\(\int e^{-\coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2})^{3/2} (e x)^m \, dx\) [908]

Optimal result

Integrand size = 29, antiderivative size = 191 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} (e x)^m \, dx=-\frac {c e^2 \sqrt {c-\frac {c}{a^2 x^2}} (e x)^{-2+m}}{a^3 (2-m) \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {c e \sqrt {c-\frac {c}{a^2 x^2}} (e x)^{-1+m}}{a^2 (1-m) \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {c \sqrt {c-\frac {c}{a^2 x^2}} (e x)^m}{a m \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {c \sqrt {c-\frac {c}{a^2 x^2}} (e x)^{1+m}}{e (1+m) \sqrt {1-\frac {1}{a^2 x^2}}} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.65 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} (e x)^m \, dx=\frac {\left (c-\frac {c}{a^2 x^2}\right )^{3/2} x^{-m} (e x)^m \left (\frac {x^{-2+m} (-1+a x)^3}{2-m}-\frac {(-a (2-m)+a (1+m)) \left (-\frac {x^{-1+m}}{1-m}-\frac {2 a x^m}{m}+\frac {a^2 x^{1+m}}{1+m}\right )}{2-m}\right )}{a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.58, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {6751, 6747, 8, 84, 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.

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

\(\Big \downarrow \) 6751

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

\(\Big \downarrow \) 6747

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

\(\Big \downarrow \) 8

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

\(\Big \downarrow \) 84

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(c*e^3*Sqrt[c - c/(a^2*x^2)]*(-((e*x)^(-2 + m)/(e*(2 - m))) + (a*(e*x)^(-1 
 + m))/(e^2*(1 - m)) - (a^2*(e*x)^m)/(e^3*m) + (a^3*(e*x)^(1 + m))/(e^4*(1 
 + m))))/(a^3*Sqrt[1 - 1/(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 84

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : 
> Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, 
 d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n + p + 2, 0 
] && GtQ[n + 2*p, 0])
 

rule 2009

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

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 [A] (verified)

Time = 0.08 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.84

method	result	size
orering	\(\frac {\left (a^{3} m^{3} x^{3}-3 a^{3} m^{2} x^{3}+2 a^{3} m \,x^{3}-a^{2} x^{2} m^{3}+2 a^{2} m^{2} x^{2}+a^{2} m \,x^{2}-a \,m^{3} x -2 a^{2} x^{2}+a x \,m^{2}+2 a m x +m^{3}-m \right ) x \left (c -\frac {c}{a^{2} x^{2}}\right )^{\frac {3}{2}} \left (e x \right )^{m} \sqrt {\frac {a x -1}{a x +1}}}{\left (1+m \right ) m \left (m -1\right ) \left (-2+m \right ) \left (a x +1\right ) \left (a x -1\right )^{2}}\)	\(161\)
gosper	\(\frac {x \left (a^{3} m^{3} x^{3}-3 a^{3} m^{2} x^{3}+2 a^{3} m \,x^{3}-a^{2} x^{2} m^{3}+2 a^{2} m^{2} x^{2}+a^{2} m \,x^{2}-a \,m^{3} x -2 a^{2} x^{2}+a x \,m^{2}+2 a m x +m^{3}-m \right ) {\left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )}^{\frac {3}{2}} \left (e x \right )^{m} \sqrt {\frac {a x -1}{a x +1}}}{\left (1+m \right ) m \left (m -1\right ) \left (-2+m \right ) \left (a x +1\right ) \left (a x -1\right )^{2}}\)	\(167\)
risch	\(\frac {\sqrt {\frac {a x -1}{a x +1}}\, \sqrt {c}\, \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{\left (a x -1\right ) \left (a x +1\right )}}\, \left (a x +1\right ) \left (a^{3} m^{3} x^{3}-3 a^{3} m^{2} x^{3}+2 a^{3} m \,x^{3}-a^{2} x^{2} m^{3}+2 a^{2} m^{2} x^{2}+a^{2} m \,x^{2}-a \,m^{3} x -2 a^{2} x^{2}+a x \,m^{2}+2 a m x +m^{3}-m \right ) \left (e x \right )^{m}}{\left (a^{2} x^{2}-1\right ) a^{2} x m \left (1+m \right ) \left (m -1\right ) \left (-2+m \right )}\)	\(204\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.09 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} (e x)^m \, dx=\frac {{\left (c m^{3} + {\left (a^{3} c m^{3} - 3 \, a^{3} c m^{2} + 2 \, a^{3} c m\right )} x^{3} - {\left (a^{2} c m^{3} - 2 \, a^{2} c m^{2} - a^{2} c m + 2 \, a^{2} c\right )} x^{2} - c m - {\left (a c m^{3} - a c m^{2} - 2 \, a c m\right )} x\right )} \left (e x\right )^{m} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{{\left (a^{3} m^{4} - 2 \, a^{3} m^{3} - a^{3} m^{2} + 2 \, a^{3} m\right )} x^{2} - {\left (a^{2} m^{4} - 2 \, a^{2} m^{3} - a^{2} m^{2} + 2 \, a^{2} m\right )} x} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.08 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} (e x)^m \, dx=\frac {{\left ({\left (m^{3} - 3 \, m^{2} + 2 \, m\right )} a^{3} c^{\frac {3}{2}} e^{m} x^{3} - {\left (m^{3} - 2 \, m^{2} - m + 2\right )} a^{2} c^{\frac {3}{2}} e^{m} x^{2} - {\left (m^{3} - m^{2} - 2 \, m\right )} a c^{\frac {3}{2}} e^{m} x + {\left (m^{3} - m\right )} c^{\frac {3}{2}} e^{m}\right )} {\left (a x + 1\right )} {\left (a x - 1\right )}^{2} x^{m}}{{\left (m^{4} - 2 \, m^{3} - m^{2} + 2 \, m\right )} a^{6} x^{5} - {\left (m^{4} - 2 \, m^{3} - m^{2} + 2 \, m\right )} a^{5} x^{4} - {\left (m^{4} - 2 \, m^{3} - m^{2} + 2 \, m\right )} a^{4} x^{3} + {\left (m^{4} - 2 \, m^{3} - m^{2} + 2 \, m\right )} a^{3} x^{2}} \] Input:

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

Output:

((m^3 - 3*m^2 + 2*m)*a^3*c^(3/2)*e^m*x^3 - (m^3 - 2*m^2 - m + 2)*a^2*c^(3/ 
2)*e^m*x^2 - (m^3 - m^2 - 2*m)*a*c^(3/2)*e^m*x + (m^3 - m)*c^(3/2)*e^m)*(a 
*x + 1)*(a*x - 1)^2*x^m/((m^4 - 2*m^3 - m^2 + 2*m)*a^6*x^5 - (m^4 - 2*m^3 
- m^2 + 2*m)*a^5*x^4 - (m^4 - 2*m^3 - m^2 + 2*m)*a^4*x^3 + (m^4 - 2*m^3 - 
m^2 + 2*m)*a^3*x^2)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.66 \[ \int e^{-\coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^{3/2} (e x)^m \, dx=\frac {x^{m} e^{m} \sqrt {c}\, c \left (a^{3} m^{3} x^{3}-3 a^{3} m^{2} x^{3}+2 a^{3} m \,x^{3}-a^{2} m^{3} x^{2}+2 a^{2} m^{2} x^{2}+a^{2} m \,x^{2}-a \,m^{3} x -2 a^{2} x^{2}+a \,m^{2} x +2 a m x +m^{3}-m \right )}{a^{3} m \,x^{2} \left (m^{3}-2 m^{2}-m +2\right )} \] Input:

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

Output:

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