3.1014 \(\int \frac{1}{\sqrt{-1+\sqrt{x}} \sqrt{1+\sqrt{x}} \sqrt{x}} \, dx\)

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

[Out]

2*ArcCosh[Sqrt[x]]

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Rubi [A]  time = 0.0106923, antiderivative size = 8, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {330, 52} \[ 2 \cosh ^{-1}\left (\sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 0.0061587, size = 26, normalized size = 3.25 \[ 4 \tanh ^{-1}\left (\frac{\sqrt{\sqrt{x}-1}}{\sqrt{\sqrt{x}+1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.002, size = 40, normalized size = 5. \begin{align*} 2\,{\frac{\sqrt{ \left ( 1+\sqrt{x} \right ) \left ( -1+\sqrt{x} \right ) }\ln \left ( \sqrt{x}+\sqrt{-1+x} \right ) }{\sqrt{-1+\sqrt{x}}\sqrt{1+\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.917757, size = 22, normalized size = 2.75 \begin{align*} 2 \, \log \left (2 \, \sqrt{x - 1} + 2 \, \sqrt{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(2*sqrt(x - 1) + 2*sqrt(x))

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Fricas [B]  time = 0.969394, size = 85, normalized size = 10.62 \begin{align*} -\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(1/x^(1/2)/(-1+x^(1/2))^(1/2)/(1+x^(1/2))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: RuntimeError