3.4.78 \(\int \frac {1+c^2 x^2}{x \sqrt {-1+c x} \sqrt {1+c x}} \, dx\) [378]

Optimal. Leaf size=40 \[ \sqrt {-1+c x} \sqrt {1+c x}+\tan ^{-1}\left (\sqrt {-1+c x} \sqrt {1+c x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {471, 94, 211} \begin {gather*} \text {ArcTan}\left (\sqrt {c x-1} \sqrt {c x+1}\right )+\sqrt {c x-1} \sqrt {c x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.06, size = 40, normalized size = 1.00 \begin {gather*} \sqrt {-1+c x} \sqrt {1+c x}+2 \tan ^{-1}\left (\sqrt {\frac {-1+c x}{1+c x}}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]
time = 0.28, size = 53, normalized size = 1.32

method	result	size
default	\(\frac {\left (\sqrt {c^{2} x^{2}-1}-\arctan \left (\frac {1}{\sqrt {c^{2} x^{2}-1}}\right )\right ) \sqrt {c x -1}\, \sqrt {c x +1}}{\sqrt {c^{2} x^{2}-1}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 23, normalized size = 0.58 \begin {gather*} \sqrt {c^{2} x^{2} - 1} - \arcsin \left (\frac {1}{c {\left | x \right |}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.19, size = 39, normalized size = 0.98 \begin {gather*} \sqrt {c x + 1} \sqrt {c x - 1} + 2 \, \arctan \left (-c x + \sqrt {c x + 1} \sqrt {c x - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c*x + 1)*sqrt(c*x - 1) + 2*arctan(-c*x + sqrt(c*x + 1)*sqrt(c*x - 1))

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Sympy [C] Result contains complex when optimal does not.
time = 26.88, size = 148, normalized size = 3.70 \begin {gather*} \frac {{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}}} - \frac {{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 {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}}} + \frac {i {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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)) - meij
erg(((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*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)) + I*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)/(c**2*x**2
))/(4*pi**(3/2))

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Giac [A]
time = 0.68, size = 40, normalized size = 1.00 \begin {gather*} \sqrt {c x + 1} \sqrt {c x - 1} - 2 \, \arctan \left (\frac {1}{2} \, {\left (\sqrt {c x + 1} - \sqrt {c x - 1}\right )}^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.65, size = 72, normalized size = 1.80 \begin {gather*} \sqrt {c\,x-1}\,\sqrt {c\,x+1}-\ln \left (\frac {{\left (\sqrt {c\,x-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {c\,x+1}-1\right )}^2}+1\right )\,1{}\mathrm {i}+\ln \left (\frac {\sqrt {c\,x-1}-\mathrm {i}}{\sqrt {c\,x+1}-1}\right )\,1{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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