\(\int \frac {(\sqrt {1-x}+\sqrt {1+x})^2}{x^2} \, dx\) [424]

Optimal result

Integrand size = 23, antiderivative size = 26 \[ \int \frac {\left (\sqrt {1-x}+\sqrt {1+x}\right )^2}{x^2} \, dx=-\frac {2}{x}-\frac {2 \sqrt {1-x^2}}{x}-2 \arcsin (x) \]

[Out]

-2/x-2*arcsin(x)-2*(-x^2+1)^(1/2)/x

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6874, 283, 222} \[ \int \frac {\left (\sqrt {1-x}+\sqrt {1+x}\right )^2}{x^2} \, dx=-2 \arcsin (x)-\frac {2 \sqrt {1-x^2}}{x}-\frac {2}{x} \]

[In]

Int[(Sqrt[1 - x] + Sqrt[1 + x])^2/x^2,x]

[Out]

-2/x - (2*Sqrt[1 - x^2])/x - 2*ArcSin[x]

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 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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 {2}{x^2}+\frac {2 \sqrt {1-x^2}}{x^2}\right ) \, dx \\ & = -\frac {2}{x}+2 \int \frac {\sqrt {1-x^2}}{x^2} \, dx \\ & = -\frac {2}{x}-\frac {2 \sqrt {1-x^2}}{x}-2 \int \frac {1}{\sqrt {1-x^2}} \, dx \\ & = -\frac {2}{x}-\frac {2 \sqrt {1-x^2}}{x}-2 \sin ^{-1}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.88 \[ \int \frac {\left (\sqrt {1-x}+\sqrt {1+x}\right )^2}{x^2} \, dx=-\frac {2 \left (1+\sqrt {1-x^2}-4 x \arctan \left (\frac {\sqrt {1+x}}{\sqrt {2}-\sqrt {1-x}}\right )\right )}{x} \]

[In]

Integrate[(Sqrt[1 - x] + Sqrt[1 + x])^2/x^2,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(49\) vs. \(2(24)=48\).

Time = 0.92 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.92

method	result	size
default	\(-\frac {2}{x}+\frac {2 \left (-\arcsin \left (x \right ) x -\sqrt {-x^{2}+1}\right ) \sqrt {1-x}\, \sqrt {x +1}}{x \sqrt {-x^{2}+1}}\)	\(50\)

[In]

int(((1-x)^(1/2)+(x+1)^(1/2))^2/x^2,x,method=_RETURNVERBOSE)

[Out]

-2/x+2*(-arcsin(x)*x-(-x^2+1)^(1/2))*(1-x)^(1/2)*(x+1)^(1/2)/x/(-x^2+1)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69 \[ \int \frac {\left (\sqrt {1-x}+\sqrt {1+x}\right )^2}{x^2} \, dx=\frac {2 \, {\left (2 \, x \arctan \left (\frac {\sqrt {x + 1} \sqrt {-x + 1} - 1}{x}\right ) - \sqrt {x + 1} \sqrt {-x + 1} - 1\right )}}{x} \]

[In]

integrate(((1-x)^(1/2)+(1+x)^(1/2))^2/x^2,x, algorithm="fricas")

[Out]

2*(2*x*arctan((sqrt(x + 1)*sqrt(-x + 1) - 1)/x) - sqrt(x + 1)*sqrt(-x + 1) - 1)/x

Sympy [F]

\[ \int \frac {\left (\sqrt {1-x}+\sqrt {1+x}\right )^2}{x^2} \, dx=\int \frac {\left (\sqrt {1 - x} + \sqrt {x + 1}\right )^{2}}{x^{2}}\, dx \]

[In]

integrate(((1-x)**(1/2)+(1+x)**(1/2))**2/x**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {\left (\sqrt {1-x}+\sqrt {1+x}\right )^2}{x^2} \, dx=-\frac {2 \, \sqrt {-x^{2} + 1}}{x} - \frac {2}{x} - 2 \, \arcsin \left (x\right ) \]

[In]

integrate(((1-x)^(1/2)+(1+x)^(1/2))^2/x^2,x, algorithm="maxima")

[Out]

-2*sqrt(-x^2 + 1)/x - 2/x - 2*arcsin(x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (24) = 48\).

Time = 0.31 (sec) , antiderivative size = 149, normalized size of antiderivative = 5.73 \[ \int \frac {\left (\sqrt {1-x}+\sqrt {1+x}\right )^2}{x^2} \, dx=-2 \, \pi - \frac {8 \, {\left (\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} - \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}\right )}}{{\left (\frac {\sqrt {2} - \sqrt {-x + 1}}{\sqrt {x + 1}} - \frac {\sqrt {x + 1}}{\sqrt {2} - \sqrt {-x + 1}}\right )}^{2} - 4} - \frac {2}{x} - 4 \, \arctan \left (\frac {\sqrt {x + 1} {\left (\frac {{\left (\sqrt {2} - \sqrt {-x + 1}\right )}^{2}}{x + 1} - 1\right )}}{2 \, {\left (\sqrt {2} - \sqrt {-x + 1}\right )}}\right ) \]

[In]

integrate(((1-x)^(1/2)+(1+x)^(1/2))^2/x^2,x, algorithm="giac")

[Out]

-2*pi - 8*((sqrt(2) - sqrt(-x + 1))/sqrt(x + 1) - sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)))/(((sqrt(2) - sqrt(-x +
1))/sqrt(x + 1) - sqrt(x + 1)/(sqrt(2) - sqrt(-x + 1)))^2 - 4) - 2/x - 4*arctan(1/2*sqrt(x + 1)*((sqrt(2) - s
qrt(-x + 1))^2/(x + 1) - 1)/(sqrt(2) - sqrt(-x + 1)))

Mupad [B] (verification not implemented)

Time = 17.69 (sec) , antiderivative size = 120, normalized size of antiderivative = 4.62 \[ \int \frac {\left (\sqrt {1-x}+\sqrt {1+x}\right )^2}{x^2} \, dx=8\,\mathrm {atan}\left (\frac {\sqrt {1-x}-1}{\sqrt {x+1}-1}\right )-\frac {\frac {5\,{\left (\sqrt {1-x}-1\right )}^2}{2\,{\left (\sqrt {x+1}-1\right )}^2}-\frac {1}{2}}{\frac {\sqrt {1-x}-1}{\sqrt {x+1}-1}-\frac {{\left (\sqrt {1-x}-1\right )}^3}{{\left (\sqrt {x+1}-1\right )}^3}}-\frac {\sqrt {1-x}-1}{2\,\left (\sqrt {x+1}-1\right )}-\frac {2}{x} \]

[In]

int(((x + 1)^(1/2) + (1 - x)^(1/2))^2/x^2,x)

[Out]

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