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

Optimal. Leaf size=31 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {5} x}{2 \sqrt {1-x^2}}\right )}{2 \sqrt {5}} \]

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {377, 203} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {5} x}{2 \sqrt {1-x^2}}\right )}{2 \sqrt {5}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[1 - x^2]*(4 + x^2)),x]

[Out]

ArcTan[(Sqrt[5]*x)/(2*Sqrt[1 - x^2])]/(2*Sqrt[5])

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 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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {1-x^2} \left (4+x^2\right )} \, dx &=\operatorname {Subst}\left (\int \frac {1}{4+5 x^2} \, dx,x,\frac {x}{\sqrt {1-x^2}}\right )\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {5} x}{2 \sqrt {1-x^2}}\right )}{2 \sqrt {5}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 31, normalized size = 1.00 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {5} x}{2 \sqrt {1-x^2}}\right )}{2 \sqrt {5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(Sqrt[5]*x)/(2*Sqrt[1 - x^2])]/(2*Sqrt[5])

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IntegrateAlgebraic [C]  time = 0.09, size = 55, normalized size = 1.77 \[ -\frac {i \tanh ^{-1}\left (\frac {x^2}{2 \sqrt {5}}+\frac {i \sqrt {1-x^2} x}{2 \sqrt {5}}+\frac {2}{\sqrt {5}}\right )}{2 \sqrt {5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2*I)*ArcTanh[2/Sqrt[5] + x^2/(2*Sqrt[5]) + ((I/2)*x*Sqrt[1 - x^2])/Sqrt[5]])/Sqrt[5]

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fricas [A]  time = 1.25, size = 23, normalized size = 0.74 \[ -\frac {1}{10} \, \sqrt {5} \arctan \left (\frac {2 \, \sqrt {5} \sqrt {-x^{2} + 1}}{5 \, x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^2+4)/(-x^2+1)^(1/2),x, algorithm="fricas")

[Out]

-1/10*sqrt(5)*arctan(2/5*sqrt(5)*sqrt(-x^2 + 1)/x)

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giac [B]  time = 0.63, size = 51, normalized size = 1.65 \[ \frac {1}{20} \, \sqrt {5} {\left (\pi \mathrm {sgn}\relax (x) + 2 \, \arctan \left (-\frac {\sqrt {5} x {\left (\frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{5 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*sqrt(5)*(pi*sgn(x) + 2*arctan(-1/5*sqrt(5)*x*((sqrt(-x^2 + 1) - 1)^2/x^2 - 1)/(sqrt(-x^2 + 1) - 1)))

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maple [A]  time = 0.29, size = 29, normalized size = 0.94

method	result	size
default	\(-\frac {\sqrt {5}\, \arctan \left (\frac {\sqrt {5}\, \sqrt {-x^{2}+1}\, x}{2 x^{2}-2}\right )}{10}\)	\(29\)
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}+5\right ) \ln \left (-\frac {9 \RootOf \left (\textit {\_Z}^{2}+5\right ) x^{2}-20 \sqrt {-x^{2}+1}\, x -4 \RootOf \left (\textit {\_Z}^{2}+5\right )}{x^{2}+4}\right )}{20}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*5^(1/2)*arctan(1/2*5^(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 \[ \int \frac {1}{{\left (x^{2} + 4\right )} \sqrt {-x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.52, size = 79, normalized size = 2.55 \[ \frac {\sqrt {5}\,\ln \left (\frac {\frac {\sqrt {5}\,\left (-1+x\,2{}\mathrm {i}\right )\,1{}\mathrm {i}}{5}-\sqrt {1-x^2}\,1{}\mathrm {i}}{x-2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{20}-\frac {\sqrt {5}\,\ln \left (\frac {\frac {\sqrt {5}\,\left (1+x\,2{}\mathrm {i}\right )\,1{}\mathrm {i}}{5}+\sqrt {1-x^2}\,1{}\mathrm {i}}{x+2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{20} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((1 - x^2)^(1/2)*(x^2 + 4)),x)

[Out]

(5^(1/2)*log(((5^(1/2)*(x*2i - 1)*1i)/5 - (1 - x^2)^(1/2)*1i)/(x - 2i))*1i)/20 - (5^(1/2)*log(((5^(1/2)*(x*2i
+ 1)*1i)/5 + (1 - x^2)^(1/2)*1i)/(x + 2i))*1i)/20

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 4\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(-(x - 1)*(x + 1))*(x**2 + 4)), x)

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