\(\int \frac {e^{\coth ^{-1}(a x)} (e x)^m}{(c-a c x)^3} \, dx\) [387]

Optimal result

Integrand size = 21, antiderivative size = 201 \[ \int \frac {e^{\coth ^{-1}(a x)} (e x)^m}{(c-a c x)^3} \, dx=-\frac {(e x)^m}{a^3 c^3 (1+m) \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^2}+\frac {8 \left (a+\frac {1}{x}\right ) (e x)^m}{5 a^4 c^3 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^2}-\frac {\left (1+8 m-8 m^2\right ) (e x)^m \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {2-m}{2},\frac {4-m}{2},\frac {1}{a^2 x^2}\right )}{5 a^3 c^3 (2-m) (1+m) x^2}-\frac {4 (1-2 m) (e x)^m \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {3-m}{2},\frac {5-m}{2},\frac {1}{a^2 x^2}\right )}{5 a^4 c^3 (3-m) x^3} \] Output:

-(e*x)^m/a^3/c^3/(1+m)/(1-1/a^2/x^2)^(3/2)/x^2+8/5*(a+1/x)*(e*x)^m/a^4/c^3 
/(1-1/a^2/x^2)^(5/2)/x^2-1/5*(-8*m^2+8*m+1)*(e*x)^m*hypergeom([5/2, 1-1/2* 
m],[2-1/2*m],1/a^2/x^2)/a^3/c^3/(2-m)/(1+m)/x^2-4/5*(1-2*m)*(e*x)^m*hyperg 
eom([5/2, 3/2-1/2*m],[5/2-1/2*m],1/a^2/x^2)/a^4/c^3/(3-m)/x^3
 

Mathematica [C] (warning: unable to verify)

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

Time = 0.35 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.60 \[ \int \frac {e^{\coth ^{-1}(a x)} (e x)^m}{(c-a c x)^3} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x (e x)^m \sqrt {1-a x} \sqrt {\frac {1+a x}{a^2}} \left (\operatorname {AppellF1}\left (m,-\frac {1}{2},\frac {5}{2},1+m,-a x,a x\right )-\operatorname {AppellF1}\left (m,-\frac {1}{2},\frac {7}{2},1+m,-a x,a x\right )\right )}{c^3 m \sqrt {-1+a x} \sqrt {1+a x} \sqrt {-\frac {1}{a^2}+x^2}} \] Input:

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

Output:

-((Sqrt[1 - 1/(a^2*x^2)]*x*(e*x)^m*Sqrt[1 - a*x]*Sqrt[(1 + a*x)/a^2]*(Appe 
llF1[m, -1/2, 5/2, 1 + m, -(a*x), a*x] - AppellF1[m, -1/2, 7/2, 1 + m, -(a 
*x), a*x]))/(c^3*m*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*Sqrt[-a^(-2) + x^2]))
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.31, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6729, 147, 2009}

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.

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

\(\Big \downarrow \) 6729

\(\displaystyle \frac {\left (\frac {1}{x}\right )^m (e x)^m \int \frac {\sqrt {1+\frac {1}{a x}} \left (\frac {1}{x}\right )^{1-m}}{\left (1-\frac {1}{a x}\right )^{7/2}}d\frac {1}{x}}{a^3 c^3}\)

\(\Big \downarrow \) 147

\(\displaystyle \frac {\left (\frac {1}{x}\right )^m (e x)^m \int \left (\frac {\left (\frac {1}{x}\right )^{1-m}}{\left (1-\frac {1}{a x}\right )^{7/2} \left (1+\frac {1}{a x}\right )^{7/2}}+\frac {4 \left (\frac {1}{x}\right )^{2-m}}{a \left (1-\frac {1}{a x}\right )^{7/2} \left (1+\frac {1}{a x}\right )^{7/2}}+\frac {6 \left (\frac {1}{x}\right )^{3-m}}{a^2 \left (1-\frac {1}{a x}\right )^{7/2} \left (1+\frac {1}{a x}\right )^{7/2}}+\frac {4 \left (\frac {1}{x}\right )^{4-m}}{a^3 \left (1-\frac {1}{a x}\right )^{7/2} \left (1+\frac {1}{a x}\right )^{7/2}}+\frac {\left (\frac {1}{x}\right )^{5-m}}{a^4 \left (1-\frac {1}{a x}\right )^{7/2} \left (1+\frac {1}{a x}\right )^{7/2}}\right )d\frac {1}{x}}{a^3 c^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (\frac {1}{x}\right )^m (e x)^m \left (\frac {\left (\frac {1}{x}\right )^{2-m} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {2-m}{2},\frac {4-m}{2},\frac {1}{a^2 x^2}\right )}{2-m}+\frac {4 \left (\frac {1}{x}\right )^{3-m} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {3-m}{2},\frac {5-m}{2},\frac {1}{a^2 x^2}\right )}{a (3-m)}+\frac {6 \left (\frac {1}{x}\right )^{4-m} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {4-m}{2},\frac {6-m}{2},\frac {1}{a^2 x^2}\right )}{a^2 (4-m)}+\frac {\left (\frac {1}{x}\right )^{6-m} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {6-m}{2},\frac {8-m}{2},\frac {1}{a^2 x^2}\right )}{a^4 (6-m)}+\frac {4 \left (\frac {1}{x}\right )^{5-m} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},\frac {5-m}{2},\frac {7-m}{2},\frac {1}{a^2 x^2}\right )}{a^3 (5-m)}\right )}{a^3 c^3}\)

Input:

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

Output:

((x^(-1))^m*(e*x)^m*(((x^(-1))^(2 - m)*Hypergeometric2F1[7/2, (2 - m)/2, ( 
4 - m)/2, 1/(a^2*x^2)])/(2 - m) + (4*(x^(-1))^(3 - m)*Hypergeometric2F1[7/ 
2, (3 - m)/2, (5 - m)/2, 1/(a^2*x^2)])/(a*(3 - m)) + (6*(x^(-1))^(4 - m)*H 
ypergeometric2F1[7/2, (4 - m)/2, (6 - m)/2, 1/(a^2*x^2)])/(a^2*(4 - m)) + 
(4*(x^(-1))^(5 - m)*Hypergeometric2F1[7/2, (5 - m)/2, (7 - m)/2, 1/(a^2*x^ 
2)])/(a^3*(5 - m)) + ((x^(-1))^(6 - m)*Hypergeometric2F1[7/2, (6 - m)/2, ( 
8 - m)/2, 1/(a^2*x^2)])/(a^4*(6 - m))))/(a^3*c^3)
 

Defintions of rubi rules used

rule 147

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6729

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(-(a*x + 1)*(e*x)^m*sqrt((a*x - 1)/(a*x + 1))/(a^4*c^3*x^4 - 4*a^3 
*c^3*x^3 + 6*a^2*c^3*x^2 - 4*a*c^3*x + c^3), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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