3.1013 \(\int \frac{\sqrt{x}}{\sqrt{-1+\sqrt{x}} \sqrt{1+\sqrt{x}}} \, dx\)

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

[Out]

Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]]*Sqrt[x] + ArcCosh[Sqrt[x]]

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Rubi [A]  time = 0.0206963, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {323, 330, 52} \[ \sqrt{\sqrt{x}-1} \sqrt{\sqrt{x}+1} \sqrt{x}+\cosh ^{-1}\left (\sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]]*Sqrt[x] + ArcCosh[Sqrt[x]]

Rule 323

Int[((c_.)*(x_))^(m_)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(2
*n - 1)*(c*x)^(m - 2*n + 1)*(a1 + b1*x^n)^(p + 1)*(a2 + b2*x^n)^(p + 1))/(b1*b2*(m + 2*n*p + 1)), x] - Dist[(a
1*a2*c^(2*n)*(m - 2*n + 1))/(b1*b2*(m + 2*n*p + 1)), Int[(c*x)^(m - 2*n)*(a1 + b1*x^n)^p*(a2 + b2*x^n)^p, x],
x] /; FreeQ[{a1, b1, a2, b2, c, p}, x] && EqQ[a2*b1 + a1*b2, 0] && IGtQ[2*n, 0] && GtQ[m, 2*n - 1] && NeQ[m +
2*n*p + 1, 0] && IntBinomialQ[a1*a2, b1*b2, c, 2*n, m, p, x]

Rule 330

Int[((c_.)*(x_))^(m_)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k =
Denominator[m]}, Dist[k/c, Subst[Int[x^(k*(m + 1) - 1)*(a1 + (b1*x^(k*n))/c^n)^p*(a2 + (b2*x^(k*n))/c^n)^p, x]
, x, (c*x)^(1/k)], x]] /; FreeQ[{a1, b1, a2, b2, c, p}, x] && EqQ[a2*b1 + a1*b2, 0] && IGtQ[2*n, 0] && Fractio
nQ[m] && IntBinomialQ[a1*a2, b1*b2, c, 2*n, m, p, x]

Rule 52

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ArcCosh[(b*x)/a]/b, x] /; FreeQ[{a,
b, c, d}, x] && EqQ[a + c, 0] && EqQ[b - d, 0] && GtQ[a, 0]

Rubi steps

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

Mathematica [A]  time = 0.010379, size = 55, normalized size = 1.57 \[ \sqrt{\sqrt{x}-1} \sqrt{\sqrt{x}+1} \sqrt{x}+2 \tanh ^{-1}\left (\frac{\sqrt{\sqrt{x}-1}}{\sqrt{\sqrt{x}+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 41, normalized size = 1.2 \begin{align*}{\sqrt{-1+\sqrt{x}}\sqrt{1+\sqrt{x}} \left ( \sqrt{x}\sqrt{-1+x}+\ln \left ( \sqrt{x}+\sqrt{-1+x} \right ) \right ){\frac{1}{\sqrt{-1+x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.923579, size = 32, normalized size = 0.91 \begin{align*} \sqrt{x - 1} \sqrt{x} + \log \left (2 \, \sqrt{x - 1} + 2 \, \sqrt{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x - 1)*sqrt(x) + log(2*sqrt(x - 1) + 2*sqrt(x))

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Fricas [A]  time = 0.966081, size = 151, normalized size = 4.31 \begin{align*} \sqrt{x} \sqrt{\sqrt{x} + 1} \sqrt{\sqrt{x} - 1} - \frac{1}{2} \, \log \left (2 \, \sqrt{x} \sqrt{\sqrt{x} + 1} \sqrt{\sqrt{x} - 1} - 2 \, x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 6.58692, size = 83, normalized size = 2.37 \begin{align*} \frac{{G_{6, 6}^{6, 2}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{4} & - \frac{1}{2}, - \frac{1}{2}, 0, 1 \\-1, - \frac{3}{4}, - \frac{1}{2}, - \frac{1}{4}, 0, 0 & \end{matrix} \middle |{\frac{1}{x}} \right )}}{2 \pi ^{\frac{3}{2}}} - \frac{i{G_{6, 6}^{2, 6}\left (\begin{matrix} - \frac{3}{2}, - \frac{5}{4}, -1, - \frac{3}{4}, - \frac{1}{2}, 1 & \\- \frac{5}{4}, - \frac{3}{4} & - \frac{3}{2}, -1, -1, 0 \end{matrix} \middle |{\frac{e^{2 i \pi }}{x}} \right )}}{2 \pi ^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

meijerg(((-3/4, -1/4), (-1/2, -1/2, 0, 1)), ((-1, -3/4, -1/2, -1/4, 0, 0), ()), 1/x)/(2*pi**(3/2)) - I*meijerg
(((-3/2, -5/4, -1, -3/4, -1/2, 1), ()), ((-5/4, -3/4), (-3/2, -1, -1, 0)), exp_polar(2*I*pi)/x)/(2*pi**(3/2))

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError