\(\int \frac {e^{3 \text {arctanh}(a x)}}{(c-\frac {c}{a x})^2} \, dx\) [493]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 129 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=-\frac {2 (1+a x)^4}{3 a c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac {20 (1+a x)}{3 a c^2 \sqrt {1-a^2 x^2}}+\frac {5 \sqrt {1-a^2 x^2}}{3 a c^2}+\frac {\left (1-a^2 x^2\right )^{5/2}}{5 a c^2 (1-a x)^5}-\frac {5 \arcsin (a x)}{a c^2} \] Output:

-2/3*(a*x+1)^4/a/c^2/(-a^2*x^2+1)^(3/2)+20/3*(a*x+1)/a/c^2/(-a^2*x^2+1)^(1 
/2)+5/3*(-a^2*x^2+1)^(1/2)/a/c^2+1/5*(-a^2*x^2+1)^(5/2)/a/c^2/(-a*x+1)^5-5 
*arcsin(a*x)/a/c^2
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.55 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {(1+a x)^{5/2} (-7+10 a x)+100 \sqrt {2} (-1+a x)^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2},\frac {1}{2},\frac {1}{2} (1-a x)\right )}{15 a c^2 (1-a x)^{5/2}} \] Input:

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

Output:

((1 + a*x)^(5/2)*(-7 + 10*a*x) + 100*Sqrt[2]*(-1 + a*x)^2*Hypergeometric2F 
1[-3/2, -1/2, 1/2, (1 - a*x)/2])/(15*a*c^2*(1 - a*x)^(5/2))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {6681, 6678, 570, 529, 27, 468, 462, 455, 223}

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.

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

\(\Big \downarrow \) 6681

\(\displaystyle \frac {a^2 \int \frac {e^{3 \text {arctanh}(a x)} x^2}{(1-a x)^2}dx}{c^2}\)

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 570

\(\displaystyle \frac {a^2 \int \frac {x^2 (a x+1)^5}{\left (1-a^2 x^2\right )^{7/2}}dx}{c^2}\)

\(\Big \downarrow \) 529

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 468

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

\(\Big \downarrow \) 462

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 223

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

Input:

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

Output:

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

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 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 455
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 462
Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2)^(3/2), x_Symbol] :> Simp 
[(-2^(n - 1))*d*c^(n - 2)*((c + d*x)/(b*Sqrt[a + b*x^2])), x] + Simp[d^2/b 
  Int[(1/Sqrt[a + b*x^2])*ExpandToSum[(2^(n - 1)*c^(n - 1) - (c + d*x)^(n - 
 1))/(c - d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 
0] && IGtQ[n, 2]
 

rule 468
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] - Simp[d^2*((n + 
p)/(b*(p + 1)))   Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1), x], x] /; Free 
Q[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[p, -1] && GtQ[n, 1] && I 
ntegerQ[2*p]
 

rule 529
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m, a*d + b*c*x, x], R = PolynomialRem 
ainder[x^m, a*d + b*c*x, x]}, Simp[(-c)*R*(c + d*x)^n*((a + b*x^2)^(p + 1)/ 
(2*a*d*(p + 1))), x] + Simp[c/(2*a*(p + 1))   Int[(c + d*x)^(n - 1)*(a + b* 
x^2)^(p + 1)*ExpandToSum[2*a*d*(p + 1)*Qx + R*(n + 2*p + 2), x], x], x]] /; 
 FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[b* 
c^2 + a*d^2, 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 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 = 0.32 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.50

method result size
risch \(-\frac {a^{2} x^{2}-1}{a \sqrt {-a^{2} x^{2}+1}\, c^{2}}-\frac {\left (\frac {5 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}+\frac {4 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a^{6} \left (x -\frac {1}{a}\right )^{3}}+\frac {52 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{15 a^{5} \left (x -\frac {1}{a}\right )^{2}}+\frac {143 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{15 a^{4} \left (x -\frac {1}{a}\right )}\right ) a^{2}}{c^{2}}\) \(193\)
default \(\frac {a^{2} \left (a \left (-\frac {x^{2}}{a^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {2}{a^{4} \sqrt {-a^{2} x^{2}+1}}\right )+\frac {25 x}{a^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {5 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{2} \sqrt {a^{2}}}+\frac {12}{a^{3} \sqrt {-a^{2} x^{2}+1}}+\frac {\frac {8}{5 a \left (x -\frac {1}{a}\right )^{2} \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}-\frac {24 a \left (\frac {1}{3 a \left (x -\frac {1}{a}\right ) \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}+\frac {-2 \left (x -\frac {1}{a}\right ) a^{2}-2 a}{3 a \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{5}}{a^{4}}+\frac {\frac {28}{3 a \left (x -\frac {1}{a}\right ) \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}+\frac {28 \left (-2 \left (x -\frac {1}{a}\right ) a^{2}-2 a \right )}{3 a \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}}{a^{3}}\right )}{c^{2}}\) \(349\)

Input:

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

Output:

-1/a*(a^2*x^2-1)/(-a^2*x^2+1)^(1/2)/c^2-(5/a^2/(a^2)^(1/2)*arctan((a^2)^(1 
/2)*x/(-a^2*x^2+1)^(1/2))+4/5/a^6/(x-1/a)^3*(-(x-1/a)^2*a^2-2*a*(x-1/a))^( 
1/2)+52/15/a^5/(x-1/a)^2*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2)+143/15/a^4/(x- 
1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2))*a^2/c^2
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.11 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {118 \, a^{3} x^{3} - 354 \, a^{2} x^{2} + 354 \, a x + 150 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (15 \, a^{3} x^{3} - 188 \, a^{2} x^{2} + 279 \, a x - 118\right )} \sqrt {-a^{2} x^{2} + 1} - 118}{15 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \] Input:

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

Output:

1/15*(118*a^3*x^3 - 354*a^2*x^2 + 354*a*x + 150*(a^3*x^3 - 3*a^2*x^2 + 3*a 
*x - 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (15*a^3*x^3 - 188*a^2*x^2 
 + 279*a*x - 118)*sqrt(-a^2*x^2 + 1) - 118)/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 
 3*a^2*c^2*x - a*c^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.40 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=-\frac {5 \, \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{c^{2} {\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a c^{2}} - \frac {2 \, {\left (\frac {440 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a^{2} x} - \frac {670 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac {360 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac {75 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 103\right )}}{15 \, c^{2} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{5} {\left | a \right |}} \] Input:

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

Output:

-5*arcsin(a*x)*sgn(a)/(c^2*abs(a)) + sqrt(-a^2*x^2 + 1)/(a*c^2) - 2/15*(44 
0*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 670*(sqrt(-a^2*x^2 + 1)*abs(a) 
 + a)^2/(a^4*x^2) + 360*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) - 75*( 
sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^8*x^4) - 103)/(c^2*((sqrt(-a^2*x^2 + 1 
)*abs(a) + a)/(a^2*x) - 1)^5*abs(a))
 

Mupad [B] (verification not implemented)

Time = 15.03 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.09 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {8\,a^2\,\sqrt {1-a^2\,x^2}}{15\,\left (a^5\,c^2\,x^2-2\,a^4\,c^2\,x+a^3\,c^2\right )}-\frac {5\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^2\,\sqrt {-a^2}}-\frac {4\,a\,\sqrt {1-a^2\,x^2}}{a^4\,c^2\,x^2-2\,a^3\,c^2\,x+a^2\,c^2}+\frac {\sqrt {1-a^2\,x^2}}{a\,c^2}+\frac {143\,\sqrt {1-a^2\,x^2}}{15\,\sqrt {-a^2}\,\left (c^2\,x\,\sqrt {-a^2}-\frac {c^2\,\sqrt {-a^2}}{a}\right )}+\frac {4\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (3\,c^2\,x\,\sqrt {-a^2}-\frac {c^2\,\sqrt {-a^2}}{a}+a^2\,c^2\,x^3\,\sqrt {-a^2}-3\,a\,c^2\,x^2\,\sqrt {-a^2}\right )} \] Input:

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

Output:

(8*a^2*(1 - a^2*x^2)^(1/2))/(15*(a^3*c^2 - 2*a^4*c^2*x + a^5*c^2*x^2)) - ( 
5*asinh(x*(-a^2)^(1/2)))/(c^2*(-a^2)^(1/2)) - (4*a*(1 - a^2*x^2)^(1/2))/(a 
^2*c^2 - 2*a^3*c^2*x + a^4*c^2*x^2) + (1 - a^2*x^2)^(1/2)/(a*c^2) + (143*( 
1 - a^2*x^2)^(1/2))/(15*(-a^2)^(1/2)*(c^2*x*(-a^2)^(1/2) - (c^2*(-a^2)^(1/ 
2))/a)) + (4*(1 - a^2*x^2)^(1/2))/(5*(-a^2)^(1/2)*(3*c^2*x*(-a^2)^(1/2) - 
(c^2*(-a^2)^(1/2))/a + a^2*c^2*x^3*(-a^2)^(1/2) - 3*a*c^2*x^2*(-a^2)^(1/2) 
))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.08 \[ \int \frac {e^{3 \text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx=\frac {-75 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a^{2} x^{2}+150 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a x -75 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )-75 \mathit {asin} \left (a x \right ) a^{3} x^{3}+225 \mathit {asin} \left (a x \right ) a^{2} x^{2}-225 \mathit {asin} \left (a x \right ) a x +75 \mathit {asin} \left (a x \right )+15 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-100 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+103 \sqrt {-a^{2} x^{2}+1}\, a x -30 \sqrt {-a^{2} x^{2}+1}-15 a^{4} x^{4}+261 a^{3} x^{3}-355 a^{2} x^{2}+103 a x +30}{15 a \,c^{2} \left (\sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-2 \sqrt {-a^{2} x^{2}+1}\, a x +\sqrt {-a^{2} x^{2}+1}+a^{3} x^{3}-3 a^{2} x^{2}+3 a x -1\right )} \] Input:

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

Output:

( - 75*sqrt( - a**2*x**2 + 1)*asin(a*x)*a**2*x**2 + 150*sqrt( - a**2*x**2 
+ 1)*asin(a*x)*a*x - 75*sqrt( - a**2*x**2 + 1)*asin(a*x) - 75*asin(a*x)*a* 
*3*x**3 + 225*asin(a*x)*a**2*x**2 - 225*asin(a*x)*a*x + 75*asin(a*x) + 15* 
sqrt( - a**2*x**2 + 1)*a**3*x**3 - 100*sqrt( - a**2*x**2 + 1)*a**2*x**2 + 
103*sqrt( - a**2*x**2 + 1)*a*x - 30*sqrt( - a**2*x**2 + 1) - 15*a**4*x**4 
+ 261*a**3*x**3 - 355*a**2*x**2 + 103*a*x + 30)/(15*a*c**2*(sqrt( - a**2*x 
**2 + 1)*a**2*x**2 - 2*sqrt( - a**2*x**2 + 1)*a*x + sqrt( - a**2*x**2 + 1) 
 + a**3*x**3 - 3*a**2*x**2 + 3*a*x - 1))