\(\int \frac {e^{\text {arctanh}(a x)}}{(c-\frac {c}{a^2 x^2})^2} \, dx\) [674]

Optimal result

Integrand size = 20, antiderivative size = 93 \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {1+a x}{3 a c^2 \left (1-a^2 x^2\right )^{3/2}}-\frac {2 (3+2 a x)}{3 a c^2 \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a c^2}+\frac {\arcsin (a x)}{a c^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {8-5 a x-7 a^2 x^2+3 a^3 x^3+3 (-1+a x) \sqrt {1-a^2 x^2} \arcsin (a x)}{3 a c^2 (-1+a x) \sqrt {1-a^2 x^2}} \] Input:

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

Output:

(8 - 5*a*x - 7*a^2*x^2 + 3*a^3*x^3 + 3*(-1 + a*x)*Sqrt[1 - a^2*x^2]*ArcSin 
[a*x])/(3*a*c^2*(-1 + a*x)*Sqrt[1 - a^2*x^2])
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6707, 6698, 529, 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.

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

\(\Big \downarrow \) 6707

\(\displaystyle \frac {a^4 \int \frac {e^{\text {arctanh}(a x)} x^4}{\left (1-a^2 x^2\right )^2}dx}{c^2}\)

\(\Big \downarrow \) 6698

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

\(\Big \downarrow \) 529

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

\(\Big \downarrow \) 2345

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 223

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

Input:

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

Output:

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

Leaf count of result is larger than twice the leaf count of optimal. \(185\) vs. \(2(83)=166\).

Time = 0.16 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.00

method	result	size
risch	\(\frac {a^{2} x^{2}-1}{a \sqrt {-a^{2} x^{2}+1}\, c^{2}}-\frac {\left (-\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{4} \sqrt {a^{2}}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{6 a^{7} \left (x -\frac {1}{a}\right )^{2}}-\frac {19 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{12 a^{6} \left (x -\frac {1}{a}\right )}+\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{6} \left (x +\frac {1}{a}\right )}\right ) a^{4}}{c^{2}}\)	\(186\)
default	\(\frac {a^{4} \left (\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{a^{4} \sqrt {a^{2}}}-\frac {\sqrt {-a^{2} x^{2}+1}}{a^{5}}+\frac {7 \sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{4 a^{6} \left (x -\frac {1}{a}\right )}-\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{6} \left (x +\frac {1}{a}\right )}+\frac {\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-\left (x -\frac {1}{a}\right )^{2} a^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}}{2 a^{6}}\right )}{c^{2}}\)	\(216\)

Input:

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

Output:

1/a*(a^2*x^2-1)/(-a^2*x^2+1)^(1/2)/c^2-(-1/a^4/(a^2)^(1/2)*arctan((a^2)^(1 
/2)*x/(-a^2*x^2+1)^(1/2))-1/6/a^7/(x-1/a)^2*(-(x-1/a)^2*a^2-2*a*(x-1/a))^( 
1/2)-19/12/a^6/(x-1/a)*(-(x-1/a)^2*a^2-2*a*(x-1/a))^(1/2)+1/4/a^6/(x+1/a)* 
(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))*a^4/c^2
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.53 \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=-\frac {8 \, a^{3} x^{3} - 8 \, a^{2} x^{2} - 8 \, a x + 6 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (3 \, a^{3} x^{3} - 7 \, a^{2} x^{2} - 5 \, a x + 8\right )} \sqrt {-a^{2} x^{2} + 1} + 8}{3 \, {\left (a^{4} c^{2} x^{3} - a^{3} c^{2} x^{2} - a^{2} c^{2} x + a c^{2}\right )}} \] Input:

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

Output:

-1/3*(8*a^3*x^3 - 8*a^2*x^2 - 8*a*x + 6*(a^3*x^3 - a^2*x^2 - a*x + 1)*arct 
an((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (3*a^3*x^3 - 7*a^2*x^2 - 5*a*x + 8)*s 
qrt(-a^2*x^2 + 1) + 8)/(a^4*c^2*x^3 - a^3*c^2*x^2 - a^2*c^2*x + a*c^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 22.50 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.02 \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {a\,\sqrt {1-a^2\,x^2}}{6\,\left (a^4\,c^2\,x^2-2\,a^3\,c^2\,x+a^2\,c^2\right )}+\frac {\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^2\,\sqrt {-a^2}}-\frac {\sqrt {1-a^2\,x^2}}{a\,c^2}+\frac {\sqrt {1-a^2\,x^2}}{4\,\sqrt {-a^2}\,\left (c^2\,x\,\sqrt {-a^2}+\frac {c^2\,\sqrt {-a^2}}{a}\right )}-\frac {19\,\sqrt {1-a^2\,x^2}}{12\,\sqrt {-a^2}\,\left (c^2\,x\,\sqrt {-a^2}-\frac {c^2\,\sqrt {-a^2}}{a}\right )} \] Input:

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

Output:

(a*(1 - a^2*x^2)^(1/2))/(6*(a^2*c^2 - 2*a^3*c^2*x + a^4*c^2*x^2)) + asinh( 
x*(-a^2)^(1/2))/(c^2*(-a^2)^(1/2)) - (1 - a^2*x^2)^(1/2)/(a*c^2) + (1 - a^ 
2*x^2)^(1/2)/(4*(-a^2)^(1/2)*(c^2*x*(-a^2)^(1/2) + (c^2*(-a^2)^(1/2))/a)) 
- (19*(1 - a^2*x^2)^(1/2))/(12*(-a^2)^(1/2)*(c^2*x*(-a^2)^(1/2) - (c^2*(-a 
^2)^(1/2))/a))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.23 \[ \int \frac {e^{\text {arctanh}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx=\frac {3 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right ) a x -3 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )-5 \sqrt {-a^{2} x^{2}+1}\, a x +5 \sqrt {-a^{2} x^{2}+1}+3 a^{3} x^{3}-7 a^{2} x^{2}-5 a x +8}{3 \sqrt {-a^{2} x^{2}+1}\, a \,c^{2} \left (a x -1\right )} \] Input:

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

Output:

(3*sqrt( - a**2*x**2 + 1)*asin(a*x)*a*x - 3*sqrt( - a**2*x**2 + 1)*asin(a* 
x) - 5*sqrt( - a**2*x**2 + 1)*a*x + 5*sqrt( - a**2*x**2 + 1) + 3*a**3*x**3 
 - 7*a**2*x**2 - 5*a*x + 8)/(3*sqrt( - a**2*x**2 + 1)*a*c**2*(a*x - 1))