\(\int \frac {e^{3 \text {arctanh}(a x)}}{(c-\frac {c}{a x}) x^3} \, dx\) [615]

Optimal result

Integrand size = 25, antiderivative size = 118 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right ) x^3} \, dx=-\frac {7 a^2 \sqrt {1-a^2 x^2}}{3 c (1-a x)^2}+\frac {a \sqrt {1-a^2 x^2}}{c x (1-a x)^2}-\frac {19 a^2 \sqrt {1-a^2 x^2}}{3 c (1-a x)}+\frac {4 a^2 \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )}{c} \] Output:

-7/3*a^2*(-a^2*x^2+1)^(1/2)/c/(-a*x+1)^2+a*(-a^2*x^2+1)^(1/2)/c/x/(-a*x+1) 
^2-19/3*a^2*(-a^2*x^2+1)^(1/2)/c/(-a*x+1)+4*a^2*arctanh((-a^2*x^2+1)^(1/2) 
)/c
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.78 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right ) x^3} \, dx=\frac {a \left (-3+23 a x+7 a^2 x^2-19 a^3 x^3+12 a x (-1+a x) \sqrt {1-a^2 x^2} \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )\right )}{3 c x (-1+a x) \sqrt {1-a^2 x^2}} \] Input:

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

Output:

(a*(-3 + 23*a*x + 7*a^2*x^2 - 19*a^3*x^3 + 12*a*x*(-1 + a*x)*Sqrt[1 - a^2* 
x^2]*ArcTanh[Sqrt[1 - a^2*x^2]]))/(3*c*x*(-1 + a*x)*Sqrt[1 - a^2*x^2])
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.84, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {2005, 6678, 27, 570, 532, 25, 2336, 27, 534, 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.

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

\(\Big \downarrow \) 2005

\(\displaystyle \int \frac {e^{3 \text {arctanh}(a x)}}{x^2 \left (c x-\frac {c}{a}\right )}dx\)

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 570

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

\(\Big \downarrow \) 532

\(\displaystyle -\frac {a \left (\frac {8 a (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}-\frac {1}{3} \int -\frac {13 a^2 x^2+12 a x+3}{x^2 \left (1-a^2 x^2\right )^{3/2}}dx\right )}{c}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {a \left (\frac {1}{3} \int \frac {13 a^2 x^2+12 a x+3}{x^2 \left (1-a^2 x^2\right )^{3/2}}dx+\frac {8 a (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}\right )}{c}\)

\(\Big \downarrow \) 2336

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 534

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle -\frac {a \left (\frac {1}{3} \left (3 \left (-4 a \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{x}\right )+\frac {4 a (4 a x+3)}{\sqrt {1-a^2 x^2}}\right )+\frac {8 a (a x+1)}{3 \left (1-a^2 x^2\right )^{3/2}}\right )}{c}\)

Input:

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

Output:

-((a*((8*a*(1 + a*x))/(3*(1 - a^2*x^2)^(3/2)) + ((4*a*(3 + 4*a*x))/Sqrt[1 
- a^2*x^2] + 3*(-(Sqrt[1 - a^2*x^2]/x) - 4*a*ArcTanh[Sqrt[1 - a^2*x^2]]))/ 
3))/c)
 

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 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 532

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[x^m 
*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), 
x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[p, 
 -1] && IntegerQ[2*p]
 

rule 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 570

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

rule 2005

Int[(Fx_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(m 
+ n*p)*(b + a/x^n)^p*Fx, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && Neg 
Q[n]
 

rule 2336

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema 
inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) 
^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* 
b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex 
pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F 
reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
 

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]
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.40

method	result	size
default	\(\frac {a \left (-\frac {a^{2} x}{\sqrt {-a^{2} x^{2}+1}}+\frac {1}{x \sqrt {-a^{2} x^{2}+1}}-4 a \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )+8 a \left (\frac {1}{3 a \left (x -\frac {1}{a}\right ) \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}+\frac {-2 \left (x -\frac {1}{a}\right ) a^{2}-2 a}{3 a \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )\right )}{c}\)	\(165\)
risch	\(-\frac {\left (a^{2} x^{2}-1\right ) a}{x \sqrt {-a^{2} x^{2}+1}\, c}-\frac {4 a^{2} \left (-\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}}{a}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )}\right )}{c}\)	\(181\)

Input:

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

Output:

a/c*(-a^2*x/(-a^2*x^2+1)^(1/2)+1/x/(-a^2*x^2+1)^(1/2)-4*a*(1/(-a^2*x^2+1)^ 
(1/2)-arctanh(1/(-a^2*x^2+1)^(1/2)))+8*a*(1/3/a/(x-1/a)/(-(x-1/a)^2*a^2-2* 
a*(x-1/a))^(1/2)+1/3/a*(-2*(x-1/a)*a^2-2*a)/(-(x-1/a)^2*a^2-2*a*(x-1/a))^( 
1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.02 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right ) x^3} \, dx=-\frac {20 \, a^{4} x^{3} - 40 \, a^{3} x^{2} + 20 \, a^{2} x + 12 \, {\left (a^{4} x^{3} - 2 \, a^{3} x^{2} + a^{2} x\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (19 \, a^{3} x^{2} - 26 \, a^{2} x + 3 \, a\right )} \sqrt {-a^{2} x^{2} + 1}}{3 \, {\left (a^{2} c x^{3} - 2 \, a c x^{2} + c x\right )}} \] Input:

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

Output:

-1/3*(20*a^4*x^3 - 40*a^3*x^2 + 20*a^2*x + 12*(a^4*x^3 - 2*a^3*x^2 + a^2*x 
)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (19*a^3*x^2 - 26*a^2*x + 3*a)*sqrt(-a^ 
2*x^2 + 1))/(a^2*c*x^3 - 2*a*c*x^2 + c*x)
 

Sympy [F]

\[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right ) x^3} \, dx=\frac {a \left (\int \frac {3 a x}{- a^{3} x^{5} \sqrt {- a^{2} x^{2} + 1} + a^{2} x^{4} \sqrt {- a^{2} x^{2} + 1} + a x^{3} \sqrt {- a^{2} x^{2} + 1} - x^{2} \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {3 a^{2} x^{2}}{- a^{3} x^{5} \sqrt {- a^{2} x^{2} + 1} + a^{2} x^{4} \sqrt {- a^{2} x^{2} + 1} + a x^{3} \sqrt {- a^{2} x^{2} + 1} - x^{2} \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a^{3} x^{3}}{- a^{3} x^{5} \sqrt {- a^{2} x^{2} + 1} + a^{2} x^{4} \sqrt {- a^{2} x^{2} + 1} + a x^{3} \sqrt {- a^{2} x^{2} + 1} - x^{2} \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {1}{- a^{3} x^{5} \sqrt {- a^{2} x^{2} + 1} + a^{2} x^{4} \sqrt {- a^{2} x^{2} + 1} + a x^{3} \sqrt {- a^{2} x^{2} + 1} - x^{2} \sqrt {- a^{2} x^{2} + 1}}\, dx\right )}{c} \] Input:

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

Output:

a*(Integral(3*a*x/(-a**3*x**5*sqrt(-a**2*x**2 + 1) + a**2*x**4*sqrt(-a**2* 
x**2 + 1) + a*x**3*sqrt(-a**2*x**2 + 1) - x**2*sqrt(-a**2*x**2 + 1)), x) + 
 Integral(3*a**2*x**2/(-a**3*x**5*sqrt(-a**2*x**2 + 1) + a**2*x**4*sqrt(-a 
**2*x**2 + 1) + a*x**3*sqrt(-a**2*x**2 + 1) - x**2*sqrt(-a**2*x**2 + 1)), 
x) + Integral(a**3*x**3/(-a**3*x**5*sqrt(-a**2*x**2 + 1) + a**2*x**4*sqrt( 
-a**2*x**2 + 1) + a*x**3*sqrt(-a**2*x**2 + 1) - x**2*sqrt(-a**2*x**2 + 1)) 
, x) + Integral(1/(-a**3*x**5*sqrt(-a**2*x**2 + 1) + a**2*x**4*sqrt(-a**2* 
x**2 + 1) + a*x**3*sqrt(-a**2*x**2 + 1) - x**2*sqrt(-a**2*x**2 + 1)), x))/ 
c
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (103) = 206\).

Time = 0.15 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.84 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right ) x^3} \, dx=\frac {4 \, a^{3} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{c {\left | a \right |}} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a}{2 \, c x {\left | a \right |}} + \frac {{\left (3 \, a^{3} - \frac {89 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a}{x} + \frac {153 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a x^{2}} - \frac {99 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{3} x^{3}}\right )} a^{2} x}{6 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{3} {\left | a \right |}} \] Input:

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

Output:

4*a^3*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c*abs 
(a)) + 1/2*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a/(c*x*abs(a)) + 1/6*(3*a^3 - 8 
9*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a/x + 153*(sqrt(-a^2*x^2 + 1)*abs(a) + a 
)^2/(a*x^2) - 99*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^3*x^3))*a^2*x/((sqrt 
(-a^2*x^2 + 1)*abs(a) + a)*c*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1) 
^3*abs(a))
 

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.17 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right ) x^3} \, dx=\frac {a\,\sqrt {1-a^2\,x^2}}{c\,x}-\frac {4\,a^4\,\sqrt {1-a^2\,x^2}}{3\,\left (c\,a^4\,x^2-2\,c\,a^3\,x+c\,a^2\right )}+\frac {16\,a^3\,\sqrt {1-a^2\,x^2}}{3\,\left (\frac {c\,\sqrt {-a^2}}{a}-c\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}-\frac {a^2\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{c} \] Input:

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

Output:

(a*(1 - a^2*x^2)^(1/2))/(c*x) - (4*a^4*(1 - a^2*x^2)^(1/2))/(3*(a^2*c + a^ 
4*c*x^2 - 2*a^3*c*x)) - (a^2*atan((1 - a^2*x^2)^(1/2)*1i)*4i)/c + (16*a^3* 
(1 - a^2*x^2)^(1/2))/(3*((c*(-a^2)^(1/2))/a - c*x*(-a^2)^(1/2))*(-a^2)^(1/ 
2))
 

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.42 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right ) x^3} \, dx=\frac {a^{2} \left (-24 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{4}+72 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-72 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+24 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-3 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{5}+39 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{4}-60 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+59 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-3\right )}{6 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right ) c \left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{3}-3 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )^{2}+3 \tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )-1\right )} \] Input:

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

Output:

(a**2*( - 24*log(tan(asin(a*x)/2))*tan(asin(a*x)/2)**4 + 72*log(tan(asin(a 
*x)/2))*tan(asin(a*x)/2)**3 - 72*log(tan(asin(a*x)/2))*tan(asin(a*x)/2)**2 
 + 24*log(tan(asin(a*x)/2))*tan(asin(a*x)/2) - 3*tan(asin(a*x)/2)**5 + 39* 
tan(asin(a*x)/2)**4 - 60*tan(asin(a*x)/2)**2 + 59*tan(asin(a*x)/2) - 3))/( 
6*tan(asin(a*x)/2)*c*(tan(asin(a*x)/2)**3 - 3*tan(asin(a*x)/2)**2 + 3*tan( 
asin(a*x)/2) - 1))