3.1.58 \(\int \frac {(1+x)^2}{x \sqrt {1-x^2}} \, dx\) [58]

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

[Out]

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

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Rubi [A]
time = 0.04, 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 = {1823, 858, 222, 272, 65, 212} \begin {gather*} 2 \text {ArcSin}(x)-\sqrt {1-x^2}-\tanh ^{-1}\left (\sqrt {1-x^2}\right ) \end {gather*}

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 65

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 212

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

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 272

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 858

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 1823

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} \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,x^2\right )\\ &=-\sqrt {1-x^2}+2 \sin ^{-1}(x)-\text {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 = 0.09, size = 52, normalized size = 1.62 \begin {gather*} -\sqrt {1-x^2}+4 \tan ^{-1}\left (\frac {x}{-1+\sqrt {1-x^2}}\right )-\log (x)+\log \left (-1+\sqrt {1-x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 29, normalized size = 0.91

method	result	size
default	\(-\sqrt {-x^{2}+1}+2 \arcsin \left (x \right )-\arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )\)	\(29\)
trager	\(-\sqrt {-x^{2}+1}+\ln \left (\frac {\sqrt {-x^{2}+1}-1}{x}\right )-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (x \RootOf \left (\textit {\_Z}^{2}+1\right )+\sqrt {-x^{2}+1}\right )\)	\(56\)
meijerg	\(\frac {-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{2}+1}}{2}\right )+\left (-2 \ln \left (2\right )+2 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }}{2 \sqrt {\pi }}+2 \arcsin \left (x \right )-\frac {-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {-x^{2}+1}}{2 \sqrt {\pi }}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 41, normalized size = 1.28 \begin {gather*} -\sqrt {-x^{2} + 1} + 2 \, \arcsin \left (x\right ) - \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \end {gather*}

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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Fricas [A]
time = 1.96, size = 46, normalized size = 1.44 \begin {gather*} -\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 ) \end {gather*}

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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Sympy [A]
time = 3.02, size = 31, normalized size = 0.97 \begin {gather*} - \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}{\left (x \right )} \end {gather*}

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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Giac [A]
time = 0.88, size = 34, normalized size = 1.06 \begin {gather*} -\sqrt {-x^{2} + 1} + 2 \, \arcsin \left (x\right ) + \log \left (-\frac {\sqrt {-x^{2} + 1} - 1}{{\left | x \right |}}\right ) \end {gather*}

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

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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