Integrand size = 111, antiderivative size = 25 \[ \int \frac {\log ^{\frac {2 x}{e^4 \left (-8 x-2 x^2\right )+(5+2 x) \log (\log (6))}}(6) \left (4 e^4 x^2 \log (\log (6))+10 \log ^2(\log (6))\right )}{e^8 \left (64 x^2+32 x^3+4 x^4\right )+e^4 \left (-80 x-52 x^2-8 x^3\right ) \log (\log (6))+\left (25+20 x+4 x^2\right ) \log ^2(\log (6))} \, dx=e^{\frac {x}{\frac {5}{2}+x-\frac {e^4 x (4+x)}{\log (\log (6))}}} \] Output:
exp(x/(x-exp(4)*x*(4+x)/ln(ln(6))+5/2))
Time = 1.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {\log ^{\frac {2 x}{e^4 \left (-8 x-2 x^2\right )+(5+2 x) \log (\log (6))}}(6) \left (4 e^4 x^2 \log (\log (6))+10 \log ^2(\log (6))\right )}{e^8 \left (64 x^2+32 x^3+4 x^4\right )+e^4 \left (-80 x-52 x^2-8 x^3\right ) \log (\log (6))+\left (25+20 x+4 x^2\right ) \log ^2(\log (6))} \, dx=\log ^{-\frac {2 x}{2 e^4 x (4+x)-(5+2 x) \log (\log (6))}}(6) \] Input:
Integrate[(Log[6]^((2*x)/(E^4*(-8*x - 2*x^2) + (5 + 2*x)*Log[Log[6]]))*(4* E^4*x^2*Log[Log[6]] + 10*Log[Log[6]]^2))/(E^8*(64*x^2 + 32*x^3 + 4*x^4) + E^4*(-80*x - 52*x^2 - 8*x^3)*Log[Log[6]] + (25 + 20*x + 4*x^2)*Log[Log[6]] ^2),x]
Output:
Log[6]^((-2*x)/(2*E^4*x*(4 + x) - (5 + 2*x)*Log[Log[6]]))
Time = 0.81 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {2463, 6, 7257}
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.
| \(\displaystyle \int \frac {\log ^{\frac {2 x}{e^4 \left (-2 x^2-8 x\right )+(2 x+5) \log (\log (6))}}(6) \left (4 e^4 x^2 \log (\log (6))+10 \log ^2(\log (6))\right )}{\left (4 x^2+20 x+25\right ) \log ^2(\log (6))+e^4 \left (-8 x^3-52 x^2-80 x\right ) \log (\log (6))+e^8 \left (4 x^4+32 x^3+64 x^2\right )} \, dx\) |
| \(\Big \downarrow \) 2463 |
| \(\displaystyle \int \frac {\log ^{\frac {2 x}{e^4 \left (-2 x^2-8 x\right )+(2 x+5) \log (\log (6))}}(6) \left (4 e^4 x^2 \log (\log (6))+10 \log ^2(\log (6))\right )}{\left (2 e^4 x^2+8 e^4 x-2 x \log (\log (6))-5 \log (\log (6))\right )^2}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {\log ^{\frac {2 x}{e^4 \left (-2 x^2-8 x\right )+(2 x+5) \log (\log (6))}}(6) \left (4 e^4 x^2 \log (\log (6))+10 \log ^2(\log (6))\right )}{\left (2 e^4 x^2+x \left (8 e^4-2 \log (\log (6))\right )-5 \log (\log (6))\right )^2}dx\) |
| \(\Big \downarrow \) 7257 |
| \(\displaystyle \log ^{-\frac {2 x}{2 e^4 \left (x^2+4 x\right )-(2 x+5) \log (\log (6))}}(6)\) |
Input:
Int[(Log[6]^((2*x)/(E^4*(-8*x - 2*x^2) + (5 + 2*x)*Log[Log[6]]))*(4*E^4*x^ 2*Log[Log[6]] + 10*Log[Log[6]]^2))/(E^8*(64*x^2 + 32*x^3 + 4*x^4) + E^4*(- 80*x - 52*x^2 - 8*x^3)*Log[Log[6]] + (25 + 20*x + 4*x^2)*Log[Log[6]]^2),x]
Output:
Log[6]^((-2*x)/(2*E^4*(4*x + x^2) - (5 + 2*x)*Log[Log[6]]))
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
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] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 1.41 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36
| method | result | size |
| gosper | \({\mathrm e}^{-\frac {2 \ln \left (\ln \left (6\right )\right ) x}{2 x^{2} {\mathrm e}^{4}+8 x \,{\mathrm e}^{4}-2 \ln \left (\ln \left (6\right )\right ) x -5 \ln \left (\ln \left (6\right )\right )}}\) | \(34\) |
| parallelrisch | \({\mathrm e}^{-\frac {2 \ln \left (\ln \left (6\right )\right ) x}{2 x^{2} {\mathrm e}^{4}+8 x \,{\mathrm e}^{4}-2 \ln \left (\ln \left (6\right )\right ) x -5 \ln \left (\ln \left (6\right )\right )}}\) | \(34\) |
| risch | \(\left (\ln \left (2\right )+\ln \left (3\right )\right )^{-\frac {2 x}{2 x^{2} {\mathrm e}^{4}+8 x \,{\mathrm e}^{4}-2 \ln \left (\ln \left (2\right )+\ln \left (3\right )\right ) x -5 \ln \left (\ln \left (2\right )+\ln \left (3\right )\right )}}\) | \(42\) |
| norman | \(\frac {\left (8 \,{\mathrm e}^{4}-2 \ln \left (\ln \left (6\right )\right )\right ) x \,{\mathrm e}^{\frac {2 x \ln \left (\ln \left (6\right )\right )}{\left (5+2 x \right ) \ln \left (\ln \left (6\right )\right )+\left (-2 x^{2}-8 x \right ) {\mathrm e}^{4}}}-5 \ln \left (\ln \left (6\right )\right ) {\mathrm e}^{\frac {2 x \ln \left (\ln \left (6\right )\right )}{\left (5+2 x \right ) \ln \left (\ln \left (6\right )\right )+\left (-2 x^{2}-8 x \right ) {\mathrm e}^{4}}}+2 x^{2} {\mathrm e}^{4} {\mathrm e}^{\frac {2 x \ln \left (\ln \left (6\right )\right )}{\left (5+2 x \right ) \ln \left (\ln \left (6\right )\right )+\left (-2 x^{2}-8 x \right ) {\mathrm e}^{4}}}}{2 x^{2} {\mathrm e}^{4}+8 x \,{\mathrm e}^{4}-2 \ln \left (\ln \left (6\right )\right ) x -5 \ln \left (\ln \left (6\right )\right )}\) | \(146\) |
Input:
int((10*ln(ln(6))^2+4*x^2*exp(4)*ln(ln(6)))*exp(2*x*ln(ln(6))/((5+2*x)*ln( ln(6))+(-2*x^2-8*x)*exp(4)))/((4*x^2+20*x+25)*ln(ln(6))^2+(-8*x^3-52*x^2-8 0*x)*exp(4)*ln(ln(6))+(4*x^4+32*x^3+64*x^2)*exp(4)^2),x,method=_RETURNVERB OSE)
Output:
exp(-2*ln(ln(6))*x/(2*x^2*exp(4)+8*x*exp(4)-2*ln(ln(6))*x-5*ln(ln(6))))
Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {\log ^{\frac {2 x}{e^4 \left (-8 x-2 x^2\right )+(5+2 x) \log (\log (6))}}(6) \left (4 e^4 x^2 \log (\log (6))+10 \log ^2(\log (6))\right )}{e^8 \left (64 x^2+32 x^3+4 x^4\right )+e^4 \left (-80 x-52 x^2-8 x^3\right ) \log (\log (6))+\left (25+20 x+4 x^2\right ) \log ^2(\log (6))} \, dx=\log \left (6\right )^{-\frac {2 \, x}{2 \, {\left (x^{2} + 4 \, x\right )} e^{4} - {\left (2 \, x + 5\right )} \log \left (\log \left (6\right )\right )}} \] Input:
integrate((10*log(log(6))^2+4*x^2*exp(4)*log(log(6)))*exp(2*x*log(log(6))/ ((5+2*x)*log(log(6))+(-2*x^2-8*x)*exp(4)))/((4*x^2+20*x+25)*log(log(6))^2+ (-8*x^3-52*x^2-80*x)*exp(4)*log(log(6))+(4*x^4+32*x^3+64*x^2)*exp(4)^2),x, algorithm="fricas")
Output:
log(6)^(-2*x/(2*(x^2 + 4*x)*e^4 - (2*x + 5)*log(log(6))))
Time = 0.59 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {\log ^{\frac {2 x}{e^4 \left (-8 x-2 x^2\right )+(5+2 x) \log (\log (6))}}(6) \left (4 e^4 x^2 \log (\log (6))+10 \log ^2(\log (6))\right )}{e^8 \left (64 x^2+32 x^3+4 x^4\right )+e^4 \left (-80 x-52 x^2-8 x^3\right ) \log (\log (6))+\left (25+20 x+4 x^2\right ) \log ^2(\log (6))} \, dx=e^{\frac {2 x \log {\left (\log {\left (6 \right )} \right )}}{\left (2 x + 5\right ) \log {\left (\log {\left (6 \right )} \right )} + \left (- 2 x^{2} - 8 x\right ) e^{4}}} \] Input:
integrate((10*ln(ln(6))**2+4*x**2*exp(4)*ln(ln(6)))*exp(2*x*ln(ln(6))/((5+ 2*x)*ln(ln(6))+(-2*x**2-8*x)*exp(4)))/((4*x**2+20*x+25)*ln(ln(6))**2+(-8*x **3-52*x**2-80*x)*exp(4)*ln(ln(6))+(4*x**4+32*x**3+64*x**2)*exp(4)**2),x)
Output:
exp(2*x*log(log(6))/((2*x + 5)*log(log(6)) + (-2*x**2 - 8*x)*exp(4)))
Time = 0.30 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.80 \[ \int \frac {\log ^{\frac {2 x}{e^4 \left (-8 x-2 x^2\right )+(5+2 x) \log (\log (6))}}(6) \left (4 e^4 x^2 \log (\log (6))+10 \log ^2(\log (6))\right )}{e^8 \left (64 x^2+32 x^3+4 x^4\right )+e^4 \left (-80 x-52 x^2-8 x^3\right ) \log (\log (6))+\left (25+20 x+4 x^2\right ) \log ^2(\log (6))} \, dx=\frac {1}{{\left (\log \left (3\right ) + \log \left (2\right )\right )}^{\frac {2 \, x}{2 \, x^{2} e^{4} + 2 \, x {\left (4 \, e^{4} - \log \left (\log \left (3\right ) + \log \left (2\right )\right )\right )} - 5 \, \log \left (\log \left (3\right ) + \log \left (2\right )\right )}}} \] Input:
integrate((10*log(log(6))^2+4*x^2*exp(4)*log(log(6)))*exp(2*x*log(log(6))/ ((5+2*x)*log(log(6))+(-2*x^2-8*x)*exp(4)))/((4*x^2+20*x+25)*log(log(6))^2+ (-8*x^3-52*x^2-80*x)*exp(4)*log(log(6))+(4*x^4+32*x^3+64*x^2)*exp(4)^2),x, algorithm="maxima")
Output:
1/((log(3) + log(2))^(2*x/(2*x^2*e^4 + 2*x*(4*e^4 - log(log(3) + log(2))) - 5*log(log(3) + log(2)))))
\[ \int \frac {\log ^{\frac {2 x}{e^4 \left (-8 x-2 x^2\right )+(5+2 x) \log (\log (6))}}(6) \left (4 e^4 x^2 \log (\log (6))+10 \log ^2(\log (6))\right )}{e^8 \left (64 x^2+32 x^3+4 x^4\right )+e^4 \left (-80 x-52 x^2-8 x^3\right ) \log (\log (6))+\left (25+20 x+4 x^2\right ) \log ^2(\log (6))} \, dx=\int { -\frac {2 \, {\left (2 \, x^{2} e^{4} \log \left (\log \left (6\right )\right ) + 5 \, \log \left (\log \left (6\right )\right )^{2}\right )} \log \left (6\right )^{-\frac {2 \, x}{2 \, {\left (x^{2} + 4 \, x\right )} e^{4} - {\left (2 \, x + 5\right )} \log \left (\log \left (6\right )\right )}}}{4 \, {\left (2 \, x^{3} + 13 \, x^{2} + 20 \, x\right )} e^{4} \log \left (\log \left (6\right )\right ) - {\left (4 \, x^{2} + 20 \, x + 25\right )} \log \left (\log \left (6\right )\right )^{2} - 4 \, {\left (x^{4} + 8 \, x^{3} + 16 \, x^{2}\right )} e^{8}} \,d x } \] Input:
integrate((10*log(log(6))^2+4*x^2*exp(4)*log(log(6)))*exp(2*x*log(log(6))/ ((5+2*x)*log(log(6))+(-2*x^2-8*x)*exp(4)))/((4*x^2+20*x+25)*log(log(6))^2+ (-8*x^3-52*x^2-80*x)*exp(4)*log(log(6))+(4*x^4+32*x^3+64*x^2)*exp(4)^2),x, algorithm="giac")
Output:
integrate(-2*(2*x^2*e^4*log(log(6)) + 5*log(log(6))^2)*log(6)^(-2*x/(2*(x^ 2 + 4*x)*e^4 - (2*x + 5)*log(log(6))))/(4*(2*x^3 + 13*x^2 + 20*x)*e^4*log( log(6)) - (4*x^2 + 20*x + 25)*log(log(6))^2 - 4*(x^4 + 8*x^3 + 16*x^2)*e^8 ), x)
Time = 4.33 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {\log ^{\frac {2 x}{e^4 \left (-8 x-2 x^2\right )+(5+2 x) \log (\log (6))}}(6) \left (4 e^4 x^2 \log (\log (6))+10 \log ^2(\log (6))\right )}{e^8 \left (64 x^2+32 x^3+4 x^4\right )+e^4 \left (-80 x-52 x^2-8 x^3\right ) \log (\log (6))+\left (25+20 x+4 x^2\right ) \log ^2(\log (6))} \, dx={\mathrm {e}}^{\frac {2\,x\,\ln \left (\ln \left (6\right )\right )}{5\,\ln \left (\ln \left (6\right )\right )-8\,x\,{\mathrm {e}}^4-2\,x^2\,{\mathrm {e}}^4+2\,x\,\ln \left (\ln \left (6\right )\right )}} \] Input:
int((exp(-(2*x*log(log(6)))/(exp(4)*(8*x + 2*x^2) - log(log(6))*(2*x + 5)) )*(10*log(log(6))^2 + 4*x^2*exp(4)*log(log(6))))/(log(log(6))^2*(20*x + 4* x^2 + 25) + exp(8)*(64*x^2 + 32*x^3 + 4*x^4) - exp(4)*log(log(6))*(80*x + 52*x^2 + 8*x^3)),x)
Output:
exp((2*x*log(log(6)))/(5*log(log(6)) - 8*x*exp(4) - 2*x^2*exp(4) + 2*x*log (log(6))))
Time = 0.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 3.40 \[ \int \frac {\log ^{\frac {2 x}{e^4 \left (-8 x-2 x^2\right )+(5+2 x) \log (\log (6))}}(6) \left (4 e^4 x^2 \log (\log (6))+10 \log ^2(\log (6))\right )}{e^8 \left (64 x^2+32 x^3+4 x^4\right )+e^4 \left (-80 x-52 x^2-8 x^3\right ) \log (\log (6))+\left (25+20 x+4 x^2\right ) \log ^2(\log (6))} \, dx=\frac {e^{\frac {2 e^{4} x^{2}+8 e^{4} x}{2 \,\mathrm {log}\left (\mathrm {log}\left (6\right )\right ) x +5 \,\mathrm {log}\left (\mathrm {log}\left (6\right )\right )-2 e^{4} x^{2}-8 e^{4} x}} e}{e^{\frac {5 \,\mathrm {log}\left (\mathrm {log}\left (6\right )\right )}{2 \,\mathrm {log}\left (\mathrm {log}\left (6\right )\right ) x +5 \,\mathrm {log}\left (\mathrm {log}\left (6\right )\right )-2 e^{4} x^{2}-8 e^{4} x}}} \] Input:
int((10*log(log(6))^2+4*x^2*exp(4)*log(log(6)))*exp(2*x*log(log(6))/((5+2* x)*log(log(6))+(-2*x^2-8*x)*exp(4)))/((4*x^2+20*x+25)*log(log(6))^2+(-8*x^ 3-52*x^2-80*x)*exp(4)*log(log(6))+(4*x^4+32*x^3+64*x^2)*exp(4)^2),x)
Output:
(e**((2*e**4*x**2 + 8*e**4*x)/(2*log(log(6))*x + 5*log(log(6)) - 2*e**4*x* *2 - 8*e**4*x))*e)/e**((5*log(log(6)))/(2*log(log(6))*x + 5*log(log(6)) - 2*e**4*x**2 - 8*e**4*x))