\(\int \frac {1}{\sqrt {3-b x} \sqrt {2+b x}} \, dx\) [1536]

Optimal result

Integrand size = 20, antiderivative size = 16 \[ \int \frac {1}{\sqrt {3-b x} \sqrt {2+b x}} \, dx=-\frac {\arcsin \left (\frac {1}{5} (1-2 b x)\right )}{b} \]

[Out]

arcsin(2/5*b*x-1/5)/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {55, 633, 222} \[ \int \frac {1}{\sqrt {3-b x} \sqrt {2+b x}} \, dx=-\frac {\arcsin \left (\frac {1}{5} (1-2 b x)\right )}{b} \]

[In]

Int[1/(Sqrt[3 - b*x]*Sqrt[2 + b*x]),x]

[Out]

-(ArcSin[(1 - 2*b*x)/5]/b)

Rule 55

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[1/Sqrt[a*c - b*(a - c)*x - b^2*x^2]
, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\sqrt {6+b x-b^2 x^2}} \, dx \\ & = -\frac {\text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{25 b^2}}} \, dx,x,b-2 b^2 x\right )}{5 b^2} \\ & = -\frac {\sin ^{-1}\left (\frac {1}{5} (1-2 b x)\right )}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.88 \[ \int \frac {1}{\sqrt {3-b x} \sqrt {2+b x}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {(3-b x) (2+b x)}}{-3+b x}\right )}{b} \]

[In]

Integrate[1/(Sqrt[3 - b*x]*Sqrt[2 + b*x]),x]

[Out]

(-2*ArcTan[Sqrt[(3 - b*x)*(2 + b*x)]/(-3 + b*x)])/b

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(64\) vs. \(2(11)=22\).

Time = 0.54 (sec) , antiderivative size = 65, normalized size of antiderivative = 4.06

method	result	size
default	\(\frac {\sqrt {\left (-b x +3\right ) \left (b x +2\right )}\, \arctan \left (\frac {\sqrt {b^{2}}\, \left (x -\frac {1}{2 b}\right )}{\sqrt {-b^{2} x^{2}+b x +6}}\right )}{\sqrt {-b x +3}\, \sqrt {b x +2}\, \sqrt {b^{2}}}\)	\(65\)

[In]

int(1/(-b*x+3)^(1/2)/(b*x+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

((-b*x+3)*(b*x+2))^(1/2)/(-b*x+3)^(1/2)/(b*x+2)^(1/2)/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x-1/2/b)/(-b^2*x^2+b*x+6
)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (11) = 22\).

Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.75 \[ \int \frac {1}{\sqrt {3-b x} \sqrt {2+b x}} \, dx=-\frac {\arctan \left (\frac {{\left (2 \, b x - 1\right )} \sqrt {b x + 2} \sqrt {-b x + 3}}{2 \, {\left (b^{2} x^{2} - b x - 6\right )}}\right )}{b} \]

[In]

integrate(1/(-b*x+3)^(1/2)/(b*x+2)^(1/2),x, algorithm="fricas")

[Out]

-arctan(1/2*(2*b*x - 1)*sqrt(b*x + 2)*sqrt(-b*x + 3)/(b^2*x^2 - b*x - 6))/b

Sympy [F]

\[ \int \frac {1}{\sqrt {3-b x} \sqrt {2+b x}} \, dx=\int \frac {1}{\sqrt {- b x + 3} \sqrt {b x + 2}}\, dx \]

[In]

integrate(1/(-b*x+3)**(1/2)/(b*x+2)**(1/2),x)

[Out]

Integral(1/(sqrt(-b*x + 3)*sqrt(b*x + 2)), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\sqrt {3-b x} \sqrt {2+b x}} \, dx=-\frac {\arcsin \left (-\frac {2 \, b^{2} x - b}{5 \, b}\right )}{b} \]

[In]

integrate(1/(-b*x+3)^(1/2)/(b*x+2)^(1/2),x, algorithm="maxima")

[Out]

-arcsin(-1/5*(2*b^2*x - b)/b)/b

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\sqrt {3-b x} \sqrt {2+b x}} \, dx=\frac {2 \, \arcsin \left (\frac {1}{5} \, \sqrt {5} \sqrt {b x + 2}\right )}{b} \]

[In]

integrate(1/(-b*x+3)^(1/2)/(b*x+2)^(1/2),x, algorithm="giac")

[Out]

2*arcsin(1/5*sqrt(5)*sqrt(b*x + 2))/b

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.75 \[ \int \frac {1}{\sqrt {3-b x} \sqrt {2+b x}} \, dx=-\frac {4\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {3}-\sqrt {3-b\,x}\right )}{\left (\sqrt {2}-\sqrt {b\,x+2}\right )\,\sqrt {b^2}}\right )}{\sqrt {b^2}} \]

[In]

int(1/((b*x + 2)^(1/2)*(3 - b*x)^(1/2)),x)

[Out]

-(4*atan((b*(3^(1/2) - (3 - b*x)^(1/2)))/((2^(1/2) - (b*x + 2)^(1/2))*(b^2)^(1/2))))/(b^2)^(1/2)