3.218 \(\int \frac{\sqrt{c-a c x} \sqrt{1-a^2 x^2}}{x^2} \, dx\)

Optimal. Leaf size=102 \[ -\frac{c^2 \left (1-a^2 x^2\right )^{3/2}}{x (c-a c x)^{3/2}}-\frac{a c \sqrt{1-a^2 x^2}}{\sqrt{c-a c x}}+a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-a^2 x^2}}{\sqrt{c-a c x}}\right ) \]

[Out]

-((a*c*Sqrt[1 - a^2*x^2])/Sqrt[c - a*c*x]) - (c^2*(1 - a^2*x^2)^(3/2))/(x*(c - a*c*x)^(3/2)) + a*Sqrt[c]*ArcTa
nh[(Sqrt[c]*Sqrt[1 - a^2*x^2])/Sqrt[c - a*c*x]]

________________________________________________________________________________________

Rubi [A]  time = 0.111614, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {879, 865, 875, 208} \[ -\frac{c^2 \left (1-a^2 x^2\right )^{3/2}}{x (c-a c x)^{3/2}}-\frac{a c \sqrt{1-a^2 x^2}}{\sqrt{c-a c x}}+a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-a^2 x^2}}{\sqrt{c-a c x}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*c*Sqrt[1 - a^2*x^2])/Sqrt[c - a*c*x]) - (c^2*(1 - a^2*x^2)^(3/2))/(x*(c - a*c*x)^(3/2)) + a*Sqrt[c]*ArcTa
nh[(Sqrt[c]*Sqrt[1 - a^2*x^2])/Sqrt[c - a*c*x]]

Rule 879

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e^2*(e*f
- d*g)*(d + e*x)^(m - 2)*(f + g*x)^(n + 1)*(a + c*x^2)^(p + 1))/(c*g*(n + 1)*(e*f + d*g)), x] - Dist[(e*(e*f*
(p + 1) - d*g*(2*n + p + 3)))/(g*(n + 1)*(e*f + d*g)), Int[(d + e*x)^(m - 1)*(f + g*x)^(n + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&
EqQ[m + p - 1, 0] && LtQ[n, -1] && IntegerQ[2*p]

Rule 865

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*
x)^m*(f + g*x)^(n + 1)*(a + c*x^2)^p)/(g*(m - n - 1)), x] - Dist[(c*m*(e*f + d*g))/(e^2*g*(m - n - 1)), Int[(d
+ e*x)^(m + 1)*(f + g*x)^n*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0
] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0] && GtQ[p, 0] && NeQ[m - n - 1, 0] &&  !IGtQ[n, 0]
&&  !(IntegerQ[n + p] && LtQ[n + p + 2, 0]) && RationalQ[n]

Rule 875

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

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\sqrt{c-a c x} \sqrt{1-a^2 x^2}}{x^2} \, dx &=-\frac{c^2 \left (1-a^2 x^2\right )^{3/2}}{x (c-a c x)^{3/2}}-\frac{1}{2} (a c) \int \frac{\sqrt{1-a^2 x^2}}{x \sqrt{c-a c x}} \, dx\\ &=-\frac{a c \sqrt{1-a^2 x^2}}{\sqrt{c-a c x}}-\frac{c^2 \left (1-a^2 x^2\right )^{3/2}}{x (c-a c x)^{3/2}}-\frac{1}{2} a \int \frac{\sqrt{c-a c x}}{x \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{a c \sqrt{1-a^2 x^2}}{\sqrt{c-a c x}}-\frac{c^2 \left (1-a^2 x^2\right )^{3/2}}{x (c-a c x)^{3/2}}-\left (a^3 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{-a^2 c+a^2 c^2 x^2} \, dx,x,\frac{\sqrt{1-a^2 x^2}}{\sqrt{c-a c x}}\right )\\ &=-\frac{a c \sqrt{1-a^2 x^2}}{\sqrt{c-a c x}}-\frac{c^2 \left (1-a^2 x^2\right )^{3/2}}{x (c-a c x)^{3/2}}+a \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-a^2 x^2}}{\sqrt{c-a c x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.101095, size = 93, normalized size = 0.91 \[ \frac{\sqrt{1-a^2 x^2} \left (a \sqrt{c} x \tanh ^{-1}\left (\sqrt{c} \sqrt{\frac{a x+1}{c}}\right )-c (2 a x+1) \sqrt{\frac{a x+1}{c}}\right )}{x \sqrt{\frac{a x+1}{c}} \sqrt{c-a c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(-(c*Sqrt[(1 + a*x)/c]*(1 + 2*a*x)) + a*Sqrt[c]*x*ArcTanh[Sqrt[c]*Sqrt[(1 + a*x)/c]]))/(x*S
qrt[(1 + a*x)/c]*Sqrt[c - a*c*x])

________________________________________________________________________________________

Maple [A]  time = 0.198, size = 95, normalized size = 0.9 \begin{align*}{\frac{1}{ \left ( ax-1 \right ) x} \left ( -{\it Artanh} \left ({\sqrt{c \left ( ax+1 \right ) }{\frac{1}{\sqrt{c}}}} \right ) xac+2\,xa\sqrt{c \left ( ax+1 \right ) }\sqrt{c}+\sqrt{c \left ( ax+1 \right ) }\sqrt{c} \right ) \sqrt{-c \left ( ax-1 \right ) }\sqrt{-{a}^{2}{x}^{2}+1}{\frac{1}{\sqrt{c \left ( ax+1 \right ) }}}{\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.59443, size = 473, normalized size = 4.64 \begin{align*} \left [\frac{{\left (a^{2} x^{2} - a x\right )} \sqrt{c} \log \left (-\frac{a^{2} c x^{2} + a c x - 2 \, \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c} \sqrt{c} - 2 \, c}{a x^{2} - x}\right ) + 2 \, \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}{\left (2 \, a x + 1\right )}}{2 \,{\left (a x^{2} - x\right )}}, \frac{{\left (a^{2} x^{2} - a x\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c} \sqrt{-c}}{a^{2} c x^{2} - c}\right ) + \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}{\left (2 \, a x + 1\right )}}{a x^{2} - x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a^2*x^2 - a*x)*sqrt(c)*log(-(a^2*c*x^2 + a*c*x - 2*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(c) - 2*c)/(
a*x^2 - x)) + 2*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(2*a*x + 1))/(a*x^2 - x), ((a^2*x^2 - a*x)*sqrt(-c)*arctan
(sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-c)/(a^2*c*x^2 - c)) + sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(2*a*x +
1))/(a*x^2 - x)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- c \left (a x - 1\right )} \sqrt{- \left (a x - 1\right ) \left (a x + 1\right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-c*(a*x - 1))*sqrt(-(a*x - 1)*(a*x + 1))/x**2, x)

________________________________________________________________________________________

Giac [A]  time = 1.14903, size = 149, normalized size = 1.46 \begin{align*} -\frac{{\left ({\left (\frac{c \arctan \left (\frac{\sqrt{a c x + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} + 2 \, \sqrt{a c x + c} + \frac{\sqrt{a c x + c}}{a x}\right )} a^{2} c^{2} - \frac{a^{2} c^{3} \arctan \left (\frac{\sqrt{2} \sqrt{c}}{\sqrt{-c}}\right ) + 3 \, \sqrt{2} a^{2} \sqrt{-c} c^{\frac{5}{2}}}{\sqrt{-c}}\right )}{\left | c \right |}}{a c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c*arctan(sqrt(a*c*x + c)/sqrt(-c))/sqrt(-c) + 2*sqrt(a*c*x + c) + sqrt(a*c*x + c)/(a*x))*a^2*c^2 - (a^2*c^3
*arctan(sqrt(2)*sqrt(c)/sqrt(-c)) + 3*sqrt(2)*a^2*sqrt(-c)*c^(5/2))/sqrt(-c))*abs(c)/(a*c^3)