\(\int \frac {e^{-\frac {1}{2} \text {arctanh}(a x)}}{x^4} \, dx\) [94]

Optimal result

Integrand size = 14, antiderivative size = 139 \[ \int \frac {e^{-\frac {1}{2} \text {arctanh}(a x)}}{x^4} \, dx=-\frac {\sqrt [4]{1-a x} (1+a x)^{3/4}}{3 x^3}+\frac {5 a \sqrt [4]{1-a x} (1+a x)^{3/4}}{12 x^2}-\frac {11 a^2 \sqrt [4]{1-a x} (1+a x)^{3/4}}{24 x}-\frac {3}{8} a^3 \arctan \left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )+\frac {3}{8} a^3 \text {arctanh}\left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right ) \] Output:

-1/3*(-a*x+1)^(1/4)*(a*x+1)^(3/4)/x^3+5/12*a*(-a*x+1)^(1/4)*(a*x+1)^(3/4)/ 
x^2-11/24*a^2*(-a*x+1)^(1/4)*(a*x+1)^(3/4)/x-3/8*a^3*arctan((a*x+1)^(1/4)/ 
(-a*x+1)^(1/4))+3/8*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 = 0.01 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.56 \[ \int \frac {e^{-\frac {1}{2} \text {arctanh}(a x)}}{x^4} \, dx=\frac {\sqrt [4]{1-a x} \left (-8+2 a x-a^2 x^2-11 a^3 x^3+18 a^3 x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},\frac {1-a x}{1+a x}\right )\right )}{24 x^3 \sqrt [4]{1+a x}} \] Input:

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

Output:

((1 - a*x)^(1/4)*(-8 + 2*a*x - a^2*x^2 - 11*a^3*x^3 + 18*a^3*x^3*Hypergeom 
etric2F1[1/4, 1, 5/4, (1 - a*x)/(1 + a*x)]))/(24*x^3*(1 + a*x)^(1/4))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {6676, 110, 27, 168, 27, 168, 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.

\(\displaystyle \int \frac {e^{-\frac {1}{2} \text {arctanh}(a x)}}{x^4} \, dx\)

\(\Big \downarrow \) 6676

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

\(\Big \downarrow \) 110

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 168

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{2} \int \frac {a (11-10 a x)}{2 x^2 (1-a x)^{3/4} \sqrt [4]{a x+1}}dx-\frac {5 \sqrt [4]{1-a x} (a x+1)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{4} a \int \frac {11-10 a x}{x^2 (1-a x)^{3/4} \sqrt [4]{a x+1}}dx-\frac {5 \sqrt [4]{1-a x} (a x+1)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 168

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{4} a \left (-\int \frac {9 a}{2 x (1-a x)^{3/4} \sqrt [4]{a x+1}}dx-\frac {11 \sqrt [4]{1-a x} (a x+1)^{3/4}}{x}\right )-\frac {5 \sqrt [4]{1-a x} (a x+1)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{4} a \left (-\frac {9}{2} a \int \frac {1}{x (1-a x)^{3/4} \sqrt [4]{a x+1}}dx-\frac {11 \sqrt [4]{1-a x} (a x+1)^{3/4}}{x}\right )-\frac {5 \sqrt [4]{1-a x} (a x+1)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{3 x^3}\)

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

\(\displaystyle -\frac {1}{6} a \left (-\frac {1}{4} a \left (-18 a \left (\frac {1}{2} \arctan \left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )\right )-\frac {11 \sqrt [4]{1-a x} (a x+1)^{3/4}}{x}\right )-\frac {5 \sqrt [4]{1-a x} (a x+1)^{3/4}}{2 x^2}\right )-\frac {\sqrt [4]{1-a x} (a x+1)^{3/4}}{3 x^3}\)

Input:

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

Output:

-1/3*((1 - a*x)^(1/4)*(1 + a*x)^(3/4))/x^3 - (a*((-5*(1 - a*x)^(1/4)*(1 + 
a*x)^(3/4))/(2*x^2) - (a*((-11*(1 - a*x)^(1/4)*(1 + a*x)^(3/4))/x - 18*a*( 
ArcTan[(1 + a*x)^(1/4)/(1 - a*x)^(1/4)]/2 - ArcTanh[(1 + a*x)^(1/4)/(1 - a 
*x)^(1/4)]/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 110

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

\[\int \frac {1}{\sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}\, x^{4}}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-\frac {1}{2} \text {arctanh}(a x)}}{x^4} \, dx=-\frac {18 \, a^{3} x^{3} \arctan \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}\right ) - 9 \, a^{3} x^{3} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 1\right ) + 9 \, a^{3} x^{3} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - 1\right ) + 2 \, {\left (11 \, a^{2} x^{2} - 10 \, a x + 8\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}}{48 \, x^{3}} \] Input:

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

Output:

-1/48*(18*a^3*x^3*arctan(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))) - 9*a^3*x^3* 
log(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) + 1) + 9*a^3*x^3*log(sqrt(-sqrt(-a 
^2*x^2 + 1)/(a*x - 1)) - 1) + 2*(11*a^2*x^2 - 10*a*x + 8)*sqrt(-a^2*x^2 + 
1)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)))/x^3
 

Sympy [F]

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

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

Output:

Integral(1/(x**4*sqrt((a*x + 1)/sqrt(-a**2*x**2 + 1))), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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