3.33.82 \(\int \frac {-405-774 x+36 e^2 x-432 x^2-60 x^3+9 x^4+2 x^5}{450-60 x^2+2 x^4} \, dx\)

Optimal. Leaf size=33 \[ \frac {3-e^2+x+(4+x)^2}{2-\frac {9}{3+\frac {1}{4} \left (-9+x^2\right )}} \]

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Rubi [A]  time = 0.04, antiderivative size = 34, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {6, 28, 1814, 1586} \begin {gather*} \frac {x^2}{2}-\frac {9 \left (9 x-e^2+34\right )}{15-x^2}+\frac {9 x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-405 - 774*x + 36*E^2*x - 432*x^2 - 60*x^3 + 9*x^4 + 2*x^5)/(450 - 60*x^2 + 2*x^4),x]

[Out]

(9*x)/2 + x^2/2 - (9*(34 - E^2 + 9*x))/(15 - x^2)

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 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 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 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 {-405+\left (-774+36 e^2\right ) x-432 x^2-60 x^3+9 x^4+2 x^5}{450-60 x^2+2 x^4} \, dx\\ &=2 \int \frac {-405+\left (-774+36 e^2\right ) x-432 x^2-60 x^3+9 x^4+2 x^5}{\left (-30+2 x^2\right )^2} \, dx\\ &=-\frac {9 \left (34-e^2+9 x\right )}{15-x^2}+\frac {1}{30} \int \frac {-4050-900 x+270 x^2+60 x^3}{-30+2 x^2} \, dx\\ &=-\frac {9 \left (34-e^2+9 x\right )}{15-x^2}+\frac {1}{30} \int (135+30 x) \, dx\\ &=\frac {9 x}{2}+\frac {x^2}{2}-\frac {9 \left (34-e^2+9 x\right )}{15-x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 28, normalized size = 0.85 \begin {gather*} \frac {1}{2} \left (9 x+x^2-\frac {18 \left (-34+e^2-9 x\right )}{-15+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-405 - 774*x + 36*E^2*x - 432*x^2 - 60*x^3 + 9*x^4 + 2*x^5)/(450 - 60*x^2 + 2*x^4),x]

[Out]

(9*x + x^2 - (18*(-34 + E^2 - 9*x))/(-15 + x^2))/2

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fricas [A]  time = 0.72, size = 31, normalized size = 0.94 \begin {gather*} \frac {x^{4} + 9 \, x^{3} - 15 \, x^{2} + 27 \, x - 18 \, e^{2} + 612}{2 \, {\left (x^{2} - 15\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((36*exp(2)*x+2*x^5+9*x^4-60*x^3-432*x^2-774*x-405)/(2*x^4-60*x^2+450),x, algorithm="fricas")

[Out]

1/2*(x^4 + 9*x^3 - 15*x^2 + 27*x - 18*e^2 + 612)/(x^2 - 15)

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giac [A]  time = 0.38, size = 27, normalized size = 0.82 \begin {gather*} \frac {1}{2} \, x^{2} + \frac {9}{2} \, x + \frac {9 \, {\left (9 \, x - e^{2} + 34\right )}}{x^{2} - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((36*exp(2)*x+2*x^5+9*x^4-60*x^3-432*x^2-774*x-405)/(2*x^4-60*x^2+450),x, algorithm="giac")

[Out]

1/2*x^2 + 9/2*x + 9*(9*x - e^2 + 34)/(x^2 - 15)

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maple [A]  time = 0.04, size = 27, normalized size = 0.82

method	result	size
risch	\(\frac {x^{2}}{2}+\frac {9 x}{2}+\frac {81 x +306-9 \,{\mathrm e}^{2}}{x^{2}-15}\)	\(27\)
default	\(\frac {x^{2}}{2}+\frac {9 x}{2}+\frac {81 x +306-9 \,{\mathrm e}^{2}}{x^{2}-15}\)	\(28\)
norman	\(\frac {\frac {27 x}{2}+\frac {9 x^{3}}{2}+\frac {x^{4}}{2}+\frac {387}{2}-9 \,{\mathrm e}^{2}}{x^{2}-15}\)	\(28\)
gosper	\(-\frac {-x^{4}-9 x^{3}+18 \,{\mathrm e}^{2}-27 x -387}{2 \left (x^{2}-15\right )}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((36*exp(2)*x+2*x^5+9*x^4-60*x^3-432*x^2-774*x-405)/(2*x^4-60*x^2+450),x,method=_RETURNVERBOSE)

[Out]

1/2*x^2+9/2*x+(81*x+306-9*exp(2))/(x^2-15)

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maxima [A]  time = 0.41, size = 27, normalized size = 0.82 \begin {gather*} \frac {1}{2} \, x^{2} + \frac {9}{2} \, x + \frac {9 \, {\left (9 \, x - e^{2} + 34\right )}}{x^{2} - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((36*exp(2)*x+2*x^5+9*x^4-60*x^3-432*x^2-774*x-405)/(2*x^4-60*x^2+450),x, algorithm="maxima")

[Out]

1/2*x^2 + 9/2*x + 9*(9*x - e^2 + 34)/(x^2 - 15)

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mupad [B]  time = 0.08, size = 26, normalized size = 0.79 \begin {gather*} \frac {9\,x}{2}+\frac {x^2}{2}+\frac {81\,x-9\,{\mathrm {e}}^2+306}{x^2-15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(774*x - 36*x*exp(2) + 432*x^2 + 60*x^3 - 9*x^4 - 2*x^5 + 405)/(2*x^4 - 60*x^2 + 450),x)

[Out]

(9*x)/2 + x^2/2 + (81*x - 9*exp(2) + 306)/(x^2 - 15)

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sympy [A]  time = 0.31, size = 24, normalized size = 0.73 \begin {gather*} \frac {x^{2}}{2} + \frac {9 x}{2} + \frac {81 x - 9 e^{2} + 306}{x^{2} - 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((36*exp(2)*x+2*x**5+9*x**4-60*x**3-432*x**2-774*x-405)/(2*x**4-60*x**2+450),x)

[Out]

x**2/2 + 9*x/2 + (81*x - 9*exp(2) + 306)/(x**2 - 15)

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