3.1412 \(\int \frac{\sqrt{-1+x}}{(1+x)^3} \, dx\)

Optimal. Leaf size=56 \[ \frac{\sqrt{x-1}}{8 (x+1)}-\frac{\sqrt{x-1}}{2 (x+1)^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{x-1}}{\sqrt{2}}\right )}{8 \sqrt{2}} \]

[Out]

-Sqrt[-1 + x]/(2*(1 + x)^2) + Sqrt[-1 + x]/(8*(1 + x)) + ArcTan[Sqrt[-1 + x]/Sqrt[2]]/(8*Sqrt[2])

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Rubi [A]  time = 0.0130528, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {47, 51, 63, 203} \[ \frac{\sqrt{x-1}}{8 (x+1)}-\frac{\sqrt{x-1}}{2 (x+1)^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{x-1}}{\sqrt{2}}\right )}{8 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + x]/(1 + x)^3,x]

[Out]

-Sqrt[-1 + x]/(2*(1 + x)^2) + Sqrt[-1 + x]/(8*(1 + x)) + ArcTan[Sqrt[-1 + x]/Sqrt[2]]/(8*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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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)^3} \, dx &=-\frac{\sqrt{-1+x}}{2 (1+x)^2}+\frac{1}{4} \int \frac{1}{\sqrt{-1+x} (1+x)^2} \, dx\\ &=-\frac{\sqrt{-1+x}}{2 (1+x)^2}+\frac{\sqrt{-1+x}}{8 (1+x)}+\frac{1}{16} \int \frac{1}{\sqrt{-1+x} (1+x)} \, dx\\ &=-\frac{\sqrt{-1+x}}{2 (1+x)^2}+\frac{\sqrt{-1+x}}{8 (1+x)}+\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{2+x^2} \, dx,x,\sqrt{-1+x}\right )\\ &=-\frac{\sqrt{-1+x}}{2 (1+x)^2}+\frac{\sqrt{-1+x}}{8 (1+x)}+\frac{\tan ^{-1}\left (\frac{\sqrt{-1+x}}{\sqrt{2}}\right )}{8 \sqrt{2}}\\ \end{align*}

Mathematica [C]  time = 0.0052132, size = 28, normalized size = 0.5 \[ \frac{1}{12} (x-1)^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{1-x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 + x]/(1 + x)^3,x]

[Out]

((-1 + x)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, (1 - x)/2])/12

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Maple [A]  time = 0.007, size = 40, normalized size = 0.7 \begin{align*} 2\,{\frac{1/16\, \left ( -1+x \right ) ^{3/2}-1/8\,\sqrt{-1+x}}{ \left ( 1+x \right ) ^{2}}}+{\frac{\sqrt{2}}{16}\arctan \left ({\frac{\sqrt{2}}{2}\sqrt{-1+x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1+x)^(1/2)/(1+x)^3,x)

[Out]

2*(1/16*(-1+x)^(3/2)-1/8*(-1+x)^(1/2))/(1+x)^2+1/16*arctan(1/2*(-1+x)^(1/2)*2^(1/2))*2^(1/2)

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Maxima [A]  time = 1.43662, size = 58, normalized size = 1.04 \begin{align*} \frac{1}{16} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} \sqrt{x - 1}\right ) + \frac{{\left (x - 1\right )}^{\frac{3}{2}} - 2 \, \sqrt{x - 1}}{8 \,{\left ({\left (x - 1\right )}^{2} + 4 \, x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)^(1/2)/(1+x)^3,x, algorithm="maxima")

[Out]

1/16*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(x - 1)) + 1/8*((x - 1)^(3/2) - 2*sqrt(x - 1))/((x - 1)^2 + 4*x)

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Fricas [A]  time = 1.83783, size = 140, normalized size = 2.5 \begin{align*} \frac{\sqrt{2}{\left (x^{2} + 2 \, x + 1\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} \sqrt{x - 1}\right ) + 2 \, \sqrt{x - 1}{\left (x - 3\right )}}{16 \,{\left (x^{2} + 2 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)^(1/2)/(1+x)^3,x, algorithm="fricas")

[Out]

1/16*(sqrt(2)*(x^2 + 2*x + 1)*arctan(1/2*sqrt(2)*sqrt(x - 1)) + 2*sqrt(x - 1)*(x - 3))/(x^2 + 2*x + 1)

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Sympy [A]  time = 2.4935, size = 167, normalized size = 2.98 \begin{align*} \begin{cases} \frac{\sqrt{2} i \operatorname{acosh}{\left (\frac{\sqrt{2}}{\sqrt{x + 1}} \right )}}{16} - \frac{i}{8 \sqrt{-1 + \frac{2}{x + 1}} \sqrt{x + 1}} + \frac{3 i}{4 \sqrt{-1 + \frac{2}{x + 1}} \left (x + 1\right )^{\frac{3}{2}}} - \frac{i}{\sqrt{-1 + \frac{2}{x + 1}} \left (x + 1\right )^{\frac{5}{2}}} & \text{for}\: \frac{2}{\left |{x + 1}\right |} > 1 \\- \frac{\sqrt{2} \operatorname{asin}{\left (\frac{\sqrt{2}}{\sqrt{x + 1}} \right )}}{16} + \frac{1}{8 \sqrt{1 - \frac{2}{x + 1}} \sqrt{x + 1}} - \frac{3}{4 \sqrt{1 - \frac{2}{x + 1}} \left (x + 1\right )^{\frac{3}{2}}} + \frac{1}{\sqrt{1 - \frac{2}{x + 1}} \left (x + 1\right )^{\frac{5}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)**(1/2)/(1+x)**3,x)

[Out]

Piecewise((sqrt(2)*I*acosh(sqrt(2)/sqrt(x + 1))/16 - I/(8*sqrt(-1 + 2/(x + 1))*sqrt(x + 1)) + 3*I/(4*sqrt(-1 +
2/(x + 1))*(x + 1)**(3/2)) - I/(sqrt(-1 + 2/(x + 1))*(x + 1)**(5/2)), 2/Abs(x + 1) > 1), (-sqrt(2)*asin(sqrt(
2)/sqrt(x + 1))/16 + 1/(8*sqrt(1 - 2/(x + 1))*sqrt(x + 1)) - 3/(4*sqrt(1 - 2/(x + 1))*(x + 1)**(3/2)) + 1/(sqr
t(1 - 2/(x + 1))*(x + 1)**(5/2)), True))

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Giac [A]  time = 1.08067, size = 50, normalized size = 0.89 \begin{align*} \frac{1}{16} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2} \sqrt{x - 1}\right ) + \frac{{\left (x - 1\right )}^{\frac{3}{2}} - 2 \, \sqrt{x - 1}}{8 \,{\left (x + 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)^(1/2)/(1+x)^3,x, algorithm="giac")

[Out]

1/16*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(x - 1)) + 1/8*((x - 1)^(3/2) - 2*sqrt(x - 1))/(x + 1)^2