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

Optimal. Leaf size=30 \[ \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )-\sin ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {402, 216, 377, 203} \[ \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )-\sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - x^2]/(1 + x^2),x]

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 30, normalized size = 1.00 \[ \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )-\sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - x^2]/(1 + x^2),x]

[Out]

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

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fricas [A]  time = 0.64, size = 42, normalized size = 1.40 \[ -\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-x^{2} + 1}}{2 \, x}\right ) + 2 \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.61, size = 95, normalized size = 3.17 \[ -\frac {1}{2} \, \pi \mathrm {sgn}\relax (x) + \frac {1}{2} \, \sqrt {2} {\left (\pi \mathrm {sgn}\relax (x) + 2 \, \arctan \left (-\frac {\sqrt {2} x {\left (\frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{4 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right )\right )} - \arctan \left (-\frac {x {\left (\frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*pi*sgn(x) + 1/2*sqrt(2)*(pi*sgn(x) + 2*arctan(-1/4*sqrt(2)*x*((sqrt(-x^2 + 1) - 1)^2/x^2 - 1)/(sqrt(-x^2
+ 1) - 1))) - arctan(-1/2*x*((sqrt(-x^2 + 1) - 1)^2/x^2 - 1)/(sqrt(-x^2 + 1) - 1))

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maple [A]  time = 0.02, size = 33, normalized size = 1.10 \[ -\arcsin \relax (x )-\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-x^{2}+1}\, x}{x^{2}-1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-x^2 + 1)/(x^2 + 1), x)

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mupad [B]  time = 0.39, size = 83, normalized size = 2.77 \[ -\mathrm {asin}\relax (x)+\frac {\sqrt {2}\,\ln \left (\frac {\frac {\sqrt {2}\,\left (-1+x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\sqrt {1-x^2}\,1{}\mathrm {i}}{x-\mathrm {i}}\right )\,1{}\mathrm {i}}{2}-\frac {\sqrt {2}\,\ln \left (\frac {\frac {\sqrt {2}\,\left (1+x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\sqrt {1-x^2}\,1{}\mathrm {i}}{x+1{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - x^2)^(1/2)/(x^2 + 1),x)

[Out]

(2^(1/2)*log(((2^(1/2)*(x*1i - 1)*1i)/2 - (1 - x^2)^(1/2)*1i)/(x - 1i))*1i)/2 - asin(x) - (2^(1/2)*log(((2^(1/
2)*(x*1i + 1)*1i)/2 + (1 - x^2)^(1/2)*1i)/(x + 1i))*1i)/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- \left (x - 1\right ) \left (x + 1\right )}}{x^{2} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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