3.155 \(\int \frac{\sqrt{2-x-x^2}}{x^2} \, dx\)

Optimal. Leaf size=68 \[ -\frac{\sqrt{-x^2-x+2}}{x}+\frac{\tanh ^{-1}\left (\frac{4-x}{2 \sqrt{2} \sqrt{-x^2-x+2}}\right )}{2 \sqrt{2}}+\sin ^{-1}\left (\frac{1}{3} (-2 x-1)\right ) \]

[Out]

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

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Rubi [A]  time = 0.0367591, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {732, 843, 619, 216, 724, 206} \[ -\frac{\sqrt{-x^2-x+2}}{x}+\frac{\tanh ^{-1}\left (\frac{4-x}{2 \sqrt{2} \sqrt{-x^2-x+2}}\right )}{2 \sqrt{2}}+\sin ^{-1}\left (\frac{1}{3} (-2 x-1)\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 - x - x^2]/x^2,x]

[Out]

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

Rule 732

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ
[2*c*d - b*e, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 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 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

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{\sqrt{2-x-x^2}}{x^2} \, dx &=-\frac{\sqrt{2-x-x^2}}{x}+\frac{1}{2} \int \frac{-1-2 x}{x \sqrt{2-x-x^2}} \, dx\\ &=-\frac{\sqrt{2-x-x^2}}{x}-\frac{1}{2} \int \frac{1}{x \sqrt{2-x-x^2}} \, dx-\int \frac{1}{\sqrt{2-x-x^2}} \, dx\\ &=-\frac{\sqrt{2-x-x^2}}{x}+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{9}}} \, dx,x,-1-2 x\right )+\operatorname{Subst}\left (\int \frac{1}{8-x^2} \, dx,x,\frac{4-x}{\sqrt{2-x-x^2}}\right )\\ &=-\frac{\sqrt{2-x-x^2}}{x}+\sin ^{-1}\left (\frac{1}{3} (-1-2 x)\right )+\frac{\tanh ^{-1}\left (\frac{4-x}{2 \sqrt{2} \sqrt{2-x-x^2}}\right )}{2 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0274292, size = 68, normalized size = 1. \[ -\frac{\sqrt{-x^2-x+2}}{x}+\frac{\tanh ^{-1}\left (\frac{4-x}{2 \sqrt{2} \sqrt{-x^2-x+2}}\right )}{2 \sqrt{2}}+\sin ^{-1}\left (\frac{1}{3} (-2 x-1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 - x - x^2]/x^2,x]

[Out]

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

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Maple [A]  time = 0.005, size = 88, normalized size = 1.3 \begin{align*} -{\frac{1}{2\,x} \left ( -{x}^{2}-x+2 \right ) ^{{\frac{3}{2}}}}-{\frac{1}{4}\sqrt{-{x}^{2}-x+2}}-\arcsin \left ({\frac{1}{3}}+{\frac{2\,x}{3}} \right ) +{\frac{\sqrt{2}}{4}{\it Artanh} \left ({\frac{ \left ( 4-x \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{-{x}^{2}-x+2}}}} \right ) }+{\frac{-1-2\,x}{4}\sqrt{-{x}^{2}-x+2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/x*(-x^2-x+2)^(3/2)-1/4*(-x^2-x+2)^(1/2)-arcsin(1/3+2/3*x)+1/4*arctanh(1/4*(4-x)*2^(1/2)/(-x^2-x+2)^(1/2))
*2^(1/2)+1/4*(-1-2*x)*(-x^2-x+2)^(1/2)

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Maxima [A]  time = 1.45172, size = 80, normalized size = 1.18 \begin{align*} \frac{1}{4} \, \sqrt{2} \log \left (\frac{2 \, \sqrt{2} \sqrt{-x^{2} - x + 2}}{{\left | x \right |}} + \frac{4}{{\left | x \right |}} - 1\right ) - \frac{\sqrt{-x^{2} - x + 2}}{x} + \arcsin \left (-\frac{2}{3} \, x - \frac{1}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*log(2*sqrt(2)*sqrt(-x^2 - x + 2)/abs(x) + 4/abs(x) - 1) - sqrt(-x^2 - x + 2)/x + arcsin(-2/3*x - 1
/3)

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Fricas [A]  time = 1.12746, size = 232, normalized size = 3.41 \begin{align*} \frac{\sqrt{2} x \log \left (-\frac{4 \, \sqrt{2} \sqrt{-x^{2} - x + 2}{\left (x - 4\right )} + 7 \, x^{2} + 16 \, x - 32}{x^{2}}\right ) + 8 \, x \arctan \left (\frac{\sqrt{-x^{2} - x + 2}{\left (2 \, x + 1\right )}}{2 \,{\left (x^{2} + x - 2\right )}}\right ) - 8 \, \sqrt{-x^{2} - x + 2}}{8 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(sqrt(2)*x*log(-(4*sqrt(2)*sqrt(-x^2 - x + 2)*(x - 4) + 7*x^2 + 16*x - 32)/x^2) + 8*x*arctan(1/2*sqrt(-x^2
- x + 2)*(2*x + 1)/(x^2 + x - 2)) - 8*sqrt(-x^2 - x + 2))/x

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- \left (x - 1\right ) \left (x + 2\right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-(x - 1)*(x + 2))/x**2, x)

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Giac [B]  time = 1.13398, size = 227, normalized size = 3.34 \begin{align*} -\frac{1}{4} \, \sqrt{2} \log \left (\frac{{\left | -4 \, \sqrt{2} + \frac{2 \,{\left (2 \, \sqrt{-x^{2} - x + 2} - 3\right )}}{2 \, x + 1} + 6 \right |}}{{\left | 4 \, \sqrt{2} + \frac{2 \,{\left (2 \, \sqrt{-x^{2} - x + 2} - 3\right )}}{2 \, x + 1} + 6 \right |}}\right ) + \frac{6 \,{\left (\frac{3 \,{\left (2 \, \sqrt{-x^{2} - x + 2} - 3\right )}}{2 \, x + 1} + 1\right )}}{\frac{6 \,{\left (2 \, \sqrt{-x^{2} - x + 2} - 3\right )}}{2 \, x + 1} + \frac{{\left (2 \, \sqrt{-x^{2} - x + 2} - 3\right )}^{2}}{{\left (2 \, x + 1\right )}^{2}} + 1} - \arcsin \left (\frac{2}{3} \, x + \frac{1}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*log(abs(-4*sqrt(2) + 2*(2*sqrt(-x^2 - x + 2) - 3)/(2*x + 1) + 6)/abs(4*sqrt(2) + 2*(2*sqrt(-x^2 -
x + 2) - 3)/(2*x + 1) + 6)) + 6*(3*(2*sqrt(-x^2 - x + 2) - 3)/(2*x + 1) + 1)/(6*(2*sqrt(-x^2 - x + 2) - 3)/(2
*x + 1) + (2*sqrt(-x^2 - x + 2) - 3)^2/(2*x + 1)^2 + 1) - arcsin(2/3*x + 1/3)