\(\int \frac {1}{\sqrt {1-x^2} (1+x^2)} \, dx\) [106]

Optimal result

Integrand size = 19, antiderivative size = 25 \[ \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )}{\sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {385, 209} \[ \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )}{\sqrt {2}} \]

[In]

Int[1/(Sqrt[1 - x^2]*(1 + x^2)),x]

[Out]

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

Rule 209

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

Rule 385

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1-x^2}}\right )}{\sqrt {2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.35 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96

method	result	size
pseudoelliptic	\(-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-x^{2}+1}}{2 x}\right )}{2}\)	\(24\)
default	\(-\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {-x^{2}+1}\, x}{x^{2}-1}\right )}{2}\)	\(28\)
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{2}+4 x \sqrt {-x^{2}+1}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )}{x^{2}+1}\right )}{4}\)	\(50\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-x^{2} + 1}}{2 \, x}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (20) = 40\).

Time = 0.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04 \[ \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx=\frac {1}{4} \, \sqrt {2} {\left (\pi \mathrm {sgn}\left (x\right ) + 2 \, \arctan \left (-\frac {\sqrt {2} x {\left (\frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{4 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right )\right )} \]

[In]

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

[Out]

1/4*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)))

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.16 \[ \int \frac {1}{\sqrt {1-x^2} \left (1+x^2\right )} \, dx=\frac {\sqrt {2}\,\ln \left (\frac {\frac {\sqrt {2}\,\left (-1+x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\sqrt {1-x^2}\,1{}\mathrm {i}}{x-\mathrm {i}}\right )\,1{}\mathrm {i}}{4}-\frac {\sqrt {2}\,\ln \left (\frac {\frac {\sqrt {2}\,\left (1+x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\sqrt {1-x^2}\,1{}\mathrm {i}}{x+1{}\mathrm {i}}\right )\,1{}\mathrm {i}}{4} \]

[In]

int(1/((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)/4 - (2^(1/2)*log(((2^(1/2)*(x*1i +
1)*1i)/2 + (1 - x^2)^(1/2)*1i)/(x + 1i))*1i)/4