\(\int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx\) [552]

Optimal result

Integrand size = 27, antiderivative size = 188 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {1888 a^3 c \sqrt {1-\frac {1}{a^2 x^2}}}{105 \sqrt {c-\frac {c}{a x}}}+\frac {472}{105} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}+\frac {59 a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}}{35 c}+\frac {2 a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{5/2}}{7 c^2}+\frac {a^3 \left (c-\frac {c}{a x}\right )^{7/2}}{c^3 \sqrt {1-\frac {1}{a^2 x^2}}} \] Output:

1888/105*a^3*c*(1-1/a^2/x^2)^(1/2)/(c-c/a/x)^(1/2)+472/105*a^3*(1-1/a^2/x^ 
2)^(1/2)*(c-c/a/x)^(1/2)+59/35*a^3*(1-1/a^2/x^2)^(1/2)*(c-c/a/x)^(3/2)/c+2 
/7*a^3*(1-1/a^2/x^2)^(1/2)*(c-c/a/x)^(5/2)/c^2+a^3*(c-c/a/x)^(7/2)/c^3/(1- 
1/a^2/x^2)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.41 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {2 a \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}} \left (-15+66 a x-167 a^2 x^2+668 a^3 x^3+1336 a^4 x^4\right )}{105 x^2 \left (-1+a^2 x^2\right )} \] Input:

Integrate[Sqrt[c - c/(a*x)]/(E^(3*ArcCoth[a*x])*x^4),x]
 

Output:

(2*a*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)]*(-15 + 66*a*x - 167*a^2*x^2 + 
 668*a^3*x^3 + 1336*a^4*x^4))/(105*x^2*(-1 + a^2*x^2))
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {6733, 581, 27, 669, 459, 459, 458}

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.

\(\displaystyle \int \frac {\sqrt {c-\frac {c}{a x}} e^{-3 \coth ^{-1}(a x)}}{x^4} \, dx\)

\(\Big \downarrow \) 6733

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

\(\Big \downarrow \) 581

\(\displaystyle -\frac {\frac {2 a^2 \int \frac {c^2 \left (9 a-\frac {2}{x}\right ) \left (c-\frac {c}{a x}\right )^{7/2}}{2 a \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}d\frac {1}{x}}{7 c^2}+\frac {2 a^3 \left (c-\frac {c}{a x}\right )^{9/2}}{7 c \sqrt {1-\frac {1}{a^2 x^2}}}}{c^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {1}{7} a \int \frac {\left (9 a-\frac {2}{x}\right ) \left (c-\frac {c}{a x}\right )^{7/2}}{\left (1-\frac {1}{a^2 x^2}\right )^{3/2}}d\frac {1}{x}+\frac {2 a^3 \left (c-\frac {c}{a x}\right )^{9/2}}{7 c \sqrt {1-\frac {1}{a^2 x^2}}}}{c^3}\)

\(\Big \downarrow \) 669

\(\displaystyle -\frac {\frac {1}{7} a \left (-\frac {59}{2} a c \int \frac {\left (c-\frac {c}{a x}\right )^{5/2}}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}-\frac {11 a^2 \left (c-\frac {c}{a x}\right )^{7/2}}{\sqrt {1-\frac {1}{a^2 x^2}}}\right )+\frac {2 a^3 \left (c-\frac {c}{a x}\right )^{9/2}}{7 c \sqrt {1-\frac {1}{a^2 x^2}}}}{c^3}\)

\(\Big \downarrow \) 459

\(\displaystyle -\frac {\frac {1}{7} a \left (-\frac {59}{2} a c \left (\frac {8}{5} c \int \frac {\left (c-\frac {c}{a x}\right )^{3/2}}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}+\frac {2}{5} a c \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}\right )-\frac {11 a^2 \left (c-\frac {c}{a x}\right )^{7/2}}{\sqrt {1-\frac {1}{a^2 x^2}}}\right )+\frac {2 a^3 \left (c-\frac {c}{a x}\right )^{9/2}}{7 c \sqrt {1-\frac {1}{a^2 x^2}}}}{c^3}\)

\(\Big \downarrow \) 459

\(\displaystyle -\frac {\frac {1}{7} a \left (-\frac {59}{2} a c \left (\frac {8}{5} c \left (\frac {4}{3} c \int \frac {\sqrt {c-\frac {c}{a x}}}{\sqrt {1-\frac {1}{a^2 x^2}}}d\frac {1}{x}+\frac {2}{3} a c \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}\right )+\frac {2}{5} a c \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}\right )-\frac {11 a^2 \left (c-\frac {c}{a x}\right )^{7/2}}{\sqrt {1-\frac {1}{a^2 x^2}}}\right )+\frac {2 a^3 \left (c-\frac {c}{a x}\right )^{9/2}}{7 c \sqrt {1-\frac {1}{a^2 x^2}}}}{c^3}\)

\(\Big \downarrow \) 458

\(\displaystyle -\frac {\frac {1}{7} a \left (-\frac {59}{2} a c \left (\frac {8}{5} c \left (\frac {8 a c^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 \sqrt {c-\frac {c}{a x}}}+\frac {2}{3} a c \sqrt {1-\frac {1}{a^2 x^2}} \sqrt {c-\frac {c}{a x}}\right )+\frac {2}{5} a c \sqrt {1-\frac {1}{a^2 x^2}} \left (c-\frac {c}{a x}\right )^{3/2}\right )-\frac {11 a^2 \left (c-\frac {c}{a x}\right )^{7/2}}{\sqrt {1-\frac {1}{a^2 x^2}}}\right )+\frac {2 a^3 \left (c-\frac {c}{a x}\right )^{9/2}}{7 c \sqrt {1-\frac {1}{a^2 x^2}}}}{c^3}\)

Input:

Int[Sqrt[c - c/(a*x)]/(E^(3*ArcCoth[a*x])*x^4),x]
 

Output:

-(((a*((-59*a*c*((8*c*((8*a*c^2*Sqrt[1 - 1/(a^2*x^2)])/(3*Sqrt[c - c/(a*x) 
]) + (2*a*c*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)])/3))/5 + (2*a*c*Sqrt[1 
 - 1/(a^2*x^2)]*(c - c/(a*x))^(3/2))/5))/2 - (11*a^2*(c - c/(a*x))^(7/2))/ 
Sqrt[1 - 1/(a^2*x^2)]))/7 + (2*a^3*(c - c/(a*x))^(9/2))/(7*c*Sqrt[1 - 1/(a 
^2*x^2)]))/c^3)
 

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 458

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] /; FreeQ[{a, b, c 
, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + p, 0]
 

rule 459

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* 
(Simplify[n + p]/(n + 2*p + 1))   Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && IGtQ[Simplif 
y[n + p], 0]
 

rule 581

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[(c + d*x)^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*d^(m - 1)*(m + n + 
2*p + 1))), x] + Simp[1/(d^m*(m + n + 2*p + 1))   Int[(c + d*x)^n*(a + b*x^ 
2)^p*ExpandToSum[d^m*(m + n + 2*p + 1)*x^m - (m + n + 2*p + 1)*(c + d*x)^m 
+ c*(c + d*x)^(m - 2)*(c*(m + n - 1) + c*(m + n + 2*p + 1) + 2*d*(m + n + p 
)*x), x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] & 
& IGtQ[m, 1] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[2*p] || ILtQ[m + n, 0] 
)
 

rule 669

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^( 
p_), x_Symbol] :> Simp[(d*g + e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d* 
(p + 1))), x] - Simp[e*((m*(d*g + e*f) + 2*e*f*(p + 1))/(2*c*d*(p + 1))) 
Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, 
g}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0]
 

rule 6733

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_S 
ymbol] :> Simp[-c^n   Subst[Int[(c + d*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^(m 
 + 2)), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] && Int 
egerQ[(n - 1)/2] && IntegerQ[m] && IntegerQ[2*p]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.40

method	result	size
orering	\(\frac {2 \left (1336 a^{4} x^{4}+668 a^{3} x^{3}-167 a^{2} x^{2}+66 a x -15\right ) \left (a x +1\right ) \sqrt {c -\frac {c}{a x}}\, \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{105 x^{3} \left (a x -1\right )^{2}}\)	\(76\)
gosper	\(\frac {2 \left (a x +1\right ) \left (1336 a^{4} x^{4}+668 a^{3} x^{3}-167 a^{2} x^{2}+66 a x -15\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{105 x^{3} \left (a x -1\right )^{2}}\)	\(78\)
default	\(\frac {2 \left (a x +1\right ) \left (1336 a^{4} x^{4}+668 a^{3} x^{3}-167 a^{2} x^{2}+66 a x -15\right ) \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{105 x^{3} \left (a x -1\right )^{2}}\)	\(78\)
risch	\(\frac {2 \left (916 a^{4} x^{4}+668 a^{3} x^{3}-167 a^{2} x^{2}+66 a x -15\right ) \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\frac {c \left (a x -1\right )}{a x}}}{105 x^{3} \left (a x -1\right )}+\frac {8 x \,a^{4} \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\frac {c \left (a x -1\right )}{a x}}}{a x -1}\)	\(117\)

Input:

int((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x,method=_RETURNVERBOSE)
 

Output:

2/105*(1336*a^4*x^4+668*a^3*x^3-167*a^2*x^2+66*a*x-15)*(a*x+1)/x^3/(a*x-1) 
^2*(c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.41 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {2 \, {\left (1336 \, a^{4} x^{4} + 668 \, a^{3} x^{3} - 167 \, a^{2} x^{2} + 66 \, a x - 15\right )} \sqrt {\frac {a x - 1}{a x + 1}} \sqrt {\frac {a c x - c}{a x}}}{105 \, {\left (a x^{4} - x^{3}\right )}} \] Input:

integrate((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x, algorithm="fricas 
")
 

Output:

2/105*(1336*a^4*x^4 + 668*a^3*x^3 - 167*a^2*x^2 + 66*a*x - 15)*sqrt((a*x - 
 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x))/(a*x^4 - x^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\text {Timed out} \] Input:

integrate((c-c/a/x)**(1/2)*((a*x-1)/(a*x+1))**(3/2)/x**4,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x, algorithm="maxima 
")
 

Output:

integrate(sqrt(c - c/(a*x))*((a*x - 1)/(a*x + 1))^(3/2)/x^4, x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,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 [B] (verification not implemented)

Time = 13.79 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.53 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {2\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (1336\,a^3\,x^3+2004\,a^2\,x^2+1837\,a\,x+1903\right )\,\sqrt {\frac {c\,\left (a\,x-1\right )}{a\,x}}}{105\,x^3}+\frac {3776\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\sqrt {\frac {c\,\left (a\,x-1\right )}{a\,x}}}{105\,x^3\,\left (a\,x-1\right )} \] Input:

int(((c - c/(a*x))^(1/2)*((a*x - 1)/(a*x + 1))^(3/2))/x^4,x)
 

Output:

(2*((a*x - 1)/(a*x + 1))^(1/2)*(1837*a*x + 2004*a^2*x^2 + 1336*a^3*x^3 + 1 
903)*((c*(a*x - 1))/(a*x))^(1/2))/(105*x^3) + (3776*((a*x - 1)/(a*x + 1))^ 
(1/2)*((c*(a*x - 1))/(a*x))^(1/2))/(105*x^3*(a*x - 1))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.44 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a x}}}{x^4} \, dx=\frac {2 \sqrt {c}\, \left (-1336 \sqrt {a x +1}\, a^{4} x^{4}+1336 \sqrt {x}\, \sqrt {a}\, a^{4} x^{4}+668 \sqrt {x}\, \sqrt {a}\, a^{3} x^{3}-167 \sqrt {x}\, \sqrt {a}\, a^{2} x^{2}+66 \sqrt {x}\, \sqrt {a}\, a x -15 \sqrt {x}\, \sqrt {a}\right )}{105 \sqrt {a x +1}\, a \,x^{4}} \] Input:

int((c-c/a/x)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x^4,x)
 

Output:

(2*sqrt(c)*( - 1336*sqrt(a*x + 1)*a**4*x**4 + 1336*sqrt(x)*sqrt(a)*a**4*x* 
*4 + 668*sqrt(x)*sqrt(a)*a**3*x**3 - 167*sqrt(x)*sqrt(a)*a**2*x**2 + 66*sq 
rt(x)*sqrt(a)*a*x - 15*sqrt(x)*sqrt(a)))/(105*sqrt(a*x + 1)*a*x**4)