3.229 \(\int \frac{1}{(-3-2 x+x^2)^{5/2}} \, dx\)

Optimal. Leaf size=43 \[ \frac{1-x}{12 \left (x^2-2 x-3\right )^{3/2}}-\frac{1-x}{24 \sqrt{x^2-2 x-3}} \]

[Out]

(1 - x)/(12*(-3 - 2*x + x^2)^(3/2)) - (1 - x)/(24*Sqrt[-3 - 2*x + x^2])

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Rubi [A]  time = 0.006261, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {614, 613} \[ \frac{1-x}{12 \left (x^2-2 x-3\right )^{3/2}}-\frac{1-x}{24 \sqrt{x^2-2 x-3}} \]

Antiderivative was successfully verified.

[In]

Int[(-3 - 2*x + x^2)^(-5/2),x]

[Out]

(1 - x)/(12*(-3 - 2*x + x^2)^(3/2)) - (1 - x)/(24*Sqrt[-3 - 2*x + x^2])

Rule 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \frac{1}{\left (-3-2 x+x^2\right )^{5/2}} \, dx &=\frac{1-x}{12 \left (-3-2 x+x^2\right )^{3/2}}-\frac{1}{6} \int \frac{1}{\left (-3-2 x+x^2\right )^{3/2}} \, dx\\ &=\frac{1-x}{12 \left (-3-2 x+x^2\right )^{3/2}}-\frac{1-x}{24 \sqrt{-3-2 x+x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0108458, size = 27, normalized size = 0.63 \[ \frac{(x-1) \left (x^2-2 x-5\right )}{24 \left (x^2-2 x-3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-3 - 2*x + x^2)^(-5/2),x]

[Out]

((-1 + x)*(-5 - 2*x + x^2))/(24*(-3 - 2*x + x^2)^(3/2))

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Maple [A]  time = 0.003, size = 32, normalized size = 0.7 \begin{align*}{\frac{ \left ( 1+x \right ) \left ( -3+x \right ) \left ({x}^{3}-3\,{x}^{2}-3\,x+5 \right ) }{24} \left ({x}^{2}-2\,x-3 \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2-2*x-3)^(5/2),x)

[Out]

1/24*(1+x)*(-3+x)*(x^3-3*x^2-3*x+5)/(x^2-2*x-3)^(5/2)

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Maxima [A]  time = 0.940605, size = 69, normalized size = 1.6 \begin{align*} \frac{x}{24 \, \sqrt{x^{2} - 2 \, x - 3}} - \frac{1}{24 \, \sqrt{x^{2} - 2 \, x - 3}} - \frac{x}{12 \,{\left (x^{2} - 2 \, x - 3\right )}^{\frac{3}{2}}} + \frac{1}{12 \,{\left (x^{2} - 2 \, x - 3\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2-2*x-3)^(5/2),x, algorithm="maxima")

[Out]

1/24*x/sqrt(x^2 - 2*x - 3) - 1/24/sqrt(x^2 - 2*x - 3) - 1/12*x/(x^2 - 2*x - 3)^(3/2) + 1/12/(x^2 - 2*x - 3)^(3
/2)

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Fricas [B]  time = 1.71169, size = 159, normalized size = 3.7 \begin{align*} \frac{x^{4} - 4 \, x^{3} - 2 \, x^{2} +{\left (x^{3} - 3 \, x^{2} - 3 \, x + 5\right )} \sqrt{x^{2} - 2 \, x - 3} + 12 \, x + 9}{24 \,{\left (x^{4} - 4 \, x^{3} - 2 \, x^{2} + 12 \, x + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2-2*x-3)^(5/2),x, algorithm="fricas")

[Out]

1/24*(x^4 - 4*x^3 - 2*x^2 + (x^3 - 3*x^2 - 3*x + 5)*sqrt(x^2 - 2*x - 3) + 12*x + 9)/(x^4 - 4*x^3 - 2*x^2 + 12*
x + 9)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (x^{2} - 2 x - 3\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**2-2*x-3)**(5/2),x)

[Out]

Integral((x**2 - 2*x - 3)**(-5/2), x)

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Giac [A]  time = 1.08948, size = 31, normalized size = 0.72 \begin{align*} \frac{{\left ({\left (x - 3\right )} x - 3\right )} x + 5}{24 \,{\left (x^{2} - 2 \, x - 3\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2-2*x-3)^(5/2),x, algorithm="giac")

[Out]

1/24*(((x - 3)*x - 3)*x + 5)/(x^2 - 2*x - 3)^(3/2)