\(\int e^{\coth ^{-1}(a x)} (e x)^m (c-a c x)^3 \, dx\) [382]

Optimal result

Integrand size = 21, antiderivative size = 145 \[ \int e^{\coth ^{-1}(a x)} (e x)^m (c-a c x)^3 \, dx=\frac {a^3 c^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^4 (e x)^m}{1+m}-\frac {a^3 c^3 (5+2 m) x^4 (e x)^m \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (-4-m),\frac {1}{2} (-2-m),\frac {1}{a^2 x^2}\right )}{(1+m) (4+m)}+\frac {2 a^2 c^3 x^3 (e x)^m \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (-3-m),\frac {1}{2} (-1-m),\frac {1}{a^2 x^2}\right )}{3+m} \] Output:

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.83 \[ \int e^{\coth ^{-1}(a x)} (e x)^m (c-a c x)^3 \, dx=a c^3 x^2 (e x)^m \left (-\frac {a^2 x^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-2-\frac {m}{2},-1-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{4+m}+\frac {2 a x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {3}{2}-\frac {m}{2},-\frac {1}{2}-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{3+m}-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-1-\frac {m}{2},-\frac {m}{2},\frac {1}{a^2 x^2}\right )}{2+m}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.07, 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.

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

\(\Big \downarrow \) 6729

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

\(\Big \downarrow \) 147

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

a^3*c^3*(x^(-1))^m*(e*x)^m*(-(((x^(-1))^(-4 - m)*Hypergeometric2F1[-1/2, ( 
-4 - m)/2, (-2 - m)/2, 1/(a^2*x^2)])/(4 + m)) + (2*(x^(-1))^(-3 - m)*Hyper 
geometric2F1[-1/2, (-3 - m)/2, (-1 - m)/2, 1/(a^2*x^2)])/(a*(3 + m)) - ((x 
^(-1))^(-2 - m)*Hypergeometric2F1[-1/2, (-2 - m)/2, -1/2*m, 1/(a^2*x^2)])/ 
(a^2*(2 + m)))
 

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]

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

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]

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

Giac [F(-2)]

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

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

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

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

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

Output:

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