\(\int \frac {e^{-\text {arctanh}(a x)}}{(c-\frac {c}{a x})^3} \, dx\) [509]

Optimal result

Integrand size = 22, antiderivative size = 98 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {\sqrt {1-a^2 x^2}}{a c^3}-\frac {\sqrt {1-a^2 x^2}}{3 a c^3 (1-a x)^2}+\frac {8 \sqrt {1-a^2 x^2}}{3 a c^3 (1-a x)}-\frac {2 \arcsin (a x)}{a c^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.52 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {\frac {\sqrt {1+a x} \left (10-14 a x+3 a^2 x^2\right )}{(1-a x)^{3/2}}-6 \arcsin (a x)}{3 a c^3} \] Input:

Integrate[1/(E^ArcTanh[a*x]*(c - c/(a*x))^3),x]
 

Output:

((Sqrt[1 + a*x]*(10 - 14*a*x + 3*a^2*x^2))/(1 - a*x)^(3/2) - 6*ArcSin[a*x] 
)/(3*a*c^3)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6681, 6678, 570, 529, 2166, 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.

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

\(\Big \downarrow \) 6681

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

\(\Big \downarrow \) 6678

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

\(\Big \downarrow \) 570

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

\(\Big \downarrow \) 529

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

\(\Big \downarrow \) 2166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 223

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

Input:

Int[1/(E^ArcTanh[a*x]*(c - c/(a*x))^3),x]
 

Output:

-((a^3*((1 + a*x)^2/(3*a^4*(1 - a^2*x^2)^(3/2)) + ((-8*(1 + a*x))/(a^4*Sqr 
t[1 - a^2*x^2]) + (3*(-(Sqrt[1 - a^2*x^2]/a) + (2*ArcSin[a*x])/a))/a^3)/3) 
)/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 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 2166

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde 
r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e 
*(p + 1))), x] + Simp[d/(2*a*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^( 
p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ 
[{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 
 0] && GtQ[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]
 

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 = 151, normalized size of antiderivative = 1.54

method	result	size
risch	\(-\frac {a^{2} x^{2}-1}{a \sqrt {-a^{2} x^{2}+1}\, c^{3}}+\frac {\left (-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a^{6} \left (x -\frac {1}{a}\right )^{2}}-\frac {8 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a^{5} \left (x -\frac {1}{a}\right )}-\frac {2 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{3} \sqrt {a^{2}}}\right ) a^{3}}{c^{3}}\)	\(151\)
default	\(\frac {a^{3} \left (\frac {\left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{6 a^{7} \left (x -\frac {1}{a}\right )^{3}}+\frac {\frac {5 \left (-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{4 a \left (x -\frac {1}{a}\right )^{2}}+\frac {5 a \left (\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}-\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}\right )}{4}}{a^{5}}+\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}}{8 a^{4}}+\frac {\frac {7 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{8}-\frac {7 a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{8 \sqrt {a^{2}}}}{a^{4}}\right )}{c^{3}}\)	\(312\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.09 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {10 \, a^{2} x^{2} - 20 \, a x + 12 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (3 \, a^{2} x^{2} - 14 \, a x + 10\right )} \sqrt {-a^{2} x^{2} + 1} + 10}{3 \, {\left (a^{3} c^{3} x^{2} - 2 \, a^{2} c^{3} x + a c^{3}\right )}} \] Input:

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

Output:

1/3*(10*a^2*x^2 - 20*a*x + 12*(a^2*x^2 - 2*a*x + 1)*arctan((sqrt(-a^2*x^2 
+ 1) - 1)/(a*x)) + (3*a^2*x^2 - 14*a*x + 10)*sqrt(-a^2*x^2 + 1) + 10)/(a^3 
*c^3*x^2 - 2*a^2*c^3*x + a*c^3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(c-c/a/x)^3,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 [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {\sqrt {1-a^2\,x^2}}{a\,c^3}-\frac {2\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^3\,\sqrt {-a^2}}-\frac {a\,\sqrt {1-a^2\,x^2}}{3\,\left (a^4\,c^3\,x^2-2\,a^3\,c^3\,x+a^2\,c^3\right )}+\frac {8\,\sqrt {1-a^2\,x^2}}{3\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.85 \[ \int \frac {e^{-\text {arctanh}(a x)}}{\left (c-\frac {c}{a x}\right )^3} \, dx=\frac {-6 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a x +6 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )-6 \mathit {asin} \left (a x \right ) a^{2} x^{2}+12 \mathit {asin} \left (a x \right ) a x -6 \mathit {asin} \left (a x \right )+3 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-8 \sqrt {-a^{2} x^{2}+1}\, a x +4 \sqrt {-a^{2} x^{2}+1}-3 a^{3} x^{3}+17 a^{2} x^{2}-8 a x -4}{3 a \,c^{3} \left (\sqrt {-a^{2} x^{2}+1}\, a x -\sqrt {-a^{2} x^{2}+1}+a^{2} x^{2}-2 a x +1\right )} \] Input:

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

Output:

( - 6*sqrt( - a**2*x**2 + 1)*asin(a*x)*a*x + 6*sqrt( - a**2*x**2 + 1)*asin 
(a*x) - 6*asin(a*x)*a**2*x**2 + 12*asin(a*x)*a*x - 6*asin(a*x) + 3*sqrt( - 
 a**2*x**2 + 1)*a**2*x**2 - 8*sqrt( - a**2*x**2 + 1)*a*x + 4*sqrt( - a**2* 
x**2 + 1) - 3*a**3*x**3 + 17*a**2*x**2 - 8*a*x - 4)/(3*a*c**3*(sqrt( - a** 
2*x**2 + 1)*a*x - sqrt( - a**2*x**2 + 1) + a**2*x**2 - 2*a*x + 1))