\(\int \frac {e^{\coth ^{-1}(a x)} (e x)^m}{\sqrt {c-a c x}} \, dx\) [391]

Optimal result

Integrand size = 23, antiderivative size = 74 \[ \int \frac {e^{\coth ^{-1}(a x)} (e x)^m}{\sqrt {c-a c x}} \, dx=\frac {2 \sqrt {1-\frac {1}{a x}} x (e x)^m \operatorname {AppellF1}\left (-\frac {1}{2}-m,1,-\frac {1}{2},\frac {1}{2}-m,\frac {1}{a x},-\frac {1}{a x}\right )}{(1+2 m) \sqrt {c-a c x}} \] Output:

2*(1-1/a/x)^(1/2)*x*(e*x)^m*AppellF1(-1/2-m,1,-1/2,1/2-m,1/a/x,-1/a/x)/(1+ 
2*m)/(-a*c*x+c)^(1/2)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6730, 27, 150}

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.

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

\(\Big \downarrow \) 6730

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 150

\(\displaystyle \frac {2 x \sqrt {1-\frac {1}{a x}} (e x)^m \operatorname {AppellF1}\left (-m-\frac {1}{2},-\frac {1}{2},1,\frac {1}{2}-m,-\frac {1}{a x},\frac {1}{a x}\right )}{(2 m+1) \sqrt {c-a c x}}\)

Input:

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

Output:

(2*Sqrt[1 - 1/(a*x)]*x*(e*x)^m*AppellF1[-1/2 - m, -1/2, 1, 1/2 - m, -(1/(a 
*x)), 1/(a*x)])/((1 + 2*m)*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 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 6730

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

Maple [F]

\[\int \frac {\left (e x \right )^{m}}{\sqrt {\frac {a x -1}{a x +1}}\, \sqrt {-a c x +c}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(-sqrt(-a*c*x + c)*(a*x + 1)*(e*x)^m*sqrt((a*x - 1)/(a*x + 1))/(a^ 
2*c*x^2 - 2*a*c*x + c), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(e**m*int((x**m*sqrt(a*x + 1))/(sqrt(a*x - 1)*sqrt( - a*x + 1)),x))/sqrt(c 
)