\(\int \frac {(1+x)^2}{\sqrt {1-x^2}} \, dx\) [57]

Optimal result

Integrand size = 17, antiderivative size = 40 \[ \int \frac {(1+x)^2}{\sqrt {1-x^2}} \, dx=-\frac {3}{2} \sqrt {1-x^2}-\frac {1}{2} (1+x) \sqrt {1-x^2}+\frac {3 \arcsin (x)}{2} \]

[Out]

3/2*arcsin(x)-3/2*(-x^2+1)^(1/2)-1/2*(1+x)*(-x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {685, 655, 222} \[ \int \frac {(1+x)^2}{\sqrt {1-x^2}} \, dx=\frac {3 \arcsin (x)}{2}-\frac {1}{2} \sqrt {1-x^2} (x+1)-\frac {3 \sqrt {1-x^2}}{2} \]

[In]

Int[(1 + x)^2/Sqrt[1 - x^2],x]

[Out]

(-3*Sqrt[1 - x^2])/2 - ((1 + x)*Sqrt[1 - x^2])/2 + (3*ArcSin[x])/2

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 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 685

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
+ 1)/(c*(m + 2*p + 1))), x] + Dist[2*c*d*((m + p)/(c*(m + 2*p + 1))), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{2} (1+x) \sqrt {1-x^2}+\frac {3}{2} \int \frac {1+x}{\sqrt {1-x^2}} \, dx \\ & = -\frac {3}{2} \sqrt {1-x^2}-\frac {1}{2} (1+x) \sqrt {1-x^2}+\frac {3}{2} \int \frac {1}{\sqrt {1-x^2}} \, dx \\ & = -\frac {3}{2} \sqrt {1-x^2}-\frac {1}{2} (1+x) \sqrt {1-x^2}+\frac {3}{2} \sin ^{-1}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.02 \[ \int \frac {(1+x)^2}{\sqrt {1-x^2}} \, dx=\frac {1}{2} (-4-x) \sqrt {1-x^2}-3 \arctan \left (\frac {\sqrt {1-x^2}}{1+x}\right ) \]

[In]

Integrate[(1 + x)^2/Sqrt[1 - x^2],x]

[Out]

((-4 - x)*Sqrt[1 - x^2])/2 - 3*ArcTan[Sqrt[1 - x^2]/(1 + x)]

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.62

method	result	size
risch	\(\frac {\left (4+x \right ) \left (x^{2}-1\right )}{2 \sqrt {-x^{2}+1}}+\frac {3 \arcsin \left (x \right )}{2}\)	\(25\)
default	\(\frac {3 \arcsin \left (x \right )}{2}-\frac {x \sqrt {-x^{2}+1}}{2}-2 \sqrt {-x^{2}+1}\)	\(29\)
trager	\(\left (-2-\frac {x}{2}\right ) \sqrt {-x^{2}+1}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {-x^{2}+1}+x \right )}{2}\)	\(44\)
meijerg	\(\arcsin \left (x \right )-\frac {-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-x^{2}+1}}{\sqrt {\pi }}+\frac {i \left (i \sqrt {\pi }\, x \sqrt {-x^{2}+1}-i \sqrt {\pi }\, \arcsin \left (x \right )\right )}{2 \sqrt {\pi }}\)	\(60\)

[In]

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

[Out]

1/2*(4+x)*(x^2-1)/(-x^2+1)^(1/2)+3/2*arcsin(x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \frac {(1+x)^2}{\sqrt {1-x^2}} \, dx=-\frac {1}{2} \, \sqrt {-x^{2} + 1} {\left (x + 4\right )} - 3 \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \]

[In]

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

[Out]

-1/2*sqrt(-x^2 + 1)*(x + 4) - 3*arctan((sqrt(-x^2 + 1) - 1)/x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.68 \[ \int \frac {(1+x)^2}{\sqrt {1-x^2}} \, dx=- \frac {x \sqrt {1 - x^{2}}}{2} - 2 \sqrt {1 - x^{2}} + \frac {3 \operatorname {asin}{\left (x \right )}}{2} \]

[In]

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

[Out]

-x*sqrt(1 - x**2)/2 - 2*sqrt(1 - x**2) + 3*asin(x)/2

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.70 \[ \int \frac {(1+x)^2}{\sqrt {1-x^2}} \, dx=-\frac {1}{2} \, \sqrt {-x^{2} + 1} x - 2 \, \sqrt {-x^{2} + 1} + \frac {3}{2} \, \arcsin \left (x\right ) \]

[In]

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

[Out]

-1/2*sqrt(-x^2 + 1)*x - 2*sqrt(-x^2 + 1) + 3/2*arcsin(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.48 \[ \int \frac {(1+x)^2}{\sqrt {1-x^2}} \, dx=-\frac {1}{2} \, \sqrt {-x^{2} + 1} {\left (x + 4\right )} + \frac {3}{2} \, \arcsin \left (x\right ) \]

[In]

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

[Out]

-1/2*sqrt(-x^2 + 1)*(x + 4) + 3/2*arcsin(x)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.52 \[ \int \frac {(1+x)^2}{\sqrt {1-x^2}} \, dx=\frac {3\,\mathrm {asin}\left (x\right )}{2}-\left (\frac {x}{2}+2\right )\,\sqrt {1-x^2} \]

[In]

int((x + 1)^2/(1 - x^2)^(1/2),x)

[Out]

(3*asin(x))/2 - (x/2 + 2)*(1 - x^2)^(1/2)