3.25.43 \(\int \frac {5+e^3 (-1-4 x)-6 x-x^2}{75+3 e^6+30 x^2+3 x^4+e^3 (-30-6 x^2)} \, dx\)

Optimal. Leaf size=27 \[ \frac {3+\frac {1}{3} \left (4+x+2 x^2\right )}{5-e^3+x^2} \]

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Rubi [A]  time = 0.08, antiderivative size = 46, normalized size of antiderivative = 1.70, number of steps used = 4, number of rules used = 4, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1994, 28, 1814, 8} \begin {gather*} \frac {\left (5-e^3\right ) x-2 e^6+7 e^3+15}{3 \left (5-e^3\right ) \left (x^2-e^3+5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + E^3*(-1 - 4*x) - 6*x - x^2)/(75 + 3*E^6 + 30*x^2 + 3*x^4 + E^3*(-30 - 6*x^2)),x]

[Out]

(15 + 7*E^3 - 2*E^6 + (5 - E^3)*x)/(3*(5 - E^3)*(5 - E^3 + x^2))

Rule 8

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

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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]

Rule 1994

Int[(Pq_)*(u_)^(p_.), x_Symbol] :> Int[Pq*ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && PolyQ[Pq, x] && TrinomialQ
[u, x] &&  !TrinomialMatchQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5+e^3 (-1-4 x)-6 x-x^2}{3 \left (5-e^3\right )^2+6 \left (5-e^3\right ) x^2+3 x^4} \, dx\\ &=3 \int \frac {5+e^3 (-1-4 x)-6 x-x^2}{\left (3 \left (5-e^3\right )+3 x^2\right )^2} \, dx\\ &=\frac {15+7 e^3-2 e^6+\left (5-e^3\right ) x}{3 \left (5-e^3\right ) \left (5-e^3+x^2\right )}-\frac {\int 0 \, dx}{2 \left (5-e^3\right )}\\ &=\frac {15+7 e^3-2 e^6+\left (5-e^3\right ) x}{3 \left (5-e^3\right ) \left (5-e^3+x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.96 \begin {gather*} -\frac {-3-2 e^3-x}{3 \left (5-e^3+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + E^3*(-1 - 4*x) - 6*x - x^2)/(75 + 3*E^6 + 30*x^2 + 3*x^4 + E^3*(-30 - 6*x^2)),x]

[Out]

-1/3*(-3 - 2*E^3 - x)/(5 - E^3 + x^2)

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fricas [A]  time = 0.56, size = 20, normalized size = 0.74 \begin {gather*} \frac {x + 2 \, e^{3} + 3}{3 \, {\left (x^{2} - e^{3} + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-1)*exp(3)-x^2-6*x+5)/(3*exp(3)^2+(-6*x^2-30)*exp(3)+3*x^4+30*x^2+75),x, algorithm="fricas")

[Out]

1/3*(x + 2*e^3 + 3)/(x^2 - e^3 + 5)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-1)*exp(3)-x^2-6*x+5)/(3*exp(3)^2+(-6*x^2-30)*exp(3)+3*x^4+30*x^2+75),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.06, size = 21, normalized size = 0.78

method	result	size
gosper	\(-\frac {3+x +2 \,{\mathrm e}^{3}}{3 \left (-x^{2}+{\mathrm e}^{3}-5\right )}\)	\(21\)
default	\(-\frac {3+x +2 \,{\mathrm e}^{3}}{3 \left (-x^{2}+{\mathrm e}^{3}-5\right )}\)	\(21\)
norman	\(\frac {-\frac {x}{3}-1-\frac {2 \,{\mathrm e}^{3}}{3}}{-x^{2}+{\mathrm e}^{3}-5}\)	\(22\)
risch	\(\frac {-\frac {x}{3}-1-\frac {2 \,{\mathrm e}^{3}}{3}}{-x^{2}+{\mathrm e}^{3}-5}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x-1)*exp(3)-x^2-6*x+5)/(3*exp(3)^2+(-6*x^2-30)*exp(3)+3*x^4+30*x^2+75),x,method=_RETURNVERBOSE)

[Out]

-1/3*(3+x+2*exp(3))/(-x^2+exp(3)-5)

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maxima [A]  time = 0.51, size = 20, normalized size = 0.74 \begin {gather*} \frac {x + 2 \, e^{3} + 3}{3 \, {\left (x^{2} - e^{3} + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-1)*exp(3)-x^2-6*x+5)/(3*exp(3)^2+(-6*x^2-30)*exp(3)+3*x^4+30*x^2+75),x, algorithm="maxima")

[Out]

1/3*(x + 2*e^3 + 3)/(x^2 - e^3 + 5)

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mupad [B]  time = 0.13, size = 21, normalized size = 0.78 \begin {gather*} \frac {\frac {x}{3}+\frac {2\,{\mathrm {e}}^3}{3}+1}{x^2-{\mathrm {e}}^3+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(6*x + x^2 + exp(3)*(4*x + 1) - 5)/(3*exp(6) - exp(3)*(6*x^2 + 30) + 30*x^2 + 3*x^4 + 75),x)

[Out]

(x/3 + (2*exp(3))/3 + 1)/(x^2 - exp(3) + 5)

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sympy [A]  time = 0.40, size = 22, normalized size = 0.81 \begin {gather*} - \frac {- x - 2 e^{3} - 3}{3 x^{2} - 3 e^{3} + 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-1)*exp(3)-x**2-6*x+5)/(3*exp(3)**2+(-6*x**2-30)*exp(3)+3*x**4+30*x**2+75),x)

[Out]

-(-x - 2*exp(3) - 3)/(3*x**2 - 3*exp(3) + 15)

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