3.1411 \(\int \frac {\sqrt {-1+x}}{(1+x)^2} \, dx\)

Optimal. Leaf size=35 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {x-1}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {\sqrt {x-1}}{x+1} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {47, 63, 203} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {x-1}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {\sqrt {x-1}}{x+1} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + x]/(1 + x)^2,x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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

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Mathematica [A]  time = 0.03, size = 51, normalized size = 1.46 \[ \frac {-2 x-\sqrt {2-2 x} (x+1) \tanh ^{-1}\left (\frac {\sqrt {1-x}}{\sqrt {2}}\right )+2}{2 \sqrt {x-1} (x+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 + x]/(1 + x)^2,x]

[Out]

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

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fricas [A]  time = 0.44, size = 33, normalized size = 0.94 \[ \frac {\sqrt {2} {\left (x + 1\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {x - 1}\right ) - 2 \, \sqrt {x - 1}}{2 \, {\left (x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.96, size = 29, normalized size = 0.83 \[ \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {x - 1}\right ) - \frac {\sqrt {x - 1}}{x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 30, normalized size = 0.86 \[ \frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {x -1}\, \sqrt {2}}{2}\right )}{2}-\frac {\sqrt {x -1}}{x +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.97, size = 29, normalized size = 0.83 \[ \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {x - 1}\right ) - \frac {\sqrt {x - 1}}{x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.06, size = 29, normalized size = 0.83 \[ \frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {x-1}}{2}\right )}{2}-\frac {\sqrt {x-1}}{x+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 1)^(1/2)/(x + 1)^2,x)

[Out]

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

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sympy [A]  time = 1.50, size = 104, normalized size = 2.97 \[ \begin {cases} \frac {\sqrt {2} i \operatorname {acosh}{\left (\frac {\sqrt {2}}{\sqrt {x + 1}} \right )}}{2} + \frac {i}{\sqrt {-1 + \frac {2}{x + 1}} \sqrt {x + 1}} - \frac {2 i}{\sqrt {-1 + \frac {2}{x + 1}} \left (x + 1\right )^{\frac {3}{2}}} & \text {for}\: \frac {2}{\left |{x + 1}\right |} > 1 \\- \frac {\sqrt {1 - \frac {2}{x + 1}}}{\sqrt {x + 1}} - \frac {\sqrt {2} \operatorname {asin}{\left (\frac {\sqrt {2}}{\sqrt {x + 1}} \right )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((sqrt(2)*I*acosh(sqrt(2)/sqrt(x + 1))/2 + I/(sqrt(-1 + 2/(x + 1))*sqrt(x + 1)) - 2*I/(sqrt(-1 + 2/(x
+ 1))*(x + 1)**(3/2)), 2/Abs(x + 1) > 1), (-sqrt(1 - 2/(x + 1))/sqrt(x + 1) - sqrt(2)*asin(sqrt(2)/sqrt(x + 1
))/2, True))

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