\(\int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx\) [31]

Optimal result

Integrand size = 24, antiderivative size = 39 \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=-\frac {\sqrt {-1+x} \sqrt {1+x}}{x}+2 a \arctan \left (\sqrt {-1+x} \sqrt {1+x}\right ) \]

[Out]

2*a*arctan((-1+x)^(1/2)*(1+x)^(1/2))-(-1+x)^(1/2)*(1+x)^(1/2)/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {156, 12, 94, 209} \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=2 a \arctan \left (\sqrt {x-1} \sqrt {x+1}\right )-\frac {\sqrt {x-1} \sqrt {x+1}}{x} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 94

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

Rule 209

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=-\frac {\sqrt {-1+x} \sqrt {1+x}}{x}+4 a \arctan \left (\sqrt {\frac {-1+x}{1+x}}\right ) \]

[In]

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

[Out]

-((Sqrt[-1 + x]*Sqrt[1 + x])/x) + 4*a*ArcTan[Sqrt[(-1 + x)/(1 + x)]]

Maple [A] (verified)

Time = 1.58 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.13

method	result	size
default	\(\frac {\left (-2 a x \arctan \left (\frac {1}{\sqrt {x^{2}-1}}\right )-\sqrt {x^{2}-1}\right ) \sqrt {-1+x}\, \sqrt {1+x}}{x \sqrt {x^{2}-1}}\)	\(44\)
risch	\(-\frac {\sqrt {-1+x}\, \sqrt {1+x}}{x}-\frac {2 a \arctan \left (\frac {1}{\sqrt {x^{2}-1}}\right ) \sqrt {\left (-1+x \right ) \left (1+x \right )}}{\sqrt {-1+x}\, \sqrt {1+x}}\)	\(47\)

[In]

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

[Out]

(-2*a*x*arctan(1/(x^2-1)^(1/2))-(x^2-1)^(1/2))*(-1+x)^(1/2)*(1+x)^(1/2)/x/(x^2-1)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.03 \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=\frac {4 \, a x \arctan \left (\sqrt {x + 1} \sqrt {x - 1} - x\right ) - \sqrt {x + 1} \sqrt {x - 1} - x}{x} \]

[In]

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

[Out]

(4*a*x*arctan(sqrt(x + 1)*sqrt(x - 1) - x) - sqrt(x + 1)*sqrt(x - 1) - x)/x

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 22.40 (sec) , antiderivative size = 117, normalized size of antiderivative = 3.00 \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=- \frac {a {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4}, 1 & 1, 1, \frac {3}{2} \\\frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2} & 0 \end {matrix} \middle | {\frac {1}{x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}}} + \frac {i a {G_{6, 6}^{2, 6}\left (\begin {matrix} 0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 1 & \\\frac {1}{4}, \frac {3}{4} & 0, \frac {1}{2}, \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{x^{2}}} \right )}}{2 \pi ^{\frac {3}{2}}} + \frac {{G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4}, 1 & \frac {3}{2}, \frac {3}{2}, 2 \\1, \frac {5}{4}, \frac {3}{2}, \frac {7}{4}, 2 & 0 \end {matrix} \middle | {\frac {1}{x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {i {G_{6, 6}^{2, 6}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4}, 1, \frac {5}{4}, \frac {3}{2}, 1 & \\\frac {3}{4}, \frac {5}{4} & \frac {1}{2}, 1, 1, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} \]

[In]

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

[Out]

-a*meijerg(((3/4, 5/4, 1), (1, 1, 3/2)), ((1/2, 3/4, 1, 5/4, 3/2), (0,)), x**(-2))/(2*pi**(3/2)) + I*a*meijerg
(((0, 1/4, 1/2, 3/4, 1, 1), ()), ((1/4, 3/4), (0, 1/2, 1/2, 0)), exp_polar(2*I*pi)/x**2)/(2*pi**(3/2)) + meije
rg(((5/4, 7/4, 1), (3/2, 3/2, 2)), ((1, 5/4, 3/2, 7/4, 2), (0,)), x**(-2))/(4*pi**(3/2)) + I*meijerg(((1/2, 3/
4, 1, 5/4, 3/2, 1), ()), ((3/4, 5/4), (1/2, 1, 1, 0)), exp_polar(2*I*pi)/x**2)/(4*pi**(3/2))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.54 \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=-2 \, a \arcsin \left (\frac {1}{{\left | x \right |}}\right ) - \frac {\sqrt {x^{2} - 1}}{x} \]

[In]

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

[Out]

-2*a*arcsin(1/abs(x)) - sqrt(x^2 - 1)/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.10 \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=-4 \, a \arctan \left (\frac {1}{2} \, {\left (\sqrt {x + 1} - \sqrt {x - 1}\right )}^{2}\right ) - \frac {8}{{\left (\sqrt {x + 1} - \sqrt {x - 1}\right )}^{4} + 4} \]

[In]

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

[Out]

-4*a*arctan(1/2*(sqrt(x + 1) - sqrt(x - 1))^2) - 8/((sqrt(x + 1) - sqrt(x - 1))^4 + 4)

Mupad [B] (verification not implemented)

Time = 4.57 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.67 \[ \int \frac {-1+2 a x}{\sqrt {-1+x} x^2 \sqrt {1+x}} \, dx=-\frac {\sqrt {x-1}\,\sqrt {x+1}}{x}-a\,\left (\ln \left (\frac {{\left (\sqrt {x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {x+1}-1\right )}^2}+1\right )-\ln \left (\frac {\sqrt {x-1}-\mathrm {i}}{\sqrt {x+1}-1}\right )\right )\,2{}\mathrm {i} \]

[In]

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

[Out]

- a*(log(((x - 1)^(1/2) - 1i)^2/((x + 1)^(1/2) - 1)^2 + 1) - log(((x - 1)^(1/2) - 1i)/((x + 1)^(1/2) - 1)))*2i
- ((x - 1)^(1/2)*(x + 1)^(1/2))/x