\(\int e^{-3 \coth ^{-1}(a x)} (c x)^m \, dx\) [147]

Optimal result

Integrand size = 14, antiderivative size = 120 \[ \int e^{-3 \coth ^{-1}(a x)} (c x)^m \, dx=-\frac {4 \left (a-\frac {1}{x}\right ) x (c x)^m}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {(5+4 m) x (c x)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1-m),\frac {1-m}{2},\frac {1}{a^2 x^2}\right )}{1+m}-\frac {(3+4 m) (c x)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{a m} \] Output:

-4*(a-1/x)*x*(c*x)^m/a/(1-1/a^2/x^2)^(1/2)+(5+4*m)*x*(c*x)^m*hypergeom([1/ 
2, -1/2-1/2*m],[1/2-1/2*m],1/a^2/x^2)/(1+m)-(3+4*m)*(c*x)^m*hypergeom([1/2 
, -1/2*m],[1-1/2*m],1/a^2/x^2)/a/m
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.32 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.61 \[ \int e^{-3 \coth ^{-1}(a x)} (c x)^m \, dx=\frac {x (c x)^m \left (-3 (1+m) \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {\frac {-1+a x}{a^2}} \operatorname {AppellF1}\left (m,-\frac {1}{2},\frac {1}{2},1+m,a x,-a x\right )+2 (1+m) \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {\frac {-1+a x}{a^2}} \operatorname {AppellF1}\left (m,-\frac {1}{2},\frac {3}{2},1+m,a x,-a x\right )+m \sqrt {1-a x} \sqrt {-\frac {1}{a^2}+x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{2}-\frac {m}{2},\frac {1}{2}-\frac {m}{2},\frac {1}{a^2 x^2}\right )\right )}{m (1+m) \sqrt {1-a x} \sqrt {-\frac {1}{a^2}+x^2}} \] Input:

Integrate[(c*x)^m/E^(3*ArcCoth[a*x]),x]
 

Output:

(x*(c*x)^m*(-3*(1 + m)*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[(-1 + a*x)/a^2]*AppellF1 
[m, -1/2, 1/2, 1 + m, a*x, -(a*x)] + 2*(1 + m)*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[ 
(-1 + a*x)/a^2]*AppellF1[m, -1/2, 3/2, 1 + m, a*x, -(a*x)] + m*Sqrt[1 - a* 
x]*Sqrt[-a^(-2) + x^2]*Hypergeometric2F1[-1/2, -1/2 - m/2, 1/2 - m/2, 1/(a 
^2*x^2)]))/(m*(1 + m)*Sqrt[1 - a*x]*Sqrt[-a^(-2) + x^2])
 

Rubi [A] (verified)

Time = 1.14 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.50, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6722, 27, 2355, 557, 278, 583, 557, 278}

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.

\(\displaystyle \int e^{-3 \coth ^{-1}(a x)} (c x)^m \, dx\)

\(\Big \downarrow \) 6722

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2355

\(\displaystyle -\frac {\left (\frac {1}{x}\right )^m (c x)^m \left (4 a^2 \int \frac {\left (\frac {1}{x}\right )^{-m-2}}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a+\frac {1}{x}\right )}d\frac {1}{x}+\int \frac {\left (\frac {1}{x}-3 a\right ) \left (\frac {1}{x}\right )^{-m-2}}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}\right )}{a}\)

\(\Big \downarrow \) 557

\(\displaystyle -\frac {\left (\frac {1}{x}\right )^m (c x)^m \left (4 a^2 \int \frac {\left (\frac {1}{x}\right )^{-m-2}}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a+\frac {1}{x}\right )}d\frac {1}{x}-3 a \int \frac {\left (\frac {1}{x}\right )^{-m-2}}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}+\int \frac {\left (\frac {1}{x}\right )^{-m-1}}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}\right )}{a}\)

\(\Big \downarrow \) 278

\(\displaystyle -\frac {\left (\frac {1}{x}\right )^m (c x)^m \left (4 a^2 \int \frac {\left (\frac {1}{x}\right )^{-m-2}}{\sqrt {1-\frac {1}{a^2 x^2}} \left (a+\frac {1}{x}\right )}d\frac {1}{x}+\frac {3 a \left (\frac {1}{x}\right )^{-m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-1),\frac {1-m}{2},\frac {1}{a^2 x^2}\right )}{m+1}-\frac {\left (\frac {1}{x}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{m}\right )}{a}\)

\(\Big \downarrow \) 583

\(\displaystyle -\frac {\left (\frac {1}{x}\right )^m (c x)^m \left (4 \int \frac {\left (a-\frac {1}{x}\right ) \left (\frac {1}{x}\right )^{-m-2}}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}d\frac {1}{x}+\frac {3 a \left (\frac {1}{x}\right )^{-m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-1),\frac {1-m}{2},\frac {1}{a^2 x^2}\right )}{m+1}-\frac {\left (\frac {1}{x}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{m}\right )}{a}\)

\(\Big \downarrow \) 557

\(\displaystyle -\frac {\left (\frac {1}{x}\right )^m (c x)^m \left (4 \left (a \int \frac {\left (\frac {1}{x}\right )^{-m-2}}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}d\frac {1}{x}-\int \frac {\left (\frac {1}{x}\right )^{-m-1}}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}d\frac {1}{x}\right )+\frac {3 a \left (\frac {1}{x}\right )^{-m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-1),\frac {1-m}{2},\frac {1}{a^2 x^2}\right )}{m+1}-\frac {\left (\frac {1}{x}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{m}\right )}{a}\)

\(\Big \downarrow \) 278

\(\displaystyle -\frac {\left (\frac {1}{x}\right )^m (c x)^m \left (\frac {3 a \left (\frac {1}{x}\right )^{-m-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-m-1),\frac {1-m}{2},\frac {1}{a^2 x^2}\right )}{m+1}-\frac {\left (\frac {1}{x}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{m}+4 \left (\frac {\left (\frac {1}{x}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {m}{2},1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{m}-\frac {a \left (\frac {1}{x}\right )^{-m-1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{2} (-m-1),\frac {1-m}{2},\frac {1}{a^2 x^2}\right )}{m+1}\right )\right )}{a}\)

Input:

Int[(c*x)^m/E^(3*ArcCoth[a*x]),x]
 

Output:

-(((x^(-1))^m*(c*x)^m*((3*a*(x^(-1))^(-1 - m)*Hypergeometric2F1[1/2, (-1 - 
 m)/2, (1 - m)/2, 1/(a^2*x^2)])/(1 + m) - Hypergeometric2F1[1/2, -1/2*m, 1 
 - m/2, 1/(a^2*x^2)]/(m*(x^(-1))^m) + 4*(-((a*(x^(-1))^(-1 - m)*Hypergeome 
tric2F1[3/2, (-1 - m)/2, (1 - m)/2, 1/(a^2*x^2)])/(1 + m)) + Hypergeometri 
c2F1[3/2, -1/2*m, 1 - m/2, 1/(a^2*x^2)]/(m*(x^(-1))^m))))/a)
 

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 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 557

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym 
bol] :> Simp[c   Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e   Int[(e*x)^( 
m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
 

rule 583

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[c^(2*n)/a^n   Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ 
n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I 
LtQ[n, 0]
 

rule 2355

Int[(Px_)*((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2) 
^(p_.), x_Symbol] :> Int[PolynomialQuotient[Px, c + d*x, x]*(e*x)^m*(c + d* 
x)^(n + 1)*(a + b*x^2)^p, x] + Simp[PolynomialRemainder[Px, c + d*x, x]   I 
nt[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p} 
, x] && PolynomialQ[Px, x] && LtQ[n, 0]
 

rule 6722

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_.)*(x_))^(m_), x_Symbol] :> Simp[(-(c 
*x)^m)*(1/x)^m   Subst[Int[(1 + x/a)^((n + 1)/2)/(x^(m + 2)*(1 - x/a)^((n - 
 1)/2)*Sqrt[1 - x^2/a^2]), x], x, 1/x], x] /; FreeQ[{a, c, m}, x] && Intege 
rQ[(n - 1)/2] &&  !IntegerQ[m]
 

Maple [F]

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

Input:

int((x*c)^m*((a*x-1)/(a*x+1))^(3/2),x)
 

Output:

int((x*c)^m*((a*x-1)/(a*x+1))^(3/2),x)
 

Fricas [F]

\[ \int e^{-3 \coth ^{-1}(a x)} (c x)^m \, dx=\int { \left (c x\right )^{m} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} \,d x } \] Input:

integrate((c*x)^m*((a*x-1)/(a*x+1))^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} (c x)^m \, dx=\text {Timed out} \] Input:

integrate((c*x)**m*((a*x-1)/(a*x+1))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int e^{-3 \coth ^{-1}(a x)} (c x)^m \, dx=\int { \left (c x\right )^{m} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} \,d x } \] Input:

integrate((c*x)^m*((a*x-1)/(a*x+1))^(3/2),x, algorithm="maxima")
 

Output:

integrate((c*x)^m*((a*x - 1)/(a*x + 1))^(3/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int e^{-3 \coth ^{-1}(a x)} (c x)^m \, dx=\text {Exception raised: TypeError} \] Input:

integrate((c*x)^m*((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int e^{-3 \coth ^{-1}(a x)} (c x)^m \, dx=\int {\left (c\,x\right )}^m\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2} \,d x \] Input:

int((c*x)^m*((a*x - 1)/(a*x + 1))^(3/2),x)
 

Output:

int((c*x)^m*((a*x - 1)/(a*x + 1))^(3/2), x)
 

Reduce [F]

\[ \int e^{-3 \coth ^{-1}(a x)} (c x)^m \, dx=c^{m} \left (\left (\int \frac {x^{m} \sqrt {a x -1}\, x}{\sqrt {a x +1}\, a x +\sqrt {a x +1}}d x \right ) a -\left (\int \frac {x^{m} \sqrt {a x -1}}{\sqrt {a x +1}\, a x +\sqrt {a x +1}}d x \right )\right ) \] Input:

int((c*x)^m*((a*x-1)/(a*x+1))^(3/2),x)
                                                                                    
                                                                                    
 

Output:

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