3.827 \(\int \frac {1}{3-3 x^2-5 \sqrt {1-x^2}-4 x \sqrt {1-x^2}} \, dx\)

Optimal. Leaf size=31 \[ \frac {\sqrt {1-x^2}}{5 x+4}+\frac {3}{5 (5 x+4)} \]

[Out]

3/5/(4+5*x)+(-x^2+1)^(1/2)/(4+5*x)

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Rubi [A]  time = 0.13, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6742, 665, 216, 733, 844, 725, 206, 735} \[ \frac {\sqrt {1-x^2}}{5 x+4}+\frac {3}{5 (5 x+4)} \]

Antiderivative was successfully verified.

[In]

Int[(3 - 3*x^2 - 5*Sqrt[1 - x^2] - 4*x*Sqrt[1 - x^2])^(-1),x]

[Out]

3/(5*(4 + 5*x)) + Sqrt[1 - x^2]/(4 + 5*x)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] - Dist[(2*c*d*p)/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[
m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 733

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 1)), x] - Dist[(2*c*p)/(e*(m + 1)), Int[x*(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c,
d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m +
2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 735

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] + Dist[(2*p)/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[a*e - c*d*x, x]*(a + c*x^2)^(p - 1),
x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {1}{3-3 x^2-5 \sqrt {1-x^2}-4 x \sqrt {1-x^2}} \, dx &=\int \left (-\frac {3}{(4+5 x)^2}+\frac {\sqrt {1-x^2}}{18 (-1+x)}-\frac {\sqrt {1-x^2}}{2 (1+x)}-\frac {5 \sqrt {1-x^2}}{(4+5 x)^2}+\frac {20 \sqrt {1-x^2}}{9 (4+5 x)}\right ) \, dx\\ &=\frac {3}{5 (4+5 x)}+\frac {1}{18} \int \frac {\sqrt {1-x^2}}{-1+x} \, dx-\frac {1}{2} \int \frac {\sqrt {1-x^2}}{1+x} \, dx+\frac {20}{9} \int \frac {\sqrt {1-x^2}}{4+5 x} \, dx-5 \int \frac {\sqrt {1-x^2}}{(4+5 x)^2} \, dx\\ &=\frac {3}{5 (4+5 x)}+\frac {\sqrt {1-x^2}}{4+5 x}-\frac {1}{18} \int \frac {1}{\sqrt {1-x^2}} \, dx+\frac {4}{9} \int \frac {5+4 x}{(4+5 x) \sqrt {1-x^2}} \, dx-\frac {1}{2} \int \frac {1}{\sqrt {1-x^2}} \, dx+\int \frac {x}{(4+5 x) \sqrt {1-x^2}} \, dx\\ &=\frac {3}{5 (4+5 x)}+\frac {\sqrt {1-x^2}}{4+5 x}-\frac {5}{9} \sin ^{-1}(x)+\frac {1}{5} \int \frac {1}{\sqrt {1-x^2}} \, dx+\frac {16}{45} \int \frac {1}{\sqrt {1-x^2}} \, dx\\ &=\frac {3}{5 (4+5 x)}+\frac {\sqrt {1-x^2}}{4+5 x}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 23, normalized size = 0.74 \[ \frac {5 \sqrt {1-x^2}+3}{25 x+20} \]

Antiderivative was successfully verified.

[In]

Integrate[(3 - 3*x^2 - 5*Sqrt[1 - x^2] - 4*x*Sqrt[1 - x^2])^(-1),x]

[Out]

(3 + 5*Sqrt[1 - x^2])/(20 + 25*x)

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fricas [A]  time = 0.40, size = 25, normalized size = 0.81 \[ \frac {25 \, x + 20 \, \sqrt {-x^{2} + 1} + 32}{20 \, {\left (5 \, x + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(25*x + 20*sqrt(-x^2 + 1) + 32)/(5*x + 4)

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giac [B]  time = 0.37, size = 68, normalized size = 2.19 \[ \frac {\frac {5 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{x} - 4}{4 \, {\left (\frac {5 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{x} - \frac {2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 2\right )}} + \frac {3}{5 \, {\left (5 \, x + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 81, normalized size = 2.61 \[ \frac {5 \sqrt {\frac {8 x}{5}-\left (x +\frac {4}{5}\right )^{2}+\frac {41}{25}}\, x}{9}+\frac {3}{5 \left (5 x +4\right )}+\frac {\sqrt {-2 x -\left (x -1\right )^{2}+2}}{18}-\frac {\sqrt {2 x -\left (x +1\right )^{2}+2}}{2}+\frac {5 \left (\frac {8 x}{5}-\left (x +\frac {4}{5}\right )^{2}+\frac {41}{25}\right )^{\frac {3}{2}}}{9 \left (x +\frac {4}{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/9*(8/5*x-(x+4/5)^2+41/25)^(1/2)*x+3/5/(5*x+4)+1/18*(-2*x-(x-1)^2+2)^(1/2)-1/2*(2*x-(x+1)^2+2)^(1/2)+5/9/(x+4
/5)*(8/5*x-(x+4/5)^2+41/25)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {1}{3 \, x^{2} + 4 \, \sqrt {-x^{2} + 1} x + 5 \, \sqrt {-x^{2} + 1} - 3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate(1/(3*x^2 + 4*sqrt(-x^2 + 1)*x + 5*sqrt(-x^2 + 1) - 3), x)

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mupad [B]  time = 0.05, size = 19, normalized size = 0.61 \[ \frac {\sqrt {1-x^2}+\frac {3}{5}}{5\,x+4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((1 - x^2)^(1/2) + 3/5)/(5*x + 4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {1}{3 x^{2} + 4 x \sqrt {1 - x^{2}} + 5 \sqrt {1 - x^{2}} - 3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(3*x**2 + 4*x*sqrt(1 - x**2) + 5*sqrt(1 - x**2) - 3), x)

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