3.24.69 \(\int \frac {\sqrt {-1-x+x^2}}{1+x} \, dx\) [2369]

Optimal. Leaf size=61 \[ \sqrt {-1-x+x^2}+\frac {3}{2} \tanh ^{-1}\left (\frac {1-2 x}{2 \sqrt {-1-x+x^2}}\right )+\tanh ^{-1}\left (\frac {1+3 x}{2 \sqrt {-1-x+x^2}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {748, 857, 635, 212, 738} \begin {gather*} \sqrt {x^2-x-1}+\frac {3}{2} \tanh ^{-1}\left (\frac {1-2 x}{2 \sqrt {x^2-x-1}}\right )+\tanh ^{-1}\left (\frac {3 x+1}{2 \sqrt {x^2-x-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 - x + x^2]/(1 + x),x]

[Out]

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

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 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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 748

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 + 2*p + 1))), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, 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] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 857

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]

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 57, normalized size = 0.93 \begin {gather*} \sqrt {-1-x+x^2}+2 \tanh ^{-1}\left (1+x-\sqrt {-1-x+x^2}\right )+\frac {3}{2} \log \left (1-2 x+2 \sqrt {-1-x+x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.90, size = 54, normalized size = 0.89

method	result	size
risch	\(\sqrt {x^{2}-x -1}-\frac {3 \ln \left (-\frac {1}{2}+x +\sqrt {x^{2}-x -1}\right )}{2}-\arctanh \left (\frac {-3 x -1}{2 \sqrt {\left (x +1\right )^{2}-3 x -2}}\right )\)	\(50\)
default	\(\sqrt {\left (x +1\right )^{2}-3 x -2}-\frac {3 \ln \left (-\frac {1}{2}+x +\sqrt {\left (x +1\right )^{2}-3 x -2}\right )}{2}-\arctanh \left (\frac {-3 x -1}{2 \sqrt {\left (x +1\right )^{2}-3 x -2}}\right )\)	\(54\)
trager	\(\sqrt {x^{2}-x -1}-\frac {\ln \left (\frac {32 \sqrt {x^{2}-x -1}\, x^{4}+32 x^{5}+96 \sqrt {x^{2}-x -1}\, x^{3}+80 x^{4}+78 \sqrt {x^{2}-x -1}\, x^{2}+10 x^{3}-16 \sqrt {x^{2}-x -1}\, x -125 x^{2}-38 \sqrt {x^{2}-x -1}-120 x -41}{\left (x +1\right )^{2}}\right )}{2}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.32, size = 62, normalized size = 1.02 \begin {gather*} \sqrt {x^{2} - x - 1} - \frac {3}{2} \, \log \left (2 \, x + 2 \, \sqrt {x^{2} - x - 1} - 1\right ) - \log \left (\frac {2 \, \sqrt {x^{2} - x - 1}}{{\left | x + 1 \right |}} + \frac {2}{{\left | x + 1 \right |}} - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.03, size = 64, normalized size = 1.05 \begin {gather*} \sqrt {x^{2} - x - 1} - \log \left (-x + \sqrt {x^{2} - x - 1}\right ) + \log \left (-x + \sqrt {x^{2} - x - 1} - 2\right ) + \frac {3}{2} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} - x - 1} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.14, size = 67, normalized size = 1.10 \begin {gather*} \sqrt {x^{2} - x - 1} - \log \left ({\left | -x + \sqrt {x^{2} - x - 1} \right |}\right ) + \log \left ({\left | -x + \sqrt {x^{2} - x - 1} - 2 \right |}\right ) + \frac {3}{2} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} - x - 1} + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - x - 1)^(1/2)/(x + 1),x)

[Out]

int((x^2 - x - 1)^(1/2)/(x + 1), x)

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