\(\int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^4} \, dx\) [127]

Optimal result

Integrand size = 14, antiderivative size = 165 \[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^4} \, dx=-\frac {287 a^3 \sqrt [4]{1-a x}}{24 \sqrt [4]{1+a x}}-\frac {\sqrt [4]{1-a x}}{3 x^3 \sqrt [4]{1+a x}}+\frac {13 a \sqrt [4]{1-a x}}{12 x^2 \sqrt [4]{1+a x}}-\frac {61 a^2 \sqrt [4]{1-a x}}{24 x \sqrt [4]{1+a x}}-\frac {55}{8} a^3 \arctan \left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )+\frac {55}{8} a^3 \text {arctanh}\left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right ) \] Output:

-287/24*a^3*(-a*x+1)^(1/4)/(a*x+1)^(1/4)-1/3*(-a*x+1)^(1/4)/x^3/(a*x+1)^(1 
/4)+13/12*a*(-a*x+1)^(1/4)/x^2/(a*x+1)^(1/4)-61/24*a^2*(-a*x+1)^(1/4)/x/(a 
*x+1)^(1/4)-55/8*a^3*arctan((a*x+1)^(1/4)/(-a*x+1)^(1/4))+55/8*a^3*arctanh 
((a*x+1)^(1/4)/(-a*x+1)^(1/4))
 

Mathematica [C] (verified)

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

Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.47 \[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^4} \, dx=\frac {\sqrt [4]{1-a x} \left (-8+26 a x-61 a^2 x^2-287 a^3 x^3+330 a^3 x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {1-a x}{1+a x}\right )\right )}{24 x^3 \sqrt [4]{1+a x}} \] Input:

Integrate[1/(E^((5*ArcTanh[a*x])/2)*x^4),x]
 

Output:

((1 - a*x)^(1/4)*(-8 + 26*a*x - 61*a^2*x^2 - 287*a^3*x^3 + 330*a^3*x^3*Hyp 
ergeometric2F1[1/4, 1, 5/4, (1 - a*x)/(1 + a*x)]))/(24*x^3*(1 + a*x)^(1/4) 
)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6676, 109, 27, 168, 27, 168, 27, 172, 27, 104, 25, 827, 216, 219}

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.

\(\displaystyle \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^4} \, dx\)

\(\Big \downarrow \) 6676

\(\displaystyle \int \frac {(1-a x)^{5/4}}{x^4 (a x+1)^{5/4}}dx\)

\(\Big \downarrow \) 109

\(\displaystyle -\frac {1}{3} \int \frac {a (13-12 a x)}{2 x^3 (1-a x)^{3/4} (a x+1)^{5/4}}dx-\frac {\sqrt [4]{1-a x}}{3 x^3 \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{6} a \int \frac {13-12 a x}{x^3 (1-a x)^{3/4} (a x+1)^{5/4}}dx-\frac {\sqrt [4]{1-a x}}{3 x^3 \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 168

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{2} \int \frac {a (61-52 a x)}{2 x^2 (1-a x)^{3/4} (a x+1)^{5/4}}dx-\frac {13 \sqrt [4]{1-a x}}{2 x^2 \sqrt [4]{a x+1}}\right )-\frac {\sqrt [4]{1-a x}}{3 x^3 \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{4} a \int \frac {61-52 a x}{x^2 (1-a x)^{3/4} (a x+1)^{5/4}}dx-\frac {13 \sqrt [4]{1-a x}}{2 x^2 \sqrt [4]{a x+1}}\right )-\frac {\sqrt [4]{1-a x}}{3 x^3 \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 168

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{4} a \left (-\int \frac {a (165-122 a x)}{2 x (1-a x)^{3/4} (a x+1)^{5/4}}dx-\frac {61 \sqrt [4]{1-a x}}{x \sqrt [4]{a x+1}}\right )-\frac {13 \sqrt [4]{1-a x}}{2 x^2 \sqrt [4]{a x+1}}\right )-\frac {\sqrt [4]{1-a x}}{3 x^3 \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{4} a \left (-\frac {1}{2} a \int \frac {165-122 a x}{x (1-a x)^{3/4} (a x+1)^{5/4}}dx-\frac {61 \sqrt [4]{1-a x}}{x \sqrt [4]{a x+1}}\right )-\frac {13 \sqrt [4]{1-a x}}{2 x^2 \sqrt [4]{a x+1}}\right )-\frac {\sqrt [4]{1-a x}}{3 x^3 \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 172

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{4} a \left (-\frac {1}{2} a \left (\frac {2 \int \frac {165 a}{2 x (1-a x)^{3/4} \sqrt [4]{a x+1}}dx}{a}+\frac {574 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )-\frac {61 \sqrt [4]{1-a x}}{x \sqrt [4]{a x+1}}\right )-\frac {13 \sqrt [4]{1-a x}}{2 x^2 \sqrt [4]{a x+1}}\right )-\frac {\sqrt [4]{1-a x}}{3 x^3 \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{4} a \left (-\frac {1}{2} a \left (165 \int \frac {1}{x (1-a x)^{3/4} \sqrt [4]{a x+1}}dx+\frac {574 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )-\frac {61 \sqrt [4]{1-a x}}{x \sqrt [4]{a x+1}}\right )-\frac {13 \sqrt [4]{1-a x}}{2 x^2 \sqrt [4]{a x+1}}\right )-\frac {\sqrt [4]{1-a x}}{3 x^3 \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{4} a \left (-\frac {1}{2} a \left (660 \int -\frac {\sqrt {a x+1}}{\sqrt {1-a x} \left (1-\frac {a x+1}{1-a x}\right )}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}+\frac {574 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )-\frac {61 \sqrt [4]{1-a x}}{x \sqrt [4]{a x+1}}\right )-\frac {13 \sqrt [4]{1-a x}}{2 x^2 \sqrt [4]{a x+1}}\right )-\frac {\sqrt [4]{1-a x}}{3 x^3 \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{4} a \left (-\frac {1}{2} a \left (\frac {574 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}-660 \int \frac {\sqrt {a x+1}}{\sqrt {1-a x} \left (1-\frac {a x+1}{1-a x}\right )}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {61 \sqrt [4]{1-a x}}{x \sqrt [4]{a x+1}}\right )-\frac {13 \sqrt [4]{1-a x}}{2 x^2 \sqrt [4]{a x+1}}\right )-\frac {\sqrt [4]{1-a x}}{3 x^3 \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 827

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{4} a \left (-\frac {1}{2} a \left (660 \left (\frac {1}{2} \int \frac {1}{\frac {\sqrt {a x+1}}{\sqrt {1-a x}}+1}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {a x+1}}{\sqrt {1-a x}}}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )+\frac {574 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )-\frac {61 \sqrt [4]{1-a x}}{x \sqrt [4]{a x+1}}\right )-\frac {13 \sqrt [4]{1-a x}}{2 x^2 \sqrt [4]{a x+1}}\right )-\frac {\sqrt [4]{1-a x}}{3 x^3 \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{4} a \left (-\frac {1}{2} a \left (660 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {a x+1}}{\sqrt {1-a x}}}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )+\frac {574 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )-\frac {61 \sqrt [4]{1-a x}}{x \sqrt [4]{a x+1}}\right )-\frac {13 \sqrt [4]{1-a x}}{2 x^2 \sqrt [4]{a x+1}}\right )-\frac {\sqrt [4]{1-a x}}{3 x^3 \sqrt [4]{a x+1}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{4} a \left (-\frac {1}{2} a \left (660 \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )\right )+\frac {574 \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )-\frac {61 \sqrt [4]{1-a x}}{x \sqrt [4]{a x+1}}\right )-\frac {13 \sqrt [4]{1-a x}}{2 x^2 \sqrt [4]{a x+1}}\right )-\frac {\sqrt [4]{1-a x}}{3 x^3 \sqrt [4]{a x+1}}\)

Input:

Int[1/(E^((5*ArcTanh[a*x])/2)*x^4),x]
 

Output:

-1/3*(1 - a*x)^(1/4)/(x^3*(1 + a*x)^(1/4)) - (a*((-13*(1 - a*x)^(1/4))/(2* 
x^2*(1 + a*x)^(1/4)) - (a*((-61*(1 - a*x)^(1/4))/(x*(1 + a*x)^(1/4)) - (a* 
((574*(1 - a*x)^(1/4))/(1 + a*x)^(1/4) + 660*(ArcTan[(1 + a*x)^(1/4)/(1 - 
a*x)^(1/4)]/2 - ArcTanh[(1 + a*x)^(1/4)/(1 - a*x)^(1/4)]/2)))/2))/4))/6
 

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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 109

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

rule 168

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
 

rule 172

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> With[{mnp = Simplify[m + n + p]}, Simp[ 
(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1) 
*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) 
 Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f 
)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g 
 - a*h)*(mnp + 3)*x, x], x], x] /; ILtQ[mnp + 2, 0] && (SumSimplerQ[m, 1] | 
| ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 1 
])))] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && NeQ[m, -1]
 

rule 216

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

rule 219

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

rule 827

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

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]
 

Maple [F]

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

Input:

int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^4,x)
 

Output:

int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^4,x)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^4} \, dx=-\frac {2 \, {\left (287 \, a^{3} x^{3} + 61 \, a^{2} x^{2} - 26 \, a x + 8\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 330 \, {\left (a^{4} x^{4} + a^{3} x^{3}\right )} \arctan \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}\right ) - 165 \, {\left (a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 1\right ) + 165 \, {\left (a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - 1\right )}{48 \, {\left (a x^{4} + x^{3}\right )}} \] Input:

integrate(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^4,x, algorithm="fricas")
 

Output:

-1/48*(2*(287*a^3*x^3 + 61*a^2*x^2 - 26*a*x + 8)*sqrt(-a^2*x^2 + 1)*sqrt(- 
sqrt(-a^2*x^2 + 1)/(a*x - 1)) + 330*(a^4*x^4 + a^3*x^3)*arctan(sqrt(-sqrt( 
-a^2*x^2 + 1)/(a*x - 1))) - 165*(a^4*x^4 + a^3*x^3)*log(sqrt(-sqrt(-a^2*x^ 
2 + 1)/(a*x - 1)) + 1) + 165*(a^4*x^4 + a^3*x^3)*log(sqrt(-sqrt(-a^2*x^2 + 
 1)/(a*x - 1)) - 1))/(a*x^4 + x^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^4} \, dx=\text {Timed out} \] Input:

integrate(1/((a*x+1)/(-a**2*x**2+1)**(1/2))**(5/2)/x**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^4} \, dx=\int { \frac {1}{x^{4} \left (\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^4,x, algorithm="maxima")
 

Output:

integrate(1/(x^4*((a*x + 1)/sqrt(-a^2*x^2 + 1))^(5/2)), x)
 

Giac [F]

\[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^4} \, dx=\int { \frac {1}{x^{4} \left (\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^4,x, algorithm="giac")
 

Output:

integrate(1/(x^4*((a*x + 1)/sqrt(-a^2*x^2 + 1))^(5/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^4} \, dx=\int \frac {1}{x^4\,{\left (\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}\right )}^{5/2}} \,d x \] Input:

int(1/(x^4*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(5/2)),x)
                                                                                    
                                                                                    
 

Output:

int(1/(x^4*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(5/2)), x)
 

Reduce [F]

\[ \int \frac {e^{-\frac {5}{2} \text {arctanh}(a x)}}{x^4} \, dx=\int \frac {1}{{\left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}^{\frac {5}{2}} x^{4}}d x \] Input:

int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^4,x)
 

Output:

int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^4,x)