}{\left (2 x^2+x^4\right ) \log ^2\left (2+x^2\right )} \, dx=(2+x) \left (x+\frac {e^x+\frac {x^2}{\log \left (2+x^2\right )}}{x}\right ) \] Output:
(2+x)*(x+(x^2/ln(x^2+2)+exp(x))/x)
Time = 0.54 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x^4-2 x^5+\left (4 x^2+4 x^3+2 x^4+2 x^5\right ) \log \left (2+x^2\right )+\left (4 x^2+4 x^3+2 x^4+2 x^5+e^x \left (-4+4 x+2 x^3+x^4\right )\right ) \log ^2\left (2+x^2\right )}{\left (2 x^2+x^4\right ) \log ^2\left (2+x^2\right )} \, dx=\frac {(2+x) \left (e^x+x^2+\frac {x^2}{\log \left (2+x^2\right )}\right )}{x} \] Input:
Integrate[(-4*x^4 - 2*x^5 + (4*x^2 + 4*x^3 + 2*x^4 + 2*x^5)*Log[2 + x^2] + (4*x^2 + 4*x^3 + 2*x^4 + 2*x^5 + E^x*(-4 + 4*x + 2*x^3 + x^4))*Log[2 + x^ 2]^2)/((2*x^2 + x^4)*Log[2 + x^2]^2),x]
Output:
((2 + x)*(E^x + x^2 + x^2/Log[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.
\(\displaystyle \int \frac {-2 x^5-4 x^4+\left (2 x^5+2 x^4+4 x^3+4 x^2+e^x \left (x^4+2 x^3+4 x-4\right )\right ) \log ^2\left (x^2+2\right )+\left (2 x^5+2 x^4+4 x^3+4 x^2\right ) \log \left (x^2+2\right )}{\left (x^4+2 x^2\right ) \log ^2\left (x^2+2\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-2 x^5-4 x^4+\left (2 x^5+2 x^4+4 x^3+4 x^2+e^x \left (x^4+2 x^3+4 x-4\right )\right ) \log ^2\left (x^2+2\right )+\left (2 x^5+2 x^4+4 x^3+4 x^2\right ) \log \left (x^2+2\right )}{x^2 \left (x^2+2\right ) \log ^2\left (x^2+2\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {e^x \left (x^2+2 x-2\right )}{x^2}-\frac {2 (x+2) x^2}{\left (x^2+2\right ) \log ^2\left (x^2+2\right )}+\frac {2 (x+1)}{\log \left (x^2+2\right )}+2 (x+1)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int \frac {1}{\log ^2\left (x^2+2\right )}dx+8 \int \frac {1}{\left (x^2+2\right ) \log ^2\left (x^2+2\right )}dx+2 \int \frac {x+1}{\log \left (x^2+2\right )}dx-\operatorname {LogIntegral}\left (x^2+2\right )+\frac {x^2+2}{\log \left (x^2+2\right )}-\frac {2}{\log \left (x^2+2\right )}+(x+1)^2+e^x+\frac {2 e^x}{x}\) |
Input:
Int[(-4*x^4 - 2*x^5 + (4*x^2 + 4*x^3 + 2*x^4 + 2*x^5)*Log[2 + x^2] + (4*x^ 2 + 4*x^3 + 2*x^4 + 2*x^5 + E^x*(-4 + 4*x + 2*x^3 + x^4))*Log[2 + x^2]^2)/ ((2*x^2 + x^4)*Log[2 + x^2]^2),x]
Output:
$Aborted
Time = 0.93 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38
\[\frac {x^{3}+2 x^{2}+{\mathrm e}^{x} x +2 \,{\mathrm e}^{x}}{x}+\frac {x \left (2+x \right )}{\ln \left (x^{2}+2\right )}\]
Input:
int((((x^4+2*x^3+4*x-4)*exp(x)+2*x^5+2*x^4+4*x^3+4*x^2)*ln(x^2+2)^2+(2*x^5 +2*x^4+4*x^3+4*x^2)*ln(x^2+2)-2*x^5-4*x^4)/(x^4+2*x^2)/ln(x^2+2)^2,x)
Output:
(x^3+2*x^2+exp(x)*x+2*exp(x))/x+x*(2+x)/ln(x^2+2)
Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {-4 x^4-2 x^5+\left (4 x^2+4 x^3+2 x^4+2 x^5\right ) \log \left (2+x^2\right )+\left (4 x^2+4 x^3+2 x^4+2 x^5+e^x \left (-4+4 x+2 x^3+x^4\right )\right ) \log ^2\left (2+x^2\right )}{\left (2 x^2+x^4\right ) \log ^2\left (2+x^2\right )} \, dx=\frac {x^{3} + 2 \, x^{2} + {\left (x^{3} + 2 \, x^{2} + {\left (x + 2\right )} e^{x}\right )} \log \left (x^{2} + 2\right )}{x \log \left (x^{2} + 2\right )} \] Input:
integrate((((x^4+2*x^3+4*x-4)*exp(x)+2*x^5+2*x^4+4*x^3+4*x^2)*log(x^2+2)^2 +(2*x^5+2*x^4+4*x^3+4*x^2)*log(x^2+2)-2*x^5-4*x^4)/(x^4+2*x^2)/log(x^2+2)^ 2,x, algorithm="fricas")
Output:
(x^3 + 2*x^2 + (x^3 + 2*x^2 + (x + 2)*e^x)*log(x^2 + 2))/(x*log(x^2 + 2))
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {-4 x^4-2 x^5+\left (4 x^2+4 x^3+2 x^4+2 x^5\right ) \log \left (2+x^2\right )+\left (4 x^2+4 x^3+2 x^4+2 x^5+e^x \left (-4+4 x+2 x^3+x^4\right )\right ) \log ^2\left (2+x^2\right )}{\left (2 x^2+x^4\right ) \log ^2\left (2+x^2\right )} \, dx=x^{2} + 2 x + \frac {x^{2} + 2 x}{\log {\left (x^{2} + 2 \right )}} + \frac {\left (x + 2\right ) e^{x}}{x} \] Input:
integrate((((x**4+2*x**3+4*x-4)*exp(x)+2*x**5+2*x**4+4*x**3+4*x**2)*ln(x** 2+2)**2+(2*x**5+2*x**4+4*x**3+4*x**2)*ln(x**2+2)-2*x**5-4*x**4)/(x**4+2*x* *2)/ln(x**2+2)**2,x)
Output:
x**2 + 2*x + (x**2 + 2*x)/log(x**2 + 2) + (x + 2)*exp(x)/x
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {-4 x^4-2 x^5+\left (4 x^2+4 x^3+2 x^4+2 x^5\right ) \log \left (2+x^2\right )+\left (4 x^2+4 x^3+2 x^4+2 x^5+e^x \left (-4+4 x+2 x^3+x^4\right )\right ) \log ^2\left (2+x^2\right )}{\left (2 x^2+x^4\right ) \log ^2\left (2+x^2\right )} \, dx=\frac {x^{3} + 2 \, x^{2} + {\left (x^{3} + 2 \, x^{2} + {\left (x + 2\right )} e^{x}\right )} \log \left (x^{2} + 2\right )}{x \log \left (x^{2} + 2\right )} \] Input:
integrate((((x^4+2*x^3+4*x-4)*exp(x)+2*x^5+2*x^4+4*x^3+4*x^2)*log(x^2+2)^2 +(2*x^5+2*x^4+4*x^3+4*x^2)*log(x^2+2)-2*x^5-4*x^4)/(x^4+2*x^2)/log(x^2+2)^ 2,x, algorithm="maxima")
Output:
(x^3 + 2*x^2 + (x^3 + 2*x^2 + (x + 2)*e^x)*log(x^2 + 2))/(x*log(x^2 + 2))
Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (25) = 50\).
Time = 0.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.38 \[ \int \frac {-4 x^4-2 x^5+\left (4 x^2+4 x^3+2 x^4+2 x^5\right ) \log \left (2+x^2\right )+\left (4 x^2+4 x^3+2 x^4+2 x^5+e^x \left (-4+4 x+2 x^3+x^4\right )\right ) \log ^2\left (2+x^2\right )}{\left (2 x^2+x^4\right ) \log ^2\left (2+x^2\right )} \, dx=\frac {x^{3} \log \left (x^{2} + 2\right ) + x^{3} + 2 \, x^{2} \log \left (x^{2} + 2\right ) + x e^{x} \log \left (x^{2} + 2\right ) + 2 \, x^{2} + 2 \, e^{x} \log \left (x^{2} + 2\right )}{x \log \left (x^{2} + 2\right )} \] Input:
integrate((((x^4+2*x^3+4*x-4)*exp(x)+2*x^5+2*x^4+4*x^3+4*x^2)*log(x^2+2)^2 +(2*x^5+2*x^4+4*x^3+4*x^2)*log(x^2+2)-2*x^5-4*x^4)/(x^4+2*x^2)/log(x^2+2)^ 2,x, algorithm="giac")
Output:
(x^3*log(x^2 + 2) + x^3 + 2*x^2*log(x^2 + 2) + x*e^x*log(x^2 + 2) + 2*x^2 + 2*e^x*log(x^2 + 2))/(x*log(x^2 + 2))
Time = 2.58 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {-4 x^4-2 x^5+\left (4 x^2+4 x^3+2 x^4+2 x^5\right ) \log \left (2+x^2\right )+\left (4 x^2+4 x^3+2 x^4+2 x^5+e^x \left (-4+4 x+2 x^3+x^4\right )\right ) \log ^2\left (2+x^2\right )}{\left (2 x^2+x^4\right ) \log ^2\left (2+x^2\right )} \, dx=2\,x+{\mathrm {e}}^x+\frac {2\,{\mathrm {e}}^x}{x}+\frac {x^2}{\ln \left (x^2+2\right )}+x^2+\frac {2\,x}{\ln \left (x^2+2\right )} \] Input:
int((log(x^2 + 2)^2*(exp(x)*(4*x + 2*x^3 + x^4 - 4) + 4*x^2 + 4*x^3 + 2*x^ 4 + 2*x^5) + log(x^2 + 2)*(4*x^2 + 4*x^3 + 2*x^4 + 2*x^5) - 4*x^4 - 2*x^5) /(log(x^2 + 2)^2*(2*x^2 + x^4)),x)
Output:
2*x + exp(x) + (2*exp(x))/x + x^2/log(x^2 + 2) + x^2 + (2*x)/log(x^2 + 2)
Time = 0.21 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.46 \[ \int \frac {-4 x^4-2 x^5+\left (4 x^2+4 x^3+2 x^4+2 x^5\right ) \log \left (2+x^2\right )+\left (4 x^2+4 x^3+2 x^4+2 x^5+e^x \left (-4+4 x+2 x^3+x^4\right )\right ) \log ^2\left (2+x^2\right )}{\left (2 x^2+x^4\right ) \log ^2\left (2+x^2\right )} \, dx=\frac {e^{x} \mathrm {log}\left (x^{2}+2\right ) x +2 e^{x} \mathrm {log}\left (x^{2}+2\right )+\mathrm {log}\left (x^{2}+2\right ) x^{3}+2 \,\mathrm {log}\left (x^{2}+2\right ) x^{2}+x^{3}+2 x^{2}}{\mathrm {log}\left (x^{2}+2\right ) x} \] Input:
int((((x^4+2*x^3+4*x-4)*exp(x)+2*x^5+2*x^4+4*x^3+4*x^2)*log(x^2+2)^2+(2*x^ 5+2*x^4+4*x^3+4*x^2)*log(x^2+2)-2*x^5-4*x^4)/(x^4+2*x^2)/log(x^2+2)^2,x)
Output:
(e**x*log(x**2 + 2)*x + 2*e**x*log(x**2 + 2) + log(x**2 + 2)*x**3 + 2*log( x**2 + 2)*x**2 + x**3 + 2*x**2)/(log(x**2 + 2)*x)