3.733 \(\int \frac {\sqrt {-1+x}}{x^2 \sqrt {1+x}} \, dx\)

Optimal. Leaf size=36 \[ \tan ^{-1}\left (\sqrt {x-1} \sqrt {x+1}\right )-\frac {\sqrt {x-1} \sqrt {x+1}}{x} \]

[Out]

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

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Rubi [A]  time = 0.00, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {94, 92, 203} \[ \tan ^{-1}\left (\sqrt {x-1} \sqrt {x+1}\right )-\frac {\sqrt {x-1} \sqrt {x+1}}{x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + x]/(x^2*Sqrt[1 + x]),x]

[Out]

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

Rule 92

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

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 203

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 50, normalized size = 1.39 \[ \frac {\sqrt {\frac {x-1}{x+1}} \left (-x^2+\sqrt {x^2-1} x \tan ^{-1}\left (\sqrt {x^2-1}\right )+1\right )}{(x-1) x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 39, normalized size = 1.08 \[ \frac {2 \, x \arctan \left (\sqrt {x + 1} \sqrt {x - 1} - x\right ) - \sqrt {x + 1} \sqrt {x - 1} - x}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*x*arctan(sqrt(x + 1)*sqrt(x - 1) - x) - sqrt(x + 1)*sqrt(x - 1) - x)/x

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giac [A]  time = 0.33, size = 42, normalized size = 1.17 \[ -\frac {8}{{\left (\sqrt {x + 1} - \sqrt {x - 1}\right )}^{4} + 4} - 2 \, \arctan \left (\frac {1}{2} \, {\left (\sqrt {x + 1} - \sqrt {x - 1}\right )}^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/((sqrt(x + 1) - sqrt(x - 1))^4 + 4) - 2*arctan(1/2*(sqrt(x + 1) - sqrt(x - 1))^2)

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maple [A]  time = 0.02, size = 43, normalized size = 1.19 \[ \frac {\left (-x \arctan \left (\frac {1}{\sqrt {x^{2}-1}}\right )-\sqrt {x^{2}-1}\right ) \sqrt {x -1}\, \sqrt {x +1}}{\sqrt {x^{2}-1}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.98, size = 20, normalized size = 0.56 \[ -\frac {\sqrt {x^{2} - 1}}{x} - \arcsin \left (\frac {1}{{\left | x \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(x^2 - 1)/x - arcsin(1/abs(x))

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mupad [B]  time = 5.09, size = 138, normalized size = 3.83 \[ -\ln \left (\frac {{\left (\sqrt {x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {x+1}-1\right )}^2}+1\right )\,1{}\mathrm {i}+\ln \left (\frac {\sqrt {x-1}-\mathrm {i}}{\sqrt {x+1}-1}\right )\,1{}\mathrm {i}-\frac {\sqrt {x-1}-\mathrm {i}}{4\,\left (\sqrt {x+1}-1\right )}-\frac {\frac {5\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {x+1}-1\right )}^2}+1}{\frac {4\,{\left (\sqrt {x-1}-\mathrm {i}\right )}^3}{{\left (\sqrt {x+1}-1\right )}^3}+\frac {4\,\left (\sqrt {x-1}-\mathrm {i}\right )}{\sqrt {x+1}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(((x - 1)^(1/2) - 1i)/((x + 1)^(1/2) - 1))*1i - log(((x - 1)^(1/2) - 1i)^2/((x + 1)^(1/2) - 1)^2 + 1)*1i -
((x - 1)^(1/2) - 1i)/(4*((x + 1)^(1/2) - 1)) - ((5*((x - 1)^(1/2) - 1i)^2)/((x + 1)^(1/2) - 1)^2 + 1)/((4*((x
- 1)^(1/2) - 1i)^3)/((x + 1)^(1/2) - 1)^3 + (4*((x - 1)^(1/2) - 1i))/((x + 1)^(1/2) - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x - 1}}{x^{2} \sqrt {x + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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