3.1.10 \(\int \frac {1+2 x}{(-1+x^2) \sqrt {-1+x+x^2}} \, dx\) [10]

Optimal. Leaf size=47 \[ -\frac {1}{2} \tan ^{-1}\left (\frac {3+x}{2 \sqrt {-1+x+x^2}}\right )+\frac {3}{2} \tanh ^{-1}\left (\frac {1-3 x}{2 \sqrt {-1+x+x^2}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {1047, 738, 212, 210} \begin {gather*} \frac {3}{2} \tanh ^{-1}\left (\frac {1-3 x}{2 \sqrt {x^2+x-1}}\right )-\frac {1}{2} \text {ArcTan}\left (\frac {x+3}{2 \sqrt {x^2+x-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 1047

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

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 37, normalized size = 0.79 \begin {gather*} -\tan ^{-1}\left (1+x-\sqrt {-1+x+x^2}\right )-3 \tanh ^{-1}\left (1-x+\sqrt {-1+x+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.17, size = 46, normalized size = 0.98

method	result	size
default	\(-\frac {3 \arctanh \left (\frac {-1+3 x}{2 \sqrt {\left (-1+x \right )^{2}-2+3 x}}\right )}{2}+\frac {\arctan \left (\frac {-3-x}{2 \sqrt {\left (1+x \right )^{2}-2-x}}\right )}{2}\)	\(46\)
trager	\(-\frac {3 \ln \left (-\frac {2 \sqrt {x^{2}+x -1}-1+3 x}{-1+x}\right )}{2}+\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {x \RootOf \left (\textit {\_Z}^{2}+1\right )+2 \sqrt {x^{2}+x -1}+3 \RootOf \left (\textit {\_Z}^{2}+1\right )}{1+x}\right )}{2}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.55, size = 65, normalized size = 1.38 \begin {gather*} -\frac {1}{2} \, \arcsin \left (\frac {2 \, \sqrt {5} x}{5 \, {\left | 2 \, x + 2 \right |}} + \frac {6 \, \sqrt {5}}{5 \, {\left | 2 \, x + 2 \right |}}\right ) - \frac {3}{2} \, \log \left (\frac {2 \, \sqrt {x^{2} + x - 1}}{{\left | 2 \, x - 2 \right |}} + \frac {2}{{\left | 2 \, x - 2 \right |}} + \frac {3}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arcsin(2/5*sqrt(5)*x/abs(2*x + 2) + 6/5*sqrt(5)/abs(2*x + 2)) - 3/2*log(2*sqrt(x^2 + x - 1)/abs(2*x - 2)
+ 2/abs(2*x - 2) + 3/2)

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Fricas [A]
time = 0.36, size = 46, normalized size = 0.98 \begin {gather*} \arctan \left (-x + \sqrt {x^{2} + x - 1} - 1\right ) - \frac {3}{2} \, \log \left (-x + \sqrt {x^{2} + x - 1} + 2\right ) + \frac {3}{2} \, \log \left (-x + \sqrt {x^{2} + x - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 4.31, size = 48, normalized size = 1.02 \begin {gather*} \arctan \left (-x + \sqrt {x^{2} + x - 1} - 1\right ) - \frac {3}{2} \, \log \left ({\left | -x + \sqrt {x^{2} + x - 1} + 2 \right |}\right ) + \frac {3}{2} \, \log \left ({\left | -x + \sqrt {x^{2} + x - 1} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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