Integrand size = 158, antiderivative size = 30 \[ \int \frac {-1+2 x+e^{-2+e^4} (-20+40 x)+e^{-4+2 e^4} \left (-158+313 x+4 x^2\right )+e^{-6+3 e^4} \left (-580+1130 x+40 x^2\right )+e^{-8+4 e^4} \left (-841+1595 x+115 x^2+2 x^3\right )}{x+20 e^{-2+e^4} x+e^{-4+2 e^4} \left (158 x+2 x^2\right )+e^{-6+3 e^4} \left (580 x+20 x^2\right )+e^{-8+4 e^4} \left (841 x+58 x^2+x^3\right )} \, dx=7+2 x-\frac {x}{4+\left (5+e^{2-e^4}\right )^2+x}-\log (x) \]
Leaf count is larger than twice the leaf count of optimal. \(97\) vs. \(2(30)=60\).
Time = 0.14 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.23 \[ \int \frac {-1+2 x+e^{-2+e^4} (-20+40 x)+e^{-4+2 e^4} \left (-158+313 x+4 x^2\right )+e^{-6+3 e^4} \left (-580+1130 x+40 x^2\right )+e^{-8+4 e^4} \left (-841+1595 x+115 x^2+2 x^3\right )}{x+20 e^{-2+e^4} x+e^{-4+2 e^4} \left (158 x+2 x^2\right )+e^{-6+3 e^4} \left (580 x+20 x^2\right )+e^{-8+4 e^4} \left (841 x+58 x^2+x^3\right )} \, dx=\frac {e^4 (1+2 x)+10 e^{2+e^4} (1+2 x)+e^{2 e^4} \left (29+58 x+2 x^2\right )-\left (e^4+10 e^{2+e^4}+e^{2 e^4} (29+x)\right ) \log (x)}{e^4+10 e^{2+e^4}+e^{2 e^4} (29+x)} \]
Integrate[(-1 + 2*x + E^(-2 + E^4)*(-20 + 40*x) + E^(-4 + 2*E^4)*(-158 + 3 13*x + 4*x^2) + E^(-6 + 3*E^4)*(-580 + 1130*x + 40*x^2) + E^(-8 + 4*E^4)*( -841 + 1595*x + 115*x^2 + 2*x^3))/(x + 20*E^(-2 + E^4)*x + E^(-4 + 2*E^4)* (158*x + 2*x^2) + E^(-6 + 3*E^4)*(580*x + 20*x^2) + E^(-8 + 4*E^4)*(841*x + 58*x^2 + x^3)),x]
(E^4*(1 + 2*x) + 10*E^(2 + E^4)*(1 + 2*x) + E^(2*E^4)*(29 + 58*x + 2*x^2) - (E^4 + 10*E^(2 + E^4) + E^(2*E^4)*(29 + x))*Log[x])/(E^4 + 10*E^(2 + E^4 ) + E^(2*E^4)*(29 + x))
Leaf count is larger than twice the leaf count of optimal. \(64\) vs. \(2(30)=60\).
Time = 0.73 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6, 2026, 2007, 2123, 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.
\(\displaystyle \int \frac {e^{2 e^4-4} \left (4 x^2+313 x-158\right )+e^{3 e^4-6} \left (40 x^2+1130 x-580\right )+e^{4 e^4-8} \left (2 x^3+115 x^2+1595 x-841\right )+2 x+e^{e^4-2} (40 x-20)-1}{e^{2 e^4-4} \left (2 x^2+158 x\right )+e^{3 e^4-6} \left (20 x^2+580 x\right )+e^{4 e^4-8} \left (x^3+58 x^2+841 x\right )+20 e^{e^4-2} x+x} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{2 e^4-4} \left (4 x^2+313 x-158\right )+e^{3 e^4-6} \left (40 x^2+1130 x-580\right )+e^{4 e^4-8} \left (2 x^3+115 x^2+1595 x-841\right )+2 x+e^{e^4-2} (40 x-20)-1}{e^{2 e^4-4} \left (2 x^2+158 x\right )+e^{3 e^4-6} \left (20 x^2+580 x\right )+e^{4 e^4-8} \left (x^3+58 x^2+841 x\right )+\left (1+20 e^{e^4-2}\right ) x}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{2 e^4-4} \left (4 x^2+313 x-158\right )+e^{3 e^4-6} \left (40 x^2+1130 x-580\right )+e^{4 e^4-8} \left (2 x^3+115 x^2+1595 x-841\right )+2 x+e^{e^4-2} (40 x-20)-1}{x \left (e^{4 e^4-8} x^2+2 e^{2 e^4-8} \left (e^4+29 e^{2 e^4}+10 e^{2+e^4}\right ) x+\frac {\left (e^4+29 e^{2 e^4}+10 e^{2+e^4}\right )^2}{e^8}\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{2 e^4-4} \left (4 x^2+313 x-158\right )+e^{3 e^4-6} \left (40 x^2+1130 x-580\right )+e^{4 e^4-8} \left (2 x^3+115 x^2+1595 x-841\right )+2 x+e^{e^4-2} (40 x-20)-1}{x \left (e^{\frac {1}{2} \left (4 e^4-8\right )} x+\frac {e^4+29 e^{2 e^4}+10 e^{2+e^4}}{e^4}\right )^2}dx\) |
\(\Big \downarrow \) 2123 |
\(\displaystyle \int \left (-\frac {e^{2 e^4} \left (e^4+29 e^{2 e^4}+10 e^{2+e^4}\right )}{\left (e^{2 e^4} x+10 e^{2+e^4}+29 e^{2 e^4}+e^4\right )^2}-\frac {1}{x}+2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 x+\frac {e^4+29 e^{2 e^4}+10 e^{2+e^4}}{e^{2 e^4} x+10 e^{2+e^4}+29 e^{2 e^4}+e^4}-\log (x)\) |
Int[(-1 + 2*x + E^(-2 + E^4)*(-20 + 40*x) + E^(-4 + 2*E^4)*(-158 + 313*x + 4*x^2) + E^(-6 + 3*E^4)*(-580 + 1130*x + 40*x^2) + E^(-8 + 4*E^4)*(-841 + 1595*x + 115*x^2 + 2*x^3))/(x + 20*E^(-2 + E^4)*x + E^(-4 + 2*E^4)*(158*x + 2*x^2) + E^(-6 + 3*E^4)*(580*x + 20*x^2) + E^(-8 + 4*E^4)*(841*x + 58*x ^2 + x^3)),x]
2*x + (E^4 + 29*E^(2*E^4) + 10*E^(2 + E^4))/(E^4 + 29*E^(2*E^4) + 10*E^(2 + E^4) + E^(2*E^4)*x) - Log[x]
3.27.33.3.1 Defintions of rubi rules used
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[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
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[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Leaf count of result is larger than twice the leaf count of optimal. \(103\) vs. \(2(28)=56\).
Time = 0.29 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.47
method | result | size |
norman | \(\frac {2 \,{\mathrm e}^{2 \,{\mathrm e}^{4}} {\mathrm e}^{-4} x^{2}-{\mathrm e}^{-2 \,{\mathrm e}^{4}} {\mathrm e}^{4} \left (1653 \,{\mathrm e}^{4 \,{\mathrm e}^{4}} {\mathrm e}^{-8}+1150 \,{\mathrm e}^{3 \,{\mathrm e}^{4}} {\mathrm e}^{-6}+315 \,{\mathrm e}^{2 \,{\mathrm e}^{4}} {\mathrm e}^{-4}+40 \,{\mathrm e}^{{\mathrm e}^{4}} {\mathrm e}^{-2}+2\right )}{{\mathrm e}^{2 \,{\mathrm e}^{4}-4} x +29 \,{\mathrm e}^{2 \,{\mathrm e}^{4}-4}+10 \,{\mathrm e}^{{\mathrm e}^{4}-2}+1}-\ln \left (x \right )\) | \(104\) |
risch | \(2 x +\frac {29 \,{\mathrm e}^{2 \,{\mathrm e}^{4}-4}}{{\mathrm e}^{2 \,{\mathrm e}^{4}-4} x +29 \,{\mathrm e}^{2 \,{\mathrm e}^{4}-4}+10 \,{\mathrm e}^{{\mathrm e}^{4}-2}+1}+\frac {10 \,{\mathrm e}^{{\mathrm e}^{4}-2}}{{\mathrm e}^{2 \,{\mathrm e}^{4}-4} x +29 \,{\mathrm e}^{2 \,{\mathrm e}^{4}-4}+10 \,{\mathrm e}^{{\mathrm e}^{4}-2}+1}+\frac {1}{{\mathrm e}^{2 \,{\mathrm e}^{4}-4} x +29 \,{\mathrm e}^{2 \,{\mathrm e}^{4}-4}+10 \,{\mathrm e}^{{\mathrm e}^{4}-2}+1}-\ln \left (x \right )\) | \(112\) |
parallelrisch | \(-\frac {\left ({\mathrm e}^{4 \,{\mathrm e}^{4}-8} \ln \left (x \right ) x -2 \,{\mathrm e}^{4 \,{\mathrm e}^{4}-8} x^{2}+29 \,{\mathrm e}^{4 \,{\mathrm e}^{4}-8} \ln \left (x \right )+1653 \,{\mathrm e}^{4 \,{\mathrm e}^{4}-8}+10 \,{\mathrm e}^{3 \,{\mathrm e}^{4}-6} \ln \left (x \right )+2+1150 \,{\mathrm e}^{3 \,{\mathrm e}^{4}-6}+\ln \left (x \right ) {\mathrm e}^{2 \,{\mathrm e}^{4}-4}+315 \,{\mathrm e}^{2 \,{\mathrm e}^{4}-4}+40 \,{\mathrm e}^{{\mathrm e}^{4}-2}\right ) {\mathrm e}^{-2 \,{\mathrm e}^{4}+4}}{{\mathrm e}^{2 \,{\mathrm e}^{4}-4} x +29 \,{\mathrm e}^{2 \,{\mathrm e}^{4}-4}+10 \,{\mathrm e}^{{\mathrm e}^{4}-2}+1}\) | \(130\) |
int(((2*x^3+115*x^2+1595*x-841)*exp(exp(4)-2)^4+(40*x^2+1130*x-580)*exp(ex p(4)-2)^3+(4*x^2+313*x-158)*exp(exp(4)-2)^2+(40*x-20)*exp(exp(4)-2)+2*x-1) /((x^3+58*x^2+841*x)*exp(exp(4)-2)^4+(20*x^2+580*x)*exp(exp(4)-2)^3+(2*x^2 +158*x)*exp(exp(4)-2)^2+20*x*exp(exp(4)-2)+x),x,method=_RETURNVERBOSE)
(2*exp(exp(4))^2*exp(-2)^2*x^2-1/exp(exp(4))^2/exp(-2)^2*(1653*exp(exp(4)) ^4*exp(-2)^4+1150*exp(exp(4))^3*exp(-2)^3+315*exp(exp(4))^2*exp(-2)^2+40*e xp(exp(4))*exp(-2)+2))/(exp(exp(4)-2)^2*x+29*exp(exp(4)-2)^2+10*exp(exp(4) -2)+1)-ln(x)
Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (28) = 56\).
Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.73 \[ \int \frac {-1+2 x+e^{-2+e^4} (-20+40 x)+e^{-4+2 e^4} \left (-158+313 x+4 x^2\right )+e^{-6+3 e^4} \left (-580+1130 x+40 x^2\right )+e^{-8+4 e^4} \left (-841+1595 x+115 x^2+2 x^3\right )}{x+20 e^{-2+e^4} x+e^{-4+2 e^4} \left (158 x+2 x^2\right )+e^{-6+3 e^4} \left (580 x+20 x^2\right )+e^{-8+4 e^4} \left (841 x+58 x^2+x^3\right )} \, dx=\frac {{\left (2 \, x^{2} + 58 \, x + 29\right )} e^{\left (2 \, e^{4} - 4\right )} + 10 \, {\left (2 \, x + 1\right )} e^{\left (e^{4} - 2\right )} - {\left ({\left (x + 29\right )} e^{\left (2 \, e^{4} - 4\right )} + 10 \, e^{\left (e^{4} - 2\right )} + 1\right )} \log \left (x\right ) + 2 \, x + 1}{{\left (x + 29\right )} e^{\left (2 \, e^{4} - 4\right )} + 10 \, e^{\left (e^{4} - 2\right )} + 1} \]
integrate(((2*x^3+115*x^2+1595*x-841)*exp(exp(4)-2)^4+(40*x^2+1130*x-580)* exp(exp(4)-2)^3+(4*x^2+313*x-158)*exp(exp(4)-2)^2+(40*x-20)*exp(exp(4)-2)+ 2*x-1)/((x^3+58*x^2+841*x)*exp(exp(4)-2)^4+(20*x^2+580*x)*exp(exp(4)-2)^3+ (2*x^2+158*x)*exp(exp(4)-2)^2+20*x*exp(exp(4)-2)+x),x, algorithm=\
((2*x^2 + 58*x + 29)*e^(2*e^4 - 4) + 10*(2*x + 1)*e^(e^4 - 2) - ((x + 29)* e^(2*e^4 - 4) + 10*e^(e^4 - 2) + 1)*log(x) + 2*x + 1)/((x + 29)*e^(2*e^4 - 4) + 10*e^(e^4 - 2) + 1)
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (22) = 44\).
Time = 0.85 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.93 \[ \int \frac {-1+2 x+e^{-2+e^4} (-20+40 x)+e^{-4+2 e^4} \left (-158+313 x+4 x^2\right )+e^{-6+3 e^4} \left (-580+1130 x+40 x^2\right )+e^{-8+4 e^4} \left (-841+1595 x+115 x^2+2 x^3\right )}{x+20 e^{-2+e^4} x+e^{-4+2 e^4} \left (158 x+2 x^2\right )+e^{-6+3 e^4} \left (580 x+20 x^2\right )+e^{-8+4 e^4} \left (841 x+58 x^2+x^3\right )} \, dx=2 x - \log {\left (x \right )} + \frac {e^{4} + 10 e^{2} e^{e^{4}} + 29 e^{2 e^{4}}}{x e^{2 e^{4}} + e^{4} + 10 e^{2} e^{e^{4}} + 29 e^{2 e^{4}}} \]
integrate(((2*x**3+115*x**2+1595*x-841)*exp(exp(4)-2)**4+(40*x**2+1130*x-5 80)*exp(exp(4)-2)**3+(4*x**2+313*x-158)*exp(exp(4)-2)**2+(40*x-20)*exp(exp (4)-2)+2*x-1)/((x**3+58*x**2+841*x)*exp(exp(4)-2)**4+(20*x**2+580*x)*exp(e xp(4)-2)**3+(2*x**2+158*x)*exp(exp(4)-2)**2+20*x*exp(exp(4)-2)+x),x)
2*x - log(x) + (exp(4) + 10*exp(2)*exp(exp(4)) + 29*exp(2*exp(4)))/(x*exp( 2*exp(4)) + exp(4) + 10*exp(2)*exp(exp(4)) + 29*exp(2*exp(4)))
Time = 0.20 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.73 \[ \int \frac {-1+2 x+e^{-2+e^4} (-20+40 x)+e^{-4+2 e^4} \left (-158+313 x+4 x^2\right )+e^{-6+3 e^4} \left (-580+1130 x+40 x^2\right )+e^{-8+4 e^4} \left (-841+1595 x+115 x^2+2 x^3\right )}{x+20 e^{-2+e^4} x+e^{-4+2 e^4} \left (158 x+2 x^2\right )+e^{-6+3 e^4} \left (580 x+20 x^2\right )+e^{-8+4 e^4} \left (841 x+58 x^2+x^3\right )} \, dx=2 \, x + \frac {e^{4} + 29 \, e^{\left (2 \, e^{4}\right )} + 10 \, e^{\left (e^{4} + 2\right )}}{x e^{\left (2 \, e^{4}\right )} + e^{4} + 29 \, e^{\left (2 \, e^{4}\right )} + 10 \, e^{\left (e^{4} + 2\right )}} - \log \left (x\right ) \]
integrate(((2*x^3+115*x^2+1595*x-841)*exp(exp(4)-2)^4+(40*x^2+1130*x-580)* exp(exp(4)-2)^3+(4*x^2+313*x-158)*exp(exp(4)-2)^2+(40*x-20)*exp(exp(4)-2)+ 2*x-1)/((x^3+58*x^2+841*x)*exp(exp(4)-2)^4+(20*x^2+580*x)*exp(exp(4)-2)^3+ (2*x^2+158*x)*exp(exp(4)-2)^2+20*x*exp(exp(4)-2)+x),x, algorithm=\
2*x + (e^4 + 29*e^(2*e^4) + 10*e^(e^4 + 2))/(x*e^(2*e^4) + e^4 + 29*e^(2*e ^4) + 10*e^(e^4 + 2)) - log(x)
Exception generated. \[ \int \frac {-1+2 x+e^{-2+e^4} (-20+40 x)+e^{-4+2 e^4} \left (-158+313 x+4 x^2\right )+e^{-6+3 e^4} \left (-580+1130 x+40 x^2\right )+e^{-8+4 e^4} \left (-841+1595 x+115 x^2+2 x^3\right )}{x+20 e^{-2+e^4} x+e^{-4+2 e^4} \left (158 x+2 x^2\right )+e^{-6+3 e^4} \left (580 x+20 x^2\right )+e^{-8+4 e^4} \left (841 x+58 x^2+x^3\right )} \, dx=\text {Exception raised: NotImplementedError} \]
integrate(((2*x^3+115*x^2+1595*x-841)*exp(exp(4)-2)^4+(40*x^2+1130*x-580)* exp(exp(4)-2)^3+(4*x^2+313*x-158)*exp(exp(4)-2)^2+(40*x-20)*exp(exp(4)-2)+ 2*x-1)/((x^3+58*x^2+841*x)*exp(exp(4)-2)^4+(20*x^2+580*x)*exp(exp(4)-2)^3+ (2*x^2+158*x)*exp(exp(4)-2)^2+20*x*exp(exp(4)-2)+x),x, algorithm=\
Exception raised: NotImplementedError >> unable to parse Giac output: -ln( abs(sageVARx))+(-58*exp(4*exp(4)-8)-20*exp(3*exp(4)-6)-2*exp(2*exp(4)-4))* 1/2/sqrt(-exp(2*exp(4)-4)^2-20*exp(2*exp(4)-4)*exp(3*exp(4)-6)+100*exp(2*e xp(4)-4)*exp(4*exp(
Time = 0.44 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.60 \[ \int \frac {-1+2 x+e^{-2+e^4} (-20+40 x)+e^{-4+2 e^4} \left (-158+313 x+4 x^2\right )+e^{-6+3 e^4} \left (-580+1130 x+40 x^2\right )+e^{-8+4 e^4} \left (-841+1595 x+115 x^2+2 x^3\right )}{x+20 e^{-2+e^4} x+e^{-4+2 e^4} \left (158 x+2 x^2\right )+e^{-6+3 e^4} \left (580 x+20 x^2\right )+e^{-8+4 e^4} \left (841 x+58 x^2+x^3\right )} \, dx=2\,x-\ln \left (x\right )+\frac {10\,{\mathrm {e}}^{2-{\mathrm {e}}^4}+{\mathrm {e}}^{4-2\,{\mathrm {e}}^4}+29}{x+10\,{\mathrm {e}}^{2-{\mathrm {e}}^4}+{\mathrm {e}}^{4-2\,{\mathrm {e}}^4}+29} \]
int((2*x + exp(2*exp(4) - 4)*(313*x + 4*x^2 - 158) + exp(3*exp(4) - 6)*(11 30*x + 40*x^2 - 580) + exp(4*exp(4) - 8)*(1595*x + 115*x^2 + 2*x^3 - 841) + exp(exp(4) - 2)*(40*x - 20) - 1)/(x + 20*x*exp(exp(4) - 2) + exp(4*exp(4 ) - 8)*(841*x + 58*x^2 + x^3) + exp(2*exp(4) - 4)*(158*x + 2*x^2) + exp(3* exp(4) - 6)*(580*x + 20*x^2)),x)