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

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 = 0.29, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {6742, 731, 725, 206, 807} \[ \frac {\sqrt {1-x^2}}{5 x+4}+\frac {3}{5 (5 x+4)} \]

Antiderivative was successfully verified.

[In]

Int[(-5 - 4*x - 3*Sqrt[1 - x^2])/((4 + 5*x)^2*Sqrt[1 - x^2]),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 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 731

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d)/(c*d^2 + a*e^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(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]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 6742

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

Rubi steps

\begin {align*} \int \frac {-5-4 x-3 \sqrt {1-x^2}}{(4+5 x)^2 \sqrt {1-x^2}} \, dx &=\int \left (-\frac {3}{(4+5 x)^2}-\frac {5}{(4+5 x)^2 \sqrt {1-x^2}}-\frac {4 x}{(4+5 x)^2 \sqrt {1-x^2}}\right ) \, dx\\ &=\frac {3}{5 (4+5 x)}-4 \int \frac {x}{(4+5 x)^2 \sqrt {1-x^2}} \, dx-5 \int \frac {1}{(4+5 x)^2 \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 = 0.15, size = 23, normalized size = 0.74 \[ \frac {5 \sqrt {1-x^2}+3}{25 x+20} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3 + 5*Sqrt[1 - x^2])/(20 + 25*x)

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fricas [A]  time = 0.42, 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((-5-4*x-3*(-x^2+1)^(1/2))/(4+5*x)^2/(-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 [C]  time = 0.59, size = 54, normalized size = 1.74 \[ \frac {\sqrt {\frac {8}{5 \, x + 4} + \frac {9}{{\left (5 \, x + 4\right )}^{2}} - 1}}{5 \, \mathrm {sgn}\left (\frac {1}{5 \, x + 4}\right )} + \frac {3}{5 \, {\left (5 \, x + 4\right )}} - \frac {1}{5} i \, \mathrm {sgn}\left (\frac {1}{5 \, x + 4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*sqrt(8/(5*x + 4) + 9/(5*x + 4)^2 - 1)/sgn(1/(5*x + 4)) + 3/5/(5*x + 4) - 1/5*I*sgn(1/(5*x + 4))

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maple [A]  time = 0.01, size = 32, normalized size = 1.03 \[ \frac {\sqrt {\frac {8 x}{5}-\left (x +\frac {4}{5}\right )^{2}+\frac {41}{25}}}{5 x +4}+\frac {3}{5 \left (5 x +4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.61, size = 25, normalized size = 0.81 \[ \frac {5 \, \sqrt {x + 1} \sqrt {-x + 1} + 3}{5 \, {\left (5 \, x + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, 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(-(4*x + 3*(1 - x^2)^(1/2) + 5)/((5*x + 4)^2*(1 - x^2)^(1/2)),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 \[ - \int \frac {4 x}{25 x^{2} \sqrt {1 - x^{2}} + 40 x \sqrt {1 - x^{2}} + 16 \sqrt {1 - x^{2}}}\, dx - \int \frac {3 \sqrt {1 - x^{2}}}{25 x^{2} \sqrt {1 - x^{2}} + 40 x \sqrt {1 - x^{2}} + 16 \sqrt {1 - x^{2}}}\, dx - \int \frac {5}{25 x^{2} \sqrt {1 - x^{2}} + 40 x \sqrt {1 - x^{2}} + 16 \sqrt {1 - x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(4*x/(25*x**2*sqrt(1 - x**2) + 40*x*sqrt(1 - x**2) + 16*sqrt(1 - x**2)), x) - Integral(3*sqrt(1 - x**
2)/(25*x**2*sqrt(1 - x**2) + 40*x*sqrt(1 - x**2) + 16*sqrt(1 - x**2)), x) - Integral(5/(25*x**2*sqrt(1 - x**2)
+ 40*x*sqrt(1 - x**2) + 16*sqrt(1 - x**2)), x)

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