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

Optimal. Leaf size=32 \[ -\sqrt {1-x^2}-\tanh ^{-1}\left (\sqrt {1-x^2}\right )+2 \sin ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1809, 844, 216, 266, 63, 206} \[ -\sqrt {1-x^2}-\tanh ^{-1}\left (\sqrt {1-x^2}\right )+2 \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)^2/(x*Sqrt[1 - x^2]),x]

[Out]

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

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])

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 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 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 32, normalized size = 1.00 \[ -\sqrt {1-x^2}-\tanh ^{-1}\left (\sqrt {1-x^2}\right )+2 \sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.86, size = 46, normalized size = 1.44 \[ -\sqrt {-x^{2} + 1} - 4 \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) + \log \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 34, normalized size = 1.06 \[ -\sqrt {-x^{2} + 1} + 2 \, \arcsin \relax (x) + \log \left (-\frac {\sqrt {-x^{2} + 1} - 1}{{\left | x \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 29, normalized size = 0.91 \[ -\arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )+2 \arcsin \relax (x )-\sqrt {-x^{2}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.98, size = 41, normalized size = 1.28 \[ -\sqrt {-x^{2} + 1} + 2 \, \arcsin \relax (x) - \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 32, normalized size = 1.00 \[ 2\,\mathrm {asin}\relax (x)+\ln \left (\sqrt {\frac {1}{x^2}-1}-\sqrt {\frac {1}{x^2}}\right )-\sqrt {1-x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)^2/(x*(1 - x^2)^(1/2)),x)

[Out]

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

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sympy [A]  time = 6.30, size = 31, normalized size = 0.97 \[ - \sqrt {1 - x^{2}} + \begin {cases} - \operatorname {acosh}{\left (\frac {1}{x} \right )} & \text {for}\: \frac {1}{\left |{x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{x} \right )} & \text {otherwise} \end {cases} + 2 \operatorname {asin}{\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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