3.9.26 \(\int \frac {1}{(-5-4 x) \sqrt {1-x^2}+3 (1-x^2)} \, dx\) [826]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {6874, 679, 222, 747, 858, 739, 212, 749} \begin {gather*} \frac {\sqrt {1-x^2}}{5 x+4}+\frac {3}{5 (5 x+4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-5 - 4*x)*Sqrt[1 - x^2] + 3*(1 - x^2))^(-1),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 222

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 679

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 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 747

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 749

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 858

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 6874

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

Rubi steps

\begin {align*} \int \frac {1}{(-5-4 x) \sqrt {1-x^2}+3 \left (1-x^2\right )} \, 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 = 0.29, size = 23, normalized size = 0.74 \begin {gather*} \frac {3+5 \sqrt {1-x^2}}{20+25 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3 + 5*Sqrt[1 - x^2])/(20 + 25*x)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(80\) vs. \(2(27)=54\).
time = 0.04, size = 81, normalized size = 2.61

method	result	size
trager	\(-\frac {3 x}{4 \left (5 x +4\right )}+\frac {\sqrt {-x^{2}+1}}{5 x +4}\)	\(29\)
default	\(\frac {3}{5 \left (5 x +4\right )}+\frac {5 \left (-\left (x +\frac {4}{5}\right )^{2}+\frac {8 x}{5}+\frac {41}{25}\right )^{\frac {3}{2}}}{9 \left (x +\frac {4}{5}\right )}+\frac {5 x \sqrt {-\left (x +\frac {4}{5}\right )^{2}+\frac {8 x}{5}+\frac {41}{25}}}{9}-\frac {\sqrt {-\left (1+x \right )^{2}+2+2 x}}{2}+\frac {\sqrt {-\left (-1+x \right )^{2}+2-2 x}}{18}\)	\(81\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.32, size = 25, normalized size = 0.81 \begin {gather*} \frac {25 \, x + 20 \, \sqrt {-x^{2} + 1} + 32}{20 \, {\left (5 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-3*x^2+3+(-5-4*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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {1}{3 x^{2} + 4 x \sqrt {1 - x^{2}} + 5 \sqrt {1 - x^{2}} - 3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-3*x**2+3+(-5-4*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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (27) = 54\).
time = 2.59, size = 68, normalized size = 2.19 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-3*x^2+3+(-5-4*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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Mupad [B]
time = 0.07, size = 19, normalized size = 0.61 \begin {gather*} \frac {\sqrt {1-x^2}+\frac {3}{5}}{5\,x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((1 - x^2)^(1/2) + 3/5)/(5*x + 4)

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