\(\int \frac {1}{\sqrt {-1+b x} \sqrt {2+b x}} \, dx\) [1533]

Optimal result

Integrand size = 19, antiderivative size = 21 \[ \int \frac {1}{\sqrt {-1+b x} \sqrt {2+b x}} \, dx=\frac {2 \text {arcsinh}\left (\frac {\sqrt {-1+b x}}{\sqrt {3}}\right )}{b} \]

[Out]

2*arcsinh(1/3*(b*x-1)^(1/2)*3^(1/2))/b

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {65, 221} \[ \int \frac {1}{\sqrt {-1+b x} \sqrt {2+b x}} \, dx=\frac {2 \text {arcsinh}\left (\frac {\sqrt {b x-1}}{\sqrt {3}}\right )}{b} \]

[In]

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

[Out]

(2*ArcSinh[Sqrt[-1 + b*x]/Sqrt[3]])/b

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 221

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 \text {Subst}\left (\int \frac {1}{\sqrt {3+x^2}} \, dx,x,\sqrt {-1+b x}\right )}{b} \\ & = \frac {2 \sinh ^{-1}\left (\frac {\sqrt {-1+b x}}{\sqrt {3}}\right )}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {1}{\sqrt {-1+b x} \sqrt {2+b x}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {2+b x}}{\sqrt {-1+b x}}\right )}{b} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(64\) vs. \(2(18)=36\).

Time = 0.54 (sec) , antiderivative size = 65, normalized size of antiderivative = 3.10

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

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.33 \[ \int \frac {1}{\sqrt {-1+b x} \sqrt {2+b x}} \, dx=-\frac {\log \left (-2 \, b x + 2 \, \sqrt {b x + 2} \sqrt {b x - 1} - 1\right )}{b} \]

[In]

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

[Out]

-log(-2*b*x + 2*sqrt(b*x + 2)*sqrt(b*x - 1) - 1)/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 = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {1}{\sqrt {-1+b x} \sqrt {2+b x}} \, dx=\frac {\log \left (2 \, b^{2} x + 2 \, \sqrt {b^{2} x^{2} + b x - 2} b + b\right )}{b} \]

[In]

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

[Out]

log(2*b^2*x + 2*sqrt(b^2*x^2 + b*x - 2)*b + b)/b

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\sqrt {-1+b x} \sqrt {2+b x}} \, dx=-\frac {2 \, \log \left (\sqrt {b x + 2} - \sqrt {b x - 1}\right )}{b} \]

[In]

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

[Out]

-2*log(sqrt(b*x + 2) - sqrt(b*x - 1))/b

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.10 \[ \int \frac {1}{\sqrt {-1+b x} \sqrt {2+b x}} \, dx=\frac {4\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {b\,x-1}-\mathrm {i}\right )}{\left (\sqrt {2}-\sqrt {b\,x+2}\right )\,\sqrt {-b^2}}\right )}{\sqrt {-b^2}} \]

[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)