\(\int \frac {e^{\coth ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx\) [231]

Optimal result

Integrand size = 18, antiderivative size = 119 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx=\frac {2 \sqrt {1-\frac {1}{a x}} \sqrt {1+\frac {1}{a x}} x}{\sqrt {c-a c x}}-\frac {2 \sqrt {2} \sqrt {1-\frac {1}{a x}} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}\right )}{a \sqrt {\frac {1}{a x}} \sqrt {c-a c x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx=\frac {2 \sqrt {1-\frac {1}{a x}} x \left (\sqrt {a} \sqrt {1+\frac {1}{a x}}-\sqrt {2} \sqrt {\frac {1}{x}} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {a} \sqrt {1+\frac {1}{a x}}}\right )\right )}{\sqrt {a} \sqrt {c-a c x}} \] Input:

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

Output:

(2*Sqrt[1 - 1/(a*x)]*x*(Sqrt[a]*Sqrt[1 + 1/(a*x)] - Sqrt[2]*Sqrt[x^(-1)]*A 
rcTanh[(Sqrt[2]*Sqrt[x^(-1)])/(Sqrt[a]*Sqrt[1 + 1/(a*x)])]))/(Sqrt[a]*Sqrt 
[c - a*c*x])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6727, 27, 105, 104, 219}

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.

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

\(\Big \downarrow \) 6727

\(\displaystyle -\frac {\sqrt {1-\frac {1}{a x}} \int \frac {a \sqrt {1+\frac {1}{a x}}}{\left (a-\frac {1}{x}\right ) \left (\frac {1}{x}\right )^{3/2}}d\frac {1}{x}}{\sqrt {\frac {1}{x}} \sqrt {c-a c x}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {a \sqrt {1-\frac {1}{a x}} \int \frac {\sqrt {1+\frac {1}{a x}}}{\left (a-\frac {1}{x}\right ) \left (\frac {1}{x}\right )^{3/2}}d\frac {1}{x}}{\sqrt {\frac {1}{x}} \sqrt {c-a c x}}\)

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 219

\(\displaystyle -\frac {a \sqrt {1-\frac {1}{a x}} \left (\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\frac {1}{x}}}{\sqrt {a} \sqrt {\frac {1}{a x}+1}}\right )}{a^{3/2}}-\frac {2 \sqrt {\frac {1}{a x}+1}}{a \sqrt {\frac {1}{x}}}\right )}{\sqrt {\frac {1}{x}} \sqrt {c-a c x}}\)

Input:

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

Output:

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

Defintions of rubi rules used

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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f)))   Int[(a 
+ b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, 
e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] 
 ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 6727

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_), x_Symbol] :> Si 
mp[(-(1/x)^p)*((c + d*x)^p/(1 + c/(d*x))^p)   Subst[Int[((1 + c*(x/d))^p*(( 
1 + x/a)^(n/2)/x^(p + 2)))/(1 - x/a)^(n/2), x], x, 1/x], x] /; FreeQ[{a, c, 
 d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] &&  !IntegerQ[p]
 

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.69

method	result	size
default	\(-\frac {2 \sqrt {-c \left (a x -1\right )}\, \left (-\sqrt {c}\, \sqrt {2}\, \arctan \left (\frac {\sqrt {-c \left (a x +1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )+\sqrt {-c \left (a x +1\right )}\right )}{\sqrt {\frac {a x -1}{a x +1}}\, \sqrt {-c \left (a x +1\right )}\, c a}\)	\(82\)
risch	\(\frac {2 a x -2}{a \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {-c \left (a x -1\right )}}+\frac {2 \sqrt {2}\, \arctan \left (\frac {\sqrt {-a c x -c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {-c \left (a x +1\right )}\, \left (a x -1\right )}{a \sqrt {c}\, \sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) \sqrt {-c \left (a x -1\right )}}\)	\(115\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.06 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx=\left [\frac {\sqrt {2} {\left (a c x - c\right )} \sqrt {-\frac {1}{c}} \log \left (-\frac {a^{2} x^{2} - 2 \, \sqrt {2} \sqrt {-a c x + c} {\left (a x + 1\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {-\frac {1}{c}} + 2 \, a x - 3}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 2 \, \sqrt {-a c x + c} {\left (a x + 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c x - a c}, -\frac {2 \, {\left (\sqrt {-a c x + c} {\left (a x + 1\right )} \sqrt {\frac {a x - 1}{a x + 1}} - \frac {\sqrt {2} {\left (a c x - c\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x + c} {\left (a x + 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{2 \, {\left (a x - 1\right )} \sqrt {c}}\right )}{\sqrt {c}}\right )}}{a^{2} c x - a c}\right ] \] Input:

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

Output:

[(sqrt(2)*(a*c*x - c)*sqrt(-1/c)*log(-(a^2*x^2 - 2*sqrt(2)*sqrt(-a*c*x + c 
)*(a*x + 1)*sqrt((a*x - 1)/(a*x + 1))*sqrt(-1/c) + 2*a*x - 3)/(a^2*x^2 - 2 
*a*x + 1)) - 2*sqrt(-a*c*x + c)*(a*x + 1)*sqrt((a*x - 1)/(a*x + 1)))/(a^2* 
c*x - a*c), -2*(sqrt(-a*c*x + c)*(a*x + 1)*sqrt((a*x - 1)/(a*x + 1)) - sqr 
t(2)*(a*c*x - c)*arctan(1/2*sqrt(2)*sqrt(-a*c*x + c)*(a*x + 1)*sqrt((a*x - 
 1)/(a*x + 1))/((a*x - 1)*sqrt(c)))/sqrt(c))/(a^2*c*x - a*c)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.79 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx=\frac {2 \, c {\left (\frac {\sqrt {2} {\left (\sqrt {c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {c}}\right ) - \sqrt {-c}\right )}}{c} - \frac {\sqrt {2} \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {-a c x - c}}{2 \, \sqrt {c}}\right ) - \sqrt {-a c x - c}}{c}\right )}}{a {\left | c \right |} \mathrm {sgn}\left (a x + 1\right )} \] Input:

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

Output:

2*c*(sqrt(2)*(sqrt(c)*arctan(sqrt(-c)/sqrt(c)) - sqrt(-c))/c - (sqrt(2)*sq 
rt(c)*arctan(1/2*sqrt(2)*sqrt(-a*c*x - c)/sqrt(c)) - sqrt(-a*c*x - c))/c)/ 
(a*abs(c)*sgn(a*x + 1))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.35 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\sqrt {c-a c x}} \, dx=\frac {2 \sqrt {c}\, i \left (\sqrt {a x +1}+\sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-a x +1}}{\sqrt {2}}\right )}{2}\right )\right )-\sqrt {2}\right )}{a c} \] Input:

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

Output:

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