3.3.64 \(\int \frac {1}{(-1+x^2) \sqrt {2 x+x^2}} \, dx\) [264]

Optimal. Leaf size=49 \[ -\frac {1}{2} \tan ^{-1}\left (\sqrt {2 x+x^2}\right )-\frac {\tanh ^{-1}\left (\frac {1+2 x}{\sqrt {3} \sqrt {2 x+x^2}}\right )}{2 \sqrt {3}} \]

[Out]

-1/2*arctan((x^2+2*x)^(1/2))-1/6*arctanh(1/3*(1+2*x)*3^(1/2)/(x^2+2*x)^(1/2))*3^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {998, 702, 210, 738, 212} \begin {gather*} -\frac {1}{2} \text {ArcTan}\left (\sqrt {x^2+2 x}\right )-\frac {\tanh ^{-1}\left (\frac {2 x+1}{\sqrt {3} \sqrt {x^2+2 x}}\right )}{2 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((-1 + x^2)*Sqrt[2*x + x^2]),x]

[Out]

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

Rule 210

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

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 702

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

Rule 738

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 998

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

Rubi steps

\begin {align*} \int \frac {1}{\left (-1+x^2\right ) \sqrt {2 x+x^2}} \, dx &=\frac {1}{2} \int \frac {1}{(-1-x) \sqrt {2 x+x^2}} \, dx+\frac {1}{2} \int \frac {1}{(-1+x) \sqrt {2 x+x^2}} \, dx\\ &=2 \text {Subst}\left (\int \frac {1}{-4-4 x^2} \, dx,x,\sqrt {2 x+x^2}\right )-\text {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {2+4 x}{\sqrt {2 x+x^2}}\right )\\ &=-\frac {1}{2} \tan ^{-1}\left (\sqrt {2 x+x^2}\right )-\frac {\tanh ^{-1}\left (\frac {2+4 x}{2 \sqrt {3} \sqrt {2 x+x^2}}\right )}{2 \sqrt {3}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 78, normalized size = 1.59 \begin {gather*} \frac {\sqrt {x} \sqrt {2+x} \left (3 \tan ^{-1}\left (1+x-\sqrt {x} \sqrt {2+x}\right )-\sqrt {3} \tanh ^{-1}\left (\frac {1-x+\sqrt {x} \sqrt {2+x}}{\sqrt {3}}\right )\right )}{3 \sqrt {x (2+x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*Sqrt[2 + x]*(3*ArcTan[1 + x - Sqrt[x]*Sqrt[2 + x]] - Sqrt[3]*ArcTanh[(1 - x + Sqrt[x]*Sqrt[2 + x])/Sq
rt[3]]))/(3*Sqrt[x*(2 + x)])

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Maple [A]
time = 0.15, size = 42, normalized size = 0.86

method	result	size
default	\(-\frac {\sqrt {3}\, \arctanh \left (\frac {\left (2+4 x \right ) \sqrt {3}}{6 \sqrt {\left (-1+x \right )^{2}-1+4 x}}\right )}{6}+\frac {\arctan \left (\frac {1}{\sqrt {\left (1+x \right )^{2}-1}}\right )}{2}\)	\(42\)
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (\frac {-2 \RootOf \left (\textit {\_Z}^{2}-3\right ) x +3 \sqrt {x^{2}+2 x}-\RootOf \left (\textit {\_Z}^{2}-3\right )}{-1+x}\right )}{6}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}+1\right )+\sqrt {x^{2}+2 x}}{1+x}\right )}{2}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*3^(1/2)*arctanh(1/6*(2+4*x)*3^(1/2)/((-1+x)^2-1+4*x)^(1/2))+1/2*arctan(1/((1+x)^2-1)^(1/2))

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Maxima [A]
time = 2.70, size = 54, normalized size = 1.10 \begin {gather*} -\frac {1}{6} \, \sqrt {3} \log \left (\frac {2 \, \sqrt {3} \sqrt {x^{2} + 2 \, x}}{{\left | 2 \, x - 2 \right |}} + \frac {6}{{\left | 2 \, x - 2 \right |}} + 2\right ) + \frac {1}{2} \, \arcsin \left (\frac {2}{{\left | 2 \, x + 2 \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.44, size = 62, normalized size = 1.27 \begin {gather*} \frac {1}{6} \, \sqrt {3} \log \left (-\frac {\sqrt {3} {\left (2 \, x + 1\right )} + \sqrt {x^{2} + 2 \, x} {\left (2 \, \sqrt {3} - 3\right )} - 4 \, x - 2}{x - 1}\right ) - \arctan \left (-x + \sqrt {x^{2} + 2 \, x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {x \left (x + 2\right )} \left (x - 1\right ) \left (x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.14, size = 71, normalized size = 1.45 \begin {gather*} \frac {1}{6} \, \sqrt {3} \log \left (\frac {{\left | -2 \, x - 2 \, \sqrt {3} + 2 \, \sqrt {x^{2} + 2 \, x} + 2 \right |}}{{\left | -2 \, x + 2 \, \sqrt {3} + 2 \, \sqrt {x^{2} + 2 \, x} + 2 \right |}}\right ) - \arctan \left (-x + \sqrt {x^{2} + 2 \, x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\sqrt {x^2+2\,x}\,\left (x^2-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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