3.22.76 \(\int \frac {e^{12} (-e^5+e^{10} (-1-x^2))}{1+e^5 (2+2 x-2 x^2)+e^{10} (1+2 x-x^2-2 x^3+x^4)} \, dx\)

Optimal. Leaf size=19 \[ \frac {e^{12}}{-1-\frac {1+\frac {1}{e^5}}{x}+x} \]

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Rubi [A]  time = 0.10, antiderivative size = 27, normalized size of antiderivative = 1.42, number of steps used = 5, number of rules used = 4, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.066, Rules used = {12, 1680, 1814, 8} \begin {gather*} -\frac {4 e^{17} x}{-e^5 (1-2 x)^2+5 e^5+4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^12*(-E^5 + E^10*(-1 - x^2)))/(1 + E^5*(2 + 2*x - 2*x^2) + E^10*(1 + 2*x - x^2 - 2*x^3 + x^4)),x]

[Out]

(-4*E^17*x)/(4 + 5*E^5 - E^5*(1 - 2*x)^2)

Rule 8

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[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} &=e^{12} \int \frac {-e^5+e^{10} \left (-1-x^2\right )}{1+e^5 \left (2+2 x-2 x^2\right )+e^{10} \left (1+2 x-x^2-2 x^3+x^4\right )} \, dx\\ &=e^{12} \operatorname {Subst}\left (\int \frac {4 e^5 \left (-4-5 e^5-4 e^5 x-4 e^5 x^2\right )}{\left (4+5 e^5-4 e^5 x^2\right )^2} \, dx,x,-\frac {1}{2}+x\right )\\ &=\left (4 e^{17}\right ) \operatorname {Subst}\left (\int \frac {-4-5 e^5-4 e^5 x-4 e^5 x^2}{\left (4+5 e^5-4 e^5 x^2\right )^2} \, dx,x,-\frac {1}{2}+x\right )\\ &=-\frac {4 e^{17} x}{4+5 e^5-e^5 (1-2 x)^2}-\frac {\left (2 e^{17}\right ) \operatorname {Subst}\left (\int 0 \, dx,x,-\frac {1}{2}+x\right )}{4+5 e^5}\\ &=-\frac {4 e^{17} x}{4+5 e^5-e^5 (1-2 x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 22, normalized size = 1.16 \begin {gather*} -\frac {e^{17} x}{1+e^5 \left (1+x-x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^12*(-E^5 + E^10*(-1 - x^2)))/(1 + E^5*(2 + 2*x - 2*x^2) + E^10*(1 + 2*x - x^2 - 2*x^3 + x^4)),x]

[Out]

-((E^17*x)/(1 + E^5*(1 + x - x^2)))

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fricas [A]  time = 0.54, size = 19, normalized size = 1.00 \begin {gather*} \frac {x e^{17}}{{\left (x^{2} - x - 1\right )} e^{5} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-1)*exp(5)^2-exp(5))*exp(12)/((x^4-2*x^3-x^2+2*x+1)*exp(5)^2+(-2*x^2+2*x+2)*exp(5)+1),x, algor
ithm="fricas")

[Out]

x*e^17/((x^2 - x - 1)*e^5 - 1)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (x^{2} + 1\right )} e^{10} + e^{5}\right )} e^{12}}{{\left (x^{4} - 2 \, x^{3} - x^{2} + 2 \, x + 1\right )} e^{10} - 2 \, {\left (x^{2} - x - 1\right )} e^{5} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-1)*exp(5)^2-exp(5))*exp(12)/((x^4-2*x^3-x^2+2*x+1)*exp(5)^2+(-2*x^2+2*x+2)*exp(5)+1),x, algor
ithm="giac")

[Out]

integrate(-((x^2 + 1)*e^10 + e^5)*e^12/((x^4 - 2*x^3 - x^2 + 2*x + 1)*e^10 - 2*(x^2 - x - 1)*e^5 + 1), x)

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maple [A]  time = 0.09, size = 24, normalized size = 1.26

method	result	size
gosper	\(\frac {{\mathrm e}^{17} x}{x^{2} {\mathrm e}^{5}-x \,{\mathrm e}^{5}-{\mathrm e}^{5}-1}\)	\(24\)
risch	\(\frac {{\mathrm e}^{17} x}{x^{2} {\mathrm e}^{5}-x \,{\mathrm e}^{5}-{\mathrm e}^{5}-1}\)	\(24\)
norman	\(\frac {{\mathrm e}^{5} x \,{\mathrm e}^{12}}{x^{2} {\mathrm e}^{5}-x \,{\mathrm e}^{5}-{\mathrm e}^{5}-1}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2-1)*exp(5)^2-exp(5))*exp(12)/((x^4-2*x^3-x^2+2*x+1)*exp(5)^2+(-2*x^2+2*x+2)*exp(5)+1),x,method=_RETU
RNVERBOSE)

[Out]

exp(17)*x/(x^2*exp(5)-x*exp(5)-exp(5)-1)

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maxima [A]  time = 0.41, size = 23, normalized size = 1.21 \begin {gather*} \frac {x e^{17}}{x^{2} e^{5} - x e^{5} - e^{5} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2-1)*exp(5)^2-exp(5))*exp(12)/((x^4-2*x^3-x^2+2*x+1)*exp(5)^2+(-2*x^2+2*x+2)*exp(5)+1),x, algor
ithm="maxima")

[Out]

x*e^17/(x^2*e^5 - x*e^5 - e^5 - 1)

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mupad [B]  time = 1.39, size = 22, normalized size = 1.16 \begin {gather*} -\frac {x\,{\mathrm {e}}^{17}}{-{\mathrm {e}}^5\,x^2+{\mathrm {e}}^5\,x+{\mathrm {e}}^5+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(12)*(exp(5) + exp(10)*(x^2 + 1)))/(exp(5)*(2*x - 2*x^2 + 2) + exp(10)*(2*x - x^2 - 2*x^3 + x^4 + 1)
+ 1),x)

[Out]

-(x*exp(17))/(exp(5) + x*exp(5) - x^2*exp(5) + 1)

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sympy [A]  time = 0.52, size = 20, normalized size = 1.05 \begin {gather*} \frac {x e^{17}}{x^{2} e^{5} - x e^{5} - e^{5} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2-1)*exp(5)**2-exp(5))*exp(12)/((x**4-2*x**3-x**2+2*x+1)*exp(5)**2+(-2*x**2+2*x+2)*exp(5)+1),x
)

[Out]

x*exp(17)/(x**2*exp(5) - x*exp(5) - exp(5) - 1)

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