\(\int e^{-3 \text {arctanh}(a x)} x^2 \, dx\) [52]

Optimal result

Integrand size = 12, antiderivative size = 99 \[ \int e^{-3 \text {arctanh}(a x)} x^2 \, dx=-\frac {4 (1-a x)}{a^3 \sqrt {1-a^2 x^2}}-\frac {5 \sqrt {1-a^2 x^2}}{a^3}+\frac {3 x \sqrt {1-a^2 x^2}}{2 a^2}+\frac {\left (1-a^2 x^2\right )^{3/2}}{3 a^3}-\frac {11 \arcsin (a x)}{2 a^3} \] Output:

(4*a*x-4)/a^3/(-a^2*x^2+1)^(1/2)-5*(-a^2*x^2+1)^(1/2)/a^3+3/2*x*(-a^2*x^2+ 
1)^(1/2)/a^2+1/3*(-a^2*x^2+1)^(3/2)/a^3-11/2*arcsin(a*x)/a^3
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59 \[ \int e^{-3 \text {arctanh}(a x)} x^2 \, dx=-\frac {\frac {\sqrt {1-a^2 x^2} \left (52+19 a x-7 a^2 x^2+2 a^3 x^3\right )}{1+a x}+33 \arcsin (a x)}{6 a^3} \] Input:

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

Output:

-1/6*((Sqrt[1 - a^2*x^2]*(52 + 19*a*x - 7*a^2*x^2 + 2*a^3*x^3))/(1 + a*x) 
+ 33*ArcSin[a*x])/a^3
 

Rubi [A] (verified)

Time = 1.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.22, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.083, Rules used = {6674, 2164, 2027, 2164, 27, 563, 2346, 25, 2346, 25, 27, 455, 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.

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

\(\Big \downarrow \) 6674

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

\(\Big \downarrow \) 2164

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

\(\Big \downarrow \) 2027

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

\(\Big \downarrow \) 2164

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 563

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

\(\Big \downarrow \) 2346

\(\displaystyle -\frac {\frac {1}{3} a x^2 \sqrt {1-a^2 x^2}-\frac {\int -\frac {9 x^2 a^4-14 x a^3+12 a^2}{\sqrt {1-a^2 x^2}}dx}{3 a^2}}{a^2}-\frac {4 \sqrt {1-a^2 x^2}}{a^3 (a x+1)}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {\int \frac {9 x^2 a^4-14 x a^3+12 a^2}{\sqrt {1-a^2 x^2}}dx}{3 a^2}+\frac {1}{3} a x^2 \sqrt {1-a^2 x^2}}{a^2}-\frac {4 \sqrt {1-a^2 x^2}}{a^3 (a x+1)}\)

\(\Big \downarrow \) 2346

\(\displaystyle -\frac {\frac {-\frac {\int -\frac {a^4 (33-28 a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {9}{2} a^2 x \sqrt {1-a^2 x^2}}{3 a^2}+\frac {1}{3} a x^2 \sqrt {1-a^2 x^2}}{a^2}-\frac {4 \sqrt {1-a^2 x^2}}{a^3 (a x+1)}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {\frac {\int \frac {a^4 (33-28 a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {9}{2} a^2 x \sqrt {1-a^2 x^2}}{3 a^2}+\frac {1}{3} a x^2 \sqrt {1-a^2 x^2}}{a^2}-\frac {4 \sqrt {1-a^2 x^2}}{a^3 (a x+1)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {1}{2} a^2 \int \frac {33-28 a x}{\sqrt {1-a^2 x^2}}dx-\frac {9}{2} a^2 x \sqrt {1-a^2 x^2}}{3 a^2}+\frac {1}{3} a x^2 \sqrt {1-a^2 x^2}}{a^2}-\frac {4 \sqrt {1-a^2 x^2}}{a^3 (a x+1)}\)

\(\Big \downarrow \) 455

\(\displaystyle -\frac {\frac {\frac {1}{2} a^2 \left (33 \int \frac {1}{\sqrt {1-a^2 x^2}}dx+\frac {28 \sqrt {1-a^2 x^2}}{a}\right )-\frac {9}{2} a^2 x \sqrt {1-a^2 x^2}}{3 a^2}+\frac {1}{3} a x^2 \sqrt {1-a^2 x^2}}{a^2}-\frac {4 \sqrt {1-a^2 x^2}}{a^3 (a x+1)}\)

\(\Big \downarrow \) 223

\(\displaystyle -\frac {\frac {\frac {1}{2} a^2 \left (\frac {28 \sqrt {1-a^2 x^2}}{a}+\frac {33 \arcsin (a x)}{a}\right )-\frac {9}{2} a^2 x \sqrt {1-a^2 x^2}}{3 a^2}+\frac {1}{3} a x^2 \sqrt {1-a^2 x^2}}{a^2}-\frac {4 \sqrt {1-a^2 x^2}}{a^3 (a x+1)}\)

Input:

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

Output:

(-4*Sqrt[1 - a^2*x^2])/(a^3*(1 + a*x)) - ((a*x^2*Sqrt[1 - a^2*x^2])/3 + (( 
-9*a^2*x*Sqrt[1 - a^2*x^2])/2 + (a^2*((28*Sqrt[1 - a^2*x^2])/a + (33*ArcSi 
n[a*x])/a))/2)/(3*a^2))/a^2
 

Defintions of rubi rules used

rule 25

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

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 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 455

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

rule 563

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[(-(-c)^(m - n - 2))*d^(2*n - m + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)* 
b^(n + 2)*(c + d*x))), x] - Simp[d^(2*n - m + 2)/b^(n + 1)   Int[(1/Sqrt[a 
+ b*x^2])*ExpandToSum[(2^(-n - 1)*(-c)^(m - n - 1) - d^m*x^m*(-c + d*x)^(-n 
 - 1))/(c + d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2 
, 0] && IGtQ[m, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
 

rule 2027

Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ 
(p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & 
& PosQ[s - r] &&  !(EqQ[p, 1] && EqQ[u, 1])
 

rule 2164

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] 
:> Simp[d*e   Int[(d + e*x)^(m - 1)*PolynomialQuotient[Pq, a*e + b*d*x, x]* 
(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, d, e, m, p}, x] && PolyQ[Pq, x] 
 && EqQ[b*d^2 + a*e^2, 0] && EqQ[PolynomialRemainder[Pq, a*e + b*d*x, x], 0 
]
 

rule 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

rule 6674

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_.)*(x_))^(m_.), x_Symbol] :> Int[(c*x 
)^m*((1 + a*x)^((n + 1)/2)/((1 - a*x)^((n - 1)/2)*Sqrt[1 - a^2*x^2])), x] / 
; FreeQ[{a, c, m}, x] && IntegerQ[(n - 1)/2]
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08

method	result	size
risch	\(\frac {\left (2 a^{2} x^{2}-9 a x +28\right ) \left (a^{2} x^{2}-1\right )}{6 a^{3} \sqrt {-a^{2} x^{2}+1}}-\frac {11 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 a^{2} \sqrt {a^{2}}}-\frac {4 \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{a^{4} \left (x +\frac {1}{a}\right )}\)	\(107\)
default	\(\frac {-\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{3}}-2 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{2}}+3 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}+a \left (-\frac {\left (-2 a^{2} \left (x +\frac {1}{a}\right )+2 a \right ) \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )}{a^{5}}+\frac {\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}+a \left (-\frac {\left (-2 a^{2} \left (x +\frac {1}{a}\right )+2 a \right ) \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 \sqrt {a^{2}}}\right )}{a^{3}}-\frac {2 \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a \left (x +\frac {1}{a}\right )^{2}}+3 a \left (\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}}{3}+a \left (-\frac {\left (-2 a^{2} \left (x +\frac {1}{a}\right )+2 a \right ) \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}{4 a^{2}}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 \sqrt {a^{2}}}\right )\right )\right )}{a^{4}}\)	\(457\)

Input:

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

Output:

1/6*(2*a^2*x^2-9*a*x+28)*(a^2*x^2-1)/a^3/(-a^2*x^2+1)^(1/2)-11/2/a^2/(a^2) 
^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))-4/a^4/(x+1/a)*(-a^2*(x+1/a 
)^2+2*a*(x+1/a))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.84 \[ \int e^{-3 \text {arctanh}(a x)} x^2 \, dx=-\frac {52 \, a x - 66 \, {\left (a x + 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (2 \, a^{3} x^{3} - 7 \, a^{2} x^{2} + 19 \, a x + 52\right )} \sqrt {-a^{2} x^{2} + 1} + 52}{6 \, {\left (a^{4} x + a^{3}\right )}} \] Input:

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

Output:

-1/6*(52*a*x - 66*(a*x + 1)*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + (2*a^ 
3*x^3 - 7*a^2*x^2 + 19*a*x + 52)*sqrt(-a^2*x^2 + 1) + 52)/(a^4*x + a^3)
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.79 \[ \int e^{-3 \text {arctanh}(a x)} x^2 \, dx=\frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{5} x^{2} + 2 \, a^{4} x + a^{3}} - \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{4} x + a^{3}} - \frac {6 \, \sqrt {-a^{2} x^{2} + 1}}{a^{4} x + a^{3}} + \frac {\sqrt {a^{2} x^{2} + 4 \, a x + 3} x}{2 \, a^{2}} + \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{3 \, a^{3}} - \frac {i \, \arcsin \left (a x + 2\right )}{2 \, a^{3}} - \frac {6 \, \arcsin \left (a x\right )}{a^{3}} + \frac {\sqrt {a^{2} x^{2} + 4 \, a x + 3}}{a^{3}} - \frac {3 \, \sqrt {-a^{2} x^{2} + 1}}{a^{3}} \] Input:

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

Output:

(-a^2*x^2 + 1)^(3/2)/(a^5*x^2 + 2*a^4*x + a^3) - (-a^2*x^2 + 1)^(3/2)/(a^4 
*x + a^3) - 6*sqrt(-a^2*x^2 + 1)/(a^4*x + a^3) + 1/2*sqrt(a^2*x^2 + 4*a*x 
+ 3)*x/a^2 + 1/3*(-a^2*x^2 + 1)^(3/2)/a^3 - 1/2*I*arcsin(a*x + 2)/a^3 - 6* 
arcsin(a*x)/a^3 + sqrt(a^2*x^2 + 4*a*x + 3)/a^3 - 3*sqrt(-a^2*x^2 + 1)/a^3
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.88 \[ \int e^{-3 \text {arctanh}(a x)} x^2 \, dx=-\frac {1}{6} \, \sqrt {-a^{2} x^{2} + 1} {\left (x {\left (\frac {2 \, x}{a} - \frac {9}{a^{2}}\right )} + \frac {28}{a^{3}}\right )} - \frac {11 \, \arcsin \left (a x\right ) \mathrm {sgn}\left (a\right )}{2 \, a^{2} {\left | a \right |}} + \frac {8}{a^{2} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} \] Input:

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

Output:

-1/6*sqrt(-a^2*x^2 + 1)*(x*(2*x/a - 9/a^2) + 28/a^3) - 11/2*arcsin(a*x)*sg 
n(a)/(a^2*abs(a)) + 8/(a^2*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 1)*a 
bs(a))
 

Mupad [B] (verification not implemented)

Time = 14.26 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.42 \[ \int e^{-3 \text {arctanh}(a x)} x^2 \, dx=\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {2}{3\,a\,\sqrt {-a^2}}-\frac {4\,\sqrt {-a^2}}{a^3}+\frac {a\,x^2}{3\,\sqrt {-a^2}}+\frac {3\,x\,\sqrt {-a^2}}{2\,a^2}\right )}{\sqrt {-a^2}}-\frac {11\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,a^2\,\sqrt {-a^2}}+\frac {4\,\sqrt {1-a^2\,x^2}}{a^2\,\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \] Input:

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

Output:

((1 - a^2*x^2)^(1/2)*(2/(3*a*(-a^2)^(1/2)) - (4*(-a^2)^(1/2))/a^3 + (a*x^2 
)/(3*(-a^2)^(1/2)) + (3*x*(-a^2)^(1/2))/(2*a^2)))/(-a^2)^(1/2) - (11*asinh 
(x*(-a^2)^(1/2)))/(2*a^2*(-a^2)^(1/2)) + (4*(1 - a^2*x^2)^(1/2))/(a^2*(x*( 
-a^2)^(1/2) + (-a^2)^(1/2)/a)*(-a^2)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.53 \[ \int e^{-3 \text {arctanh}(a x)} x^2 \, dx=\frac {-33 \sqrt {-a^{2} x^{2}+1}\, \mathit {asin} \left (a x \right )+33 \mathit {asin} \left (a x \right ) a x +33 \mathit {asin} \left (a x \right )+2 \sqrt {-a^{2} x^{2}+1}\, a^{3} x^{3}-7 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+19 \sqrt {-a^{2} x^{2}+1}\, a x +66 \sqrt {-a^{2} x^{2}+1}+2 a^{4} x^{4}-9 a^{3} x^{3}+26 a^{2} x^{2}+19 a x -66}{6 a^{3} \left (\sqrt {-a^{2} x^{2}+1}-a x -1\right )} \] Input:

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

Output:

( - 33*sqrt( - a**2*x**2 + 1)*asin(a*x) + 33*asin(a*x)*a*x + 33*asin(a*x) 
+ 2*sqrt( - a**2*x**2 + 1)*a**3*x**3 - 7*sqrt( - a**2*x**2 + 1)*a**2*x**2 
+ 19*sqrt( - a**2*x**2 + 1)*a*x + 66*sqrt( - a**2*x**2 + 1) + 2*a**4*x**4 
- 9*a**3*x**3 + 26*a**2*x**2 + 19*a*x - 66)/(6*a**3*(sqrt( - a**2*x**2 + 1 
) - a*x - 1))