\(\int e^{3 \text {arctanh}(a+b x)} x^2 \, dx\) [873]

Optimal result

Integrand size = 14, antiderivative size = 168 \[ \int e^{3 \text {arctanh}(a+b x)} x^2 \, dx=\frac {\left (11-18 a+6 a^2\right ) \sqrt {1-a-b x} \sqrt {1+a+b x}}{2 b^3}+\frac {\left (11-18 a+6 a^2\right ) \sqrt {1-a-b x} (1+a+b x)^{3/2}}{6 b^3}+\frac {(1-a)^2 (1+a+b x)^{5/2}}{b^3 \sqrt {1-a-b x}}+\frac {\sqrt {1-a-b x} (1+a+b x)^{5/2}}{3 b^3}-\frac {\left (11-18 a+6 a^2\right ) \arcsin (a+b x)}{2 b^3} \] Output:

1/2*(6*a^2-18*a+11)*(-b*x-a+1)^(1/2)*(b*x+a+1)^(1/2)/b^3+1/6*(6*a^2-18*a+1 
1)*(-b*x-a+1)^(1/2)*(b*x+a+1)^(3/2)/b^3+(1-a)^2*(b*x+a+1)^(5/2)/b^3/(-b*x- 
a+1)^(1/2)+1/3*(-b*x-a+1)^(1/2)*(b*x+a+1)^(5/2)/b^3-1/2*(6*a^2-18*a+11)*ar 
csin(b*x+a)/b^3
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.01 \[ \int e^{3 \text {arctanh}(a+b x)} x^2 \, dx=\frac {-\frac {\sqrt {b} \sqrt {1+a+b x} \left (-52-53 a^2+2 a^3+19 b x+7 b^2 x^2+2 b^3 x^3+a (103-16 b x)\right )}{\sqrt {1-a-b x}}+108 a \sqrt {-b} \text {arcsinh}\left (\frac {\sqrt {-b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {b}}\right )+6 \left (11+6 a^2\right ) \sqrt {-b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {1-a-b x}}{\sqrt {2} \sqrt {-b}}\right )}{6 b^{7/2}} \] Input:

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

Output:

(-((Sqrt[b]*Sqrt[1 + a + b*x]*(-52 - 53*a^2 + 2*a^3 + 19*b*x + 7*b^2*x^2 + 
 2*b^3*x^3 + a*(103 - 16*b*x)))/Sqrt[1 - a - b*x]) + 108*a*Sqrt[-b]*ArcSin 
h[(Sqrt[-b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[b])] + 6*(11 + 6*a^2)*Sqrt[-b 
]*ArcSinh[(Sqrt[b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[-b])])/(6*b^(7/2))
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {6713, 100, 27, 90, 60, 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.

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

\(\Big \downarrow \) 6713

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

\(\Big \downarrow \) 100

\(\displaystyle \frac {(1-a)^2 (a+b x+1)^{5/2}}{b^3 \sqrt {-a-b x+1}}-\frac {\int \frac {b ((3-2 a) (1-a)+b x) (a+b x+1)^{3/2}}{\sqrt {-a-b x+1}}dx}{b^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(1-a)^2 (a+b x+1)^{5/2}}{b^3 \sqrt {-a-b x+1}}-\frac {\int \frac {((3-2 a) (1-a)+b x) (a+b x+1)^{3/2}}{\sqrt {-a-b x+1}}dx}{b^2}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {(1-a)^2 (a+b x+1)^{5/2}}{b^3 \sqrt {-a-b x+1}}-\frac {\frac {1}{3} \left (6 a^2-18 a+11\right ) \int \frac {(a+b x+1)^{3/2}}{\sqrt {-a-b x+1}}dx-\frac {\sqrt {-a-b x+1} (a+b x+1)^{5/2}}{3 b}}{b^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(1-a)^2 (a+b x+1)^{5/2}}{b^3 \sqrt {-a-b x+1}}-\frac {\frac {1}{3} \left (6 a^2-18 a+11\right ) \left (\frac {3}{2} \int \frac {\sqrt {a+b x+1}}{\sqrt {-a-b x+1}}dx-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2}}{2 b}\right )-\frac {\sqrt {-a-b x+1} (a+b x+1)^{5/2}}{3 b}}{b^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(1-a)^2 (a+b x+1)^{5/2}}{b^3 \sqrt {-a-b x+1}}-\frac {\frac {1}{3} \left (6 a^2-18 a+11\right ) \left (\frac {3}{2} \left (\int \frac {1}{\sqrt {-a-b x+1} \sqrt {a+b x+1}}dx-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2}}{2 b}\right )-\frac {\sqrt {-a-b x+1} (a+b x+1)^{5/2}}{3 b}}{b^2}\)

\(\Big \downarrow \) 62

\(\displaystyle \frac {(1-a)^2 (a+b x+1)^{5/2}}{b^3 \sqrt {-a-b x+1}}-\frac {\frac {1}{3} \left (6 a^2-18 a+11\right ) \left (\frac {3}{2} \left (\int \frac {1}{\sqrt {-b^2 x^2-2 a b x+(1-a) (a+1)}}dx-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2}}{2 b}\right )-\frac {\sqrt {-a-b x+1} (a+b x+1)^{5/2}}{3 b}}{b^2}\)

\(\Big \downarrow \) 1090

\(\displaystyle \frac {(1-a)^2 (a+b x+1)^{5/2}}{b^3 \sqrt {-a-b x+1}}-\frac {\frac {1}{3} \left (6 a^2-18 a+11\right ) \left (\frac {3}{2} \left (-\frac {\int \frac {1}{\sqrt {1-\frac {\left (-2 x b^2-2 a b\right )^2}{4 b^2}}}d\left (-2 x b^2-2 a b\right )}{2 b^2}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2}}{2 b}\right )-\frac {\sqrt {-a-b x+1} (a+b x+1)^{5/2}}{3 b}}{b^2}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {(1-a)^2 (a+b x+1)^{5/2}}{b^3 \sqrt {-a-b x+1}}-\frac {\frac {1}{3} \left (6 a^2-18 a+11\right ) \left (\frac {3}{2} \left (-\frac {\arcsin \left (\frac {-2 a b-2 b^2 x}{2 b}\right )}{b}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{b}\right )-\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2}}{2 b}\right )-\frac {\sqrt {-a-b x+1} (a+b x+1)^{5/2}}{3 b}}{b^2}\)

Input:

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

Output:

((1 - a)^2*(1 + a + b*x)^(5/2))/(b^3*Sqrt[1 - a - b*x]) - (-1/3*(Sqrt[1 - 
a - b*x]*(1 + a + b*x)^(5/2))/b + ((11 - 18*a + 6*a^2)*(-1/2*(Sqrt[1 - a - 
 b*x]*(1 + a + b*x)^(3/2))/b + (3*(-((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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 100

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

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 = 0.62 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.60

method	result	size
risch	\(-\frac {\left (2 b^{2} x^{2}-2 a b x +2 a^{2}+9 b x -27 a +28\right ) \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}{6 b^{3} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {\frac {\left (8 a^{2}-16 a +8\right ) \sqrt {-\left (x +\frac {-1+a}{b}\right )^{2} b^{2}-2 \left (x +\frac {-1+a}{b}\right ) b}}{b^{2} \left (x +\frac {-1+a}{b}\right )}+\frac {11 \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}}-\frac {18 a \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}}+\frac {6 a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{\sqrt {b^{2}}}}{2 b^{2}}\)	\(268\)
default	\(\text {Expression too large to display}\)	\(1910\)

Input:

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

Output:

-1/6*(2*b^2*x^2-2*a*b*x+2*a^2+9*b*x-27*a+28)*(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*((8*a^2-16*a+8)/b^2/(x+(-1+a)/b)*(-( 
x+(-1+a)/b)^2*b^2-2*(x+(-1+a)/b)*b)^(1/2)+11/(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))-18*a/(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))+6*a^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)))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.95 \[ \int e^{3 \text {arctanh}(a+b x)} x^2 \, dx=\frac {3 \, {\left (6 \, a^{3} + {\left (6 \, a^{2} - 18 \, a + 11\right )} b x - 24 \, a^{2} + 29 \, a - 11\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + {\left (2 \, b^{3} x^{3} + 7 \, b^{2} x^{2} + 2 \, a^{3} - {\left (16 \, a - 19\right )} b x - 53 \, a^{2} + 103 \, a - 52\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{6 \, {\left (b^{4} x + {\left (a - 1\right )} b^{3}\right )}} \] Input:

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

Output:

1/6*(3*(6*a^3 + (6*a^2 - 18*a + 11)*b*x - 24*a^2 + 29*a - 11)*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^3*x^3 + 7*b^2*x^2 + 2*a^3 - (16*a - 19)*b*x - 53*a^2 + 103*a - 52)*sqr 
t(-b^2*x^2 - 2*a*b*x - a^2 + 1))/(b^4*x + (a - 1)*b^3)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1645 vs. \(2 (142) = 284\).

Time = 0.13 (sec) , antiderivative size = 1645, normalized size of antiderivative = 9.79 \[ \int e^{3 \text {arctanh}(a+b x)} x^2 \, dx=\text {Too large to display} \] Input:

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

Output:

-35*a^5*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) - 
 1/3*b*x^4/sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1) + 265/6*(a^2 - 1)*a^3*x/((a^ 
2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) + 7/6*a*x^3/sqr 
t(-b^2*x^2 - 2*a*b*x - a^2 + 1) - 35/6*(a^2 - 1)*a^4/((a^2*b^2 - (a^2 - 1) 
*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b) - 61/6*(a^2 - 1)^2*a*x/((a^2*b 
^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) + 2*(a^3 + 3*a^2 + 
 3*a + 1)*a^2*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 
 1)) + 45*(a*b^2 + b^2)*a^4*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2 
*a*b*x - a^2 + 1)*b^2) - 18*(a^2*b + 2*a*b + b)*a^3*x/((a^2*b^2 - (a^2 - 1 
)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b) - 35/6*a^2*x^2/(sqrt(-b^2*x^2 
 - 2*a*b*x - a^2 + 1)*b) + 29/6*(a^2 - 1)^2*a^2/((a^2*b^2 - (a^2 - 1)*b^2) 
*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b) - (a^3 + 3*a^2 + 3*a + 1)*(a^2 - 1) 
*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) - 93/2*( 
a*b^2 + b^2)*(a^2 - 1)*a^2*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2* 
a*b*x - a^2 + 1)*b^2) + 15*(a^2*b + 2*a*b + b)*(a^2 - 1)*a*x/((a^2*b^2 - ( 
a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b) + 4/3*(a^2 - 1)*x^2/(s 
qrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b) - 3/2*(a*b^2 + b^2)*x^3/(sqrt(-b^2*x^ 
2 - 2*a*b*x - a^2 + 1)*b^2) - 35/2*a^3*arcsin(-(b^2*x + a*b)/sqrt(a^2*b^2 
- (a^2 - 1)*b^2))/b^3 + 15/2*(a*b^2 + b^2)*(a^2 - 1)*a^3/((a^2*b^2 - (a^2 
- 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*b^3) - 3*(a^2*b + 2*a*b + ...
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.99 \[ \int e^{3 \text {arctanh}(a+b x)} x^2 \, dx=\frac {1}{6} \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (x {\left (\frac {2 \, x}{b} - \frac {2 \, a b^{6} - 9 \, b^{6}}{b^{8}}\right )} + \frac {2 \, a^{2} b^{5} - 27 \, a b^{5} + 28 \, b^{5}}{b^{8}}\right )} + \frac {{\left (6 \, a^{2} - 18 \, a + 11\right )} \arcsin \left (-b x - a\right ) \mathrm {sgn}\left (b\right )}{2 \, b^{2} {\left | b \right |}} + \frac {8 \, {\left (a^{2} - 2 \, a + 1\right )}}{b^{2} {\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} - 1\right )} {\left | b \right |}} \] Input:

integrate((b*x+a+1)^3/(1-(b*x+a)^2)^(3/2)*x^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^6 - 9*b^6)/b^8) 
+ (2*a^2*b^5 - 27*a*b^5 + 28*b^5)/b^8) + 1/2*(6*a^2 - 18*a + 11)*arcsin(-b 
*x - a)*sgn(b)/(b^2*abs(b)) + 8*(a^2 - 2*a + 1)/(b^2*((sqrt(-b^2*x^2 - 2*a 
*b*x - a^2 + 1)*abs(b) + b)/(b^2*x + a*b) - 1)*abs(b))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 491, normalized size of antiderivative = 2.92 \[ \int e^{3 \text {arctanh}(a+b x)} x^2 \, dx=\frac {66-89 a -18 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \mathit {asin} \left (b x +a \right ) a^{2}+54 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \mathit {asin} \left (b x +a \right ) a -33 \mathit {asin} \left (b x +a \right ) b x +9 a \,b^{2} x^{2}-33 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, \mathit {asin} \left (b x +a \right )+51 a^{2} b x -2 a \,b^{3} x^{3}-16 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a b x -26 b^{2} x^{2}-18 \mathit {asin} \left (b x +a \right ) a^{3}+72 \mathit {asin} \left (b x +a \right ) a^{2}-87 \mathit {asin} \left (b x +a \right ) a +2 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a^{3}-71 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a^{2}+127 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a -2 b^{4} x^{4}-9 b^{3} x^{3}-2 a^{3} b x -8 a^{2}+33 a^{3}+2 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b^{3} x^{3}+7 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b^{2} x^{2}+19 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b x -18 \mathit {asin} \left (b x +a \right ) a^{2} b x +54 \mathit {asin} \left (b x +a \right ) a b x -2 a^{4}+33 \mathit {asin} \left (b x +a \right )-66 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}-82 a b x +19 b x}{6 b^{3} \left (\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}+a +b x -1\right )} \] Input:

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

Output:

( - 18*sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)*a**2 + 54*sqr 
t( - a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)*a - 33*sqrt( - a**2 - 2 
*a*b*x - b**2*x**2 + 1)*asin(a + b*x) - 18*asin(a + b*x)*a**3 - 18*asin(a 
+ b*x)*a**2*b*x + 72*asin(a + b*x)*a**2 + 54*asin(a + b*x)*a*b*x - 87*asin 
(a + b*x)*a - 33*asin(a + b*x)*b*x + 33*asin(a + b*x) + 2*sqrt( - a**2 - 2 
*a*b*x - b**2*x**2 + 1)*a**3 - 71*sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)* 
a**2 - 16*sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*a*b*x + 127*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**3*x**3 + 7*sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1)*b**2*x**2 + 19*sqrt( 
 - a**2 - 2*a*b*x - b**2*x**2 + 1)*b*x - 66*sqrt( - a**2 - 2*a*b*x - b**2* 
x**2 + 1) - 2*a**4 - 2*a**3*b*x + 33*a**3 + 51*a**2*b*x - 8*a**2 - 2*a*b** 
3*x**3 + 9*a*b**2*x**2 - 82*a*b*x - 89*a - 2*b**4*x**4 - 9*b**3*x**3 - 26* 
b**2*x**2 + 19*b*x + 66)/(6*b**3*(sqrt( - a**2 - 2*a*b*x - b**2*x**2 + 1) 
+ a + b*x - 1))