3.8.58 \(\int \frac {1}{x \sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx\) [758]

Optimal. Leaf size=54 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {1+a+b x}}{\sqrt {1+a} \sqrt {1-a-b x}}\right )}{\sqrt {1-a^2}} \]

[Out]

-2*arctanh((1-a)^(1/2)*(b*x+a+1)^(1/2)/(1+a)^(1/2)/(-b*x-a+1)^(1/2))/(-a^2+1)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {95, 214} \begin {gather*} -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{\sqrt {1-a^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 214

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

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 88, normalized size = 1.63 \begin {gather*} \frac {2 (-1+a+b x)^{3/2} (1+a+b x)^{3/2} \tan ^{-1}\left (\frac {\sqrt {1-a^2} \sqrt {\frac {-1+a+b x}{1+a+b x}}}{-1+a}\right )}{\sqrt {1-a^2} (-((-1+a+b x) (1+a+b x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-1 + a + b*x)^(3/2)*(1 + a + b*x)^(3/2)*ArcTan[(Sqrt[1 - a^2]*Sqrt[(-1 + a + b*x)/(1 + a + b*x)])/(-1 + a)
])/(Sqrt[1 - a^2]*(-((-1 + a + b*x)*(1 + a + b*x)))^(3/2))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.09, size = 114, normalized size = 2.11

method	result	size
default	\(\frac {\sqrt {-b x -a +1}\, \sqrt {b x +a +1}\, \mathrm {csgn}\left (b \right )^{2} \ln \left (-\frac {2 \left (a b x +a^{2}-\sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}-1\right )}{x}\right ) \sqrt {-a^{2}+1}}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \left (1+a \right ) \left (a -1\right )}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a-1>0)', see `assume?` for mor
e details)Is

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Fricas [A]
time = 0.99, size = 180, normalized size = 3.33 \begin {gather*} \left [-\frac {\sqrt {-a^{2} + 1} \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x + 2 \, {\left (a b x + a^{2} - 1\right )} \sqrt {-a^{2} + 1} \sqrt {b x + a + 1} \sqrt {-b x - a + 1} - 4 \, a^{2} + 2}{x^{2}}\right )}{2 \, {\left (a^{2} - 1\right )}}, \frac {\arctan \left (\frac {{\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1} \sqrt {b x + a + 1} \sqrt {-b x - a + 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \, {\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right )}{\sqrt {a^{2} - 1}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(-a^2 + 1)*log(((2*a^2 - 1)*b^2*x^2 + 2*a^4 + 4*(a^3 - a)*b*x + 2*(a*b*x + a^2 - 1)*sqrt(-a^2 + 1)*s
qrt(b*x + a + 1)*sqrt(-b*x - a + 1) - 4*a^2 + 2)/x^2)/(a^2 - 1), arctan((a*b*x + a^2 - 1)*sqrt(a^2 - 1)*sqrt(b
*x + a + 1)*sqrt(-b*x - a + 1)/((a^2 - 1)*b^2*x^2 + a^4 + 2*(a^3 - a)*b*x - 2*a^2 + 1))/sqrt(a^2 - 1)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \sqrt {- a - b x + 1} \sqrt {a + b x + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Undef/Unsigned Inf encountered in limitLimit: Max order reached or unable to make series expansion Error: B
ad Argument

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Mupad [B]
time = 3.22, size = 260, normalized size = 4.81 \begin {gather*} \frac {\ln \left (\frac {2\,a\,\left (\sqrt {1-a}-\sqrt {1-b\,x-a}\right )}{\sqrt {a+1}-\sqrt {a+b\,x+1}}-\sqrt {1-a}\,\sqrt {a+1}+\frac {\sqrt {1-a}\,\sqrt {a+1}\,{\left (\sqrt {1-a}-\sqrt {1-b\,x-a}\right )}^2}{{\left (\sqrt {a+1}-\sqrt {a+b\,x+1}\right )}^2}\right )-\ln \left (\frac {8\,\left (\sqrt {1-a}-\sqrt {1-b\,x-a}\right )}{\sqrt {a+1}-\sqrt {a+b\,x+1}}-a\,\sqrt {1-a}\,\sqrt {a+1}+a\,{\left (1-a\right )}^{3/2}\,{\left (a+1\right )}^{3/2}-\frac {8\,a^2\,\left (\sqrt {1-a}-\sqrt {1-b\,x-a}\right )}{\sqrt {a+1}-\sqrt {a+b\,x+1}}+a^3\,\sqrt {1-a}\,\sqrt {a+1}\right )}{\sqrt {1-a}\,\sqrt {a+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((2*a*((1 - a)^(1/2) - (1 - b*x - a)^(1/2)))/((a + 1)^(1/2) - (a + b*x + 1)^(1/2)) - (1 - a)^(1/2)*(a + 1)
^(1/2) + ((1 - a)^(1/2)*(a + 1)^(1/2)*((1 - a)^(1/2) - (1 - b*x - a)^(1/2))^2)/((a + 1)^(1/2) - (a + b*x + 1)^
(1/2))^2) - log((8*((1 - a)^(1/2) - (1 - b*x - a)^(1/2)))/((a + 1)^(1/2) - (a + b*x + 1)^(1/2)) - a*(1 - a)^(1
/2)*(a + 1)^(1/2) + a*(1 - a)^(3/2)*(a + 1)^(3/2) - (8*a^2*((1 - a)^(1/2) - (1 - b*x - a)^(1/2)))/((a + 1)^(1/
2) - (a + b*x + 1)^(1/2)) + a^3*(1 - a)^(1/2)*(a + 1)^(1/2)))/((1 - a)^(1/2)*(a + 1)^(1/2))

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