Integrand size = 130, antiderivative size = 26 \[ \int \frac {e^{\frac {\log ^2(2 x)}{25 x^4-10 x^6+x^8}} \left (-375 x^4-250 x^5+225 x^6+150 x^7-45 x^8-30 x^9+3 x^{10}+2 x^{11}+\left (-30-10 x+6 x^2+2 x^3\right ) \log (2 x)+\left (60+20 x-24 x^2-8 x^3\right ) \log ^2(2 x)\right )}{-125 x^4+75 x^6-15 x^8+x^{10}} \, dx=e^{\frac {\log ^2(2 x)}{x^4 \left (5-x^2\right )^2}} x (3+x) \]
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {\log ^2(2 x)}{25 x^4-10 x^6+x^8}} \left (-375 x^4-250 x^5+225 x^6+150 x^7-45 x^8-30 x^9+3 x^{10}+2 x^{11}+\left (-30-10 x+6 x^2+2 x^3\right ) \log (2 x)+\left (60+20 x-24 x^2-8 x^3\right ) \log ^2(2 x)\right )}{-125 x^4+75 x^6-15 x^8+x^{10}} \, dx=e^{\frac {\log ^2(2 x)}{x^4 \left (-5+x^2\right )^2}} x (3+x) \]
Integrate[(E^(Log[2*x]^2/(25*x^4 - 10*x^6 + x^8))*(-375*x^4 - 250*x^5 + 22 5*x^6 + 150*x^7 - 45*x^8 - 30*x^9 + 3*x^10 + 2*x^11 + (-30 - 10*x + 6*x^2 + 2*x^3)*Log[2*x] + (60 + 20*x - 24*x^2 - 8*x^3)*Log[2*x]^2))/(-125*x^4 + 75*x^6 - 15*x^8 + x^10),x]
Leaf count is larger than twice the leaf count of optimal. \(149\) vs. \(2(26)=52\).
Time = 1.25 (sec) , antiderivative size = 149, normalized size of antiderivative = 5.73, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {2026, 2070, 2726}
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.
\(\displaystyle \int \frac {e^{\frac {\log ^2(2 x)}{x^8-10 x^6+25 x^4}} \left (2 x^{11}+3 x^{10}-30 x^9-45 x^8+150 x^7+225 x^6-250 x^5-375 x^4+\left (-8 x^3-24 x^2+20 x+60\right ) \log ^2(2 x)+\left (2 x^3+6 x^2-10 x-30\right ) \log (2 x)\right )}{x^{10}-15 x^8+75 x^6-125 x^4} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {\log ^2(2 x)}{x^8-10 x^6+25 x^4}} \left (2 x^{11}+3 x^{10}-30 x^9-45 x^8+150 x^7+225 x^6-250 x^5-375 x^4+\left (-8 x^3-24 x^2+20 x+60\right ) \log ^2(2 x)+\left (2 x^3+6 x^2-10 x-30\right ) \log (2 x)\right )}{x^4 \left (x^6-15 x^4+75 x^2-125\right )}dx\) |
\(\Big \downarrow \) 2070 |
\(\displaystyle \int \frac {e^{\frac {\log ^2(2 x)}{x^8-10 x^6+25 x^4}} \left (2 x^{11}+3 x^{10}-30 x^9-45 x^8+150 x^7+225 x^6-250 x^5-375 x^4+\left (-8 x^3-24 x^2+20 x+60\right ) \log ^2(2 x)+\left (2 x^3+6 x^2-10 x-30\right ) \log (2 x)\right )}{x^4 \left (x^2-5\right )^3}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {e^{\frac {\log ^2(2 x)}{x^8-10 x^6+25 x^4}} \left (\left (-x^3-3 x^2+5 x+15\right ) \log (2 x)-2 \left (-2 x^3-6 x^2+5 x+15\right ) \log ^2(2 x)\right )}{x^4 \left (5-x^2\right )^3 \left (\frac {\log (2 x)}{x \left (x^8-10 x^6+25 x^4\right )}-\frac {2 \left (2 x^7-15 x^5+25 x^3\right ) \log ^2(2 x)}{\left (x^8-10 x^6+25 x^4\right )^2}\right )}\) |
Int[(E^(Log[2*x]^2/(25*x^4 - 10*x^6 + x^8))*(-375*x^4 - 250*x^5 + 225*x^6 + 150*x^7 - 45*x^8 - 30*x^9 + 3*x^10 + 2*x^11 + (-30 - 10*x + 6*x^2 + 2*x^ 3)*Log[2*x] + (60 + 20*x - 24*x^2 - 8*x^3)*Log[2*x]^2))/(-125*x^4 + 75*x^6 - 15*x^8 + x^10),x]
(E^(Log[2*x]^2/(25*x^4 - 10*x^6 + x^8))*((15 + 5*x - 3*x^2 - x^3)*Log[2*x] - 2*(15 + 5*x - 6*x^2 - 2*x^3)*Log[2*x]^2))/(x^4*(5 - x^2)^3*(Log[2*x]/(x *(25*x^4 - 10*x^6 + x^8)) - (2*(25*x^3 - 15*x^5 + 2*x^7)*Log[2*x]^2)/(25*x ^4 - 10*x^6 + x^8)^2))
3.5.3.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x^2, 0], Expon[Px , x^2]], b = Rt[Coeff[Px, x^2, Expon[Px, x^2]], Expon[Px, x^2]]}, Int[u*(a + b*x^2)^(Expon[Px, x^2]*p), x] /; EqQ[Px, (a + b*x^2)^Expon[Px, x^2]]] /; IntegerQ[p] && PolyQ[Px, x^2] && GtQ[Expon[Px, x^2], 1] && NeQ[Coeff[Px, x^ 2, 0], 0]
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 322.32 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92
method | result | size |
risch | \(\left (3+x \right ) x \,{\mathrm e}^{\frac {\ln \left (2 x \right )^{2}}{x^{4} \left (x^{2}-5\right )^{2}}}\) | \(24\) |
parallelrisch | \(x^{2} {\mathrm e}^{\frac {\ln \left (2 x \right )^{2}}{x^{4} \left (x^{4}-10 x^{2}+25\right )}}+3 x \,{\mathrm e}^{\frac {\ln \left (2 x \right )^{2}}{x^{4} \left (x^{4}-10 x^{2}+25\right )}}\) | \(55\) |
int(((-8*x^3-24*x^2+20*x+60)*ln(2*x)^2+(2*x^3+6*x^2-10*x-30)*ln(2*x)+2*x^1 1+3*x^10-30*x^9-45*x^8+150*x^7+225*x^6-250*x^5-375*x^4)*exp(ln(2*x)^2/(x^8 -10*x^6+25*x^4))/(x^10-15*x^8+75*x^6-125*x^4),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{\frac {\log ^2(2 x)}{25 x^4-10 x^6+x^8}} \left (-375 x^4-250 x^5+225 x^6+150 x^7-45 x^8-30 x^9+3 x^{10}+2 x^{11}+\left (-30-10 x+6 x^2+2 x^3\right ) \log (2 x)+\left (60+20 x-24 x^2-8 x^3\right ) \log ^2(2 x)\right )}{-125 x^4+75 x^6-15 x^8+x^{10}} \, dx={\left (x^{2} + 3 \, x\right )} e^{\left (\frac {\log \left (2 \, x\right )^{2}}{x^{8} - 10 \, x^{6} + 25 \, x^{4}}\right )} \]
integrate(((-8*x^3-24*x^2+20*x+60)*log(2*x)^2+(2*x^3+6*x^2-10*x-30)*log(2* x)+2*x^11+3*x^10-30*x^9-45*x^8+150*x^7+225*x^6-250*x^5-375*x^4)*exp(log(2* x)^2/(x^8-10*x^6+25*x^4))/(x^10-15*x^8+75*x^6-125*x^4),x, algorithm=\
Timed out. \[ \int \frac {e^{\frac {\log ^2(2 x)}{25 x^4-10 x^6+x^8}} \left (-375 x^4-250 x^5+225 x^6+150 x^7-45 x^8-30 x^9+3 x^{10}+2 x^{11}+\left (-30-10 x+6 x^2+2 x^3\right ) \log (2 x)+\left (60+20 x-24 x^2-8 x^3\right ) \log ^2(2 x)\right )}{-125 x^4+75 x^6-15 x^8+x^{10}} \, dx=\text {Timed out} \]
integrate(((-8*x**3-24*x**2+20*x+60)*ln(2*x)**2+(2*x**3+6*x**2-10*x-30)*ln (2*x)+2*x**11+3*x**10-30*x**9-45*x**8+150*x**7+225*x**6-250*x**5-375*x**4) *exp(ln(2*x)**2/(x**8-10*x**6+25*x**4))/(x**10-15*x**8+75*x**6-125*x**4),x )
Leaf count of result is larger than twice the leaf count of optimal. 157 vs. \(2 (23) = 46\).
Time = 0.49 (sec) , antiderivative size = 157, normalized size of antiderivative = 6.04 \[ \int \frac {e^{\frac {\log ^2(2 x)}{25 x^4-10 x^6+x^8}} \left (-375 x^4-250 x^5+225 x^6+150 x^7-45 x^8-30 x^9+3 x^{10}+2 x^{11}+\left (-30-10 x+6 x^2+2 x^3\right ) \log (2 x)+\left (60+20 x-24 x^2-8 x^3\right ) \log ^2(2 x)\right )}{-125 x^4+75 x^6-15 x^8+x^{10}} \, dx={\left (x^{2} + 3 \, x\right )} e^{\left (\frac {\log \left (2\right )^{2}}{25 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} - \frac {2 \, \log \left (2\right )^{2}}{125 \, {\left (x^{2} - 5\right )}} + \frac {2 \, \log \left (2\right ) \log \left (x\right )}{25 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} - \frac {4 \, \log \left (2\right ) \log \left (x\right )}{125 \, {\left (x^{2} - 5\right )}} + \frac {\log \left (x\right )^{2}}{25 \, {\left (x^{4} - 10 \, x^{2} + 25\right )}} - \frac {2 \, \log \left (x\right )^{2}}{125 \, {\left (x^{2} - 5\right )}} + \frac {2 \, \log \left (2\right )^{2}}{125 \, x^{2}} + \frac {4 \, \log \left (2\right ) \log \left (x\right )}{125 \, x^{2}} + \frac {2 \, \log \left (x\right )^{2}}{125 \, x^{2}} + \frac {\log \left (2\right )^{2}}{25 \, x^{4}} + \frac {2 \, \log \left (2\right ) \log \left (x\right )}{25 \, x^{4}} + \frac {\log \left (x\right )^{2}}{25 \, x^{4}}\right )} \]
integrate(((-8*x^3-24*x^2+20*x+60)*log(2*x)^2+(2*x^3+6*x^2-10*x-30)*log(2* x)+2*x^11+3*x^10-30*x^9-45*x^8+150*x^7+225*x^6-250*x^5-375*x^4)*exp(log(2* x)^2/(x^8-10*x^6+25*x^4))/(x^10-15*x^8+75*x^6-125*x^4),x, algorithm=\
(x^2 + 3*x)*e^(1/25*log(2)^2/(x^4 - 10*x^2 + 25) - 2/125*log(2)^2/(x^2 - 5 ) + 2/25*log(2)*log(x)/(x^4 - 10*x^2 + 25) - 4/125*log(2)*log(x)/(x^2 - 5) + 1/25*log(x)^2/(x^4 - 10*x^2 + 25) - 2/125*log(x)^2/(x^2 - 5) + 2/125*lo g(2)^2/x^2 + 4/125*log(2)*log(x)/x^2 + 2/125*log(x)^2/x^2 + 1/25*log(2)^2/ x^4 + 2/25*log(2)*log(x)/x^4 + 1/25*log(x)^2/x^4)
\[ \int \frac {e^{\frac {\log ^2(2 x)}{25 x^4-10 x^6+x^8}} \left (-375 x^4-250 x^5+225 x^6+150 x^7-45 x^8-30 x^9+3 x^{10}+2 x^{11}+\left (-30-10 x+6 x^2+2 x^3\right ) \log (2 x)+\left (60+20 x-24 x^2-8 x^3\right ) \log ^2(2 x)\right )}{-125 x^4+75 x^6-15 x^8+x^{10}} \, dx=\int { \frac {{\left (2 \, x^{11} + 3 \, x^{10} - 30 \, x^{9} - 45 \, x^{8} + 150 \, x^{7} + 225 \, x^{6} - 250 \, x^{5} - 375 \, x^{4} - 4 \, {\left (2 \, x^{3} + 6 \, x^{2} - 5 \, x - 15\right )} \log \left (2 \, x\right )^{2} + 2 \, {\left (x^{3} + 3 \, x^{2} - 5 \, x - 15\right )} \log \left (2 \, x\right )\right )} e^{\left (\frac {\log \left (2 \, x\right )^{2}}{x^{8} - 10 \, x^{6} + 25 \, x^{4}}\right )}}{x^{10} - 15 \, x^{8} + 75 \, x^{6} - 125 \, x^{4}} \,d x } \]
integrate(((-8*x^3-24*x^2+20*x+60)*log(2*x)^2+(2*x^3+6*x^2-10*x-30)*log(2* x)+2*x^11+3*x^10-30*x^9-45*x^8+150*x^7+225*x^6-250*x^5-375*x^4)*exp(log(2* x)^2/(x^8-10*x^6+25*x^4))/(x^10-15*x^8+75*x^6-125*x^4),x, algorithm=\
integrate((2*x^11 + 3*x^10 - 30*x^9 - 45*x^8 + 150*x^7 + 225*x^6 - 250*x^5 - 375*x^4 - 4*(2*x^3 + 6*x^2 - 5*x - 15)*log(2*x)^2 + 2*(x^3 + 3*x^2 - 5* x - 15)*log(2*x))*e^(log(2*x)^2/(x^8 - 10*x^6 + 25*x^4))/(x^10 - 15*x^8 + 75*x^6 - 125*x^4), x)
Time = 8.76 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.73 \[ \int \frac {e^{\frac {\log ^2(2 x)}{25 x^4-10 x^6+x^8}} \left (-375 x^4-250 x^5+225 x^6+150 x^7-45 x^8-30 x^9+3 x^{10}+2 x^{11}+\left (-30-10 x+6 x^2+2 x^3\right ) \log (2 x)+\left (60+20 x-24 x^2-8 x^3\right ) \log ^2(2 x)\right )}{-125 x^4+75 x^6-15 x^8+x^{10}} \, dx=x\,x^{\frac {2\,\ln \left (2\right )}{x^8-10\,x^6+25\,x^4}}\,{\mathrm {e}}^{\frac {{\ln \left (x\right )}^2}{x^8-10\,x^6+25\,x^4}}\,{\mathrm {e}}^{\frac {{\ln \left (2\right )}^2}{x^8-10\,x^6+25\,x^4}}\,\left (x+3\right ) \]
int((exp(log(2*x)^2/(25*x^4 - 10*x^6 + x^8))*(log(2*x)*(10*x - 6*x^2 - 2*x ^3 + 30) - log(2*x)^2*(20*x - 24*x^2 - 8*x^3 + 60) + 375*x^4 + 250*x^5 - 2 25*x^6 - 150*x^7 + 45*x^8 + 30*x^9 - 3*x^10 - 2*x^11))/(125*x^4 - 75*x^6 + 15*x^8 - x^10),x)