3.8.49 \(\int \frac {x}{(1+x) \sqrt {-1+\frac {2}{1+x}}} \, dx\) [749]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {526, 528, 382, 75} \begin {gather*} -\left ((x+1) \sqrt {\frac {2}{x+1}-1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 + x)*Sqrt[-1 + 2/(1 + x)])

Rule 75

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]

Rule 382

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 526

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 528

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])

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 19, normalized size = 1.06 \begin {gather*} \frac {-1+x}{\sqrt {\frac {1-x}{1+x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1 + x)/Sqrt[(1 - x)/(1 + x)]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(36\) vs. \(2(16)=32\).
time = 0.41, size = 37, normalized size = 2.06

method	result	size
gosper	\(\frac {-1+x}{\sqrt {-\frac {-1+x}{1+x}}}\)	\(17\)
risch	\(\frac {-1+x}{\sqrt {-\frac {-1+x}{1+x}}}\)	\(17\)
trager	\(\left (-1-x \right ) \sqrt {-\frac {-1+x}{1+x}}\)	\(19\)
default	\(-\frac {\sqrt {-\frac {-1+x}{1+x}}\, \left (1+x \right ) \sqrt {-x^{2}+1}}{\sqrt {-\left (1+x \right ) \left (-1+x \right )}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 16, normalized size = 0.89 \begin {gather*} \frac {\sqrt {x + 1} {\left (x - 1\right )}}{\sqrt {-x + 1}} \end {gather*}

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 = 0.35, size = 17, normalized size = 0.94 \begin {gather*} -{\left (x + 1\right )} \sqrt {-\frac {x - 1}{x + 1}} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {- \frac {x - 1}{x + 1}} \left (x + 1\right )}\, dx \end {gather*}

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 = 4.41, size = 17, normalized size = 0.94 \begin {gather*} -\frac {\sqrt {-x^{2} + 1}}{\mathrm {sgn}\left (x + 1\right )} \end {gather*}

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]

-sqrt(-x^2 + 1)/sgn(x + 1)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int \frac {x}{\left (x+1\right )\,\sqrt {\frac {2}{x+1}-1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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