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

Optimal. Leaf size=18 \[ -(x+1) \sqrt{\frac{2}{x+1}-1} \]

[Out]

-((1 + x)*Sqrt[-1 + 2/(1 + x)])

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Rubi [A]  time = 0.0306565, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {512, 514, 375, 74} \[ -(x+1) \sqrt{\frac{2}{x+1}-1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 + x)*Sqrt[-1 + 2/(1 + x)])

Rule 512

Int[((a_.) + (b_.)*(v_)^(n_))^(p_.)*((c_.) + (d_.)*(v_)^(n_))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/Coefficien
t[v, x, 1]^(m + 1), Subst[Int[SimplifyIntegrand[(x - Coefficient[v, x, 0])^m*(a + b*x^n)^p*(c + d*x^n)^q, x],
x], x, v], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && LinearQ[v, x] && IntegerQ[m] && NeQ[v, x]

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 74

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

Rubi steps

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

Mathematica [A]  time = 0.0067792, size = 17, normalized size = 0.94 \[ \frac{x-1}{\sqrt{\frac{2}{x+1}-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1 + x)/Sqrt[-1 + 2/(1 + x)]

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Maple [A]  time = 0.002, size = 17, normalized size = 0.9 \begin{align*}{(x-1){\frac{1}{\sqrt{-{\frac{x-1}{1+x}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02281, size = 22, normalized size = 1.22 \begin{align*} \frac{\sqrt{x + 1}{\left (x - 1\right )}}{\sqrt{-x + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x + 1)*(x - 1)/sqrt(-x + 1)

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Fricas [A]  time = 1.74462, size = 45, normalized size = 2.5 \begin{align*} -{\left (x + 1\right )} \sqrt{-\frac{x - 1}{x + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x + 1)*sqrt(-(x - 1)/(x + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23699, size = 39, normalized size = 2.17 \begin{align*} -\frac{2}{\sqrt{-\frac{x - 1}{x + 1}} + \frac{1}{\sqrt{-\frac{x - 1}{x + 1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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