\(\int e^{-\text {arctanh}(a+b x)} x^2 \, dx\) [881]

Optimal result

Integrand size = 14, antiderivative size = 130 \[ \int e^{-\text {arctanh}(a+b x)} x^2 \, dx=\frac {\left (1+2 a+2 a^2\right ) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^3}+\frac {(1+4 a) (1-a-b x)^{3/2} \sqrt {1+a+b x}}{6 b^3}-\frac {x (1-a-b x)^{3/2} \sqrt {1+a+b x}}{3 b^2}+\frac {\left (1+2 a+2 a^2\right ) \arcsin (a+b x)}{2 b^3} \] Output:

1/2*(2*a^2+2*a+1)*(-b*x-a+1)^(1/2)*(b*x+a+1)^(1/2)/b^3+1/6*(1+4*a)*(-b*x-a 
+1)^(3/2)*(b*x+a+1)^(1/2)/b^3-1/3*x*(-b*x-a+1)^(3/2)*(b*x+a+1)^(1/2)/b^2+1 
/2*(2*a^2+2*a+1)*arcsin(b*x+a)/b^3
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.97 \[ \int e^{-\text {arctanh}(a+b x)} x^2 \, dx=-\frac {\sqrt {1+a+b x} \left (-4+7 a^2+2 a^3+7 b x-5 b^2 x^2+2 b^3 x^3+a (-5+8 b x)\right )}{6 b^3 \sqrt {1-a-b x}}+\frac {\left (1+2 a+2 a^2\right ) \sqrt {-b} \text {arcsinh}\left (\frac {\sqrt {-b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {b}}\right )}{b^{7/2}} \] Input:

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

Output:

-1/6*(Sqrt[1 + a + b*x]*(-4 + 7*a^2 + 2*a^3 + 7*b*x - 5*b^2*x^2 + 2*b^3*x^ 
3 + a*(-5 + 8*b*x)))/(b^3*Sqrt[1 - a - b*x]) + ((1 + 2*a + 2*a^2)*Sqrt[-b] 
*ArcSinh[(Sqrt[-b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[b])])/b^(7/2)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6713, 101, 25, 90, 60, 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.

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

\(\Big \downarrow \) 6713

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

\(\Big \downarrow \) 101

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 62

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

\(\Big \downarrow \) 1090

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

\(\Big \downarrow \) 223

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

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 90

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

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
 

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 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 = 0.31 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.99

method	result	size
risch	\(-\frac {\left (2 b^{2} x^{2}-2 a b x +2 a^{2}-3 b x +9 a +4\right ) \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}{6 b^{3} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {\left (2 a^{2}+2 a +1\right ) \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{2 b^{2} \sqrt {b^{2}}}\)	\(129\)
default	\(\frac {\left (a^{2}+2 a +1\right ) \left (\sqrt {-\left (x +\frac {a +1}{b}\right )^{2} b^{2}+2 \left (x +\frac {a +1}{b}\right ) b}+\frac {b \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a +1}{b}-\frac {1}{b}\right )}{\sqrt {-\left (x +\frac {a +1}{b}\right )^{2} b^{2}+2 \left (x +\frac {a +1}{b}\right ) b}}\right )}{\sqrt {b^{2}}}\right )}{b^{3}}-\frac {-\frac {\left (-2 b^{2} x -2 a b \right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{4 b^{2}}-\frac {\left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{8 b^{2} \sqrt {b^{2}}}+a \left (-\frac {\left (-2 b^{2} x -2 a b \right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{4 b^{2}}-\frac {\left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{8 b^{2} \sqrt {b^{2}}}\right )-b \left (-\frac {\left (-b^{2} x^{2}-2 a b x -a^{2}+1\right )^{\frac {3}{2}}}{3 b^{2}}-\frac {a \left (-\frac {\left (-2 b^{2} x -2 a b \right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{4 b^{2}}-\frac {\left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{b}\right )}{b^{2}}\)	\(467\)

Input:

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

Output:

-1/6*(2*b^2*x^2-2*a*b*x+2*a^2-3*b*x+9*a+4)*(b^2*x^2+2*a*b*x+a^2-1)/b^3/(-b 
^2*x^2-2*a*b*x-a^2+1)^(1/2)+1/2/b^2*(2*a^2+2*a+1)/(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))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.90 \[ \int e^{-\text {arctanh}(a+b x)} x^2 \, dx=-\frac {3 \, {\left (2 \, a^{2} + 2 \, a + 1\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - {\left (2 \, b^{2} x^{2} - {\left (2 \, a + 3\right )} b x + 2 \, a^{2} + 9 \, a + 4\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{6 \, b^{3}} \] Input:

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

Output:

-1/6*(3*(2*a^2 + 2*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)) - (2*b^2*x^2 - (2*a + 3)*b*x + 2*a^2 + 
9*a + 4)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1))/b^3
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.34 \[ \int e^{-\text {arctanh}(a+b x)} x^2 \, dx=-\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a x}{b^{2}} + \frac {a^{2} \arcsin \left (b x + a\right )}{b^{3}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} x}{2 \, b^{2}} + \frac {a \arcsin \left (b x + a\right )}{b^{3}} - \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{3 \, b^{3}} + \frac {3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{2 \, b^{3}} + \frac {\arcsin \left (b x + a\right )}{2 \, b^{3}} + \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{3}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.82 \[ \int e^{-\text {arctanh}(a+b x)} x^2 \, dx=\frac {1}{6} \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (x {\left (\frac {2 \, x}{b} - \frac {2 \, a b^{3} + 3 \, b^{3}}{b^{5}}\right )} + \frac {2 \, a^{2} b^{2} + 9 \, a b^{2} + 4 \, b^{2}}{b^{5}}\right )} - \frac {{\left (2 \, a^{2} + 2 \, a + 1\right )} \arcsin \left (-b x - a\right ) \mathrm {sgn}\left (b\right )}{2 \, b^{2} {\left | b \right |}} \] Input:

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

Output:

1/6*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(x*(2*x/b - (2*a*b^3 + 3*b^3)/b^5) 
+ (2*a^2*b^2 + 9*a*b^2 + 4*b^2)/b^5) - 1/2*(2*a^2 + 2*a + 1)*arcsin(-b*x - 
 a)*sgn(b)/(b^2*abs(b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.51 \[ \int e^{-\text {arctanh}(a+b x)} x^2 \, dx=\frac {6 \mathit {asin} \left (b x +a \right ) a^{2}+6 \mathit {asin} \left (b x +a \right ) a +3 \mathit {asin} \left (b x +a \right )+2 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a^{2}-2 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a b x +9 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a +2 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b^{2} x^{2}-3 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b x +4 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}-6 a^{2}-12 a -4}{6 b^{3}} \] Input:

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

Output:

(6*asin(a + b*x)*a**2 + 6*asin(a + b*x)*a + 3*asin(a + b*x) + 2*sqrt( - a* 
*2 - 2*a*b*x - b**2*x**2 + 1)*a**2 - 2*sqrt( - a**2 - 2*a*b*x - b**2*x**2 
+ 1)*a*b*x + 9*sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*a + 2*sqrt( - a**2 
- 2*a*b*x - b**2*x**2 + 1)*b**2*x**2 - 3*sqrt( - a**2 - 2*a*b*x - b**2*x** 
2 + 1)*b*x + 4*sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1) - 6*a**2 - 12*a - 4 
)/(6*b**3)