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

Optimal. Leaf size=40 \[ -\frac{1}{2} \sqrt{1-x^2} (x+1)-\frac{3 \sqrt{1-x^2}}{2}+\frac{3}{2} \sin ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0122981, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {671, 641, 216} \[ -\frac{1}{2} \sqrt{1-x^2} (x+1)-\frac{3 \sqrt{1-x^2}}{2}+\frac{3}{2} \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[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 671

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
]

Rule 641

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 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]

Rubi steps

\begin{align*} \int \frac{(1+x)^2}{\sqrt{1-x^2}} \, dx &=-\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]  time = 0.0150907, size = 25, normalized size = 0.62 \[ \frac{1}{2} \left (3 \sin ^{-1}(x)-(x+4) \sqrt{1-x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)^2/Sqrt[1 - x^2],x]

[Out]

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

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Maple [A]  time = 0.057, size = 29, normalized size = 0.7 \begin{align*} -{\frac{x}{2}\sqrt{-{x}^{2}+1}}+{\frac{3\,\arcsin \left ( x \right ) }{2}}-2\,\sqrt{-{x}^{2}+1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.50275, size = 38, normalized size = 0.95 \begin{align*} -\frac{1}{2} \, \sqrt{-x^{2} + 1} x - 2 \, \sqrt{-x^{2} + 1} + \frac{3}{2} \, \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.02853, size = 86, normalized size = 2.15 \begin{align*} -\frac{1}{2} \, \sqrt{-x^{2} + 1}{\left (x + 4\right )} - 3 \, \arctan \left (\frac{\sqrt{-x^{2} + 1} - 1}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.208012, size = 27, normalized size = 0.68 \begin{align*} - \frac{x \sqrt{1 - x^{2}}}{2} - 2 \sqrt{1 - x^{2}} + \frac{3 \operatorname{asin}{\left (x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.10593, size = 26, normalized size = 0.65 \begin{align*} -\frac{1}{2} \, \sqrt{-x^{2} + 1}{\left (x + 4\right )} + \frac{3}{2} \, \arcsin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)