\(\int \frac {e^{-\text {arctanh}(a x)}}{(c-\frac {c}{a x})^{5/2}} \, dx\) [574]

Optimal result

Integrand size = 24, antiderivative size = 208 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\frac {(1-a x)^{3/2} \sqrt {1+a x}}{2 a^2 \left (c-\frac {c}{a x}\right )^{5/2} x}+\frac {3 (1-a x)^{5/2} \sqrt {1+a x}}{2 a^3 \left (c-\frac {c}{a x}\right )^{5/2} x^2}+\frac {3 (1-a x)^{5/2} \text {arcsinh}\left (\sqrt {a} \sqrt {x}\right )}{a^{7/2} \left (c-\frac {c}{a x}\right )^{5/2} x^{5/2}}-\frac {9 (1-a x)^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {1+a x}}\right )}{2 \sqrt {2} a^{7/2} \left (c-\frac {c}{a x}\right )^{5/2} x^{5/2}} \] Output:

1/2*(-a*x+1)^(3/2)*(a*x+1)^(1/2)/a^2/(c-c/a/x)^(5/2)/x+3/2*(-a*x+1)^(5/2)* 
(a*x+1)^(1/2)/a^3/(c-c/a/x)^(5/2)/x^2+3*(-a*x+1)^(5/2)*arcsinh(a^(1/2)*x^( 
1/2))/a^(7/2)/(c-c/a/x)^(5/2)/x^(5/2)-9/4*(-a*x+1)^(5/2)*arctanh(2^(1/2)*a 
^(1/2)*x^(1/2)/(a*x+1)^(1/2))*2^(1/2)/a^(7/2)/(c-c/a/x)^(5/2)/x^(5/2)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.61 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\frac {2 \sqrt {a} \sqrt {x} (3-2 a x) \sqrt {1+a x}-12 (-1+a x) \text {arcsinh}\left (\sqrt {a} \sqrt {x}\right )+9 \sqrt {2} (-1+a x) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {1+a x}}\right )}{4 a^{3/2} c^2 \sqrt {c-\frac {c}{a x}} \sqrt {x} \sqrt {1-a x}} \] Input:

Integrate[1/(E^ArcTanh[a*x]*(c - c/(a*x))^(5/2)),x]
 

Output:

(2*Sqrt[a]*Sqrt[x]*(3 - 2*a*x)*Sqrt[1 + a*x] - 12*(-1 + a*x)*ArcSinh[Sqrt[ 
a]*Sqrt[x]] + 9*Sqrt[2]*(-1 + a*x)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[x])/Sqrt[ 
1 + a*x]])/(4*a^(3/2)*c^2*Sqrt[c - c/(a*x)]*Sqrt[x]*Sqrt[1 - a*x])
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.72, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6684, 6679, 109, 27, 171, 25, 27, 175, 63, 104, 219, 222}

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.

\(\displaystyle \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 6684

\(\displaystyle \frac {(1-a x)^{5/2} \int \frac {e^{-\text {arctanh}(a x)} x^{5/2}}{(1-a x)^{5/2}}dx}{x^{5/2} \left (c-\frac {c}{a x}\right )^{5/2}}\)

\(\Big \downarrow \) 6679

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 171

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 175

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

\(\Big \downarrow \) 63

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 222

\(\displaystyle \frac {(1-a x)^{5/2} \left (\frac {x^{3/2} \sqrt {a x+1}}{2 a^2 (1-a x)}-\frac {3 \left (\frac {\frac {3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {x}}{\sqrt {a x+1}}\right )}{\sqrt {a}}-\frac {4 \text {arcsinh}\left (\sqrt {a} \sqrt {x}\right )}{\sqrt {a}}}{a}-\frac {2 \sqrt {x} \sqrt {a x+1}}{a}\right )}{4 a^2}\right )}{x^{5/2} \left (c-\frac {c}{a x}\right )^{5/2}}\)

Input:

Int[1/(E^ArcTanh[a*x]*(c - c/(a*x))^(5/2)),x]
 

Output:

((1 - a*x)^(5/2)*((x^(3/2)*Sqrt[1 + a*x])/(2*a^2*(1 - a*x)) - (3*((-2*Sqrt 
[x]*Sqrt[1 + a*x])/a + ((-4*ArcSinh[Sqrt[a]*Sqrt[x]])/Sqrt[a] + (3*Sqrt[2] 
*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[x])/Sqrt[1 + a*x]])/Sqrt[a])/a))/(4*a^2)))/ 
((c - c/(a*x))^(5/2)*x^(5/2))
 

Defintions of rubi rules used

rule 25

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

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 63

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2/b   S 
ubst[Int[1/Sqrt[c + d*(x^2/b)], x], x, Sqrt[b*x]], x] /; FreeQ[{b, c, d}, x 
] && GtQ[c, 0]
 

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f 
*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) 
 Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) 
+ c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) 
 + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || 
IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 171

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) 
  Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 
) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) 
+ h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] 
 && IntegersQ[2*m, 2*n, 2*p]
 

rule 175

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ 
)))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b   Int[(c + d*x)^n*(e + f*x)^p, x] 
, x] + Simp[(b*g - a*h)/b   Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] 
 /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 6679

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

rule 6684

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_), x_Symbol] 
 :> Simp[x^p*((c + d/x)^p/(1 + c*(x/d))^p)   Int[u*(1 + c*(x/d))^p*(E^(n*Ar 
cTanh[a*x])/x^p), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c^2 - a^2*d^2, 
 0] &&  !IntegerQ[p]
 

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.33

method	result	size
default	\(\frac {\sqrt {\frac {c \left (a x -1\right )}{a x}}\, x \sqrt {-a^{2} x^{2}+1}\, \left (4 \sqrt {-x \left (a x +1\right )}\, a^{\frac {5}{2}} \sqrt {2}\, \sqrt {-\frac {1}{a}}\, x -6 a^{2} \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-x \left (a x +1\right )}}\right ) \sqrt {2}\, \sqrt {-\frac {1}{a}}\, x -6 \sqrt {-x \left (a x +1\right )}\, a^{\frac {3}{2}} \sqrt {2}\, \sqrt {-\frac {1}{a}}+6 \arctan \left (\frac {2 a x +1}{2 \sqrt {a}\, \sqrt {-x \left (a x +1\right )}}\right ) a \sqrt {2}\, \sqrt {-\frac {1}{a}}+9 a^{\frac {3}{2}} \ln \left (\frac {2 \sqrt {2}\, \sqrt {-\frac {1}{a}}\, \sqrt {-x \left (a x +1\right )}\, a -3 a x -1}{a x -1}\right ) x -9 \ln \left (\frac {2 \sqrt {2}\, \sqrt {-\frac {1}{a}}\, \sqrt {-x \left (a x +1\right )}\, a -3 a x -1}{a x -1}\right ) \sqrt {a}\right ) \sqrt {2}}{8 a^{\frac {3}{2}} c^{3} \left (a x -1\right )^{2} \sqrt {-x \left (a x +1\right )}\, \sqrt {-\frac {1}{a}}}\)	\(276\)
risch	\(-\frac {\left (a x +1\right ) \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}\, \left (a x -1\right )}{a \sqrt {-\left (a x +1\right ) a c x}\, \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {-a^{2} x^{2}+1}\, c^{2}}-\frac {\left (\frac {3 \arctan \left (\frac {\sqrt {a^{2} c}\, \left (x +\frac {1}{2 a}\right )}{\sqrt {-a^{2} c \,x^{2}-a c x}}\right )}{2 a^{3} \sqrt {a^{2} c}}+\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -3 \left (x -\frac {1}{a}\right ) a c -2 c}}{2 a^{5} c \left (x -\frac {1}{a}\right )}-\frac {9 \ln \left (\frac {-4 c -3 \left (x -\frac {1}{a}\right ) a c +2 \sqrt {-2 c}\, \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2} c -3 \left (x -\frac {1}{a}\right ) a c -2 c}}{x -\frac {1}{a}}\right )}{4 a^{4} \sqrt {-2 c}}\right ) a^{2} \sqrt {\frac {c a x \left (-a^{2} x^{2}+1\right )}{a x -1}}\, \left (a x -1\right )}{x \sqrt {\frac {c \left (a x -1\right )}{a x}}\, \sqrt {-a^{2} x^{2}+1}\, c^{2}}\)	\(318\)

Input:

int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/8*(c*(a*x-1)/a/x)^(1/2)*x*(-a^2*x^2+1)^(1/2)*(4*(-x*(a*x+1))^(1/2)*a^(5/ 
2)*2^(1/2)*(-1/a)^(1/2)*x-6*a^2*arctan(1/2/a^(1/2)*(2*a*x+1)/(-x*(a*x+1))^ 
(1/2))*2^(1/2)*(-1/a)^(1/2)*x-6*(-x*(a*x+1))^(1/2)*a^(3/2)*2^(1/2)*(-1/a)^ 
(1/2)+6*arctan(1/2/a^(1/2)*(2*a*x+1)/(-x*(a*x+1))^(1/2))*a*2^(1/2)*(-1/a)^ 
(1/2)+9*a^(3/2)*ln((2*2^(1/2)*(-1/a)^(1/2)*(-x*(a*x+1))^(1/2)*a-3*a*x-1)/( 
a*x-1))*x-9*ln((2*2^(1/2)*(-1/a)^(1/2)*(-x*(a*x+1))^(1/2)*a-3*a*x-1)/(a*x- 
1))*a^(1/2))*2^(1/2)/a^(3/2)/c^3/(a*x-1)^2/(-x*(a*x+1))^(1/2)/(-1/a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 528, normalized size of antiderivative = 2.54 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\left [-\frac {9 \, \sqrt {2} {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {-c} \log \left (-\frac {17 \, a^{3} c x^{3} - 3 \, a^{2} c x^{2} - 13 \, a c x - 4 \, \sqrt {2} {\left (3 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1}\right ) + 12 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {-c} \log \left (-\frac {8 \, a^{3} c x^{3} - 7 \, a c x + 4 \, {\left (2 \, a^{2} x^{2} + a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-c} \sqrt {\frac {a c x - c}{a x}} - c}{a x - 1}\right ) - 8 \, {\left (2 \, a^{2} x^{2} - 3 \, a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{16 \, {\left (a^{3} c^{3} x^{2} - 2 \, a^{2} c^{3} x + a c^{3}\right )}}, -\frac {9 \, \sqrt {2} {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{3 \, a^{2} c x^{2} - 2 \, a c x - c}\right ) - 12 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \sqrt {c} \arctan \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1} a \sqrt {c} x \sqrt {\frac {a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) - 4 \, {\left (2 \, a^{2} x^{2} - 3 \, a x\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a c x - c}{a x}}}{8 \, {\left (a^{3} c^{3} x^{2} - 2 \, a^{2} c^{3} x + a c^{3}\right )}}\right ] \] Input:

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^(5/2),x, algorithm="frica 
s")
 

Output:

[-1/16*(9*sqrt(2)*(a^2*x^2 - 2*a*x + 1)*sqrt(-c)*log(-(17*a^3*c*x^3 - 3*a^ 
2*c*x^2 - 13*a*c*x - 4*sqrt(2)*(3*a^2*x^2 + a*x)*sqrt(-a^2*x^2 + 1)*sqrt(- 
c)*sqrt((a*c*x - c)/(a*x)) - c)/(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)) + 12*(a 
^2*x^2 - 2*a*x + 1)*sqrt(-c)*log(-(8*a^3*c*x^3 - 7*a*c*x + 4*(2*a^2*x^2 + 
a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-c)*sqrt((a*c*x - c)/(a*x)) - c)/(a*x - 1)) - 
 8*(2*a^2*x^2 - 3*a*x)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x)))/(a^3*c^ 
3*x^2 - 2*a^2*c^3*x + a*c^3), -1/8*(9*sqrt(2)*(a^2*x^2 - 2*a*x + 1)*sqrt(c 
)*arctan(2*sqrt(2)*sqrt(-a^2*x^2 + 1)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x))/ 
(3*a^2*c*x^2 - 2*a*c*x - c)) - 12*(a^2*x^2 - 2*a*x + 1)*sqrt(c)*arctan(2*s 
qrt(-a^2*x^2 + 1)*a*sqrt(c)*x*sqrt((a*c*x - c)/(a*x))/(2*a^2*c*x^2 - a*c*x 
 - c)) - 4*(2*a^2*x^2 - 3*a*x)*sqrt(-a^2*x^2 + 1)*sqrt((a*c*x - c)/(a*x))) 
/(a^3*c^3*x^2 - 2*a^2*c^3*x + a*c^3)]
 

Sympy [F]

\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{\left (- c \left (-1 + \frac {1}{a x}\right )\right )^{\frac {5}{2}} \left (a x + 1\right )}\, dx \] Input:

integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(c-c/a/x)**(5/2),x)
 

Output:

Integral(sqrt(-(a*x - 1)*(a*x + 1))/((-c*(-1 + 1/(a*x)))**(5/2)*(a*x + 1)) 
, x)
 

Maxima [F]

\[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\int { \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} {\left (c - \frac {c}{a x}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^(5/2),x, algorithm="maxim 
a")
 

Output:

integrate(sqrt(-a^2*x^2 + 1)/((a*x + 1)*(c - c/(a*x))^(5/2)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^(5/2),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. \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\int \frac {\sqrt {1-a^2\,x^2}}{{\left (c-\frac {c}{a\,x}\right )}^{5/2}\,\left (a\,x+1\right )} \,d x \] Input:

int((1 - a^2*x^2)^(1/2)/((c - c/(a*x))^(5/2)*(a*x + 1)),x)
 

Output:

int((1 - a^2*x^2)^(1/2)/((c - c/(a*x))^(5/2)*(a*x + 1)), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.70 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-9 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, \sqrt {2}\, i}{a x +1}\right ) a x +9 \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, \sqrt {2}\, i}{a x +1}\right )+12 \mathit {atan} \left (\frac {\sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, i}{a x +1}\right ) a x -12 \mathit {atan} \left (\frac {\sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, i}{a x +1}\right )+4 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, a i x -6 \sqrt {x}\, \sqrt {a}\, \sqrt {a x +1}\, i \right )}{4 a \,c^{3} \left (a x -1\right )} \] Input:

int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^(5/2),x)
 

Output:

(sqrt(c)*( - 9*sqrt(2)*atan((sqrt(x)*sqrt(a)*sqrt(a*x + 1)*sqrt(2)*i)/(a*x 
 + 1))*a*x + 9*sqrt(2)*atan((sqrt(x)*sqrt(a)*sqrt(a*x + 1)*sqrt(2)*i)/(a*x 
 + 1)) + 12*atan((sqrt(x)*sqrt(a)*sqrt(a*x + 1)*i)/(a*x + 1))*a*x - 12*ata 
n((sqrt(x)*sqrt(a)*sqrt(a*x + 1)*i)/(a*x + 1)) + 4*sqrt(x)*sqrt(a)*sqrt(a* 
x + 1)*a*i*x - 6*sqrt(x)*sqrt(a)*sqrt(a*x + 1)*i))/(4*a*c**3*(a*x - 1))