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

Optimal. Leaf size=33 \[ -\frac{\sqrt{1-x^2}}{x}-2 \tanh ^{-1}\left (\sqrt{1-x^2}\right )+\sin ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0625413, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1807, 844, 216, 266, 63, 206} \[ -\frac{\sqrt{1-x^2}}{x}-2 \tanh ^{-1}\left (\sqrt{1-x^2}\right )+\sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)^2/(x^2*Sqrt[1 - x^2]),x]

[Out]

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

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

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]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.0150139, size = 33, normalized size = 1. \[ -\frac{\sqrt{1-x^2}}{x}-2 \tanh ^{-1}\left (\sqrt{1-x^2}\right )+\sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.054, size = 30, normalized size = 0.9 \begin{align*} \arcsin \left ( x \right ) -2\,{\it Artanh} \left ({\frac{1}{\sqrt{-{x}^{2}+1}}} \right ) -{\frac{1}{x}\sqrt{-{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.4913, size = 57, normalized size = 1.73 \begin{align*} -\frac{\sqrt{-x^{2} + 1}}{x} + \arcsin \left (x\right ) - 2 \, \log \left (\frac{2 \, \sqrt{-x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.16486, size = 124, normalized size = 3.76 \begin{align*} -\frac{2 \, x \arctan \left (\frac{\sqrt{-x^{2} + 1} - 1}{x}\right ) - 2 \, x \log \left (\frac{\sqrt{-x^{2} + 1} - 1}{x}\right ) + \sqrt{-x^{2} + 1}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 3.96754, size = 51, normalized size = 1.55 \begin{align*} \begin{cases} - \frac{i \sqrt{x^{2} - 1}}{x} & \text{for}\: \left |{x^{2}}\right | > 1 \\- \frac{\sqrt{1 - x^{2}}}{x} & \text{otherwise} \end{cases} + 2 \left (\begin{cases} - \operatorname{acosh}{\left (\frac{1}{x} \right )} & \text{for}\: \frac{1}{\left |{x^{2}}\right |} > 1 \\i \operatorname{asin}{\left (\frac{1}{x} \right )} & \text{otherwise} \end{cases}\right ) + \operatorname{asin}{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*sqrt(x**2 - 1)/x, Abs(x**2) > 1), (-sqrt(1 - x**2)/x, True)) + 2*Piecewise((-acosh(1/x), 1/Abs(x
**2) > 1), (I*asin(1/x), True)) + asin(x)

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Giac [A]  time = 1.12314, size = 74, normalized size = 2.24 \begin{align*} \frac{x}{2 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}} - \frac{\sqrt{-x^{2} + 1} - 1}{2 \, x} + \arcsin \left (x\right ) + 2 \, \log \left (-\frac{\sqrt{-x^{2} + 1} - 1}{{\left | x \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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