\(\int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx\) [698]

Optimal result

Integrand size = 27, antiderivative size = 75 \[ \int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=\sqrt {c-a^2 c x^2}+2 \sqrt {c} \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )+\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=\sqrt {c-a^2 c x^2}-2 \sqrt {c} \arctan \left (\frac {a x \sqrt {c-a^2 c x^2}}{\sqrt {c} \left (-1+a^2 x^2\right )}\right )-\sqrt {c} \log (x)+\sqrt {c} \log \left (c+\sqrt {c} \sqrt {c-a^2 c x^2}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {6717, 6702, 541, 25, 27, 538, 224, 216, 243, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 6717

\(\displaystyle -\int \frac {e^{-2 \text {arctanh}(a x)} \sqrt {c-a^2 c x^2}}{x}dx\)

\(\Big \downarrow \) 6702

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

\(\Big \downarrow \) 541

\(\displaystyle -c \left (-\frac {\int -\frac {a^2 c (1-2 a x)}{x \sqrt {c-a^2 c x^2}}dx}{a^2 c}-\frac {\sqrt {c-a^2 c x^2}}{c}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle -c \left (\frac {\int \frac {a^2 c (1-2 a x)}{x \sqrt {c-a^2 c x^2}}dx}{a^2 c}-\frac {\sqrt {c-a^2 c x^2}}{c}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle -c \left (\int \frac {1-2 a x}{x \sqrt {c-a^2 c x^2}}dx-\frac {\sqrt {c-a^2 c x^2}}{c}\right )\)

\(\Big \downarrow \) 538

\(\displaystyle -c \left (-2 a \int \frac {1}{\sqrt {c-a^2 c x^2}}dx+\int \frac {1}{x \sqrt {c-a^2 c x^2}}dx-\frac {\sqrt {c-a^2 c x^2}}{c}\right )\)

\(\Big \downarrow \) 224

\(\displaystyle -c \left (\int \frac {1}{x \sqrt {c-a^2 c x^2}}dx-2 a \int \frac {1}{\frac {a^2 c x^2}{c-a^2 c x^2}+1}d\frac {x}{\sqrt {c-a^2 c x^2}}-\frac {\sqrt {c-a^2 c x^2}}{c}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle -c \left (\int \frac {1}{x \sqrt {c-a^2 c x^2}}dx-\frac {2 \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{\sqrt {c}}-\frac {\sqrt {c-a^2 c x^2}}{c}\right )\)

\(\Big \downarrow \) 243

\(\displaystyle -c \left (\frac {1}{2} \int \frac {1}{x^2 \sqrt {c-a^2 c x^2}}dx^2-\frac {2 \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{\sqrt {c}}-\frac {\sqrt {c-a^2 c x^2}}{c}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle -c \left (-\frac {\int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2 c}}d\sqrt {c-a^2 c x^2}}{a^2 c}-\frac {2 \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{\sqrt {c}}-\frac {\sqrt {c-a^2 c x^2}}{c}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle -c \left (-\frac {2 \arctan \left (\frac {a \sqrt {c} x}{\sqrt {c-a^2 c x^2}}\right )}{\sqrt {c}}-\frac {\text {arctanh}\left (\frac {\sqrt {c-a^2 c x^2}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {\sqrt {c-a^2 c x^2}}{c}\right )\)

Input:

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

Output:

-(c*(-(Sqrt[c - a^2*c*x^2]/c) - (2*ArcTan[(a*Sqrt[c]*x)/Sqrt[c - a^2*c*x^2 
]])/Sqrt[c] - ArcTanh[Sqrt[c - a^2*c*x^2]/Sqrt[c]]/Sqrt[c]))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 216

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

rule 221

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

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 538

Int[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp 
[c   Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d   Int[1/Sqrt[a + b*x^2], x] 
, x] /; FreeQ[{a, b, c, d}, x]
 

rule 541

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[d^n*x^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*(m + n + 2*p + 1))), x 
] + Simp[1/(b*(m + n + 2*p + 1))   Int[x^m*(a + b*x^2)^p*ExpandToSum[b*(m + 
 n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1)*x^n - a*d^n*(m + n - 1) 
*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, m, p}, x] && IGtQ[n, 1] && IGt 
Q[m, -2] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 6702

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Simp[1/c^(n/2)   Int[x^m*((c + d*x^2)^(p + n/2)/(1 - a*x)^n), x] 
, x] /; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || 
 GtQ[c, 0]) && ILtQ[n/2, 0]
 

rule 6717

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Simp[(-1)^(n/2)   Int[ 
u*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[a, x] && IntegerQ[n/2]
 

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.61

method	result	size
default	\(-\sqrt {-a^{2} c \,x^{2}+c}+\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {-a^{2} c \,x^{2}+c}}{x}\right )+2 \sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 \left (x +\frac {1}{a}\right ) a c}+\frac {2 a c \arctan \left (\frac {\sqrt {a^{2} c}\, x}{\sqrt {-\left (x +\frac {1}{a}\right )^{2} a^{2} c +2 \left (x +\frac {1}{a}\right ) a c}}\right )}{\sqrt {a^{2} c}}\)	\(121\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.41 \[ \int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=\left [-2 \, \sqrt {c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} a \sqrt {c} x}{a^{2} c x^{2} - c}\right ) + \frac {1}{2} \, \sqrt {c} \log \left (-\frac {a^{2} c x^{2} - 2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c} - 2 \, c}{x^{2}}\right ) + \sqrt {-a^{2} c x^{2} + c}, -\sqrt {-c} \arctan \left (\frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-c}}{c}\right ) + \sqrt {-c} \log \left (2 \, a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c x^{2} + c} a \sqrt {-c} x - c\right ) + \sqrt {-a^{2} c x^{2} + c}\right ] \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=\int \frac {\sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )} \left (a x - 1\right )}{x \left (a x + 1\right )}\, dx \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.15 \[ \int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=a^{2} {\left (\frac {\sqrt {c} \arcsin \left (a x\right )}{a^{2}} + \frac {\sqrt {-a^{2} c x^{2} + c}}{a^{2}}\right )} + a {\left (\frac {\sqrt {c} \arcsin \left (a x\right )}{a} + \frac {\sqrt {c} \log \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {c}}{{\left | x \right |}} + \frac {2 \, c}{{\left | x \right |}}\right )}{a}\right )} \] Input:

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

Output:

a^2*(sqrt(c)*arcsin(a*x)/a^2 + sqrt(-a^2*c*x^2 + c)/a^2) + a*(sqrt(c)*arcs 
in(a*x)/a + sqrt(c)*log(2*sqrt(-a^2*c*x^2 + c)*sqrt(c)/abs(x) + 2*c/abs(x) 
)/a)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.27 \[ \int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=-\frac {2 \, c \arctan \left (-\frac {\sqrt {-a^{2} c} x - \sqrt {-a^{2} c x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {2 \, a \sqrt {-c} \log \left ({\left | -\sqrt {-a^{2} c} x + \sqrt {-a^{2} c x^{2} + c} \right |}\right )}{{\left | a \right |}} + \sqrt {-a^{2} c x^{2} + c} \] Input:

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

Output:

-2*c*arctan(-(sqrt(-a^2*c)*x - sqrt(-a^2*c*x^2 + c))/sqrt(-c))/sqrt(-c) + 
2*a*sqrt(-c)*log(abs(-sqrt(-a^2*c)*x + sqrt(-a^2*c*x^2 + c)))/abs(a) + sqr 
t(-a^2*c*x^2 + c)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=\int \frac {\sqrt {c-a^2\,c\,x^2}\,\left (a\,x-1\right )}{x\,\left (a\,x+1\right )} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.43 \[ \int \frac {e^{-2 \coth ^{-1}(a x)} \sqrt {c-a^2 c x^2}}{x} \, dx=\sqrt {c}\, \left (2 \mathit {asin} \left (a x \right )+\sqrt {-a^{2} x^{2}+1}-\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right )-1\right ) \] Input:

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

Output:

sqrt(c)*(2*asin(a*x) + sqrt( - a**2*x**2 + 1) - log(tan(asin(a*x)/2)) - 1)