\(\int \frac {e^{-\coth ^{-1}(a x)} (e x)^m}{(c-\frac {c}{a^2 x^2})^{3/2}} \, dx\) [911]

Optimal result

Integrand size = 29, antiderivative size = 150 \[ \int \frac {e^{-\coth ^{-1}(a x)} (e x)^m}{\left (c-\frac {c}{a^2 x^2}\right )^{3/2}} \, dx=-\frac {a^3 \sqrt {1-\frac {1}{a^2 x^2}} (e x)^{4+m} \operatorname {Hypergeometric2F1}\left (2,\frac {4+m}{2},\frac {6+m}{2},a^2 x^2\right )}{c e^4 (4+m) \sqrt {c-\frac {c}{a^2 x^2}}}+\frac {a^4 \sqrt {1-\frac {1}{a^2 x^2}} (e x)^{5+m} \operatorname {Hypergeometric2F1}\left (2,\frac {5+m}{2},\frac {7+m}{2},a^2 x^2\right )}{c e^5 (5+m) \sqrt {c-\frac {c}{a^2 x^2}}} \] Output:

-a^3*(1-1/a^2/x^2)^(1/2)*(e*x)^(4+m)*hypergeom([2, 2+1/2*m],[3+1/2*m],a^2* 
x^2)/c/e^4/(4+m)/(c-c/a^2/x^2)^(1/2)+a^4*(1-1/a^2/x^2)^(1/2)*(e*x)^(5+m)*h 
ypergeom([2, 5/2+1/2*m],[7/2+1/2*m],a^2*x^2)/c/e^5/(5+m)/(c-c/a^2/x^2)^(1/ 
2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-\coth ^{-1}(a x)} (e x)^m}{\left (c-\frac {c}{a^2 x^2}\right )^{3/2}} \, dx=\frac {a^3 \sqrt {1-\frac {1}{a^2 x^2}} x^4 (e x)^m \left (-\left ((5+m) \operatorname {Hypergeometric2F1}\left (2,2+\frac {m}{2},3+\frac {m}{2},a^2 x^2\right )\right )+a (4+m) x \operatorname {Hypergeometric2F1}\left (2,\frac {5+m}{2},\frac {7+m}{2},a^2 x^2\right )\right )}{c (4+m) (5+m) \sqrt {c-\frac {c}{a^2 x^2}}} \] Input:

Integrate[(e*x)^m/(E^ArcCoth[a*x]*(c - c/(a^2*x^2))^(3/2)),x]
 

Output:

(a^3*Sqrt[1 - 1/(a^2*x^2)]*x^4*(e*x)^m*(-((5 + m)*Hypergeometric2F1[2, 2 + 
 m/2, 3 + m/2, a^2*x^2]) + a*(4 + m)*x*Hypergeometric2F1[2, (5 + m)/2, (7 
+ m)/2, a^2*x^2]))/(c*(4 + m)*(5 + m)*Sqrt[c - c/(a^2*x^2)])
 

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.81, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {6751, 6747, 8, 25, 92, 82, 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.

\(\displaystyle \int \frac {e^{-\coth ^{-1}(a x)} (e x)^m}{\left (c-\frac {c}{a^2 x^2}\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 6751

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

\(\Big \downarrow \) 6747

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

\(\Big \downarrow \) 8

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 92

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

\(\Big \downarrow \) 82

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

\(\Big \downarrow \) 278

\(\displaystyle -\frac {a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (\frac {(e x)^{m+4} \operatorname {Hypergeometric2F1}\left (2,\frac {m+4}{2},\frac {m+6}{2},a^2 x^2\right )}{e (m+4)}-\frac {a (e x)^{m+5} \operatorname {Hypergeometric2F1}\left (2,\frac {m+5}{2},\frac {m+7}{2},a^2 x^2\right )}{e^2 (m+5)}\right )}{c e^3 \sqrt {c-\frac {c}{a^2 x^2}}}\)

Input:

Int[(e*x)^m/(E^ArcCoth[a*x]*(c - c/(a^2*x^2))^(3/2)),x]
 

Output:

-((a^3*Sqrt[1 - 1/(a^2*x^2)]*(((e*x)^(4 + m)*Hypergeometric2F1[2, (4 + m)/ 
2, (6 + m)/2, a^2*x^2])/(e*(4 + m)) - (a*(e*x)^(5 + m)*Hypergeometric2F1[2 
, (5 + m)/2, (7 + m)/2, a^2*x^2])/(e^2*(5 + m))))/(c*e^3*Sqrt[c - c/(a^2*x 
^2)]))
 

Defintions of rubi rules used

rule 8

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_))^(p_), x_Symbol] :> Simp[1/a^m   Int[u*(a* 
x)^(m + p), x], x] /; FreeQ[{a, m, p}, x] && IntegerQ[m]
 

rule 25

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

rule 82

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

rule 92

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

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 6747

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

rule 6751

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((e*x)^m*a^4*x^4*sqrt((a*x - 1)/(a*x + 1))*sqrt((a^2*c*x^2 - c)/(a 
^2*x^2))/(a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-\coth ^{-1}(a x)} (e x)^m}{\left (c-\frac {c}{a^2 x^2}\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

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

Output:

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

Reduce [F]

\[ \int \frac {e^{-\coth ^{-1}(a x)} (e x)^m}{\left (c-\frac {c}{a^2 x^2}\right )^{3/2}} \, dx=\frac {e^{m} \sqrt {c}\, \left (x^{m} a m x -x^{m} m -x^{m}-\left (\int \frac {x^{m}}{a^{3} x^{4}+a^{2} x^{3}-a \,x^{2}-x}d x \right ) m^{2}-\left (\int \frac {x^{m}}{a^{3} x^{4}+a^{2} x^{3}-a \,x^{2}-x}d x \right ) m +2 \left (\int \frac {x^{m} x}{a^{3} x^{3}+a^{2} x^{2}-a x -1}d x \right ) a^{2} m^{2}+2 \left (\int \frac {x^{m} x}{a^{3} x^{3}+a^{2} x^{2}-a x -1}d x \right ) a^{2} m \right )}{a \,c^{2} m \left (m +1\right )} \] Input:

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

Output:

(e**m*sqrt(c)*(x**m*a*m*x - x**m*m - x**m - int(x**m/(a**3*x**4 + a**2*x** 
3 - a*x**2 - x),x)*m**2 - int(x**m/(a**3*x**4 + a**2*x**3 - a*x**2 - x),x) 
*m + 2*int((x**m*x)/(a**3*x**3 + a**2*x**2 - a*x - 1),x)*a**2*m**2 + 2*int 
((x**m*x)/(a**3*x**3 + a**2*x**2 - a*x - 1),x)*a**2*m))/(a*c**2*m*(m + 1))