\(\int \frac {1}{\sqrt {-1-b x} \sqrt {2+b x}} \, dx\) [1540]

Optimal result

Integrand size = 20, antiderivative size = 11 \[ \int \frac {1}{\sqrt {-1-b x} \sqrt {2+b x}} \, dx=\frac {\arcsin (3+2 b x)}{b} \]

[Out]

arcsin(2*b*x+3)/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, 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} \[ \int \frac {1}{\sqrt {-1-b x} \sqrt {2+b x}} \, dx=\frac {\arcsin (2 b x+3)}{b} \]

[In]

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

[Out]

ArcSin[3 + 2*b*x]/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 {-2-3 b x-b^2 x^2}} \, dx \\ & = -\frac {\text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{b^2}}} \, dx,x,-3 b-2 b^2 x\right )}{b^2} \\ & = \frac {\sin ^{-1}(3+2 b x)}{b} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(59\) vs. \(2(11)=22\).

Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 5.36 \[ \int \frac {1}{\sqrt {-1-b x} \sqrt {2+b x}} \, dx=\frac {2 \sqrt {1+b x} \sqrt {2+b x} \text {arctanh}\left (\frac {\sqrt {2+b x}}{\sqrt {1+b x}}\right )}{b \sqrt {-((1+b x) (2+b x))}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(65\) vs. \(2(11)=22\).

Time = 0.54 (sec) , antiderivative size = 66, normalized size of antiderivative = 6.00

method	result	size
default	\(\frac {\sqrt {\left (-b x -1\right ) \left (b x +2\right )}\, \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {3}{2 b}\right )}{\sqrt {-b^{2} x^{2}-3 b x -2}}\right )}{\sqrt {-b x -1}\, \sqrt {b x +2}\, \sqrt {b^{2}}}\)	\(66\)

[In]

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

[Out]

((-b*x-1)*(b*x+2))^(1/2)/(-b*x-1)^(1/2)/(b*x+2)^(1/2)/(b^2)^(1/2)*arctan((b^2)^(1/2)*(x+3/2/b)/(-b^2*x^2-3*b*x
-2)^(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 = 4.00 \[ \int \frac {1}{\sqrt {-1-b x} \sqrt {2+b x}} \, dx=-\frac {\arctan \left (\frac {{\left (2 \, b x + 3\right )} \sqrt {b x + 2} \sqrt {-b x - 1}}{2 \, {\left (b^{2} x^{2} + 3 \, b x + 2\right )}}\right )}{b} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.91 \[ \int \frac {1}{\sqrt {-1-b x} \sqrt {2+b x}} \, dx=-\frac {\arcsin \left (-\frac {2 \, b^{2} x + 3 \, b}{b}\right )}{b} \]

[In]

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

[Out]

-arcsin(-(2*b^2*x + 3*b)/b)/b

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\sqrt {-1-b x} \sqrt {2+b x}} \, dx=\frac {2 \, \arcsin \left (\sqrt {b x + 2}\right )}{b} \]

[In]

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

[Out]

2*arcsin(sqrt(b*x + 2))/b

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 41, normalized size of antiderivative = 3.73 \[ \int \frac {1}{\sqrt {-1-b x} \sqrt {2+b x}} \, dx=\frac {4\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {-b\,x-1}-\mathrm {i}\right )}{\left (\sqrt {2}-\sqrt {b\,x+2}\right )\,\sqrt {b^2}}\right )}{\sqrt {b^2}} \]

[In]

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

[Out]

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