Integrand size = 88, antiderivative size = 31 \[ \int \frac {-8 e^3 x-32 x^3+8 e^3 x \log \left (-x^2\right )+\left (-e^6+48 x^2-16 x^4+e^3 \left (-12-8 x^2\right )\right ) \log ^2\left (-x^2\right )}{\left (e^6+8 e^3 x^2+16 x^4\right ) \log ^2\left (-x^2\right )} \, dx=-x+\frac {-3+\frac {x}{\log \left (-x^2\right )}}{\frac {e^3}{4 x}+x} \] Output:
1/(1/4*exp(3)/x+x)*(x/ln(-x^2)-3)-x
Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {-8 e^3 x-32 x^3+8 e^3 x \log \left (-x^2\right )+\left (-e^6+48 x^2-16 x^4+e^3 \left (-12-8 x^2\right )\right ) \log ^2\left (-x^2\right )}{\left (e^6+8 e^3 x^2+16 x^4\right ) \log ^2\left (-x^2\right )} \, dx=x \left (-1-\frac {12}{e^3+4 x^2}+\frac {4 x}{\left (e^3+4 x^2\right ) \log \left (-x^2\right )}\right ) \] Input:
Integrate[(-8*E^3*x - 32*x^3 + 8*E^3*x*Log[-x^2] + (-E^6 + 48*x^2 - 16*x^4 + E^3*(-12 - 8*x^2))*Log[-x^2]^2)/((E^6 + 8*E^3*x^2 + 16*x^4)*Log[-x^2]^2 ),x]
Output:
x*(-1 - 12/(E^3 + 4*x^2) + (4*x)/((E^3 + 4*x^2)*Log[-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.
\(\displaystyle \int \frac {-32 x^3+8 e^3 x \log \left (-x^2\right )+\left (-16 x^4+48 x^2+e^3 \left (-8 x^2-12\right )-e^6\right ) \log ^2\left (-x^2\right )-8 e^3 x}{\left (16 x^4+8 e^3 x^2+e^6\right ) \log ^2\left (-x^2\right )} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle 16 \int -\frac {32 x^3-8 e^3 \log \left (-x^2\right ) x+8 e^3 x+\left (16 x^4-48 x^2+4 e^3 \left (2 x^2+3\right )+e^6\right ) \log ^2\left (-x^2\right )}{16 \left (4 x^2+e^3\right )^2 \log ^2\left (-x^2\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {32 x^3-8 e^3 \log \left (-x^2\right ) x+8 e^3 x+\left (16 x^4-48 x^2+4 e^3 \left (2 x^2+3\right )+e^6\right ) \log ^2\left (-x^2\right )}{\left (4 x^2+e^3\right )^2 \log ^2\left (-x^2\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (-\frac {8 e^3 x}{\left (4 x^2+e^3\right )^2 \log \left (-x^2\right )}+\frac {8 x}{\left (4 x^2+e^3\right ) \log ^2\left (-x^2\right )}+\frac {16 x^4-8 \left (6-e^3\right ) x^2+e^3 \left (12+e^3\right )}{\left (4 x^2+e^3\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \text {Subst}\left (\int \frac {1}{\left (4 x+e^3\right ) \log ^2(-x)}dx,x,x^2\right )+4 e^3 \text {Subst}\left (\int \frac {1}{\left (4 x+e^3\right )^2 \log (-x)}dx,x,x^2\right )-\frac {12 x}{4 x^2+e^3}-x\) |
Input:
Int[(-8*E^3*x - 32*x^3 + 8*E^3*x*Log[-x^2] + (-E^6 + 48*x^2 - 16*x^4 + E^3 *(-12 - 8*x^2))*Log[-x^2]^2)/((E^6 + 8*E^3*x^2 + 16*x^4)*Log[-x^2]^2),x]
Output:
$Aborted
Time = 2.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52
method | result | size |
risch | \(-\frac {x \left (4 x^{2}+{\mathrm e}^{3}+12\right )}{4 x^{2}+{\mathrm e}^{3}}+\frac {4 x^{2}}{\left (4 x^{2}+{\mathrm e}^{3}\right ) \ln \left (-x^{2}\right )}\) | \(47\) |
default | \(-x -\frac {12 x}{4 x^{2}+{\mathrm e}^{3}}+\frac {1}{\ln \left (-x^{2}\right )}-\frac {{\mathrm e}^{3}}{\ln \left (-x^{2}\right ) \left (4 x^{2}+{\mathrm e}^{3}\right )}\) | \(48\) |
parts | \(-x -\frac {12 x}{4 x^{2}+{\mathrm e}^{3}}+\frac {1}{\ln \left (-x^{2}\right )}-\frac {{\mathrm e}^{3}}{\ln \left (-x^{2}\right ) \left (4 x^{2}+{\mathrm e}^{3}\right )}\) | \(48\) |
norman | \(\frac {\left (-{\mathrm e}^{3}-12\right ) x \ln \left (-x^{2}\right )+4 x^{2}-4 x^{3} \ln \left (-x^{2}\right )}{\left (4 x^{2}+{\mathrm e}^{3}\right ) \ln \left (-x^{2}\right )}\) | \(51\) |
parallelrisch | \(\frac {-16 x^{3} \ln \left (-x^{2}\right )-4 x \,{\mathrm e}^{3} \ln \left (-x^{2}\right )+16 x^{2}-48 x \ln \left (-x^{2}\right )}{4 \ln \left (-x^{2}\right ) \left (4 x^{2}+{\mathrm e}^{3}\right )}\) | \(58\) |
Input:
int(((-exp(3)^2+(-8*x^2-12)*exp(3)-16*x^4+48*x^2)*ln(-x^2)^2+8*x*exp(3)*ln (-x^2)-8*x*exp(3)-32*x^3)/(exp(3)^2+8*x^2*exp(3)+16*x^4)/ln(-x^2)^2,x,meth od=_RETURNVERBOSE)
Output:
-x*(4*x^2+exp(3)+12)/(4*x^2+exp(3))+4*x^2/(4*x^2+exp(3))/ln(-x^2)
Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {-8 e^3 x-32 x^3+8 e^3 x \log \left (-x^2\right )+\left (-e^6+48 x^2-16 x^4+e^3 \left (-12-8 x^2\right )\right ) \log ^2\left (-x^2\right )}{\left (e^6+8 e^3 x^2+16 x^4\right ) \log ^2\left (-x^2\right )} \, dx=\frac {4 \, x^{2} - {\left (4 \, x^{3} + x e^{3} + 12 \, x\right )} \log \left (-x^{2}\right )}{{\left (4 \, x^{2} + e^{3}\right )} \log \left (-x^{2}\right )} \] Input:
integrate(((-exp(3)^2+(-8*x^2-12)*exp(3)-16*x^4+48*x^2)*log(-x^2)^2+8*x*ex p(3)*log(-x^2)-8*x*exp(3)-32*x^3)/(exp(3)^2+8*x^2*exp(3)+16*x^4)/log(-x^2) ^2,x, algorithm="fricas")
Output:
(4*x^2 - (4*x^3 + x*e^3 + 12*x)*log(-x^2))/((4*x^2 + e^3)*log(-x^2))
Time = 0.13 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {-8 e^3 x-32 x^3+8 e^3 x \log \left (-x^2\right )+\left (-e^6+48 x^2-16 x^4+e^3 \left (-12-8 x^2\right )\right ) \log ^2\left (-x^2\right )}{\left (e^6+8 e^3 x^2+16 x^4\right ) \log ^2\left (-x^2\right )} \, dx=\frac {4 x^{2}}{\left (4 x^{2} + e^{3}\right ) \log {\left (- x^{2} \right )}} - x - \frac {12 x}{4 x^{2} + e^{3}} \] Input:
integrate(((-exp(3)**2+(-8*x**2-12)*exp(3)-16*x**4+48*x**2)*ln(-x**2)**2+8 *x*exp(3)*ln(-x**2)-8*x*exp(3)-32*x**3)/(exp(3)**2+8*x**2*exp(3)+16*x**4)/ ln(-x**2)**2,x)
Output:
4*x**2/((4*x**2 + exp(3))*log(-x**2)) - x - 12*x/(4*x**2 + exp(3))
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.16 \[ \int \frac {-8 e^3 x-32 x^3+8 e^3 x \log \left (-x^2\right )+\left (-e^6+48 x^2-16 x^4+e^3 \left (-12-8 x^2\right )\right ) \log ^2\left (-x^2\right )}{\left (e^6+8 e^3 x^2+16 x^4\right ) \log ^2\left (-x^2\right )} \, dx=-\frac {4 i \, \pi x^{3} + {\left (12 i \, \pi + i \, \pi e^{3}\right )} x - 4 \, x^{2} + 2 \, {\left (4 \, x^{3} + x {\left (e^{3} + 12\right )}\right )} \log \left (x\right )}{4 i \, \pi x^{2} + i \, \pi e^{3} + 2 \, {\left (4 \, x^{2} + e^{3}\right )} \log \left (x\right )} \] Input:
integrate(((-exp(3)^2+(-8*x^2-12)*exp(3)-16*x^4+48*x^2)*log(-x^2)^2+8*x*ex p(3)*log(-x^2)-8*x*exp(3)-32*x^3)/(exp(3)^2+8*x^2*exp(3)+16*x^4)/log(-x^2) ^2,x, algorithm="maxima")
Output:
-(4*I*pi*x^3 + (12*I*pi + I*pi*e^3)*x - 4*x^2 + 2*(4*x^3 + x*(e^3 + 12))*l og(x))/(4*I*pi*x^2 + I*pi*e^3 + 2*(4*x^2 + e^3)*log(x))
Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (30) = 60\).
Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.97 \[ \int \frac {-8 e^3 x-32 x^3+8 e^3 x \log \left (-x^2\right )+\left (-e^6+48 x^2-16 x^4+e^3 \left (-12-8 x^2\right )\right ) \log ^2\left (-x^2\right )}{\left (e^6+8 e^3 x^2+16 x^4\right ) \log ^2\left (-x^2\right )} \, dx=-\frac {4 \, x^{3} \log \left (-x^{2}\right ) + x e^{3} \log \left (-x^{2}\right ) - 4 \, x^{2} + 24 \, x \log \left (-x^{2}\right )}{4 \, x^{2} \log \left (-x^{2}\right ) + e^{3} \log \left (-x^{2}\right )} \] Input:
integrate(((-exp(3)^2+(-8*x^2-12)*exp(3)-16*x^4+48*x^2)*log(-x^2)^2+8*x*ex p(3)*log(-x^2)-8*x*exp(3)-32*x^3)/(exp(3)^2+8*x^2*exp(3)+16*x^4)/log(-x^2) ^2,x, algorithm="giac")
Output:
-(4*x^3*log(-x^2) + x*e^3*log(-x^2) - 4*x^2 + 24*x*log(-x^2))/(4*x^2*log(- x^2) + e^3*log(-x^2))
Time = 0.99 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.71 \[ \int \frac {-8 e^3 x-32 x^3+8 e^3 x \log \left (-x^2\right )+\left (-e^6+48 x^2-16 x^4+e^3 \left (-12-8 x^2\right )\right ) \log ^2\left (-x^2\right )}{\left (e^6+8 e^3 x^2+16 x^4\right ) \log ^2\left (-x^2\right )} \, dx=-\frac {x\,\left (12\,\ln \left (-x^2\right )-4\,x+{\mathrm {e}}^3\,\ln \left (-x^2\right )+4\,x^2\,\ln \left (-x^2\right )\right )}{\ln \left (-x^2\right )\,\left (4\,x^2+{\mathrm {e}}^3\right )} \] Input:
int(-(8*x*exp(3) + 32*x^3 + log(-x^2)^2*(exp(6) + exp(3)*(8*x^2 + 12) - 48 *x^2 + 16*x^4) - 8*x*exp(3)*log(-x^2))/(log(-x^2)^2*(exp(6) + 8*x^2*exp(3) + 16*x^4)),x)
Output:
-(x*(12*log(-x^2) - 4*x + exp(3)*log(-x^2) + 4*x^2*log(-x^2)))/(log(-x^2)* (exp(3) + 4*x^2))
Time = 0.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.77 \[ \int \frac {-8 e^3 x-32 x^3+8 e^3 x \log \left (-x^2\right )+\left (-e^6+48 x^2-16 x^4+e^3 \left (-12-8 x^2\right )\right ) \log ^2\left (-x^2\right )}{\left (e^6+8 e^3 x^2+16 x^4\right ) \log ^2\left (-x^2\right )} \, dx=\frac {x \left (-\mathrm {log}\left (-x^{2}\right ) e^{3}-4 \,\mathrm {log}\left (-x^{2}\right ) x^{2}-12 \,\mathrm {log}\left (-x^{2}\right )+4 x \right )}{\mathrm {log}\left (-x^{2}\right ) \left (e^{3}+4 x^{2}\right )} \] Input:
int(((-exp(3)^2+(-8*x^2-12)*exp(3)-16*x^4+48*x^2)*log(-x^2)^2+8*x*exp(3)*l og(-x^2)-8*x*exp(3)-32*x^3)/(exp(3)^2+8*x^2*exp(3)+16*x^4)/log(-x^2)^2,x)
Output:
(x*( - log( - x**2)*e**3 - 4*log( - x**2)*x**2 - 12*log( - x**2) + 4*x))/( log( - x**2)*(e**3 + 4*x**2))