\(\int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} (e x)^m \, dx\) [554]

Optimal result
Mathematica [A] (verified)
Rubi [F]
Maple [F]
Fricas [F]
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 27, antiderivative size = 62 \[ \int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} (e x)^m \, dx=\frac {c \sqrt {1-\frac {1}{a x}} x (e x)^m \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-1-m,-m,-\frac {1}{a x}\right )}{(1+m) \sqrt {c-\frac {c}{a x}}} \] Output:

c*(1-1/a/x)^(1/2)*x*(e*x)^m*hypergeom([-1/2, -1-m],[-m],-1/a/x)/(1+m)/(c-c 
/a/x)^(1/2)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.98 \[ \int e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}} (e x)^m \, dx=\frac {\sqrt {c-\frac {c}{a x}} x (e x)^m \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-1-m,-m,-\frac {1}{a x}\right )}{(1+m) \sqrt {1-\frac {1}{a x}}} \] Input:

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

Output:

(Sqrt[c - c/(a*x)]*x*(e*x)^m*Hypergeometric2F1[-1/2, -1 - m, -m, -(1/(a*x) 
)])/((1 + m)*Sqrt[1 - 1/(a*x)])
 

Rubi [F]

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.

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

\(\Big \downarrow \) 6736

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

\(\Big \downarrow \) 7268

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

\(\Big \downarrow \) 2044

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

\(\Big \downarrow \) 7299

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 2044
Int[(u_.)*((c_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Simp[S 
imp[(c*(a + b*x^n)^q)^p/(a + b*x^n)^(p*q)]   Int[u*(a + b*x^n)^(p*q), x], x 
] /; FreeQ[{a, b, c, n, p, q}, x] && GeQ[a, 0]
 

rule 6736
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_), x_Symbol] 
 :> Simp[(c + d/x)^p/(1 + d/(c*x))^p   Int[u*(1 + d/(c*x))^p*E^(n*ArcCoth[a 
*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !Int 
egerQ[n/2] &&  !(IntegerQ[p] || GtQ[c, 0])
 

rule 7268
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfQuotientOfLinears 
[u, x]}, Simp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/ls 
t[[2]])], x] /;  !FalseQ[lst]]
 

rule 7299
Int[u_, x_] :> CannotIntegrate[u, x]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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