\(\int \frac {-5-4 x-3 \sqrt {1-x^2}}{(4+5 x)^2 \sqrt {1-x^2}} \, dx\) [825]

Optimal result

Integrand size = 37, antiderivative size = 31 \[ \int \frac {-5-4 x-3 \sqrt {1-x^2}}{(4+5 x)^2 \sqrt {1-x^2}} \, dx=\frac {3}{5 (4+5 x)}+\frac {\sqrt {1-x^2}}{4+5 x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {6874, 745, 739, 212, 821} \[ \int \frac {-5-4 x-3 \sqrt {1-x^2}}{(4+5 x)^2 \sqrt {1-x^2}} \, dx=\frac {\sqrt {1-x^2}}{5 x+4}+\frac {3}{5 (5 x+4)} \]

[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 212

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

Rule 739

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 745

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 821

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 6874

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

Rubi steps \begin{align*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74 \[ \int \frac {-5-4 x-3 \sqrt {1-x^2}}{(4+5 x)^2 \sqrt {1-x^2}} \, dx=\frac {3+5 \sqrt {1-x^2}}{20+25 x} \]

[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)

Maple [A] (verified)

Time = 1.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94

method	result	size
trager	\(-\frac {3 x}{4 \left (4+5 x \right )}+\frac {\sqrt {-x^{2}+1}}{4+5 x}\)	\(29\)
default	\(\frac {\sqrt {-\left (\frac {4}{5}+x \right )^{2}+\frac {8 x}{5}+\frac {41}{25}}}{4+5 x}+\frac {3}{5 \left (4+5 x \right )}\)	\(32\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {-5-4 x-3 \sqrt {1-x^2}}{(4+5 x)^2 \sqrt {1-x^2}} \, dx=\frac {25 \, x + 20 \, \sqrt {-x^{2} + 1} + 32}{20 \, {\left (5 \, x + 4\right )}} \]

[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)

Sympy [F]

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

[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)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {-5-4 x-3 \sqrt {1-x^2}}{(4+5 x)^2 \sqrt {1-x^2}} \, dx=\frac {5 \, \sqrt {x + 1} \sqrt {-x + 1} + 3}{5 \, {\left (5 \, x + 4\right )}} \]

[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)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74 \[ \int \frac {-5-4 x-3 \sqrt {1-x^2}}{(4+5 x)^2 \sqrt {1-x^2}} \, dx=\frac {\sqrt {\frac {8}{5 \, x + 4} + \frac {9}{{\left (5 \, x + 4\right )}^{2}} - 1}}{5 \, \mathrm {sgn}\left (\frac {1}{5 \, x + 4}\right )} + \frac {3}{5 \, {\left (5 \, x + 4\right )}} - \frac {1}{5} i \, \mathrm {sgn}\left (\frac {1}{5 \, x + 4}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.61 \[ \int \frac {-5-4 x-3 \sqrt {1-x^2}}{(4+5 x)^2 \sqrt {1-x^2}} \, dx=\frac {\sqrt {1-x^2}+\frac {3}{5}}{5\,x+4} \]

[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)