3.1.59 \(\int \frac {(1+x)^2}{x^2 \sqrt {1-x^2}} \, dx\) [59]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 33, 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 = {1821, 858, 222, 272, 65, 212} \begin {gather*} \text {ArcSin}(x)-\frac {\sqrt {1-x^2}}{x}-2 \tanh ^{-1}\left (\sqrt {1-x^2}\right ) \end {gather*}

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 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 1821

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])

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)+\text {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {1-x^2}}{x}+\sin ^{-1}(x)-2 \text {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*}

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Mathematica [A]
time = 0.09, size = 57, normalized size = 1.73 \begin {gather*} -\frac {\sqrt {1-x^2}}{x}+2 \tan ^{-1}\left (\frac {x}{-1+\sqrt {1-x^2}}\right )-2 \log (x)+2 \log \left (-1+\sqrt {1-x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 30, normalized size = 0.91

method	result	size
default	\(\arcsin \left (x \right )-\frac {\sqrt {-x^{2}+1}}{x}-2 \arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )\)	\(30\)
risch	\(\frac {x^{2}-1}{x \sqrt {-x^{2}+1}}+\arcsin \left (x \right )-2 \arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )\)	\(34\)
meijerg	\(-\frac {\sqrt {-x^{2}+1}}{x}+\frac {-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{2}+1}}{2}\right )+\left (-2 \ln \left (2\right )+2 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }}{\sqrt {\pi }}+\arcsin \left (x \right )\)	\(59\)
trager	\(-\frac {\sqrt {-x^{2}+1}}{x}-\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (x \RootOf \left (\textit {\_Z}^{2}+1\right )+\sqrt {-x^{2}+1}\right )-2 \ln \left (\frac {\sqrt {-x^{2}+1}+1}{x}\right )\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 42, normalized size = 1.27 \begin {gather*} -\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 {gather*}

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 = 1.73, size = 53, normalized size = 1.61 \begin {gather*} -\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 {gather*}

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] Result contains complex when optimal does not.
time = 2.25, size = 51, normalized size = 1.55 \begin {gather*} \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 {gather*}

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 = 0.80, size = 55, normalized size = 1.67 \begin {gather*} \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 {gather*}

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))

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Mupad [B]
time = 0.08, size = 35, normalized size = 1.06 \begin {gather*} \mathrm {asin}\left (x\right )+2\,\ln \left (\sqrt {\frac {1}{x^2}-1}-\sqrt {\frac {1}{x^2}}\right )-\frac {\sqrt {1-x^2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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