Integrand size = 104, antiderivative size = 29 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=\frac {12}{-1+\frac {2}{x-\frac {2 x^2}{e^4}}}+\log \left (2-x^2\right ) \] Output:
4/(2/3/(x-2*x^2/exp(4))-1/3)+ln(-x^2+2)
Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=-\frac {24 e^4}{e^4 (-2+x)-2 x^2}+\log \left (2-x^2\right ) \] Input:
Integrate[(8*x^5 + E^8*(-48 + 8*x + 16*x^2 + 2*x^3) + E^4*(192*x - 80*x^3 - 8*x^4))/(-8*x^4 + 4*x^6 + E^8*(-8 + 8*x + 2*x^2 - 4*x^3 + x^4) + E^4*(-1 6*x^2 + 8*x^3 + 8*x^4 - 4*x^5)),x]
Output:
(-24*E^4)/(E^4*(-2 + x) - 2*x^2) + Log[2 - x^2]
Time = 0.34 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {2462, 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.
\(\displaystyle \int \frac {8 x^5+e^4 \left (-8 x^4-80 x^3+192 x\right )+e^8 \left (2 x^3+16 x^2+8 x-48\right )}{4 x^6-8 x^4+e^8 \left (x^4-4 x^3+2 x^2+8 x-8\right )+e^4 \left (-4 x^5+8 x^4+8 x^3-16 x^2\right )} \, dx\) |
\(\Big \downarrow \) 2462 |
\(\displaystyle \int \left (\frac {2 x}{x^2-2}+\frac {24 \left (e^8-4 e^4 x\right )}{\left (-2 x^2+e^4 x-2 e^4\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {24 e^4}{2 x^2-e^4 x+2 e^4}+\log \left (2-x^2\right )\) |
Input:
Int[(8*x^5 + E^8*(-48 + 8*x + 16*x^2 + 2*x^3) + E^4*(192*x - 80*x^3 - 8*x^ 4))/(-8*x^4 + 4*x^6 + E^8*(-8 + 8*x + 2*x^2 - 4*x^3 + x^4) + E^4*(-16*x^2 + 8*x^3 + 8*x^4 - 4*x^5)),x]
Output:
(24*E^4)/(2*E^4 - E^4*x + 2*x^2) + Log[2 - x^2]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u*Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && GtQ [Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0 ] && RationalFunctionQ[u, x]
Time = 0.69 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97
method | result | size |
norman | \(-\frac {24 \,{\mathrm e}^{4}}{x \,{\mathrm e}^{4}-2 x^{2}-2 \,{\mathrm e}^{4}}+\ln \left (x^{2}-2\right )\) | \(28\) |
risch | \(-\frac {24 \,{\mathrm e}^{4}}{x \,{\mathrm e}^{4}-2 x^{2}-2 \,{\mathrm e}^{4}}+\ln \left (x^{2}-2\right )\) | \(28\) |
parallelrisch | \(\frac {2 \,{\mathrm e}^{4} \ln \left (x^{2}-2\right ) x -4 \ln \left (x^{2}-2\right ) x^{2}-4 \,{\mathrm e}^{4} \ln \left (x^{2}-2\right )-48 \,{\mathrm e}^{4}}{2 x \,{\mathrm e}^{4}-4 x^{2}-4 \,{\mathrm e}^{4}}\) | \(56\) |
default | \(-12 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (4 \textit {\_Z}^{4}-4 \textit {\_Z}^{3} {\mathrm e}^{4}+\left (8 \,{\mathrm e}^{4}+{\mathrm e}^{8}\right ) \textit {\_Z}^{2}-4 \,{\mathrm e}^{8} \textit {\_Z} +4 \,{\mathrm e}^{8}\right )}{\sum }\frac {\left (-4 \textit {\_R} \,{\mathrm e}^{4}+{\mathrm e}^{8}\right ) \ln \left (x -\textit {\_R} \right )}{6 \textit {\_R}^{2} {\mathrm e}^{4}-8 \textit {\_R}^{3}-8 \textit {\_R} \,{\mathrm e}^{4}-{\mathrm e}^{8} \textit {\_R} +2 \,{\mathrm e}^{8}}\right )+\ln \left (x^{2}-2\right )\) | \(91\) |
Input:
int(((2*x^3+16*x^2+8*x-48)*exp(4)^2+(-8*x^4-80*x^3+192*x)*exp(4)+8*x^5)/(( x^4-4*x^3+2*x^2+8*x-8)*exp(4)^2+(-4*x^5+8*x^4+8*x^3-16*x^2)*exp(4)+4*x^6-8 *x^4),x,method=_RETURNVERBOSE)
Output:
-24*exp(4)/(x*exp(4)-2*x^2-2*exp(4))+ln(x^2-2)
Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=\frac {{\left (2 \, x^{2} - {\left (x - 2\right )} e^{4}\right )} \log \left (x^{2} - 2\right ) + 24 \, e^{4}}{2 \, x^{2} - {\left (x - 2\right )} e^{4}} \] Input:
integrate(((2*x^3+16*x^2+8*x-48)*exp(4)^2+(-8*x^4-80*x^3+192*x)*exp(4)+8*x ^5)/((x^4-4*x^3+2*x^2+8*x-8)*exp(4)^2+(-4*x^5+8*x^4+8*x^3-16*x^2)*exp(4)+4 *x^6-8*x^4),x, algorithm="fricas")
Output:
((2*x^2 - (x - 2)*e^4)*log(x^2 - 2) + 24*e^4)/(2*x^2 - (x - 2)*e^4)
Time = 1.36 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=\log {\left (x^{2} - 2 \right )} + \frac {24 e^{4}}{2 x^{2} - x e^{4} + 2 e^{4}} \] Input:
integrate(((2*x**3+16*x**2+8*x-48)*exp(4)**2+(-8*x**4-80*x**3+192*x)*exp(4 )+8*x**5)/((x**4-4*x**3+2*x**2+8*x-8)*exp(4)**2+(-4*x**5+8*x**4+8*x**3-16* x**2)*exp(4)+4*x**6-8*x**4),x)
Output:
log(x**2 - 2) + 24*exp(4)/(2*x**2 - x*exp(4) + 2*exp(4))
Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=\frac {24 \, e^{4}}{2 \, x^{2} - x e^{4} + 2 \, e^{4}} + \log \left (x^{2} - 2\right ) \] Input:
integrate(((2*x^3+16*x^2+8*x-48)*exp(4)^2+(-8*x^4-80*x^3+192*x)*exp(4)+8*x ^5)/((x^4-4*x^3+2*x^2+8*x-8)*exp(4)^2+(-4*x^5+8*x^4+8*x^3-16*x^2)*exp(4)+4 *x^6-8*x^4),x, algorithm="maxima")
Output:
24*e^4/(2*x^2 - x*e^4 + 2*e^4) + log(x^2 - 2)
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=\frac {24 \, e^{4}}{2 \, x^{2} - x e^{4} + 2 \, e^{4}} + \log \left ({\left | x^{2} - 2 \right |}\right ) \] Input:
integrate(((2*x^3+16*x^2+8*x-48)*exp(4)^2+(-8*x^4-80*x^3+192*x)*exp(4)+8*x ^5)/((x^4-4*x^3+2*x^2+8*x-8)*exp(4)^2+(-4*x^5+8*x^4+8*x^3-16*x^2)*exp(4)+4 *x^6-8*x^4),x, algorithm="giac")
Output:
24*e^4/(2*x^2 - x*e^4 + 2*e^4) + log(abs(x^2 - 2))
Time = 0.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=\ln \left (x^2-2\right )+\frac {24\,{\mathrm {e}}^4}{2\,x^2-{\mathrm {e}}^4\,x+2\,{\mathrm {e}}^4} \] Input:
int((exp(8)*(8*x + 16*x^2 + 2*x^3 - 48) - exp(4)*(80*x^3 - 192*x + 8*x^4) + 8*x^5)/(exp(8)*(8*x + 2*x^2 - 4*x^3 + x^4 - 8) - 8*x^4 + 4*x^6 - exp(4)* (16*x^2 - 8*x^3 - 8*x^4 + 4*x^5)),x)
Output:
log(x^2 - 2) + (24*exp(4))/(2*exp(4) - x*exp(4) + 2*x^2)
Time = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 3.14 \[ \int \frac {8 x^5+e^8 \left (-48+8 x+16 x^2+2 x^3\right )+e^4 \left (192 x-80 x^3-8 x^4\right )}{-8 x^4+4 x^6+e^8 \left (-8+8 x+2 x^2-4 x^3+x^4\right )+e^4 \left (-16 x^2+8 x^3+8 x^4-4 x^5\right )} \, dx=\frac {\mathrm {log}\left (-\sqrt {2}+x \right ) e^{4} x -2 \,\mathrm {log}\left (-\sqrt {2}+x \right ) e^{4}-2 \,\mathrm {log}\left (-\sqrt {2}+x \right ) x^{2}+\mathrm {log}\left (\sqrt {2}+x \right ) e^{4} x -2 \,\mathrm {log}\left (\sqrt {2}+x \right ) e^{4}-2 \,\mathrm {log}\left (\sqrt {2}+x \right ) x^{2}-24 e^{4}}{e^{4} x -2 e^{4}-2 x^{2}} \] Input:
int(((2*x^3+16*x^2+8*x-48)*exp(4)^2+(-8*x^4-80*x^3+192*x)*exp(4)+8*x^5)/(( x^4-4*x^3+2*x^2+8*x-8)*exp(4)^2+(-4*x^5+8*x^4+8*x^3-16*x^2)*exp(4)+4*x^6-8 *x^4),x)
Output:
(log( - sqrt(2) + x)*e**4*x - 2*log( - sqrt(2) + x)*e**4 - 2*log( - sqrt(2 ) + x)*x**2 + log(sqrt(2) + x)*e**4*x - 2*log(sqrt(2) + x)*e**4 - 2*log(sq rt(2) + x)*x**2 - 24*e**4)/(e**4*x - 2*e**4 - 2*x**2)