\(\int \frac {e^{\text {arctanh}(a+b x)}}{x (1-a^2-2 a b x-b^2 x^2)} \, dx\) [903]

Optimal result

Integrand size = 34, antiderivative size = 93 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x \left (1-a^2-2 a b x-b^2 x^2\right )} \, dx=\frac {\sqrt {1+a+b x}}{(1-a) \sqrt {1-a-b x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {1-a} \sqrt {1+a+b x}}{\sqrt {1+a} \sqrt {1-a-b x}}\right )}{(1-a) \sqrt {1-a^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.27 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x \left (1-a^2-2 a b x-b^2 x^2\right )} \, dx=-\frac {-\frac {\sqrt {1-a^2-2 a b x-b^2 x^2}}{-1+a+b x}+\frac {\log (x)}{\sqrt {1-a^2}}-\frac {\log \left (1-a^2-a b x+\sqrt {1-a^2} \sqrt {1-a^2-2 a b x-b^2 x^2}\right )}{\sqrt {1-a^2}}}{-1+a} \] Input:

Integrate[E^ArcTanh[a + b*x]/(x*(1 - a^2 - 2*a*b*x - b^2*x^2)),x]
 

Output:

-((-(Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]/(-1 + a + b*x)) + Log[x]/Sqrt[1 - a 
^2] - Log[1 - a^2 - a*b*x + Sqrt[1 - a^2]*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2 
]]/Sqrt[1 - a^2])/(-1 + a))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6714, 107, 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.

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

\(\Big \downarrow \) 6714

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

\(\Big \downarrow \) 107

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

\(\Big \downarrow \) 104

\(\displaystyle \frac {2 \int \frac {1}{-a+\frac {(1-a) (a+b x+1)}{-a-b x+1}-1}d\frac {\sqrt {a+b x+1}}{\sqrt {-a-b x+1}}}{1-a}+\frac {\sqrt {a+b x+1}}{(1-a) \sqrt {-a-b x+1}}\)

\(\Big \downarrow \) 221

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

Input:

Int[E^ArcTanh[a + b*x]/(x*(1 - a^2 - 2*a*b*x - b^2*x^2)),x]
 

Output:

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

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 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 
 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[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 6714

Int[E^(ArcTanh[(a_) + (b_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_) + (e_.)*( 
x_)^2)^(p_.), x_Symbol] :> Simp[(c/(1 - a^2))^p   Int[u*(1 - a - b*x)^(p - 
n/2)*(1 + a + b*x)^(p + n/2), x], x] /; FreeQ[{a, b, c, d, e, n, p}, x] && 
EqQ[b*d - 2*a*e, 0] && EqQ[b^2*c + e*(1 - a^2), 0] && (IntegerQ[p] || GtQ[c 
/(1 - a^2), 0])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(227\) vs. \(2(79)=158\).

Time = 0.56 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.45

method	result	size
default	\(\frac {2 b \left (-2 b^{2} x -2 a b \right )}{\left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\left (a +1\right ) \left (\frac {1}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {2 a b \left (-2 b^{2} x -2 a b \right )}{\left (-a^{2}+1\right ) \left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {\ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\left (-a^{2}+1\right )^{\frac {3}{2}}}\right )\)	\(228\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 316, normalized size of antiderivative = 3.40 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x \left (1-a^2-2 a b x-b^2 x^2\right )} \, dx=\left [-\frac {\sqrt {-a^{2} + 1} {\left (b x + a - 1\right )} \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {-a^{2} + 1} - 4 \, a^{2} + 2}{x^{2}}\right ) - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )}}{2 \, {\left (a^{4} - 2 \, a^{3} + {\left (a^{3} - a^{2} - a + 1\right )} b x + 2 \, a - 1\right )}}, -\frac {\sqrt {a^{2} - 1} {\left (b x + a - 1\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \, {\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) - \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )}}{a^{4} - 2 \, a^{3} + {\left (a^{3} - a^{2} - a + 1\right )} b x + 2 \, a - 1}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x \left (1-a^2-2 a b x-b^2 x^2\right )} \, dx=-\frac {2 \, b \arctan \left (\frac {\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt {a^{2} - 1}}\right )}{\sqrt {a^{2} - 1} {\left (a {\left | b \right |} - {\left | b \right |}\right )}} - \frac {2 \, b}{{\left (a {\left | b \right |} - {\left | b \right |}\right )} {\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} - 1\right )}} \] Input:

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

Output:

-2*b*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))/(sqrt(a^2 - 1)*(a*abs(b) - abs(b))) - 2*b/((a*abs(b 
) - abs(b))*((sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*abs(b) + b)/(b^2*x + a*b) 
 - 1))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.70 \[ \int \frac {e^{\text {arctanh}(a+b x)}}{x \left (1-a^2-2 a b x-b^2 x^2\right )} \, dx=\frac {2 \sqrt {a^{2}-1}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right ) a -1}{\sqrt {a^{2}-1}}\right ) \tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right )-2 \sqrt {a^{2}-1}\, \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right ) a -1}{\sqrt {a^{2}-1}}\right )+2 \tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right ) a^{2}-2 \tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right )}{\tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right ) a^{3}-\tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right ) a^{2}-\tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right ) a +\tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right )-a^{3}+a^{2}+a -1} \] Input:

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

Output:

(2*(sqrt(a**2 - 1)*atan((tan(asin(a + b*x)/2)*a - 1)/sqrt(a**2 - 1))*tan(a 
sin(a + b*x)/2) - sqrt(a**2 - 1)*atan((tan(asin(a + b*x)/2)*a - 1)/sqrt(a* 
*2 - 1)) + tan(asin(a + b*x)/2)*a**2 - tan(asin(a + b*x)/2)))/(tan(asin(a 
+ b*x)/2)*a**3 - tan(asin(a + b*x)/2)*a**2 - tan(asin(a + b*x)/2)*a + tan( 
asin(a + b*x)/2) - a**3 + a**2 + a - 1)