3.1007 \(\int \frac {\sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}}}{x^{5/2}} \, dx\)

Optimal. Leaf size=31 \[ \frac {2 \left (\sqrt {x}-1\right )^{3/2} \left (\sqrt {x}+1\right )^{3/2}}{3 x^{3/2}} \]

[Out]

2/3*(-1+x^(1/2))^(3/2)*(1+x^(1/2))^(3/2)/x^(3/2)

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Rubi [A]  time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {265} \[ \frac {2 \left (\sqrt {x}-1\right )^{3/2} \left (\sqrt {x}+1\right )^{3/2}}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]])/x^(5/2),x]

[Out]

(2*(-1 + Sqrt[x])^(3/2)*(1 + Sqrt[x])^(3/2))/(3*x^(3/2))

Rule 265

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

Rubi steps

\begin {align*} \int \frac {\sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}}}{x^{5/2}} \, dx &=\frac {2 \left (-1+\sqrt {x}\right )^{3/2} \left (1+\sqrt {x}\right )^{3/2}}{3 x^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 31, normalized size = 1.00 \[ \frac {2 \left (\sqrt {x}-1\right )^{3/2} \left (\sqrt {x}+1\right )^{3/2}}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]])/x^(5/2),x]

[Out]

(2*(-1 + Sqrt[x])^(3/2)*(1 + Sqrt[x])^(3/2))/(3*x^(3/2))

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fricas [A]  time = 0.70, size = 30, normalized size = 0.97 \[ \frac {2 \, {\left ({\left (x - 1\right )} \sqrt {x} \sqrt {\sqrt {x} + 1} \sqrt {\sqrt {x} - 1} + x^{2}\right )}}{3 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x^(1/2))^(1/2)*(1+x^(1/2))^(1/2)/x^(5/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.23, size = 48, normalized size = 1.55 \[ \frac {16 \, {\left (3 \, {\left (\sqrt {\sqrt {x} + 1} - \sqrt {\sqrt {x} - 1}\right )}^{8} + 16\right )}}{3 \, {\left ({\left (\sqrt {\sqrt {x} + 1} - \sqrt {\sqrt {x} - 1}\right )}^{4} + 4\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x^(1/2))^(1/2)*(1+x^(1/2))^(1/2)/x^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 23, normalized size = 0.74 \[ \frac {2 \sqrt {\sqrt {x}-1}\, \sqrt {\sqrt {x}+1}\, \left (x -1\right )}{3 x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(1/2)-1)^(1/2)*(x^(1/2)+1)^(1/2)/x^(5/2),x)

[Out]

2/3*(x^(1/2)-1)^(1/2)*(x^(1/2)+1)^(1/2)*(x-1)/x^(3/2)

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maxima [A]  time = 1.32, size = 10, normalized size = 0.32 \[ \frac {2 \, {\left (x - 1\right )}^{\frac {3}{2}}}{3 \, x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x^(1/2))^(1/2)*(1+x^(1/2))^(1/2)/x^(5/2),x, algorithm="maxima")

[Out]

2/3*(x - 1)^(3/2)/x^(3/2)

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mupad [B]  time = 5.26, size = 31, normalized size = 1.00 \[ \frac {\sqrt {\sqrt {x}-1}\,\left (\frac {2\,x\,\sqrt {\sqrt {x}+1}}{3}-\frac {2\,\sqrt {\sqrt {x}+1}}{3}\right )}{x^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^(1/2) - 1)^(1/2)*(x^(1/2) + 1)^(1/2))/x^(5/2),x)

[Out]

((x^(1/2) - 1)^(1/2)*((2*x*(x^(1/2) + 1)^(1/2))/3 - (2*(x^(1/2) + 1)^(1/2))/3))/x^(3/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\sqrt {x} - 1} \sqrt {\sqrt {x} + 1}}{x^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x**(1/2))**(1/2)*(1+x**(1/2))**(1/2)/x**(5/2),x)

[Out]

Integral(sqrt(sqrt(x) - 1)*sqrt(sqrt(x) + 1)/x**(5/2), x)

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