3.1.92 \(\int e^{-\frac {1}{2} \text {arctanh}(a x)} x^2 \, dx\) [92]

3.1.92.1 Optimal result

Integrand size = 14, antiderivative size = 282 \[ \int e^{-\frac {1}{2} \text {arctanh}(a x)} x^2 \, dx=\frac {3 \sqrt [4]{1-a x} (1+a x)^{3/4}}{8 a^3}+\frac {(1-a x)^{5/4} (1+a x)^{3/4}}{12 a^3}-\frac {x (1-a x)^{5/4} (1+a x)^{3/4}}{3 a^2}+\frac {3 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 \sqrt {2} a^3}-\frac {3 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 \sqrt {2} a^3}+\frac {3 \log \left (1+\frac {\sqrt {1-a x}}{\sqrt {1+a x}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{16 \sqrt {2} a^3}-\frac {3 \log \left (1+\frac {\sqrt {1-a x}}{\sqrt {1+a x}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{16 \sqrt {2} a^3} \]

output

3/8*(-a*x+1)^(1/4)*(a*x+1)^(3/4)/a^3+1/12*(-a*x+1)^(5/4)*(a*x+1)^(3/4)/a^3 
-1/3*x*(-a*x+1)^(5/4)*(a*x+1)^(3/4)/a^2-3/16*arctan(-1+(-a*x+1)^(1/4)*2^(1 
/2)/(a*x+1)^(1/4))/a^3*2^(1/2)-3/16*arctan(1+(-a*x+1)^(1/4)*2^(1/2)/(a*x+1 
)^(1/4))/a^3*2^(1/2)+3/32*ln(1-(-a*x+1)^(1/4)*2^(1/2)/(a*x+1)^(1/4)+(-a*x+ 
1)^(1/2)/(a*x+1)^(1/2))/a^3*2^(1/2)-3/32*ln(1+(-a*x+1)^(1/4)*2^(1/2)/(a*x+ 
1)^(1/4)+(-a*x+1)^(1/2)/(a*x+1)^(1/2))/a^3*2^(1/2)
 

3.1.92.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.22 \[ \int e^{-\frac {1}{2} \text {arctanh}(a x)} x^2 \, dx=-\frac {(1-a x)^{5/4} \left (5 (1+a x)^{3/4} (-1+4 a x)+9\ 2^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {5}{4},\frac {9}{4},\frac {1}{2} (1-a x)\right )\right )}{60 a^3} \]

input

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

output

-1/60*((1 - a*x)^(5/4)*(5*(1 + a*x)^(3/4)*(-1 + 4*a*x) + 9*2^(3/4)*Hyperge 
ometric2F1[1/4, 5/4, 9/4, (1 - a*x)/2]))/a^3
 

3.1.92.3 Rubi [A] (warning: unable to verify)

Time = 0.46 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.96, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.071, Rules used = {6676, 101, 27, 90, 60, 73, 770, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 6676

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

\(\Big \downarrow \) 101

\(\displaystyle -\frac {\int -\frac {\sqrt [4]{1-a x} (2-a x)}{2 \sqrt [4]{a x+1}}dx}{3 a^2}-\frac {x (a x+1)^{3/4} (1-a x)^{5/4}}{3 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt [4]{1-a x} (2-a x)}{\sqrt [4]{a x+1}}dx}{6 a^2}-\frac {x (1-a x)^{5/4} (a x+1)^{3/4}}{3 a^2}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {9}{4} \int \frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}dx+\frac {(a x+1)^{3/4} (1-a x)^{5/4}}{2 a}}{6 a^2}-\frac {x (1-a x)^{5/4} (a x+1)^{3/4}}{3 a^2}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\frac {9}{4} \left (\frac {1}{2} \int \frac {1}{(1-a x)^{3/4} \sqrt [4]{a x+1}}dx+\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}\right )+\frac {(a x+1)^{3/4} (1-a x)^{5/4}}{2 a}}{6 a^2}-\frac {x (1-a x)^{5/4} (a x+1)^{3/4}}{3 a^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {9}{4} \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \int \frac {1}{\sqrt [4]{a x+1}}d\sqrt [4]{1-a x}}{a}\right )+\frac {(a x+1)^{3/4} (1-a x)^{5/4}}{2 a}}{6 a^2}-\frac {x (1-a x)^{5/4} (a x+1)^{3/4}}{3 a^2}\)

\(\Big \downarrow \) 770

\(\displaystyle \frac {\frac {9}{4} \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \int \frac {1}{2-a x}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{a}\right )+\frac {(a x+1)^{3/4} (1-a x)^{5/4}}{2 a}}{6 a^2}-\frac {x (1-a x)^{5/4} (a x+1)^{3/4}}{3 a^2}\)

\(\Big \downarrow \) 755

\(\displaystyle \frac {\frac {9}{4} \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \left (\frac {1}{2} \int \frac {1-\sqrt {1-a x}}{2-a x}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+\frac {1}{2} \int \frac {\sqrt {1-a x}+1}{2-a x}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{a}\right )+\frac {(a x+1)^{3/4} (1-a x)^{5/4}}{2 a}}{6 a^2}-\frac {x (1-a x)^{5/4} (a x+1)^{3/4}}{3 a^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {\frac {9}{4} \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \left (\frac {1}{2} \int \frac {1-\sqrt {1-a x}}{2-a x}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-a x}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+\frac {1}{2} \int \frac {1}{\sqrt {1-a x}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )\right )}{a}\right )+\frac {(a x+1)^{3/4} (1-a x)^{5/4}}{2 a}}{6 a^2}-\frac {x (1-a x)^{5/4} (a x+1)^{3/4}}{3 a^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\frac {9}{4} \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-\sqrt {1-a x}-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\sqrt {1-a x}-1}d\left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-\sqrt {1-a x}}{2-a x}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{a}\right )+\frac {(a x+1)^{3/4} (1-a x)^{5/4}}{2 a}}{6 a^2}-\frac {x (1-a x)^{5/4} (a x+1)^{3/4}}{3 a^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {9}{4} \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \left (\frac {1}{2} \int \frac {1-\sqrt {1-a x}}{2-a x}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )\right )}{a}\right )+\frac {(a x+1)^{3/4} (1-a x)^{5/4}}{2 a}}{6 a^2}-\frac {x (1-a x)^{5/4} (a x+1)^{3/4}}{3 a^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\frac {9}{4} \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-\frac {2 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{\sqrt {1-a x}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {1-a x}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )\right )}{a}\right )+\frac {(a x+1)^{3/4} (1-a x)^{5/4}}{2 a}}{6 a^2}-\frac {x (1-a x)^{5/4} (a x+1)^{3/4}}{3 a^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {9}{4} \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{\sqrt {1-a x}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {1-a x}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )\right )}{a}\right )+\frac {(a x+1)^{3/4} (1-a x)^{5/4}}{2 a}}{6 a^2}-\frac {x (1-a x)^{5/4} (a x+1)^{3/4}}{3 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {9}{4} \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-\frac {2 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{\sqrt {1-a x}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}{\sqrt {1-a x}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1}d\frac {\sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )\right )}{a}\right )+\frac {(a x+1)^{3/4} (1-a x)^{5/4}}{2 a}}{6 a^2}-\frac {x (1-a x)^{5/4} (a x+1)^{3/4}}{3 a^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {9}{4} \left (\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{a}-\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\sqrt {1-a x}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {1-a x}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{2 \sqrt {2}}\right )\right )}{a}\right )+\frac {(a x+1)^{3/4} (1-a x)^{5/4}}{2 a}}{6 a^2}-\frac {x (1-a x)^{5/4} (a x+1)^{3/4}}{3 a^2}\)

input

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

output

-1/3*(x*(1 - a*x)^(5/4)*(1 + a*x)^(3/4))/a^2 + (((1 - a*x)^(5/4)*(1 + a*x) 
^(3/4))/(2*a) + (9*(((1 - a*x)^(1/4)*(1 + a*x)^(3/4))/a - (2*((-(ArcTan[1 
- (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/Sqrt[2]) + ArcTan[1 + (Sqrt[2 
]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/Sqrt[2])/2 + (-1/2*Log[1 + Sqrt[1 - a* 
x] - (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/Sqrt[2] + Log[1 + Sqrt[1 - 
 a*x] + (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)]/(2*Sqrt[2]))/2))/a))/4) 
/(6*a^2)
 

3.1.92.3.1 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 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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 770

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n)   Subst[In 
t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, 
 b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 
/n]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 6676

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

3.1.92.4 Maple [F]

\[\int \frac {x^{2}}{\sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}}d x\]

input

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

output

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

3.1.92.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.89 \[ \int e^{-\frac {1}{2} \text {arctanh}(a x)} x^2 \, dx=\frac {9 \, a^{3} \left (-\frac {1}{a^{12}}\right )^{\frac {1}{4}} \log \left (a^{9} \left (-\frac {1}{a^{12}}\right )^{\frac {3}{4}} + \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}\right ) - 9 i \, a^{3} \left (-\frac {1}{a^{12}}\right )^{\frac {1}{4}} \log \left (i \, a^{9} \left (-\frac {1}{a^{12}}\right )^{\frac {3}{4}} + \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}\right ) + 9 i \, a^{3} \left (-\frac {1}{a^{12}}\right )^{\frac {1}{4}} \log \left (-i \, a^{9} \left (-\frac {1}{a^{12}}\right )^{\frac {3}{4}} + \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}\right ) - 9 \, a^{3} \left (-\frac {1}{a^{12}}\right )^{\frac {1}{4}} \log \left (-a^{9} \left (-\frac {1}{a^{12}}\right )^{\frac {3}{4}} + \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}\right ) + 2 \, {\left (8 \, a^{2} x^{2} - 10 \, a x + 11\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}}{48 \, a^{3}} \]

input

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

output

1/48*(9*a^3*(-1/a^12)^(1/4)*log(a^9*(-1/a^12)^(3/4) + sqrt(-sqrt(-a^2*x^2 
+ 1)/(a*x - 1))) - 9*I*a^3*(-1/a^12)^(1/4)*log(I*a^9*(-1/a^12)^(3/4) + sqr 
t(-sqrt(-a^2*x^2 + 1)/(a*x - 1))) + 9*I*a^3*(-1/a^12)^(1/4)*log(-I*a^9*(-1 
/a^12)^(3/4) + sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))) - 9*a^3*(-1/a^12)^(1/4 
)*log(-a^9*(-1/a^12)^(3/4) + sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))) + 2*(8*a 
^2*x^2 - 10*a*x + 11)*sqrt(-a^2*x^2 + 1)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1 
)))/a^3
 

3.1.92.6 Sympy [F]

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

input

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

output

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

3.1.92.7 Maxima [F]

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

input

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

output

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

3.1.92.8 Giac [F(-2)]

Exception generated. \[ \int e^{-\frac {1}{2} \text {arctanh}(a x)} x^2 \, dx=\text {Exception raised: TypeError} \]

input

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

output

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

3.1.92.9 Mupad [F(-1)]

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

input

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

output

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