3.26.4 \(\int \frac {e^4 (-4-2 x)-11 x^2-8 e^{15} x^2-e^{20} x^2-8 x^3-x^4+e^8 (4+2 x)+e^5 (-4 e^4+4 e^8-28 x^2-8 x^3)+e^{10} (-e^4+e^8-23 x^2-2 x^3)}{16 x^2+8 e^{15} x^2+e^{20} x^2+8 x^3+x^4+e^{10} (24 x^2+2 x^3)+e^5 (32 x^2+8 x^3)} \, dx\)

Optimal. Leaf size=34 \[ -x+\frac {\frac {e^4-e^8-x}{x}+x}{\left (2+e^5\right )^2+x} \]

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Rubi [A]  time = 0.29, antiderivative size = 45, normalized size of antiderivative = 1.32, number of steps used = 8, number of rules used = 5, integrand size = 160, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {6, 1680, 1814, 21, 8} \begin {gather*} \frac {\left (-5-4 e^5-e^{10}\right ) x+e^4 \left (1-e^4\right )}{x \left (x+\left (2+e^5\right )^2\right )}-x \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

-x + (E^4*(1 - E^4) + (-5 - 4*E^5 - E^10)*x)/(x*((2 + E^5)^2 + x))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^4 (-4-2 x)-e^{20} x^2+\left (-11-8 e^{15}\right ) 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\\ &=\int \frac {e^4 (-4-2 x)+\left (-11-8 e^{15}-e^{20}\right ) 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\\ &=\int \frac {e^4 (-4-2 x)+\left (-11-8 e^{15}-e^{20}\right ) 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 )}{e^{20} x^2+\left (16+8 e^{15}\right ) 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\\ &=\int \frac {e^4 (-4-2 x)+\left (-11-8 e^{15}-e^{20}\right ) 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 )}{\left (16+8 e^{15}+e^{20}\right ) 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\\ &=\operatorname {Subst}\left (\int \frac {\left (2+e^5\right )^4 \left (4-16 e^5-20 e^{10}-8 e^{15}-e^{20}\right )-16 \left (20+2 e^4+36 e^5-2 e^8+25 e^{10}+8 e^{15}+e^{20}\right ) x+8 \left (26+40 e^5+26 e^{10}+8 e^{15}+e^{20}\right ) x^2-16 x^4}{\left (16+32 e^5+24 e^{10}+8 e^{15}+e^{20}-4 x^2\right )^2} \, dx,x,\frac {1}{4} \left (8+8 e^5+2 e^{10}\right )+x\right )\\ &=\frac {e^4 \left (1-e^4\right )+\left (-5-4 e^5-e^{10}\right ) x}{x \left (\left (2+e^5\right )^2+x\right )}-\frac {\operatorname {Subst}\left (\int \frac {2 \left (2+e^5\right )^8-8 \left (2+e^5\right )^4 x^2}{16+32 e^5+24 e^{10}+8 e^{15}+e^{20}-4 x^2} \, dx,x,\frac {1}{4} \left (8+8 e^5+2 e^{10}\right )+x\right )}{2 \left (2+e^5\right )^4}\\ &=\frac {e^4 \left (1-e^4\right )+\left (-5-4 e^5-e^{10}\right ) x}{x \left (\left (2+e^5\right )^2+x\right )}-\operatorname {Subst}\left (\int 1 \, dx,x,\frac {1}{4} \left (8+8 e^5+2 e^{10}\right )+x\right )\\ &=-x+\frac {e^4 \left (1-e^4\right )+\left (-5-4 e^5-e^{10}\right ) x}{x \left (\left (2+e^5\right )^2+x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 54, normalized size = 1.59 \begin {gather*} -\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 )} \end {gather*}

Antiderivative was successfully verified.

[In]

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*
(24*x^2 + 2*x^3) + E^5*(32*x^2 + 8*x^3)),x]

[Out]

-((-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)))

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fricas [A]  time = 0.96, size = 55, normalized size = 1.62 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)+(2*x+4)*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="fricas")

[Out]

-(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)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)+(2*x+4)*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="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -sageVARx-(exp(20)*exp(8)*exp(10)+4*exp(
20)*exp(8)*exp(5)+4*exp(20)*exp(8)-exp(20)*exp(4)*exp(10)-4*exp(20)*exp(4)*exp(5)-4*exp(20)*exp(4)+8*exp(8)*ex
p(15)*exp(10)+32*exp(

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maple [A]  time = 0.21, size = 39, normalized size = 1.15

method	result	size
risch	\(-x +\frac {\left (-{\mathrm e}^{10}-4 \,{\mathrm e}^{5}-5\right ) x -{\mathrm e}^{8}+{\mathrm e}^{4}}{\left ({\mathrm e}^{10}+4 \,{\mathrm e}^{5}+x +4\right ) x}\)	\(39\)
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	\(\frac {x \,{\mathrm e}^{20}+8 x \,{\mathrm e}^{15}+23 x \,{\mathrm e}^{10}-x^{3}+28 x \,{\mathrm e}^{5}-{\mathrm e}^{8}+{\mathrm e}^{4}+11 x}{x \left ({\mathrm e}^{10}+4 \,{\mathrm e}^{5}+x +4\right )}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)*e
xp(5)+(2*x+4)*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)

[Out]

-x+((-exp(10)-4*exp(5)-5)*x-exp(8)+exp(4))/(exp(10)+4*exp(5)+x+4)/x

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maxima [A]  time = 0.41, size = 39, normalized size = 1.15 \begin {gather*} -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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)+(2*x+4)*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="maxima")

[Out]

-x - (x*(e^10 + 4*e^5 + 5) + e^8 - e^4)/(x^2 + x*(e^10 + 4*e^5 + 4))

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mupad [B]  time = 1.53, size = 37, normalized size = 1.09 \begin {gather*} -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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

- x - (exp(8) - exp(4) + x*(4*exp(5) + exp(10) + 5))/(x*(x + 4*exp(5) + exp(10) + 4))

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sympy [A]  time = 5.55, size = 36, normalized size = 1.06 \begin {gather*} - 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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)+(2*x+4)*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)

[Out]

-x - (x*(5 + 4*exp(5) + exp(10)) - exp(4) + exp(8))/(x**2 + x*(4 + 4*exp(5) + exp(10)))

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