\(\int \frac {e^{\text {arctanh}(a x)}}{(c-\frac {c}{a^2 x^2})^3} \, dx\) [660]

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

Optimal result

Integrand size = 20, antiderivative size = 150 \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=-\frac {1+a x}{5 a c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {1}{a c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {11 x}{15 c^3 \left (1-a^2 x^2\right )^{3/2}}-\frac {3}{a c^3 \sqrt {1-a^2 x^2}}-\frac {23 x}{15 c^3 \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a c^3}+\frac {\arcsin (a x)}{a c^3} \] Output:

-1/5*(a*x+1)/a/c^3/(-a^2*x^2+1)^(5/2)+1/a/c^3/(-a^2*x^2+1)^(3/2)+11/15*x/c 
^3/(-a^2*x^2+1)^(3/2)-3/a/c^3/(-a^2*x^2+1)^(1/2)-23/15*x/c^3/(-a^2*x^2+1)^ 
(1/2)-(-a^2*x^2+1)^(1/2)/a/c^3+arcsin(a*x)/a/c^3
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.72 \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {-48+33 a x+87 a^2 x^2-52 a^3 x^3-38 a^4 x^4+15 a^5 x^5+15 (-1+a x)^2 (1+a x) \sqrt {1-a^2 x^2} \arcsin (a x)}{15 a c^3 (-1+a x)^2 (1+a x) \sqrt {1-a^2 x^2}} \] Input:

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

Output:

(-48 + 33*a*x + 87*a^2*x^2 - 52*a^3*x^3 - 38*a^4*x^4 + 15*a^5*x^5 + 15*(-1 
 + a*x)^2*(1 + a*x)*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(15*a*c^3*(-1 + a*x)^2* 
(1 + a*x)*Sqrt[1 - a^2*x^2])
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.86, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6707, 6698, 529, 2345, 2345, 27, 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.

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

\(\Big \downarrow \) 6707

\(\displaystyle -\frac {a^6 \int \frac {e^{\text {arctanh}(a x)} x^6}{\left (1-a^2 x^2\right )^3}dx}{c^3}\)

\(\Big \downarrow \) 6698

\(\displaystyle -\frac {a^6 \int \frac {x^6 (a x+1)}{\left (1-a^2 x^2\right )^{7/2}}dx}{c^3}\)

\(\Big \downarrow \) 529

\(\displaystyle -\frac {a^6 \left (\frac {a x+1}{5 a^7 \left (1-a^2 x^2\right )^{5/2}}-\frac {1}{5} \int \frac {\frac {5 x^5}{a}+\frac {5 x^4}{a^2}+\frac {5 x^3}{a^3}+\frac {5 x^2}{a^4}+\frac {5 x}{a^5}+\frac {1}{a^6}}{\left (1-a^2 x^2\right )^{5/2}}dx\right )}{c^3}\)

\(\Big \downarrow \) 2345

\(\displaystyle -\frac {a^6 \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {\frac {15 x^3}{a^3}+\frac {15 x^2}{a^4}+\frac {30 x}{a^5}+\frac {8}{a^6}}{\left (1-a^2 x^2\right )^{3/2}}dx-\frac {11 a x+15}{3 a^7 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {a x+1}{5 a^7 \left (1-a^2 x^2\right )^{5/2}}\right )}{c^3}\)

\(\Big \downarrow \) 2345

\(\displaystyle -\frac {a^6 \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {23 a x+45}{a^7 \sqrt {1-a^2 x^2}}-\int \frac {15 (a x+1)}{a^6 \sqrt {1-a^2 x^2}}dx\right )-\frac {11 a x+15}{3 a^7 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {a x+1}{5 a^7 \left (1-a^2 x^2\right )^{5/2}}\right )}{c^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {a^6 \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {23 a x+45}{a^7 \sqrt {1-a^2 x^2}}-\frac {15 \int \frac {a x+1}{\sqrt {1-a^2 x^2}}dx}{a^6}\right )-\frac {11 a x+15}{3 a^7 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {a x+1}{5 a^7 \left (1-a^2 x^2\right )^{5/2}}\right )}{c^3}\)

\(\Big \downarrow \) 455

\(\displaystyle -\frac {a^6 \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {23 a x+45}{a^7 \sqrt {1-a^2 x^2}}-\frac {15 \left (\int \frac {1}{\sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{a}\right )}{a^6}\right )-\frac {11 a x+15}{3 a^7 \left (1-a^2 x^2\right )^{3/2}}\right )+\frac {a x+1}{5 a^7 \left (1-a^2 x^2\right )^{5/2}}\right )}{c^3}\)

\(\Big \downarrow \) 223

\(\displaystyle -\frac {a^6 \left (\frac {a x+1}{5 a^7 \left (1-a^2 x^2\right )^{5/2}}+\frac {1}{5} \left (\frac {1}{3} \left (\frac {23 a x+45}{a^7 \sqrt {1-a^2 x^2}}-\frac {15 \left (\frac {\arcsin (a x)}{a}-\frac {\sqrt {1-a^2 x^2}}{a}\right )}{a^6}\right )-\frac {11 a x+15}{3 a^7 \left (1-a^2 x^2\right )^{3/2}}\right )\right )}{c^3}\)

Input:

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

Output:

-((a^6*((1 + a*x)/(5*a^7*(1 - a^2*x^2)^(5/2)) + (-1/3*(15 + 11*a*x)/(a^7*( 
1 - a^2*x^2)^(3/2)) + ((45 + 23*a*x)/(a^7*Sqrt[1 - a^2*x^2]) - (15*(-(Sqrt 
[1 - a^2*x^2]/a) + ArcSin[a*x]/a))/a^6)/3)/5))/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 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 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 2345
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[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))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

rule 6698
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[c^p   Int[x^m*(1 - a^2*x^2)^(p - n/2)*(1 + a*x)^n, x], x] 
/; FreeQ[{a, c, d, m, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 
 0]) && IGtQ[(n + 1)/2, 0] &&  !IntegerQ[p - n/2]
 

rule 6707
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb 
ol] :> Simp[d^p   Int[(u/x^(2*p))*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x 
] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]
 
Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.73

method result size
risch \(\frac {a^{2} x^{2}-1}{a \sqrt {-a^{2} x^{2}+1}\, c^{3}}+\frac {\left (\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{6} \sqrt {a^{2}}}+\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{20 a^{10} \left (x -\frac {1}{a}\right )^{3}}+\frac {23 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{60 a^{9} \left (x -\frac {1}{a}\right )^{2}}+\frac {493 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{240 a^{8} \left (x -\frac {1}{a}\right )}+\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{24 a^{9} \left (x +\frac {1}{a}\right )^{2}}-\frac {25 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{48 a^{8} \left (x +\frac {1}{a}\right )}\right ) a^{6}}{c^{3}}\) \(260\)
default \(\frac {a^{6} \left (\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{6} \sqrt {a^{2}}}-\frac {\sqrt {-a^{2} x^{2}+1}}{a^{7}}-\frac {9 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{16 a^{8} \left (x +\frac {1}{a}\right )}+\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a \left (x -\frac {1}{a}\right )^{3}}-\frac {2 a \left (\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{5}}{4 a^{9}}+\frac {\frac {5 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{12 a \left (x -\frac {1}{a}\right )^{2}}-\frac {5 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{12 \left (x -\frac {1}{a}\right )}}{a^{8}}+\frac {39 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{16 a^{8} \left (x -\frac {1}{a}\right )}-\frac {-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{3 a \left (x +\frac {1}{a}\right )^{2}}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{3 \left (x +\frac {1}{a}\right )}}{8 a^{8}}\right )}{c^{3}}\) \(419\)

Input:

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

Output:

1/a*(a^2*x^2-1)/(-a^2*x^2+1)^(1/2)/c^3+(1/a^6/(a^2)^(1/2)*arctan((a^2)^(1/ 
2)*x/(-a^2*x^2+1)^(1/2))+1/20/a^10/(x-1/a)^3*(-(x-1/a)^2*a^2-2*a*(x-1/a))^ 
(1/2)+23/60/a^9/(x-1/a)^2*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2)+493/240/a^8/( 
x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2)+1/24/a^9/(x+1/a)^2*(-a^2*(x+1/a) 
^2+2*a*(x+1/a))^(1/2)-25/48/a^8/(x+1/a)*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2) 
)*a^6/c^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.41 \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=-\frac {48 \, a^{5} x^{5} - 48 \, a^{4} x^{4} - 96 \, a^{3} x^{3} + 96 \, a^{2} x^{2} + 48 \, a x + 30 \, {\left (a^{5} x^{5} - a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} + a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (15 \, a^{5} x^{5} - 38 \, a^{4} x^{4} - 52 \, a^{3} x^{3} + 87 \, a^{2} x^{2} + 33 \, a x - 48\right )} \sqrt {-a^{2} x^{2} + 1} - 48}{15 \, {\left (a^{6} c^{3} x^{5} - a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{3} c^{3} x^{2} + a^{2} c^{3} x - a c^{3}\right )}} \] Input:

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

Output:

-1/15*(48*a^5*x^5 - 48*a^4*x^4 - 96*a^3*x^3 + 96*a^2*x^2 + 48*a*x + 30*(a^ 
5*x^5 - a^4*x^4 - 2*a^3*x^3 + 2*a^2*x^2 + a*x - 1)*arctan((sqrt(-a^2*x^2 + 
 1) - 1)/(a*x)) + (15*a^5*x^5 - 38*a^4*x^4 - 52*a^3*x^3 + 87*a^2*x^2 + 33* 
a*x - 48)*sqrt(-a^2*x^2 + 1) - 48)/(a^6*c^3*x^5 - a^5*c^3*x^4 - 2*a^4*c^3* 
x^3 + 2*a^3*c^3*x^2 + a^2*c^3*x - a*c^3)
 

Sympy [F]

\[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {a^{6} \int \frac {x^{6}}{a^{5} x^{5} \sqrt {- a^{2} x^{2} + 1} - a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} - 2 a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} + 2 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + a x \sqrt {- a^{2} x^{2} + 1} - \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{3}} \] Input:

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

Output:

a**6*Integral(x**6/(a**5*x**5*sqrt(-a**2*x**2 + 1) - a**4*x**4*sqrt(-a**2* 
x**2 + 1) - 2*a**3*x**3*sqrt(-a**2*x**2 + 1) + 2*a**2*x**2*sqrt(-a**2*x**2 
 + 1) + a*x*sqrt(-a**2*x**2 + 1) - sqrt(-a**2*x**2 + 1)), x)/c**3
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.45 \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {5\,a\,\sqrt {1-a^2\,x^2}}{12\,\left (a^4\,c^3\,x^2-2\,a^3\,c^3\,x+a^2\,c^3\right )}+\frac {a\,\sqrt {1-a^2\,x^2}}{24\,\left (a^4\,c^3\,x^2+2\,a^3\,c^3\,x+a^2\,c^3\right )}+\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^3\,\sqrt {-a^2}}-\frac {a^6\,\sqrt {1-a^2\,x^2}}{30\,\left (a^9\,c^3\,x^2-2\,a^8\,c^3\,x+a^7\,c^3\right )}-\frac {\sqrt {1-a^2\,x^2}}{a\,c^3}+\frac {25\,\sqrt {1-a^2\,x^2}}{48\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}+\frac {c^3\,\sqrt {-a^2}}{a}\right )}-\frac {493\,\sqrt {1-a^2\,x^2}}{240\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}\right )}-\frac {\sqrt {1-a^2\,x^2}}{20\,\sqrt {-a^2}\,\left (3\,c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}+a^2\,c^3\,x^3\,\sqrt {-a^2}-3\,a\,c^3\,x^2\,\sqrt {-a^2}\right )} \] Input:

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

Output:

(5*a*(1 - a^2*x^2)^(1/2))/(12*(a^2*c^3 - 2*a^3*c^3*x + a^4*c^3*x^2)) + (a* 
(1 - a^2*x^2)^(1/2))/(24*(a^2*c^3 + 2*a^3*c^3*x + a^4*c^3*x^2)) + asinh(x* 
(-a^2)^(1/2))/(c^3*(-a^2)^(1/2)) - (a^6*(1 - a^2*x^2)^(1/2))/(30*(a^7*c^3 
- 2*a^8*c^3*x + a^9*c^3*x^2)) - (1 - a^2*x^2)^(1/2)/(a*c^3) + (25*(1 - a^2 
*x^2)^(1/2))/(48*(-a^2)^(1/2)*(c^3*x*(-a^2)^(1/2) + (c^3*(-a^2)^(1/2))/a)) 
 - (493*(1 - a^2*x^2)^(1/2))/(240*(-a^2)^(1/2)*(c^3*x*(-a^2)^(1/2) - (c^3* 
(-a^2)^(1/2))/a)) - (1 - a^2*x^2)^(1/2)/(20*(-a^2)^(1/2)*(3*c^3*x*(-a^2)^( 
1/2) - (c^3*(-a^2)^(1/2))/a + a^2*c^3*x^3*(-a^2)^(1/2) - 3*a*c^3*x^2*(-a^2 
)^(1/2)))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.53 \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {15 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a^{3} x^{3}-15 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a^{2} x^{2}-15 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a x +15 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )-33 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}+33 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+33 \sqrt {-a^{2} x^{2}+1}\, a x -33 \sqrt {-a^{2} x^{2}+1}+15 a^{5} x^{5}-38 a^{4} x^{4}-52 a^{3} x^{3}+87 a^{2} x^{2}+33 a x -48}{15 \sqrt {-a^{2} x^{2}+1}\, a \,c^{3} \left (a^{3} x^{3}-a^{2} x^{2}-a x +1\right )} \] Input:

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

Output:

(15*sqrt( - a**2*x**2 + 1)*asin(a*x)*a**3*x**3 - 15*sqrt( - a**2*x**2 + 1) 
*asin(a*x)*a**2*x**2 - 15*sqrt( - a**2*x**2 + 1)*asin(a*x)*a*x + 15*sqrt( 
- a**2*x**2 + 1)*asin(a*x) - 33*sqrt( - a**2*x**2 + 1)*a**3*x**3 + 33*sqrt 
( - a**2*x**2 + 1)*a**2*x**2 + 33*sqrt( - a**2*x**2 + 1)*a*x - 33*sqrt( - 
a**2*x**2 + 1) + 15*a**5*x**5 - 38*a**4*x**4 - 52*a**3*x**3 + 87*a**2*x**2 
 + 33*a*x - 48)/(15*sqrt( - a**2*x**2 + 1)*a*c**3*(a**3*x**3 - a**2*x**2 - 
 a*x + 1))