∫ 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 = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {41, 222} \begin {gather*} \frac {\sin ^{-1}\left (\frac {b x}{2}\right )}{b} \end {gather*}

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b

, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}

\begin {align*} \int \frac {1}{\sqrt {2-b x} \sqrt {2+b x}} \, dx &=\int \frac {1}{\sqrt {4-b^2 x^2}} \, dx\\ &=\frac {\sin ^{-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. \(39\) vs. \(2(11)=22\).
time = 0.03, size = 39, normalized size = 3.55 \begin {gather*} -\frac {\log \left (-\sqrt {-b^2} x+\sqrt {4-b^2 x^2}\right )}{\sqrt {-b^2}} \end {gather*}

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

-(Log[-(Sqrt[-b^2]*x) + Sqrt[4 - b^2*x^2]]/Sqrt[-b^2])

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 15.09, size = 69, normalized size = 6.27 \begin {gather*} \frac {-I \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}{b^2 x^2}\right ]+\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 \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 / 4, 3 / 4}, {1 / 2, 1 / 2, 1, 1}}, {{0, 1 / 4, 1 / 2, 3 / 4, 1, 0}, {}}, 4 / (b ^ 2 x ^ 2)] +

meijerg[{{-1 / 2, -1 / 4, 0, 1 / 4, 1 / 2, 1}, {}}, {{-1 / 4, 1 / 4}, {-1 / 2, 0, 0, 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. \(55\) vs. \(2(9)=18\).
time = 0.16, size = 56, normalized size = 5.09


method	result	size

default	\(\frac {\sqrt {\left (-b x +2\right ) \left (b x +2\right )}\, \arctan \left (\frac {\sqrt {b^{2}}\, x}{\sqrt {-x^{2} b^{2}+4}}\right )}{\sqrt {-b x +2}\, \sqrt {b x +2}\, \sqrt {b^{2}}}\)	\(56\)

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

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Maxima [A]
time = 0.35, size = 9, normalized size = 0.82 \begin {gather*} \frac {\arcsin \left (\frac {1}{2} \, b x\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. 31 vs. \(2 (9) = 18\).
time = 0.29, size = 31, normalized size = 2.82 \begin {gather*} -\frac {2 \, \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b x + 2} - 2}{b x}\right )}{b} \end {gather*}

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

-2*arctan((sqrt(b*x + 2)*sqrt(-b*x + 2) - 2)/(b*x))/b

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Sympy [C] Result contains complex when optimal does not.
time = 17.17, size = 76, normalized size = 6.91 \begin {gather*} - \frac {i {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}{b^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} b} + \frac {{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 e^{- 2 i \pi }}{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)

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

ijerg(((-1/2, -1/4, 0, 1/4, 1/2, 1), ()), ((-1/4, 1/4), (-1/2, 0, 0, 0)), 4*exp_polar(-2*I*pi)/(b**2*x**2))/(4

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Giac [A]
time = 0.00, size = 19, normalized size = 1.73 \begin {gather*} -\frac {2 \arcsin \left (\frac {\sqrt {-b x+2}}{2}\right )}{b} \end {gather*}

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Mupad [B]
time = 0.08, size = 44, normalized size = 4.00 \begin {gather*} -\frac {4\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {2}-\sqrt {2-b\,x}\right )}{\left (\sqrt {2}-\sqrt {b\,x+2}\right )\,\sqrt {b^2}}\right )}{\sqrt {b^2}} \end {gather*}

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

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3.16.37 \(\int \frac {1}{\sqrt {2-b x} \sqrt {2+b x}} \, dx\) [1537]