\(\int e^{-\text {arctanh}(a x)} (c-\frac {c}{a x})^3 \, dx\) [504]

Optimal result

Integrand size = 22, antiderivative size = 111 \[ \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=\frac {c^3 \sqrt {1-a^2 x^2}}{a}+\frac {c^3 \sqrt {1-a^2 x^2}}{2 a^3 x^2}-\frac {4 c^3 \sqrt {1-a^2 x^2}}{a^2 x}+\frac {4 c^3 \arcsin (a x)}{a}+\frac {13 c^3 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{2 a} \] Output:

c^3*(-a^2*x^2+1)^(1/2)/a+1/2*c^3*(-a^2*x^2+1)^(1/2)/a^3/x^2-4*c^3*(-a^2*x^ 
2+1)^(1/2)/a^2/x+4*c^3*arcsin(a*x)/a+13/2*c^3*arctanh((-a^2*x^2+1)^(1/2))/ 
a
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.42 \[ \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=\frac {c^3 \left (1-8 a x+a^2 x^2+8 a^3 x^3-2 a^4 x^4+13 a^2 x^2 \sqrt {1-a^2 x^2} \arcsin (a x)+10 a^2 x^2 \sqrt {1-a^2 x^2} \arcsin \left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )+13 a^2 x^2 \sqrt {1-a^2 x^2} \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right )}{2 a^3 x^2 \sqrt {1-a^2 x^2}} \] Input:

Integrate[(c - c/(a*x))^3/E^ArcTanh[a*x],x]
 

Output:

(c^3*(1 - 8*a*x + a^2*x^2 + 8*a^3*x^3 - 2*a^4*x^4 + 13*a^2*x^2*Sqrt[1 - a^ 
2*x^2]*ArcSin[a*x] + 10*a^2*x^2*Sqrt[1 - a^2*x^2]*ArcSin[Sqrt[1 - a*x]/Sqr 
t[2]] + 13*a^2*x^2*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]]))/(2*a^3*x 
^2*Sqrt[1 - a^2*x^2])
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.92, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {6681, 6678, 540, 2338, 2340, 25, 27, 538, 223, 243, 73, 221}

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.

\(\displaystyle \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx\)

\(\Big \downarrow \) 6681

\(\displaystyle -\frac {c^3 \int \frac {e^{-\text {arctanh}(a x)} (1-a x)^3}{x^3}dx}{a^3}\)

\(\Big \downarrow \) 6678

\(\displaystyle -\frac {c^3 \int \frac {(1-a x)^4}{x^3 \sqrt {1-a^2 x^2}}dx}{a^3}\)

\(\Big \downarrow \) 540

\(\displaystyle -\frac {c^3 \left (-\frac {1}{2} \int \frac {-2 x^3 a^4+8 x^2 a^3-13 x a^2+8 a}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )}{a^3}\)

\(\Big \downarrow \) 2338

\(\displaystyle -\frac {c^3 \left (\frac {1}{2} \left (\int \frac {2 x^2 a^4-8 x a^3+13 a^2}{x \sqrt {1-a^2 x^2}}dx+\frac {8 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )}{a^3}\)

\(\Big \downarrow \) 2340

\(\displaystyle -\frac {c^3 \left (\frac {1}{2} \left (-\frac {\int -\frac {a^4 (13-8 a x)}{x \sqrt {1-a^2 x^2}}dx}{a^2}-2 a^2 \sqrt {1-a^2 x^2}+\frac {8 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )}{a^3}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {c^3 \left (\frac {1}{2} \left (\frac {\int \frac {a^4 (13-8 a x)}{x \sqrt {1-a^2 x^2}}dx}{a^2}-2 a^2 \sqrt {1-a^2 x^2}+\frac {8 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )}{a^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {c^3 \left (\frac {1}{2} \left (a^2 \int \frac {13-8 a x}{x \sqrt {1-a^2 x^2}}dx-2 a^2 \sqrt {1-a^2 x^2}+\frac {8 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )}{a^3}\)

\(\Big \downarrow \) 538

\(\displaystyle -\frac {c^3 \left (\frac {1}{2} \left (a^2 \left (13 \int \frac {1}{x \sqrt {1-a^2 x^2}}dx-8 a \int \frac {1}{\sqrt {1-a^2 x^2}}dx\right )-2 a^2 \sqrt {1-a^2 x^2}+\frac {8 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )}{a^3}\)

\(\Big \downarrow \) 223

\(\displaystyle -\frac {c^3 \left (\frac {1}{2} \left (a^2 \left (13 \int \frac {1}{x \sqrt {1-a^2 x^2}}dx-8 \arcsin (a x)\right )-2 a^2 \sqrt {1-a^2 x^2}+\frac {8 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )}{a^3}\)

\(\Big \downarrow \) 243

\(\displaystyle -\frac {c^3 \left (\frac {1}{2} \left (a^2 \left (\frac {13}{2} \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2-8 \arcsin (a x)\right )-2 a^2 \sqrt {1-a^2 x^2}+\frac {8 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )}{a^3}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {c^3 \left (\frac {1}{2} \left (a^2 \left (-\frac {13 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a^2}-8 \arcsin (a x)\right )-2 a^2 \sqrt {1-a^2 x^2}+\frac {8 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )}{a^3}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {c^3 \left (\frac {1}{2} \left (a^2 \left (-13 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )-8 \arcsin (a x)\right )-2 a^2 \sqrt {1-a^2 x^2}+\frac {8 a \sqrt {1-a^2 x^2}}{x}\right )-\frac {\sqrt {1-a^2 x^2}}{2 x^2}\right )}{a^3}\)

Input:

Int[(c - c/(a*x))^3/E^ArcTanh[a*x],x]
 

Output:

-((c^3*(-1/2*Sqrt[1 - a^2*x^2]/x^2 + (-2*a^2*Sqrt[1 - a^2*x^2] + (8*a*Sqrt 
[1 - a^2*x^2])/x + a^2*(-8*ArcSin[a*x] - 13*ArcTanh[Sqrt[1 - a^2*x^2]]))/2 
))/a^3)
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 223

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 538

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

rule 540

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain 
der[(c + d*x)^n, x, x]}, Simp[R*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))) 
, x] + Simp[1/(a*(m + 1))   Int[x^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*(m + 
1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG 
tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 2338

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ 
Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( 
m + 1))   Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( 
m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

rule 2340

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 
)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m 
+ q + 2*p + 1))   Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) 
*Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; 
GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ 
[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
 

rule 6678

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

rule 6681

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

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.07

method	result	size
risch	\(\frac {\left (8 a^{3} x^{3}-a^{2} x^{2}-8 a x +1\right ) c^{3}}{2 x^{2} \sqrt {-a^{2} x^{2}+1}\, a^{3}}+\frac {\left (\frac {13 a^{2} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}+\frac {4 a^{3} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}+a^{2} \sqrt {-a^{2} x^{2}+1}\right ) c^{3}}{a^{3}}\)	\(119\)
default	\(\frac {c^{3} \left (\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{2 x^{2}}-\frac {13 a^{2} \left (\sqrt {-a^{2} x^{2}+1}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )}{2}+4 a \left (-\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{x}-2 a^{2} \left (\frac {x \sqrt {-a^{2} x^{2}+1}}{2}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}\right )\right )+8 a^{2} \left (\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}\right )\right )}{a^{3}}\)	\(195\)

Input:

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

Output:

1/2*(8*a^3*x^3-a^2*x^2-8*a*x+1)/x^2/(-a^2*x^2+1)^(1/2)*c^3/a^3+(13/2*a^2*a 
rctanh(1/(-a^2*x^2+1)^(1/2))+4*a^3/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2* 
x^2+1)^(1/2))+a^2*(-a^2*x^2+1)^(1/2))*c^3/a^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.07 \[ \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=-\frac {16 \, a^{2} c^{3} x^{2} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + 13 \, a^{2} c^{3} x^{2} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - 2 \, a^{2} c^{3} x^{2} - {\left (2 \, a^{2} c^{3} x^{2} - 8 \, a c^{3} x + c^{3}\right )} \sqrt {-a^{2} x^{2} + 1}}{2 \, a^{3} x^{2}} \] Input:

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

Output:

-1/2*(16*a^2*c^3*x^2*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + 13*a^2*c^3*x 
^2*log((sqrt(-a^2*x^2 + 1) - 1)/x) - 2*a^2*c^3*x^2 - (2*a^2*c^3*x^2 - 8*a* 
c^3*x + c^3)*sqrt(-a^2*x^2 + 1))/(a^3*x^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (99) = 198\).

Time = 0.13 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.86 \[ \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=-\frac {{\left (c^{3} - \frac {16 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c^{3}}{a^{2} x}\right )} a^{4} x^{2}}{8 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} {\left | a \right |}} + \frac {4 \, c^{3} \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{{\left | a \right |}} + \frac {13 \, c^{3} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{2 \, {\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1} c^{3}}{a} - \frac {\frac {16 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c^{3} {\left | a \right |}}{a^{2} x} - \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c^{3} {\left | a \right |}}{a^{4} x^{2}}}{8 \, a^{2}} \] Input:

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

Output:

-1/8*(c^3 - 16*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*c^3/(a^2*x))*a^4*x^2/((sqrt 
(-a^2*x^2 + 1)*abs(a) + a)^2*abs(a)) + 4*c^3*arcsin(a*x)*sgn(a)/abs(a) + 1 
3/2*c^3*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs( 
a) + sqrt(-a^2*x^2 + 1)*c^3/a - 1/8*(16*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*c^ 
3*abs(a)/(a^2*x) - (sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c^3*abs(a)/(a^4*x^2)) 
/a^2
 

Mupad [B] (verification not implemented)

Time = 15.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.02 \[ \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=\frac {4\,c^3\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{\sqrt {-a^2}}+\frac {c^3\,\sqrt {1-a^2\,x^2}}{a}-\frac {4\,c^3\,\sqrt {1-a^2\,x^2}}{a^2\,x}+\frac {c^3\,\sqrt {1-a^2\,x^2}}{2\,a^3\,x^2}-\frac {c^3\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,13{}\mathrm {i}}{2\,a} \] Input:

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

Output:

(4*c^3*asinh(x*(-a^2)^(1/2)))/(-a^2)^(1/2) - (c^3*atan((1 - a^2*x^2)^(1/2) 
*1i)*13i)/(2*a) + (c^3*(1 - a^2*x^2)^(1/2))/a - (4*c^3*(1 - a^2*x^2)^(1/2) 
)/(a^2*x) + (c^3*(1 - a^2*x^2)^(1/2))/(2*a^3*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.86 \[ \int e^{-\text {arctanh}(a x)} \left (c-\frac {c}{a x}\right )^3 \, dx=\frac {c^{3} \left (32 \mathit {asin} \left (a x \right ) a^{2} x^{2}+8 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-32 \sqrt {-a^{2} x^{2}+1}\, a x +4 \sqrt {-a^{2} x^{2}+1}-52 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{2} x^{2}-9 a^{2} x^{2}\right )}{8 a^{3} x^{2}} \] Input:

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

Output:

(c**3*(32*asin(a*x)*a**2*x**2 + 8*sqrt( - a**2*x**2 + 1)*a**2*x**2 - 32*sq 
rt( - a**2*x**2 + 1)*a*x + 4*sqrt( - a**2*x**2 + 1) - 52*log(tan(asin(a*x) 
/2))*a**2*x**2 - 9*a**2*x**2))/(8*a**3*x**2)