\(\int \frac {e^{\text {arctanh}(a+b x)} x}{1-a^2-2 a b x-b^2 x^2} \, dx\) [901]

Optimal result

Integrand size = 32, antiderivative size = 44 \[ \int \frac {e^{\text {arctanh}(a+b x)} x}{1-a^2-2 a b x-b^2 x^2} \, dx=\frac {(1-a) \sqrt {1+a+b x}}{b^2 \sqrt {1-a-b x}}-\frac {\arcsin (a+b x)}{b^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\text {arctanh}(a+b x)} x}{1-a^2-2 a b x-b^2 x^2} \, dx=-\frac {-\frac {(-1+a) \sqrt {1-a^2-2 a b x-b^2 x^2}}{-1+a+b x}+\arcsin (a+b x)}{b^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.27, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {6714, 87, 62, 1090, 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.

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

\(\Big \downarrow \) 6714

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 62

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

\(\Big \downarrow \) 1090

\(\displaystyle \frac {\int \frac {1}{\sqrt {1-\frac {\left (-2 x b^2-2 a b\right )^2}{4 b^2}}}d\left (-2 x b^2-2 a b\right )}{2 b^3}+\frac {(1-a) \sqrt {a+b x+1}}{b^2 \sqrt {-a-b x+1}}\)

\(\Big \downarrow \) 223

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 62

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Int[ 
1/Sqrt[a*c - b*(a - c)*x - b^2*x^2], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b 
+ d, 0] && GtQ[a + c, 0]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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 1090

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* 
(c/(b^2 - 4*a*c)))^p)   Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, 
b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
 

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. \(265\) vs. \(2(40)=80\).

Time = 0.51 (sec) , antiderivative size = 266, normalized size of antiderivative = 6.05

method	result	size
default	\(b \left (\frac {x}{b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {a \left (\frac {1}{b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {2 a \left (-2 b^{2} x -2 a b \right )}{b \left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{b}-\frac {\arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{b^{2} \sqrt {b^{2}}}\right )+\left (a +1\right ) \left (\frac {1}{b^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {2 a \left (-2 b^{2} x -2 a b \right )}{b \left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )\)	\(266\)

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:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (39) = 78\).

Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.23 \[ \int \frac {e^{\text {arctanh}(a+b x)} x}{1-a^2-2 a b x-b^2 x^2} \, dx=\frac {{\left (b x + a - 1\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a - 1\right )}}{b^{3} x + {\left (a - 1\right )} b^{2}} \] 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:

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

Sympy [F]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (39) = 78\).

Time = 0.15 (sec) , antiderivative size = 120, normalized size of antiderivative = 2.73 \[ \int \frac {e^{\text {arctanh}(a+b x)} x}{1-a^2-2 a b x-b^2 x^2} \, dx=\frac {b^{2} {\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{b^{4} x + a b^{3} - b^{3}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{4} x + a b^{3} - b^{3}} - \frac {\arcsin \left (b x + a\right )}{b^{3}}\right )}}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}} \] 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:

b^2*(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a/(b^4*x + a*b^3 - b^3) - sqrt(-b^ 
2*x^2 - 2*a*b*x - a^2 + 1)/(b^4*x + a*b^3 - b^3) - arcsin(b*x + a)/b^3)/sq 
rt(a^2*b^2 - (a^2 - 1)*b^2)
                                                                                    
                                                                                    
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.70 \[ \int \frac {e^{\text {arctanh}(a+b x)} x}{1-a^2-2 a b x-b^2 x^2} \, dx=\frac {\arcsin \left (-b x - a\right ) \mathrm {sgn}\left (b\right )}{b {\left | b \right |}} - \frac {2 \, {\left (a - 1\right )}}{b {\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} - 1\right )} {\left | b \right |}} \] 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:

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

Mupad [B] (verification not implemented)

Time = 16.95 (sec) , antiderivative size = 229, normalized size of antiderivative = 5.20 \[ \int \frac {e^{\text {arctanh}(a+b x)} x}{1-a^2-2 a b x-b^2 x^2} \, dx=\frac {\left (\frac {a^2\,x}{b}+\frac {a\,\left (a^2-1\right )}{b^2}\right )\,\sqrt {1-{\left (a+b\,x\right )}^2}}{a^2+2\,a\,b\,x+b^2\,x^2-1}+\frac {b\,\ln \left (\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}-\frac {x\,b^2+a\,b}{\sqrt {-b^2}}\right )}{{\left (-b^2\right )}^{3/2}}+\frac {\left (\frac {a^2-1}{b^2}+\frac {a\,x}{b}\right )\,\sqrt {1-{\left (a+b\,x\right )}^2}}{a^2+2\,a\,b\,x+b^2\,x^2-1}+\frac {x\,\left (b^2\,\left (a^2-1\right )-2\,a^2\,b^2\right )-a\,b\,\left (a^2-1\right )}{b\,\left (b^2\,\left (a^2-1\right )-a^2\,b^2\right )\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.45 \[ \int \frac {e^{\text {arctanh}(a+b x)} x}{1-a^2-2 a b x-b^2 x^2} \, dx=\frac {-\mathit {asin} \left (b x +a \right ) \tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right )+\mathit {asin} \left (b x +a \right )+2 \tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right ) a -2 \tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right )}{b^{2} \left (\tan \left (\frac {\mathit {asin} \left (b x +a \right )}{2}\right )-1\right )} \] 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:

( - asin(a + b*x)*tan(asin(a + b*x)/2) + asin(a + b*x) + 2*tan(asin(a + b* 
x)/2)*a - 2*tan(asin(a + b*x)/2))/(b**2*(tan(asin(a + b*x)/2) - 1))