\(\int e^{-\text {arctanh}(a+b x)} x^3 \, dx\) [880]

Optimal result

Integrand size = 14, antiderivative size = 179 \[ \int e^{-\text {arctanh}(a+b x)} x^3 \, dx=-\frac {\left (3+12 a+12 a^2+8 a^3\right ) \sqrt {1-a-b x} \sqrt {1+a+b x}}{8 b^4}-\frac {5 \left (1+6 a^2\right ) (1-a-b x)^{3/2} \sqrt {1+a+b x}}{24 b^4}-\frac {x^2 (1-a-b x)^{3/2} \sqrt {1+a+b x}}{4 b^2}-\frac {(1+6 a) (1-a-b x)^{5/2} \sqrt {1+a+b x}}{12 b^4}-\frac {\left (3+12 a+12 a^2+8 a^3\right ) \arcsin (a+b x)}{8 b^4} \] Output:

-1/8*(8*a^3+12*a^2+12*a+3)*(-b*x-a+1)^(1/2)*(b*x+a+1)^(1/2)/b^4-5/24*(6*a^ 
2+1)*(-b*x-a+1)^(3/2)*(b*x+a+1)^(1/2)/b^4-1/4*x^2*(-b*x-a+1)^(3/2)*(b*x+a+ 
1)^(1/2)/b^2-1/12*(1+6*a)*(-b*x-a+1)^(5/2)*(b*x+a+1)^(1/2)/b^4-1/8*(8*a^3+ 
12*a^2+12*a+3)*arcsin(b*x+a)/b^4
 

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.89 \[ \int e^{-\text {arctanh}(a+b x)} x^3 \, dx=\frac {\frac {\sqrt {1+a+b x} \left (-16+38 a^3+6 a^4+25 b x-17 b^2 x^2+14 b^3 x^3-6 b^4 x^4+5 a^2 (-1+6 b x)+a \left (-23+50 b x-18 b^2 x^2\right )\right )}{\sqrt {1-a-b x}}+\frac {6 \left (3+12 a+12 a^2+8 a^3\right ) \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {-b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {b}}\right )}{\sqrt {-b}}}{24 b^4} \] Input:

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

Output:

((Sqrt[1 + a + b*x]*(-16 + 38*a^3 + 6*a^4 + 25*b*x - 17*b^2*x^2 + 14*b^3*x 
^3 - 6*b^4*x^4 + 5*a^2*(-1 + 6*b*x) + a*(-23 + 50*b*x - 18*b^2*x^2)))/Sqrt 
[1 - a - b*x] + (6*(3 + 12*a + 12*a^2 + 8*a^3)*Sqrt[b]*ArcSinh[(Sqrt[-b]*S 
qrt[1 - a - b*x])/(Sqrt[2]*Sqrt[b])])/Sqrt[-b])/(24*b^4)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.92, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6713, 111, 25, 164, 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.

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

\(\Big \downarrow \) 6713

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

\(\Big \downarrow \) 111

\(\displaystyle -\frac {\int -\frac {x \sqrt {-a-b x+1} \left (2 \left (1-a^2\right )-(6 a+1) b x\right )}{\sqrt {a+b x+1}}dx}{4 b^2}-\frac {x^2 (-a-b x+1)^{3/2} \sqrt {a+b x+1}}{4 b^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {x \sqrt {-a-b x+1} \left (2 \left (1-a^2\right )-(6 a+1) b x\right )}{\sqrt {a+b x+1}}dx}{4 b^2}-\frac {x^2 (-a-b x+1)^{3/2} \sqrt {a+b x+1}}{4 b^2}\)

\(\Big \downarrow \) 164

\(\displaystyle \frac {-\frac {\left (8 a^3+12 a^2+12 a+3\right ) \int \frac {\sqrt {-a-b x+1}}{\sqrt {a+b x+1}}dx}{2 b}-\frac {\sqrt {a+b x+1} \left (18 a^2-2 (6 a+1) b x+10 a+7\right ) (-a-b x+1)^{3/2}}{6 b^2}}{4 b^2}-\frac {x^2 (-a-b x+1)^{3/2} \sqrt {a+b x+1}}{4 b^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {-\frac {\left (8 a^3+12 a^2+12 a+3\right ) \left (\int \frac {1}{\sqrt {-a-b x+1} \sqrt {a+b x+1}}dx+\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )}{2 b}-\frac {\sqrt {a+b x+1} \left (18 a^2-2 (6 a+1) b x+10 a+7\right ) (-a-b x+1)^{3/2}}{6 b^2}}{4 b^2}-\frac {x^2 (-a-b x+1)^{3/2} \sqrt {a+b x+1}}{4 b^2}\)

\(\Big \downarrow \) 62

\(\displaystyle \frac {-\frac {\left (8 a^3+12 a^2+12 a+3\right ) \left (\int \frac {1}{\sqrt {-b^2 x^2-2 a b x+(1-a) (a+1)}}dx+\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )}{2 b}-\frac {\sqrt {a+b x+1} \left (18 a^2-2 (6 a+1) b x+10 a+7\right ) (-a-b x+1)^{3/2}}{6 b^2}}{4 b^2}-\frac {x^2 (-a-b x+1)^{3/2} \sqrt {a+b x+1}}{4 b^2}\)

\(\Big \downarrow \) 1090

\(\displaystyle \frac {-\frac {\left (8 a^3+12 a^2+12 a+3\right ) \left (\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}-\frac {\int \frac {1}{\sqrt {1-\frac {\left (-2 x b^2-2 a b\right )^2}{4 b^2}}}d\left (-2 x b^2-2 a b\right )}{2 b^2}\right )}{2 b}-\frac {\sqrt {a+b x+1} \left (18 a^2-2 (6 a+1) b x+10 a+7\right ) (-a-b x+1)^{3/2}}{6 b^2}}{4 b^2}-\frac {x^2 (-a-b x+1)^{3/2} \sqrt {a+b x+1}}{4 b^2}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {-\frac {\sqrt {a+b x+1} \left (18 a^2-2 (6 a+1) b x+10 a+7\right ) (-a-b x+1)^{3/2}}{6 b^2}-\frac {\left (8 a^3+12 a^2+12 a+3\right ) \left (\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}-\frac {\arcsin \left (\frac {-2 a b-2 b^2 x}{2 b}\right )}{b}\right )}{2 b}}{4 b^2}-\frac {x^2 (-a-b x+1)^{3/2} \sqrt {a+b x+1}}{4 b^2}\)

Input:

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

Output:

-1/4*(x^2*(1 - a - b*x)^(3/2)*Sqrt[1 + a + b*x])/b^2 + (-1/6*((1 - a - b*x 
)^(3/2)*Sqrt[1 + a + b*x]*(7 + 10*a + 18*a^2 - 2*(1 + 6*a)*b*x))/b^2 - ((3 
 + 12*a + 12*a^2 + 8*a^3)*((Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/b - ArcSi 
n[(-2*a*b - 2*b^2*x)/(2*b)]/b))/(2*b))/(4*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 111

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 
)/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1))   Int[(a + b*x) 
^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m 
 - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m 
 + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & 
& GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
 

rule 164

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ 
))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - 
 b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( 
c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h 
*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 
3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + 
d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3))   Int[( 
a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] 
&& NeQ[m + n + 2, 0] && NeQ[m + n + 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 = 0.36 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.91

method	result	size
risch	\(\frac {\left (-6 b^{3} x^{3}+6 a \,b^{2} x^{2}-6 a^{2} b x +8 b^{2} x^{2}+6 a^{3}-20 a b x +44 a^{2}-9 b x +39 a +16\right ) \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}{24 b^{4} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {\left (8 a^{3}+12 a^{2}+12 a +3\right ) \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{8 b^{3} \sqrt {b^{2}}}\)	\(163\)
default	\(\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^{2} \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 (a +1\right ) \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} \left (-\frac {x \left (-b^{2} x^{2}-2 a b x -a^{2}+1\right )^{\frac {3}{2}}}{4 b^{2}}-\frac {5 a \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 )}{4 b}+\frac {\left (-a^{2}+1\right ) \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 )}{4 b^{2}}\right )+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^{3}}-\frac {\left (a^{3}+3 a^{2}+3 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^{4}}\)	\(883\)

Input:

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

Output:

1/24*(-6*b^3*x^3+6*a*b^2*x^2-6*a^2*b*x+8*b^2*x^2+6*a^3-20*a*b*x+44*a^2-9*b 
*x+39*a+16)*(b^2*x^2+2*a*b*x+a^2-1)/b^4/(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)-1/8 
/b^3*(8*a^3+12*a^2+12*a+3)/(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 = 0.10 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.80 \[ \int e^{-\text {arctanh}(a+b x)} x^3 \, dx=\frac {3 \, {\left (8 \, a^{3} + 12 \, a^{2} + 12 \, a + 3\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + {\left (6 \, b^{3} x^{3} - 2 \, {\left (3 \, a + 4\right )} b^{2} x^{2} - 6 \, a^{3} + {\left (6 \, a^{2} + 20 \, a + 9\right )} b x - 44 \, a^{2} - 39 \, a - 16\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{24 \, b^{4}} \] Input:

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

Output:

1/24*(3*(8*a^3 + 12*a^2 + 12*a + 3)*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)) + (6*b^3*x^3 - 2*(3*a + 4)*b^ 
2*x^2 - 6*a^3 + (6*a^2 + 20*a + 9)*b*x - 44*a^2 - 39*a - 16)*sqrt(-b^2*x^2 
 - 2*a*b*x - a^2 + 1))/b^4
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 338 vs. \(2 (153) = 306\).

Time = 0.12 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.89 \[ \int e^{-\text {arctanh}(a+b x)} x^3 \, dx=\frac {3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{2} x}{2 \, b^{3}} - \frac {a^{3} \arcsin \left (b x + a\right )}{b^{4}} + \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{3}}{2 \, b^{4}} - \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} x}{4 \, b^{3}} + \frac {3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a x}{2 \, b^{3}} - \frac {3 \, a^{2} \arcsin \left (b x + a\right )}{2 \, b^{4}} + \frac {3 \, {\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}} a}{4 \, b^{4}} - \frac {3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{2}}{2 \, b^{4}} + \frac {5 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} x}{8 \, b^{3}} - \frac {3 \, a \arcsin \left (b x + a\right )}{2 \, b^{4}} + \frac {{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}{3 \, b^{4}} - \frac {19 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{8 \, b^{4}} - \frac {3 \, \arcsin \left (b x + a\right )}{8 \, b^{4}} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{4}} \] Input:

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

Output:

3/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a^2*x/b^3 - a^3*arcsin(b*x + a)/b^4 
 + 1/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a^3/b^4 - 1/4*(-b^2*x^2 - 2*a*b* 
x - a^2 + 1)^(3/2)*x/b^3 + 3/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a*x/b^3 
- 3/2*a^2*arcsin(b*x + a)/b^4 + 3/4*(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2)*a 
/b^4 - 3/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a^2/b^4 + 5/8*sqrt(-b^2*x^2 
- 2*a*b*x - a^2 + 1)*x/b^3 - 3/2*a*arcsin(b*x + a)/b^4 + 1/3*(-b^2*x^2 - 2 
*a*b*x - a^2 + 1)^(3/2)/b^4 - 19/8*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*a/b^ 
4 - 3/8*arcsin(b*x + a)/b^4 - sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)/b^4
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.83 \[ \int e^{-\text {arctanh}(a+b x)} x^3 \, dx=\frac {1}{24} \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left ({\left (2 \, x {\left (\frac {3 \, x}{b} - \frac {3 \, a b^{5} + 4 \, b^{5}}{b^{7}}\right )} + \frac {6 \, a^{2} b^{4} + 20 \, a b^{4} + 9 \, b^{4}}{b^{7}}\right )} x - \frac {6 \, a^{3} b^{3} + 44 \, a^{2} b^{3} + 39 \, a b^{3} + 16 \, b^{3}}{b^{7}}\right )} + \frac {{\left (8 \, a^{3} + 12 \, a^{2} + 12 \, a + 3\right )} \arcsin \left (-b x - a\right ) \mathrm {sgn}\left (b\right )}{8 \, b^{3} {\left | b \right |}} \] Input:

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

Output:

1/24*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*((2*x*(3*x/b - (3*a*b^5 + 4*b^5)/b 
^7) + (6*a^2*b^4 + 20*a*b^4 + 9*b^4)/b^7)*x - (6*a^3*b^3 + 44*a^2*b^3 + 39 
*a*b^3 + 16*b^3)/b^7) + 1/8*(8*a^3 + 12*a^2 + 12*a + 3)*arcsin(-b*x - a)*s 
gn(b)/(b^3*abs(b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.82 \[ \int e^{-\text {arctanh}(a+b x)} x^3 \, dx=\frac {-24 \mathit {asin} \left (b x +a \right ) a^{3}-36 \mathit {asin} \left (b x +a \right ) a^{2}-36 \mathit {asin} \left (b x +a \right ) a -9 \mathit {asin} \left (b x +a \right )-6 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a^{3}+6 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a^{2} b x -44 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a^{2}-6 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a \,b^{2} x^{2}+20 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a b x -39 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a +6 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b^{3} x^{3}-8 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b^{2} x^{2}+9 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b x -16 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}+24 a^{3}+72 a^{2}+48 a +16}{24 b^{4}} \] Input:

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

Output:

( - 24*asin(a + b*x)*a**3 - 36*asin(a + b*x)*a**2 - 36*asin(a + b*x)*a - 9 
*asin(a + b*x) - 6*sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*a**3 + 6*sqrt( 
- a**2 - 2*a*b*x - b**2*x**2 + 1)*a**2*b*x - 44*sqrt( - a**2 - 2*a*b*x - b 
**2*x**2 + 1)*a**2 - 6*sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*a*b**2*x**2 
 + 20*sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*a*b*x - 39*sqrt( - a**2 - 2* 
a*b*x - b**2*x**2 + 1)*a + 6*sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*b**3* 
x**3 - 8*sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*b**2*x**2 + 9*sqrt( - a** 
2 - 2*a*b*x - b**2*x**2 + 1)*b*x - 16*sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 
 1) + 24*a**3 + 72*a**2 + 48*a + 16)/(24*b**4)