\(\int \frac {x}{1-x^2+\sqrt {1-x^2}} \, dx\) [296]

Optimal result

Integrand size = 22, antiderivative size = 16 \[ \int \frac {x}{1-x^2+\sqrt {1-x^2}} \, dx=-\log \left (1+\sqrt {1-x^2}\right ) \]

[Out]

-ln(1+(-x^2+1)^(1/2))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2186, 31} \[ \int \frac {x}{1-x^2+\sqrt {1-x^2}} \, dx=-\log \left (\sqrt {1-x^2}+1\right ) \]

[In]

Int[x/(1 - x^2 + Sqrt[1 - x^2]),x]

[Out]

-Log[1 + Sqrt[1 - x^2]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2186

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {x}{1-x^2+\sqrt {1-x^2}} \, dx=-\log \left (1+\sqrt {1-x^2}\right ) \]

[In]

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

[Out]

-Log[1 + Sqrt[1 - x^2]]

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
trager	\(-\ln \left (1+\sqrt {-x^{2}+1}\right )\)	\(15\)
default	\(-\ln \left (x \right )+\sqrt {-x^{2}+1}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-x^{2}+1}}\right )-\frac {\sqrt {-\left (1+x \right )^{2}+2 x +2}}{2}-\frac {\sqrt {-\left (-1+x \right )^{2}-2 x +2}}{2}\)	\(59\)

[In]

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

[Out]

-ln(1+(-x^2+1)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {x}{1-x^2+\sqrt {1-x^2}} \, dx=-\log \left (x\right ) + \log \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \]

[In]

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

[Out]

-log(x) + log((sqrt(-x^2 + 1) - 1)/x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (12) = 24\).

Time = 1.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 3.00 \[ \int \frac {x}{1-x^2+\sqrt {1-x^2}} \, dx=\frac {\log {\left (2 \sqrt {1 - x^{2}} \right )}}{2} - \frac {\log {\left (2 \sqrt {1 - x^{2}} + 2 \right )}}{2} - \frac {\log {\left (2 x^{2} - 2 \sqrt {1 - x^{2}} - 2 \right )}}{2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {x}{1-x^2+\sqrt {1-x^2}} \, dx=-\log \left (\sqrt {-x^{2} + 1} + 1\right ) \]

[In]

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

[Out]

-log(sqrt(-x^2 + 1) + 1)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {x}{1-x^2+\sqrt {1-x^2}} \, dx=-\log \left (\sqrt {-x^{2} + 1} + 1\right ) \]

[In]

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

[Out]

-log(sqrt(-x^2 + 1) + 1)

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {x}{1-x^2+\sqrt {1-x^2}} \, dx=\ln \left (\sqrt {\frac {1}{x^2}-1}-\sqrt {\frac {1}{x^2}}\right )-\ln \left (x\right ) \]

[In]

int(x/((1 - x^2)^(1/2) - x^2 + 1),x)

[Out]

log((1/x^2 - 1)^(1/2) - (1/x^2)^(1/2)) - log(x)