\(\int \frac {e^{-\coth ^{-1}(a x)}}{x^5} \, dx\) [41]

Optimal result

Integrand size = 12, antiderivative size = 88 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{x^5} \, dx=\frac {1}{24} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (16 a-\frac {9}{x}\right )-\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}+\frac {3}{8} a^4 \csc ^{-1}(a x) \]

[Out]

3/8*a^4*arccsc(a*x)+1/24*a^3*(16*a-9/x)*(1-1/a^2/x^2)^(1/2)-1/4*a*(1-1/a^2/x^2)^(1/2)/x^3+1/3*a^2*(1-1/a^2/x^2
)^(1/2)/x^2

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6304, 847, 794, 222} \[ \int \frac {e^{-\coth ^{-1}(a x)}}{x^5} \, dx=\frac {3}{8} a^4 \csc ^{-1}(a x)+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}-\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}+\frac {1}{24} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (16 a-\frac {9}{x}\right ) \]

[In]

Int[1/(E^ArcCoth[a*x]*x^5),x]

[Out]

(a^3*Sqrt[1 - 1/(a^2*x^2)]*(16*a - 9/x))/24 - (a*Sqrt[1 - 1/(a^2*x^2)])/(4*x^3) + (a^2*Sqrt[1 - 1/(a^2*x^2)])/
(3*x^2) + (3*a^4*ArcCsc[a*x])/8

Rule 222

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 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 847

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^
m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 6304

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

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {x^3 \left (1-\frac {x}{a}\right )}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}+\frac {1}{4} a^2 \text {Subst}\left (\int \frac {x^2 \left (\frac {3}{a}-\frac {4 x}{a^2}\right )}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}-\frac {1}{12} a^4 \text {Subst}\left (\int \frac {x \left (\frac {8}{a^2}-\frac {9 x}{a^3}\right )}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{24} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (16 a-\frac {9}{x}\right )-\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}+\frac {1}{8} \left (3 a^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{24} a^3 \sqrt {1-\frac {1}{a^2 x^2}} \left (16 a-\frac {9}{x}\right )-\frac {a \sqrt {1-\frac {1}{a^2 x^2}}}{4 x^3}+\frac {a^2 \sqrt {1-\frac {1}{a^2 x^2}}}{3 x^2}+\frac {3}{8} a^4 \csc ^{-1}(a x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.67 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{x^5} \, dx=\frac {1}{24} a \left (\frac {\sqrt {1-\frac {1}{a^2 x^2}} \left (-6+8 a x-9 a^2 x^2+16 a^3 x^3\right )}{x^3}+9 a^3 \arcsin \left (\frac {1}{a x}\right )\right ) \]

[In]

Integrate[1/(E^ArcCoth[a*x]*x^5),x]

[Out]

(a*((Sqrt[1 - 1/(a^2*x^2)]*(-6 + 8*a*x - 9*a^2*x^2 + 16*a^3*x^3))/x^3 + 9*a^3*ArcSin[1/(a*x)]))/24

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.15

method	result	size
risch	\(\frac {\left (a x +1\right ) \left (16 a^{3} x^{3}-9 a^{2} x^{2}+8 a x -6\right ) \sqrt {\frac {a x -1}{a x +1}}}{24 x^{4}}+\frac {3 a^{4} \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right ) \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{8 \left (a x -1\right )}\)	\(101\)
default	\(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) \left (-24 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{5} x^{5}+24 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, a^{3} x^{3}-9 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a^{4} x^{4}-9 a^{4} x^{4} \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )+24 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a^{5} x^{4}+24 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a^{4} x^{4}-24 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{\sqrt {a^{2}}}\right ) a^{5} x^{4}-15 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, a^{2} x^{2}+8 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, a x -6 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\right )}{24 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, x^{4} \sqrt {a^{2}}}\)	\(308\)

[In]

int(((a*x-1)/(a*x+1))^(1/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

1/24*(a*x+1)*(16*a^3*x^3-9*a^2*x^2+8*a*x-6)/x^4*((a*x-1)/(a*x+1))^(1/2)+3/8*a^4*arctan(1/(a^2*x^2-1)^(1/2))*((
a*x-1)/(a*x+1))^(1/2)*((a*x-1)*(a*x+1))^(1/2)/(a*x-1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{x^5} \, dx=-\frac {18 \, a^{4} x^{4} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) - {\left (16 \, a^{4} x^{4} + 7 \, a^{3} x^{3} - a^{2} x^{2} + 2 \, a x - 6\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{24 \, x^{4}} \]

[In]

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

[Out]

-1/24*(18*a^4*x^4*arctan(sqrt((a*x - 1)/(a*x + 1))) - (16*a^4*x^4 + 7*a^3*x^3 - a^2*x^2 + 2*a*x - 6)*sqrt((a*x
- 1)/(a*x + 1)))/x^4

Sympy [F]

\[ \int \frac {e^{-\coth ^{-1}(a x)}}{x^5} \, dx=\int \frac {\sqrt {\frac {a x - 1}{a x + 1}}}{x^{5}}\, dx \]

[In]

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

[Out]

Integral(sqrt((a*x - 1)/(a*x + 1))/x**5, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (74) = 148\).

Time = 0.31 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.97 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{x^5} \, dx=-\frac {1}{12} \, {\left (9 \, a^{3} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) - \frac {39 \, a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} + 31 \, a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} + 49 \, a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + 9 \, a^{3} \sqrt {\frac {a x - 1}{a x + 1}}}{\frac {4 \, {\left (a x - 1\right )}}{a x + 1} + \frac {6 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {4 \, {\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} + \frac {{\left (a x - 1\right )}^{4}}{{\left (a x + 1\right )}^{4}} + 1}\right )} a \]

[In]

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

[Out]

-1/12*(9*a^3*arctan(sqrt((a*x - 1)/(a*x + 1))) - (39*a^3*((a*x - 1)/(a*x + 1))^(7/2) + 31*a^3*((a*x - 1)/(a*x
+ 1))^(5/2) + 49*a^3*((a*x - 1)/(a*x + 1))^(3/2) + 9*a^3*sqrt((a*x - 1)/(a*x + 1)))/(4*(a*x - 1)/(a*x + 1) + 6
*(a*x - 1)^2/(a*x + 1)^2 + 4*(a*x - 1)^3/(a*x + 1)^3 + (a*x - 1)^4/(a*x + 1)^4 + 1))*a

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (74) = 148\).

Time = 0.29 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.93 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{x^5} \, dx=-\frac {3}{4} \, a^{4} \arctan \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1}\right ) \mathrm {sgn}\left (a x + 1\right ) + \frac {9 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{7} a^{4} \mathrm {sgn}\left (a x + 1\right ) + 33 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{5} a^{4} \mathrm {sgn}\left (a x + 1\right ) + 48 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{4} a^{3} {\left | a \right |} \mathrm {sgn}\left (a x + 1\right ) - 33 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{3} a^{4} \mathrm {sgn}\left (a x + 1\right ) + 64 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{2} a^{3} {\left | a \right |} \mathrm {sgn}\left (a x + 1\right ) - 9 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )} a^{4} \mathrm {sgn}\left (a x + 1\right ) + 16 \, a^{3} {\left | a \right |} \mathrm {sgn}\left (a x + 1\right )}{12 \, {\left ({\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{2} + 1\right )}^{4}} \]

[In]

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

[Out]

-3/4*a^4*arctan(-x*abs(a) + sqrt(a^2*x^2 - 1))*sgn(a*x + 1) + 1/12*(9*(x*abs(a) - sqrt(a^2*x^2 - 1))^7*a^4*sgn
(a*x + 1) + 33*(x*abs(a) - sqrt(a^2*x^2 - 1))^5*a^4*sgn(a*x + 1) + 48*(x*abs(a) - sqrt(a^2*x^2 - 1))^4*a^3*abs
(a)*sgn(a*x + 1) - 33*(x*abs(a) - sqrt(a^2*x^2 - 1))^3*a^4*sgn(a*x + 1) + 64*(x*abs(a) - sqrt(a^2*x^2 - 1))^2*
a^3*abs(a)*sgn(a*x + 1) - 9*(x*abs(a) - sqrt(a^2*x^2 - 1))*a^4*sgn(a*x + 1) + 16*a^3*abs(a)*sgn(a*x + 1))/((x*
abs(a) - sqrt(a^2*x^2 - 1))^2 + 1)^4

Mupad [B] (verification not implemented)

Time = 4.08 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.47 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{x^5} \, dx=\frac {2\,a^4\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{3}-\frac {\sqrt {\frac {a\,x-1}{a\,x+1}}}{4\,x^4}-\frac {3\,a^4\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{4}-\frac {a^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{24\,x^2}+\frac {7\,a^3\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{24\,x}+\frac {a\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{12\,x^3} \]

[In]

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

[Out]

(2*a^4*((a*x - 1)/(a*x + 1))^(1/2))/3 - ((a*x - 1)/(a*x + 1))^(1/2)/(4*x^4) - (3*a^4*atan(((a*x - 1)/(a*x + 1)
)^(1/2)))/4 - (a^2*((a*x - 1)/(a*x + 1))^(1/2))/(24*x^2) + (7*a^3*((a*x - 1)/(a*x + 1))^(1/2))/(24*x) + (a*((a
*x - 1)/(a*x + 1))^(1/2))/(12*x^3)