3.4.53 \(\int \frac {a+b x^2}{x \sqrt {-1+c x} \sqrt {1+c x}} \, dx\) [353]

Optimal. Leaf size=46 \[ \frac {b \sqrt {-1+c x} \sqrt {1+c x}}{c^2}+a \tan ^{-1}\left (\sqrt {-1+c x} \sqrt {1+c x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {471, 94, 211} \begin {gather*} a \text {ArcTan}\left (\sqrt {c x-1} \sqrt {c x+1}\right )+\frac {b \sqrt {c x-1} \sqrt {c x+1}}{c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/c^2 + a*ArcTan[Sqrt[-1 + c*x]*Sqrt[1 + c*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 211

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

Rule 471

Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^(p_.)*((c_) + (d_.)
*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*(a1 + b1*x^(n/2))^(p + 1)*((a2 + b2*x^(n/2))^(p + 1)/(b1*b2*e*(
m + n*(p + 1) + 1))), x] - Dist[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(b1*b2*(m + n*(p + 1) + 1)), I
nt[(e*x)^m*(a1 + b1*x^(n/2))^p*(a2 + b2*x^(n/2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, m, n, p}, x] &&
EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && NeQ[m + n*(p + 1) + 1, 0]

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 45, normalized size = 0.98 \begin {gather*} \frac {b \sqrt {-1+c x} \sqrt {1+c x}}{c^2}+2 a \tan ^{-1}\left (\sqrt {\frac {-1+c x}{1+c x}}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]
time = 0.27, size = 62, normalized size = 1.35

method	result	size
default	\(\frac {\left (-\arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right ) a \,c^{2}+b \sqrt {c^{2} x^{2}-1}\right ) \sqrt {c x -1}\, \sqrt {c x +1}}{\sqrt {c^{2} x^{2}-1}\, c^{2}}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.60, size = 29, normalized size = 0.63 \begin {gather*} -a \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) + \frac {\sqrt {c^{2} x^{2} - 1} b}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 4.06, size = 48, normalized size = 1.04 \begin {gather*} \frac {2 \, a c^{2} \arctan \left (-c x + \sqrt {c x + 1} \sqrt {c x - 1}\right ) + \sqrt {c x + 1} \sqrt {c x - 1} b}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 38.99, size = 162, normalized size = 3.52 \begin {gather*} - \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}{c^{2} x^{2}}} \right )}}{4 \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 }}{c^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}}} + \frac {b {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & 0, 0, \frac {1}{2}, 1 \\- \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {1}{c^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} c^{2}} + \frac {i b {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 1 & \\- \frac {3}{4}, - \frac {1}{4} & -1, - \frac {1}{2}, - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {e^{2 i \pi }}{c^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)/x/(c*x-1)**(1/2)/(c*x+1)**(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,)), 1/(c**2*x**2))/(4*pi**(3/2)) + I*a*m
eijerg(((0, 1/4, 1/2, 3/4, 1, 1), ()), ((1/4, 3/4), (0, 1/2, 1/2, 0)), exp_polar(2*I*pi)/(c**2*x**2))/(4*pi**(
3/2)) + b*meijerg(((-1/4, 1/4), (0, 0, 1/2, 1)), ((-1/2, -1/4, 0, 1/4, 1/2, 0), ()), 1/(c**2*x**2))/(4*pi**(3/
2)*c**2) + I*b*meijerg(((-1, -3/4, -1/2, -1/4, 0, 1), ()), ((-3/4, -1/4), (-1, -1/2, -1/2, 0)), exp_polar(2*I*
pi)/(c**2*x**2))/(4*pi**(3/2)*c**2)

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Giac [A]
time = 0.61, size = 45, normalized size = 0.98 \begin {gather*} -2 \, a \arctan \left (\frac {1}{2} \, {\left (\sqrt {c x + 1} - \sqrt {c x - 1}\right )}^{2}\right ) + \frac {\sqrt {c x + 1} \sqrt {c x - 1} b}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.86, size = 77, normalized size = 1.67 \begin {gather*} \frac {b\,\sqrt {c\,x-1}\,\sqrt {c\,x+1}}{c^2}-a\,\left (\ln \left (\frac {{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {c\,x+1}-1\right )}^2}+1\right )-\ln \left (\frac {\sqrt {c\,x-1}-\mathrm {i}}{\sqrt {c\,x+1}-1}\right )\right )\,1{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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