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]

-c^2*(-a^2*x^2+1)^(3/2)/x/(-a*c*x+c)^(3/2)+a*arctanh(c^(1/2)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(1/2))*c^(1/2)-a*c*
(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 102, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.10, 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])

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fricas [A]  time = 0.90, size = 217, normalized size = 2.13 \[ \left [\frac {{\left (a^{2} x^{2} - a x\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} + a c x - 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {c} - 2 \, c}{a x^{2} - x}\right ) + 2 \, \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (2 \, a x + 1\right )}}{2 \, {\left (a x^{2} - x\right )}}, \frac {{\left (a^{2} x^{2} - a x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} \sqrt {-c}}{a^{2} c x^{2} - c}\right ) + \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c} {\left (2 \, a x + 1\right )}}{a x^{2} - x}\right ] \]

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)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.05, size = 95, normalized size = 0.93 \[ \frac {\left (-a c x \arctanh \left (\frac {\sqrt {\left (a x +1\right ) c}}{\sqrt {c}}\right )+2 \sqrt {\left (a x +1\right ) c}\, a \sqrt {c}\, x +\sqrt {\left (a x +1\right ) c}\, \sqrt {c}\right ) \sqrt {-\left (a x -1\right ) c}\, \sqrt {-a^{2} x^{2}+1}}{\left (a x -1\right ) \sqrt {\left (a x +1\right ) c}\, \sqrt {c}\, x} \]

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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{x^{2}}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {1-a^2\,x^2}\,\sqrt {c-a\,c\,x}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- c \left (a x - 1\right )} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{x^{2}}\, dx \]

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)

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