\(\int \frac {e^{3 \text {arctanh}(a+b x)}}{x^2} \, dx\) [877]

Optimal result

Integrand size = 14, antiderivative size = 134 \[ \int \frac {e^{3 \text {arctanh}(a+b x)}}{x^2} \, dx=\frac {6 b \sqrt {1+a+b x}}{(1-a)^2 \sqrt {1-a-b x}}-\frac {(1+a+b x)^{3/2}}{(1-a) x \sqrt {1-a-b x}}-\frac {6 (1+a) b \text {arctanh}\left (\frac {\sqrt {1-a} \sqrt {1+a+b x}}{\sqrt {1+a} \sqrt {1-a-b x}}\right )}{(1-a)^2 \sqrt {1-a^2}} \] Output:

6*b*(b*x+a+1)^(1/2)/(1-a)^2/(-b*x-a+1)^(1/2)-(b*x+a+1)^(3/2)/(1-a)/x/(-b*x 
-a+1)^(1/2)-6*(1+a)*b*arctanh((1-a)^(1/2)*(b*x+a+1)^(1/2)/(1+a)^(1/2)/(-b* 
x-a+1)^(1/2))/(1-a)^2/(-a^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.79 \[ \int \frac {e^{3 \text {arctanh}(a+b x)}}{x^2} \, dx=\frac {\sqrt {1+a+b x} \left (-1+a^2+5 b x+a b x\right )}{(-1+a)^2 x \sqrt {1-a-b x}}-\frac {6 \sqrt {-1-a} b \text {arctanh}\left (\frac {\sqrt {-1-a} \sqrt {1-a-b x}}{\sqrt {-1+a} \sqrt {1+a+b x}}\right )}{(-1+a)^{5/2}} \] Input:

Integrate[E^(3*ArcTanh[a + b*x])/x^2,x]
 

Output:

(Sqrt[1 + a + b*x]*(-1 + a^2 + 5*b*x + a*b*x))/((-1 + a)^2*x*Sqrt[1 - a - 
b*x]) - (6*Sqrt[-1 - a]*b*ArcTanh[(Sqrt[-1 - a]*Sqrt[1 - a - b*x])/(Sqrt[- 
1 + a]*Sqrt[1 + a + b*x])])/(-1 + a)^(5/2)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6713, 105, 105, 104, 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.

\(\displaystyle \int \frac {e^{3 \text {arctanh}(a+b x)}}{x^2} \, dx\)

\(\Big \downarrow \) 6713

\(\displaystyle \int \frac {(a+b x+1)^{3/2}}{x^2 (-a-b x+1)^{3/2}}dx\)

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 105

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

Int[E^(3*ArcTanh[a + b*x])/x^2,x]
 

Output:

-((1 + a + b*x)^(3/2)/((1 - a)*x*Sqrt[1 - a - b*x])) + (3*b*((2*Sqrt[1 + a 
 + b*x])/((1 - a)*Sqrt[1 - a - b*x]) - (2*(1 + a)*ArcTanh[(Sqrt[1 - a]*Sqr 
t[1 + a + b*x])/(Sqrt[1 + a]*Sqrt[1 - a - b*x])])/((1 - a)*Sqrt[1 - a^2])) 
)/(1 - a)
 

Defintions of rubi rules used

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 105

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

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 6713

Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.) 
, x_Symbol] :> Int[(d + e*x)^m*((1 + a*c + b*c*x)^(n/2)/(1 - a*c - b*c*x)^( 
n/2)), x] /; FreeQ[{a, b, c, d, e, m, n}, x]
 

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.30

method	result	size
risch	\(\frac {\left (a +1\right ) \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}{\left (-1+a \right )^{2} x \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {b \left (-\frac {\left (3 a +3\right ) \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\sqrt {-a^{2}+1}}-\frac {4 \sqrt {-\left (x +\frac {-1+a}{b}\right )^{2} b^{2}-2 \left (x +\frac {-1+a}{b}\right ) b}}{b \left (x +\frac {-1+a}{b}\right )}\right )}{\left (-1+a \right )^{2}}\)	\(174\)
default	\(b^{3} \left (\frac {1}{b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {2 a \left (-2 b^{2} x -2 a b \right )}{b \left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )+\left (a^{3}+3 a^{2}+3 a +1\right ) \left (-\frac {1}{\left (-a^{2}+1\right ) x \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {3 a b \left (\frac {1}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {2 a b \left (-2 b^{2} x -2 a b \right )}{\left (-a^{2}+1\right ) \left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {\ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\left (-a^{2}+1\right )^{\frac {3}{2}}}\right )}{-a^{2}+1}+\frac {4 b^{2} \left (-2 b^{2} x -2 a b \right )}{\left (-a^{2}+1\right ) \left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )+\frac {6 b^{2} \left (-2 b^{2} x -2 a b \right )}{\left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {6 a \,b^{2} \left (-2 b^{2} x -2 a b \right )}{\left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+3 b \left (a^{2}+2 a +1\right ) \left (\frac {1}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {2 a b \left (-2 b^{2} x -2 a b \right )}{\left (-a^{2}+1\right ) \left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {\ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\left (-a^{2}+1\right )^{\frac {3}{2}}}\right )\)	\(689\)

Input:

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

Output:

(a+1)/(-1+a)^2*(b^2*x^2+2*a*b*x+a^2-1)/x/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+b/ 
(-1+a)^2*(-(3*a+3)/(-a^2+1)^(1/2)*ln((-2*a^2+2-2*a*b*x+2*(-a^2+1)^(1/2)*(- 
b^2*x^2-2*a*b*x-a^2+1)^(1/2))/x)-4/b/(x+(-1+a)/b)*(-(x+(-1+a)/b)^2*b^2-2*( 
x+(-1+a)/b)*b)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 370, normalized size of antiderivative = 2.76 \[ \int \frac {e^{3 \text {arctanh}(a+b x)}}{x^2} \, dx=\left [\frac {3 \, {\left (b^{2} x^{2} + {\left (a - 1\right )} b x\right )} \sqrt {-\frac {a + 1}{a - 1}} \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 4 \, a^{2} - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{3} + {\left (a^{2} - a\right )} b x - a^{2} - a + 1\right )} \sqrt {-\frac {a + 1}{a - 1}} + 2}{x^{2}}\right ) - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left ({\left (a + 5\right )} b x + a^{2} - 1\right )}}{2 \, {\left ({\left (a^{2} - 2 \, a + 1\right )} b x^{2} + {\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} x\right )}}, \frac {3 \, {\left (b^{2} x^{2} + {\left (a - 1\right )} b x\right )} \sqrt {\frac {a + 1}{a - 1}} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {\frac {a + 1}{a - 1}}}{{\left (a + 1\right )} b^{2} x^{2} + a^{3} + 2 \, {\left (a^{2} + a\right )} b x + a^{2} - a - 1}\right ) - \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left ({\left (a + 5\right )} b x + a^{2} - 1\right )}}{{\left (a^{2} - 2 \, a + 1\right )} b x^{2} + {\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} x}\right ] \] Input:

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

Output:

[1/2*(3*(b^2*x^2 + (a - 1)*b*x)*sqrt(-(a + 1)/(a - 1))*log(((2*a^2 - 1)*b^ 
2*x^2 + 2*a^4 + 4*(a^3 - a)*b*x - 4*a^2 - 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 
+ 1)*(a^3 + (a^2 - a)*b*x - a^2 - a + 1)*sqrt(-(a + 1)/(a - 1)) + 2)/x^2) 
- 2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*((a + 5)*b*x + a^2 - 1))/((a^2 - 2* 
a + 1)*b*x^2 + (a^3 - 3*a^2 + 3*a - 1)*x), (3*(b^2*x^2 + (a - 1)*b*x)*sqrt 
((a + 1)/(a - 1))*arctan(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(a*b*x + a^2 - 
 1)*sqrt((a + 1)/(a - 1))/((a + 1)*b^2*x^2 + a^3 + 2*(a^2 + a)*b*x + a^2 - 
 a - 1)) - sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*((a + 5)*b*x + a^2 - 1))/((a 
^2 - 2*a + 1)*b*x^2 + (a^3 - 3*a^2 + 3*a - 1)*x)]
 

Sympy [F]

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

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

Output:

Integral((a + b*x + 1)**3/(x**2*(-(a + b*x - 1)*(a + b*x + 1))**(3/2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{3 \text {arctanh}(a+b x)}}{x^2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a-1>0)', see `assume?` for more 
details)Is
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 606 vs. \(2 (109) = 218\).

Time = 0.15 (sec) , antiderivative size = 606, normalized size of antiderivative = 4.52 \[ \int \frac {e^{3 \text {arctanh}(a+b x)}}{x^2} \, dx=\frac {6 \, {\left (a b^{2} + b^{2}\right )} \arctan \left (\frac {\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt {a^{2} - 1}}\right )}{{\left (a^{2} {\left | b \right |} - 2 \, a {\left | b \right |} + {\left | b \right |}\right )} \sqrt {a^{2} - 1}} - \frac {2 \, {\left (\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a^{2} b^{2}}{b^{2} x + a b} - \frac {4 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2} a^{2} b^{2}}{{\left (b^{2} x + a b\right )}^{2}} - 5 \, a^{2} b^{2} + \frac {10 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a b^{2}}{b^{2} x + a b} - \frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2} a b^{2}}{{\left (b^{2} x + a b\right )}^{2}} - a b^{2} + \frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} b^{2}}{b^{2} x + a b} - \frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2} b^{2}}{{\left (b^{2} x + a b\right )}^{2}}\right )}}{{\left (a^{3} {\left | b \right |} - 2 \, a^{2} {\left | b \right |} + a {\left | b \right |}\right )} {\left (\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a}{b^{2} x + a b} - \frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2} a}{{\left (b^{2} x + a b\right )}^{2}} + \frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{3} a}{{\left (b^{2} x + a b\right )}^{3}} - a + \frac {2 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}}{b^{2} x + a b} - \frac {2 \, {\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )}^{2}}{{\left (b^{2} x + a b\right )}^{2}}\right )}} \] Input:

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

Output:

6*(a*b^2 + b^2)*arctan(((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a/ 
(b^2*x + a*b) - 1)/sqrt(a^2 - 1))/((a^2*abs(b) - 2*a*abs(b) + abs(b))*sqrt 
(a^2 - 1)) - 2*((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a^2*b^2/(b 
^2*x + a*b) - 4*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a^2*b^2/ 
(b^2*x + a*b)^2 - 5*a^2*b^2 + 10*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b 
) + b)*a*b^2/(b^2*x + a*b) - (sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + 
b)^2*a*b^2/(b^2*x + a*b)^2 - a*b^2 + (sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a 
bs(b) + b)*b^2/(b^2*x + a*b) - (sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) 
+ b)^2*b^2/(b^2*x + a*b)^2)/((a^3*abs(b) - 2*a^2*abs(b) + a*abs(b))*((sqrt 
(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)*a/(b^2*x + a*b) - (sqrt(-b^2*x^ 
2 - 2*a*b*x - a^2 + 1)*abs(b) + b)^2*a/(b^2*x + a*b)^2 + (sqrt(-b^2*x^2 - 
2*a*b*x - a^2 + 1)*abs(b) + b)^3*a/(b^2*x + a*b)^3 - a + 2*(sqrt(-b^2*x^2 
- 2*a*b*x - a^2 + 1)*abs(b) + b)/(b^2*x + a*b) - 2*(sqrt(-b^2*x^2 - 2*a*b* 
x - a^2 + 1)*abs(b) + b)^2/(b^2*x + a*b)^2))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 788, normalized size of antiderivative = 5.88 \[ \int \frac {e^{3 \text {arctanh}(a+b x)}}{x^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*b*( - 3*sqrt(a**2 - 1)*atan((tan(asin(a + b*x)/2)*a - 1)/sqrt(a**2 - 1) 
)*tan(asin(a + b*x)/2)**3*a**2 - 6*sqrt(a**2 - 1)*atan((tan(asin(a + b*x)/ 
2)*a - 1)/sqrt(a**2 - 1))*tan(asin(a + b*x)/2)**3*a + 3*sqrt(a**2 - 1)*ata 
n((tan(asin(a + b*x)/2)*a - 1)/sqrt(a**2 - 1))*tan(asin(a + b*x)/2)**2*a** 
2 + 12*sqrt(a**2 - 1)*atan((tan(asin(a + b*x)/2)*a - 1)/sqrt(a**2 - 1))*ta 
n(asin(a + b*x)/2)**2*a + 12*sqrt(a**2 - 1)*atan((tan(asin(a + b*x)/2)*a - 
 1)/sqrt(a**2 - 1))*tan(asin(a + b*x)/2)**2 - 3*sqrt(a**2 - 1)*atan((tan(a 
sin(a + b*x)/2)*a - 1)/sqrt(a**2 - 1))*tan(asin(a + b*x)/2)*a**2 - 12*sqrt 
(a**2 - 1)*atan((tan(asin(a + b*x)/2)*a - 1)/sqrt(a**2 - 1))*tan(asin(a + 
b*x)/2)*a - 12*sqrt(a**2 - 1)*atan((tan(asin(a + b*x)/2)*a - 1)/sqrt(a**2 
- 1))*tan(asin(a + b*x)/2) + 3*sqrt(a**2 - 1)*atan((tan(asin(a + b*x)/2)*a 
 - 1)/sqrt(a**2 - 1))*a**2 + 6*sqrt(a**2 - 1)*atan((tan(asin(a + b*x)/2)*a 
 - 1)/sqrt(a**2 - 1))*a - 4*tan(asin(a + b*x)/2)**3*a**3 + 3*tan(asin(a + 
b*x)/2)**3*a**2 + tan(asin(a + b*x)/2)**3 - 3*tan(asin(a + b*x)/2)*a**3 + 
6*tan(asin(a + b*x)/2)*a**2 + 15*tan(asin(a + b*x)/2)*a - 18*tan(asin(a + 
b*x)/2) - a**3 - 9*a**2 + 9*a + 1))/(tan(asin(a + b*x)/2)**3*a**5 - tan(as 
in(a + b*x)/2)**3*a**4 - 3*tan(asin(a + b*x)/2)**3*a**3 + 5*tan(asin(a + b 
*x)/2)**3*a**2 - 2*tan(asin(a + b*x)/2)**3*a - tan(asin(a + b*x)/2)**2*a** 
5 - tan(asin(a + b*x)/2)**2*a**4 + 5*tan(asin(a + b*x)/2)**2*a**3 + tan(as 
in(a + b*x)/2)**2*a**2 - 8*tan(asin(a + b*x)/2)**2*a + 4*tan(asin(a + b...