Integrand size = 160, antiderivative size = 34 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=-x+\frac {\frac {e^4-e^8-x}{x}+x}{\left (2+e^5\right )^2+x} \]
Time = 0.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.59 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=-\frac {-e^4+e^8+4 e^5 x (1+x)+e^{10} x (1+x)+x \left (5+4 x+x^2\right )}{x \left (4+4 e^5+e^{10}+x\right )} \]
Integrate[(E^4*(-4 - 2*x) - 11*x^2 - 8*E^15*x^2 - E^20*x^2 - 8*x^3 - x^4 + E^8*(4 + 2*x) + E^5*(-4*E^4 + 4*E^8 - 28*x^2 - 8*x^3) + E^10*(-E^4 + E^8 - 23*x^2 - 2*x^3))/(16*x^2 + 8*E^15*x^2 + E^20*x^2 + 8*x^3 + x^4 + E^10*(2 4*x^2 + 2*x^3) + E^5*(32*x^2 + 8*x^3)),x]
-((-E^4 + E^8 + 4*E^5*x*(1 + x) + E^10*x*(1 + x) + x*(5 + 4*x + x^2))/(x*( 4 + 4*E^5 + E^10 + x)))
Leaf count is larger than twice the leaf count of optimal. \(73\) vs. \(2(34)=68\).
Time = 0.47 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.15, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {6, 6, 6, 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.
\(\displaystyle \int \frac {-x^4-8 x^3-e^{20} x^2-8 e^{15} x^2-11 x^2+e^5 \left (-8 x^3-28 x^2+4 e^8-4 e^4\right )+e^{10} \left (-2 x^3-23 x^2+e^8-e^4\right )+e^4 (-2 x-4)+e^8 (2 x+4)}{x^4+8 x^3+e^{20} x^2+8 e^{15} x^2+16 x^2+e^{10} \left (2 x^3+24 x^2\right )+e^5 \left (8 x^3+32 x^2\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-x^4-8 x^3+\left (-11-8 e^{15}\right ) x^2-e^{20} x^2+e^5 \left (-8 x^3-28 x^2+4 e^8-4 e^4\right )+e^{10} \left (-2 x^3-23 x^2+e^8-e^4\right )+e^4 (-2 x-4)+e^8 (2 x+4)}{x^4+8 x^3+e^{20} x^2+8 e^{15} x^2+16 x^2+e^{10} \left (2 x^3+24 x^2\right )+e^5 \left (8 x^3+32 x^2\right )}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-x^4-8 x^3+\left (-11-8 e^{15}-e^{20}\right ) x^2+e^5 \left (-8 x^3-28 x^2+4 e^8-4 e^4\right )+e^{10} \left (-2 x^3-23 x^2+e^8-e^4\right )+e^4 (-2 x-4)+e^8 (2 x+4)}{x^4+8 x^3+e^{20} x^2+8 e^{15} x^2+16 x^2+e^{10} \left (2 x^3+24 x^2\right )+e^5 \left (8 x^3+32 x^2\right )}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-x^4-8 x^3+\left (-11-8 e^{15}-e^{20}\right ) x^2+e^5 \left (-8 x^3-28 x^2+4 e^8-4 e^4\right )+e^{10} \left (-2 x^3-23 x^2+e^8-e^4\right )+e^4 (-2 x-4)+e^8 (2 x+4)}{x^4+8 x^3+\left (16+8 e^{15}\right ) x^2+e^{20} x^2+e^{10} \left (2 x^3+24 x^2\right )+e^5 \left (8 x^3+32 x^2\right )}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-x^4-8 x^3+\left (-11-8 e^{15}-e^{20}\right ) x^2+e^5 \left (-8 x^3-28 x^2+4 e^8-4 e^4\right )+e^{10} \left (-2 x^3-23 x^2+e^8-e^4\right )+e^4 (-2 x-4)+e^8 (2 x+4)}{x^4+8 x^3+\left (16+8 e^{15}+e^{20}\right ) x^2+e^{10} \left (2 x^3+24 x^2\right )+e^5 \left (8 x^3+32 x^2\right )}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-x^4-8 x^3+\left (-11-8 e^{15}-e^{20}\right ) x^2+e^5 \left (-8 x^3-28 x^2+4 e^8-4 e^4\right )+e^{10} \left (-2 x^3-23 x^2+e^8-e^4\right )+e^4 (-2 x-4)+e^8 (2 x+4)}{x^2 \left (x^2+2 \left (2+e^5\right )^2 x+\left (2+e^5\right )^4\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {-x^4-8 x^3+\left (-11-8 e^{15}-e^{20}\right ) x^2+e^5 \left (-8 x^3-28 x^2+4 e^8-4 e^4\right )+e^{10} \left (-2 x^3-23 x^2+e^8-e^4\right )+e^4 (-2 x-4)+e^8 (2 x+4)}{x^2 \left (x+\left (2+e^5\right )^2\right )^2}dx\) |
\(\Big \downarrow \) 2123 |
\(\displaystyle \int \left (\frac {e^4 \left (e^4-1\right )}{\left (2+e^5\right )^2 x^2}+\frac {20+e^4+36 e^5-e^8+25 e^{10}+8 e^{15}+e^{20}}{\left (2+e^5\right )^2 \left (x+e^{10}+4 e^5+4\right )^2}-1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -x-\frac {20+e^4+36 e^5-e^8+25 e^{10}+8 e^{15}+e^{20}}{\left (2+e^5\right )^2 \left (x+\left (2+e^5\right )^2\right )}+\frac {e^4 \left (1-e^4\right )}{\left (2+e^5\right )^2 x}\) |
Int[(E^4*(-4 - 2*x) - 11*x^2 - 8*E^15*x^2 - E^20*x^2 - 8*x^3 - x^4 + E^8*( 4 + 2*x) + E^5*(-4*E^4 + 4*E^8 - 28*x^2 - 8*x^3) + E^10*(-E^4 + E^8 - 23*x ^2 - 2*x^3))/(16*x^2 + 8*E^15*x^2 + E^20*x^2 + 8*x^3 + x^4 + E^10*(24*x^2 + 2*x^3) + E^5*(32*x^2 + 8*x^3)),x]
(E^4*(1 - E^4))/((2 + E^5)^2*x) - x - (20 + E^4 + 36*E^5 - E^8 + 25*E^10 + 8*E^15 + E^20)/((2 + E^5)^2*((2 + E^5)^2 + x))
3.26.4.3.1 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])
Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15
method | result | size |
risch | \(-x +\frac {\left (-4 \,{\mathrm e}^{5}-{\mathrm e}^{10}-5\right ) x -{\mathrm e}^{8}+{\mathrm e}^{4}}{\left ({\mathrm e}^{10}+4 \,{\mathrm e}^{5}+x +4\right ) x}\) | \(39\) |
norman | \(\frac {\left ({\mathrm e}^{20}+8 \,{\mathrm e}^{15}+23 \,{\mathrm e}^{10}+28 \,{\mathrm e}^{5}+11\right ) x -x^{3}-{\mathrm e}^{8}+{\mathrm e}^{4}}{x \left ({\mathrm e}^{10}+4 \,{\mathrm e}^{5}+x +4\right )}\) | \(56\) |
gosper | \(\frac {x \,{\mathrm e}^{20}+8 x \,{\mathrm e}^{15}+23 x \,{\mathrm e}^{10}-x^{3}+28 x \,{\mathrm e}^{5}-{\mathrm e}^{8}+{\mathrm e}^{4}+11 x}{x \left ({\mathrm e}^{10}+4 \,{\mathrm e}^{5}+x +4\right )}\) | \(60\) |
parallelrisch | \(\frac {x \,{\mathrm e}^{20}+8 x \,{\mathrm e}^{15}+23 x \,{\mathrm e}^{10}-x^{3}+28 x \,{\mathrm e}^{5}-{\mathrm e}^{8}+{\mathrm e}^{4}+11 x}{x \left ({\mathrm e}^{10}+4 \,{\mathrm e}^{5}+x +4\right )}\) | \(60\) |
int((-x^2*exp(5)^4-8*x^2*exp(5)^3+(exp(4)^2-exp(4)-2*x^3-23*x^2)*exp(5)^2+ (4*exp(4)^2-4*exp(4)-8*x^3-28*x^2)*exp(5)+(4+2*x)*exp(4)^2+(-2*x-4)*exp(4) -x^4-8*x^3-11*x^2)/(x^2*exp(5)^4+8*x^2*exp(5)^3+(2*x^3+24*x^2)*exp(5)^2+(8 *x^3+32*x^2)*exp(5)+x^4+8*x^3+16*x^2),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.62 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=-\frac {x^{3} + 4 \, x^{2} + {\left (x^{2} + x\right )} e^{10} + 4 \, {\left (x^{2} + x\right )} e^{5} + 5 \, x + e^{8} - e^{4}}{x^{2} + x e^{10} + 4 \, x e^{5} + 4 \, x} \]
integrate((-x^2*exp(5)^4-8*x^2*exp(5)^3+(exp(4)^2-exp(4)-2*x^3-23*x^2)*exp (5)^2+(4*exp(4)^2-4*exp(4)-8*x^3-28*x^2)*exp(5)+(4+2*x)*exp(4)^2+(-2*x-4)* exp(4)-x^4-8*x^3-11*x^2)/(x^2*exp(5)^4+8*x^2*exp(5)^3+(2*x^3+24*x^2)*exp(5 )^2+(8*x^3+32*x^2)*exp(5)+x^4+8*x^3+16*x^2),x, algorithm=\
-(x^3 + 4*x^2 + (x^2 + x)*e^10 + 4*(x^2 + x)*e^5 + 5*x + e^8 - e^4)/(x^2 + x*e^10 + 4*x*e^5 + 4*x)
Time = 3.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=- x - \frac {x \left (5 + 4 e^{5} + e^{10}\right ) - e^{4} + e^{8}}{x^{2} + x \left (4 + 4 e^{5} + e^{10}\right )} \]
integrate((-x**2*exp(5)**4-8*x**2*exp(5)**3+(exp(4)**2-exp(4)-2*x**3-23*x* *2)*exp(5)**2+(4*exp(4)**2-4*exp(4)-8*x**3-28*x**2)*exp(5)+(4+2*x)*exp(4)* *2+(-2*x-4)*exp(4)-x**4-8*x**3-11*x**2)/(x**2*exp(5)**4+8*x**2*exp(5)**3+( 2*x**3+24*x**2)*exp(5)**2+(8*x**3+32*x**2)*exp(5)+x**4+8*x**3+16*x**2),x)
Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=-x - \frac {x {\left (e^{10} + 4 \, e^{5} + 5\right )} + e^{8} - e^{4}}{x^{2} + x {\left (e^{10} + 4 \, e^{5} + 4\right )}} \]
integrate((-x^2*exp(5)^4-8*x^2*exp(5)^3+(exp(4)^2-exp(4)-2*x^3-23*x^2)*exp (5)^2+(4*exp(4)^2-4*exp(4)-8*x^3-28*x^2)*exp(5)+(4+2*x)*exp(4)^2+(-2*x-4)* exp(4)-x^4-8*x^3-11*x^2)/(x^2*exp(5)^4+8*x^2*exp(5)^3+(2*x^3+24*x^2)*exp(5 )^2+(8*x^3+32*x^2)*exp(5)+x^4+8*x^3+16*x^2),x, algorithm=\
Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=-x - \frac {x e^{10} + 4 \, x e^{5} + 5 \, x + e^{8} - e^{4}}{x^{2} + x e^{10} + 4 \, x e^{5} + 4 \, x} \]
integrate((-x^2*exp(5)^4-8*x^2*exp(5)^3+(exp(4)^2-exp(4)-2*x^3-23*x^2)*exp (5)^2+(4*exp(4)^2-4*exp(4)-8*x^3-28*x^2)*exp(5)+(4+2*x)*exp(4)^2+(-2*x-4)* exp(4)-x^4-8*x^3-11*x^2)/(x^2*exp(5)^4+8*x^2*exp(5)^3+(2*x^3+24*x^2)*exp(5 )^2+(8*x^3+32*x^2)*exp(5)+x^4+8*x^3+16*x^2),x, algorithm=\
Time = 9.83 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 \left (-4 e^4+4 e^8-28 x^2-8 x^3\right )+e^{10} \left (-e^4+e^8-23 x^2-2 x^3\right )}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} \left (24 x^2+2 x^3\right )+e^5 \left (32 x^2+8 x^3\right )} \, dx=-x-\frac {{\mathrm {e}}^8-{\mathrm {e}}^4+x\,\left (4\,{\mathrm {e}}^5+{\mathrm {e}}^{10}+5\right )}{x\,\left (x+4\,{\mathrm {e}}^5+{\mathrm {e}}^{10}+4\right )} \]
int(-(exp(5)*(4*exp(4) - 4*exp(8) + 28*x^2 + 8*x^3) + 8*x^2*exp(15) + x^2* exp(20) + exp(10)*(exp(4) - exp(8) + 23*x^2 + 2*x^3) + 11*x^2 + 8*x^3 + x^ 4 + exp(4)*(2*x + 4) - exp(8)*(2*x + 4))/(exp(10)*(24*x^2 + 2*x^3) + exp(5 )*(32*x^2 + 8*x^3) + 8*x^2*exp(15) + x^2*exp(20) + 16*x^2 + 8*x^3 + x^4),x )