∫ 1_________√ −2 + bx√___

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Rubi [A]
time = 0.00, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {54} \begin {gather*} \frac {\cosh ^{-1}\left (\frac {b x}{2}\right )}{b} \end {gather*}

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]

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(25\) vs. \(2(11)=22\).
time = 0.00, size = 25, normalized size = 2.27 \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {2+b x}}{\sqrt {-2+b x}}\right )}{b} \end {gather*}

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

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 15.02, size = 69, normalized size = 6.27 \begin {gather*} \frac {I \text {meijerg}\left [\left \{\left \{-\frac {1}{2},-\frac {1}{4},0,\frac {1}{4},\frac {1}{2},1\right \},\left \{\right \}\right \},\left \{\left \{-\frac {1}{4},\frac {1}{4}\right \},\left \{-\frac {1}{2},0,0,0\right \}\right \},\frac {4}{b^2 x^2}\right ]+\text {meijerg}\left [\left \{\left \{\frac {1}{4},\frac {3}{4}\right \},\left \{\frac {1}{2},\frac {1}{2},1,1\right \}\right \},\left \{\left \{0,\frac {1}{4},\frac {1}{2},\frac {3}{4},1,0\right \},\left \{\right \}\right \},\frac {4 \text {exp\_polar}\left [2 I \text {Pi}\right ]}{b^2 x^2}\right ]}{4 \text {Pi}^{\frac {3}{2}} b} \end {gather*}

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

(I meijerg[{{-1 / 2, -1 / 4, 0, 1 / 4, 1 / 2, 1}, {}}, {{-1 / 4, 1 / 4}, {-1 / 2, 0, 0, 0}}, 4 / (b ^ 2 x ^ 2)

] + meijerg[{{1 / 4, 3 / 4}, {1 / 2, 1 / 2, 1, 1}}, {{0, 1 / 4, 1 / 2, 3 / 4, 1, 0}, {}}, 4 exp_polar[2 I Pi]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(56\) vs. \(2(9)=18\).
time = 0.15, size = 57, normalized size = 5.18


method	result	size

default	\(\frac {\sqrt {\left (b x -2\right ) \left (b x +2\right )}\, \ln \left (\frac {b^{2} x}{\sqrt {b^{2}}}+\sqrt {x^{2} b^{2}-4}\right )}{\sqrt {b x -2}\, \sqrt {b x +2}\, \sqrt {b^{2}}}\)	\(57\)

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

((b*x-2)*(b*x+2))^(1/2)/(b*x-2)^(1/2)/(b*x+2)^(1/2)*ln(b^2*x/(b^2)^(1/2)+(b^2*x^2-4)^(1/2))/(b^2)^(1/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (9) = 18\).
time = 0.28, size = 26, normalized size = 2.36 \begin {gather*} \frac {\log \left (2 \, b^{2} x + 2 \, \sqrt {b^{2} x^{2} - 4} b\right )}{b} \end {gather*}

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 26 vs. \(2 (9) = 18\).
time = 0.30, size = 26, normalized size = 2.36 \begin {gather*} -\frac {\log \left (-b x + \sqrt {b x + 2} \sqrt {b x - 2}\right )}{b} \end {gather*}

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

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Sympy [C] Result contains complex when optimal does not.
time = 16.78, size = 75, normalized size = 6.82 \begin {gather*} \frac {{G_{6, 6}^{6, 2}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} & \frac {1}{2}, \frac {1}{2}, 1, 1 \\0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 0 & \end {matrix} \middle | {\frac {4 e^{2 i \pi }}{b^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} b} + \frac {i {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1 & \\- \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, 0, 0, 0 \end {matrix} \middle | {\frac {4}{b^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} b} \end {gather*}

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

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

**(3/2)*b) + I*meijerg(((-1/2, -1/4, 0, 1/4, 1/2, 1), ()), ((-1/4, 1/4), (-1/2, 0, 0, 0)), 4/(b**2*x**2))/(4*p

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 23 vs. \(2 (9) = 18\).
time = 0.00, size = 26, normalized size = 2.36 \begin {gather*} -\frac {2 \ln \left (\sqrt {b x+2}-\sqrt {b x-2}\right )}{b} \end {gather*}

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Mupad [B]
time = 0.00, size = 50, normalized size = 4.55 \begin {gather*} -\frac {4\,\mathrm {atan}\left (\frac {b\,\left (-\sqrt {b\,x-2}+\sqrt {2}\,1{}\mathrm {i}\right )}{\left (\sqrt {2}-\sqrt {b\,x+2}\right )\,\sqrt {-b^2}}\right )}{\sqrt {-b^2}} \end {gather*}

-(4*atan((b*(2^(1/2)*1i - (b*x - 2)^(1/2)))/((2^(1/2) - (b*x + 2)^(1/2))*(-b^2)^(1/2))))/(-b^2)^(1/2)

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3.16.34 \(\int \frac {1}{\sqrt {-2+b x} \sqrt {2+b x}} \, dx\) [1534]