\(\int \frac {e^{\frac {1}{2} \coth ^{-1}(a x)}}{x^3} \, dx\) [66]

Optimal result

Integrand size = 14, antiderivative size = 319 \[ \int \frac {e^{\frac {1}{2} \coth ^{-1}(a x)}}{x^3} \, dx=\frac {1}{4} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{2} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}-\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{4 \sqrt {2}}+\frac {a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{4 \sqrt {2}}+\frac {a^2 \log \left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}}-\frac {a^2 \log \left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {6306, 81, 52, 65, 338, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {e^{\frac {1}{2} \coth ^{-1}(a x)}}{x^3} \, dx=-\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{4 \sqrt {2}}+\frac {a^2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{4 \sqrt {2}}+\frac {1}{2} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (\frac {1}{a x}+1\right )^{5/4}+\frac {1}{4} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}+\frac {a^2 \log \left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {\frac {1}{a x}+1}}-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{8 \sqrt {2}}-\frac {a^2 \log \left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {\frac {1}{a x}+1}}+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{8 \sqrt {2}} \]

[In]

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

[Out]

(a^2*(1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(1/4))/4 + (a^2*(1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(5/4))/2 - (a^2*ArcTa
n[1 - (Sqrt[2]*(1 - 1/(a*x))^(1/4))/(1 + 1/(a*x))^(1/4)])/(4*Sqrt[2]) + (a^2*ArcTan[1 + (Sqrt[2]*(1 - 1/(a*x))
^(1/4))/(1 + 1/(a*x))^(1/4)])/(4*Sqrt[2]) + (a^2*Log[1 + Sqrt[1 - 1/(a*x)]/Sqrt[1 + 1/(a*x)] - (Sqrt[2]*(1 - 1
/(a*x))^(1/4))/(1 + 1/(a*x))^(1/4)])/(8*Sqrt[2]) - (a^2*Log[1 + Sqrt[1 - 1/(a*x)]/Sqrt[1 + 1/(a*x)] + (Sqrt[2]
*(1 - 1/(a*x))^(1/4))/(1 + 1/(a*x))^(1/4)])/(8*Sqrt[2])

Rule 52

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] + Dist[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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(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 210

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 303

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

Rule 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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 1179

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

Rule 6306

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

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {x \sqrt [4]{1+\frac {x}{a}}}{\sqrt [4]{1-\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{2} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}-\frac {1}{4} a \text {Subst}\left (\int \frac {\sqrt [4]{1+\frac {x}{a}}}{\sqrt [4]{1-\frac {x}{a}}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{4} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{2} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}-\frac {1}{8} a \text {Subst}\left (\int \frac {1}{\sqrt [4]{1-\frac {x}{a}} \left (1+\frac {x}{a}\right )^{3/4}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{4} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{2} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}+\frac {1}{2} a^2 \text {Subst}\left (\int \frac {x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-\frac {1}{a x}}\right ) \\ & = \frac {1}{4} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{2} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}+\frac {1}{2} a^2 \text {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right ) \\ & = \frac {1}{4} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{2} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}-\frac {1}{4} a^2 \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )+\frac {1}{4} a^2 \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right ) \\ & = \frac {1}{4} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{2} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}+\frac {1}{8} a^2 \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )+\frac {1}{8} a^2 \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )+\frac {a^2 \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}}+\frac {a^2 \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}} \\ & = \frac {1}{4} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{2} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}+\frac {a^2 \log \left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}}-\frac {a^2 \log \left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}}+\frac {a^2 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{4 \sqrt {2}}-\frac {a^2 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{4 \sqrt {2}} \\ & = \frac {1}{4} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{2} a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}-\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{4 \sqrt {2}}+\frac {a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{4 \sqrt {2}}+\frac {a^2 \log \left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}}-\frac {a^2 \log \left (1+\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {1+\frac {1}{a x}}}+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.54 \[ \int \frac {e^{\frac {1}{2} \coth ^{-1}(a x)}}{x^3} \, dx=\frac {1}{16} a^2 \left (-\frac {32 e^{\frac {1}{2} \coth ^{-1}(a x)}}{\left (1+e^{2 \coth ^{-1}(a x)}\right )^2}+\frac {40 e^{\frac {1}{2} \coth ^{-1}(a x)}}{1+e^{2 \coth ^{-1}(a x)}}+2 \sqrt {2} \arctan \left (1-\sqrt {2} e^{\frac {1}{2} \coth ^{-1}(a x)}\right )-2 \sqrt {2} \arctan \left (1+\sqrt {2} e^{\frac {1}{2} \coth ^{-1}(a x)}\right )+\sqrt {2} \log \left (1-\sqrt {2} e^{\frac {1}{2} \coth ^{-1}(a x)}+e^{\coth ^{-1}(a x)}\right )-\sqrt {2} \log \left (1+\sqrt {2} e^{\frac {1}{2} \coth ^{-1}(a x)}+e^{\coth ^{-1}(a x)}\right )\right ) \]

[In]

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

[Out]

(a^2*((-32*E^(ArcCoth[a*x]/2))/(1 + E^(2*ArcCoth[a*x]))^2 + (40*E^(ArcCoth[a*x]/2))/(1 + E^(2*ArcCoth[a*x])) +
2*Sqrt[2]*ArcTan[1 - Sqrt[2]*E^(ArcCoth[a*x]/2)] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*E^(ArcCoth[a*x]/2)] + Sqrt[2]
*Log[1 - Sqrt[2]*E^(ArcCoth[a*x]/2) + E^ArcCoth[a*x]] - Sqrt[2]*Log[1 + Sqrt[2]*E^(ArcCoth[a*x]/2) + E^ArcCoth
[a*x]]))/16

Maple [F]

\[\int \frac {1}{\left (\frac {a x -1}{a x +1}\right )^{\frac {1}{4}} x^{3}}d x\]

[In]

int(1/((a*x-1)/(a*x+1))^(1/4)/x^3,x)

[Out]

int(1/((a*x-1)/(a*x+1))^(1/4)/x^3,x)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.63 \[ \int \frac {e^{\frac {1}{2} \coth ^{-1}(a x)}}{x^3} \, dx=\frac {\left (-a^{8}\right )^{\frac {1}{4}} x^{2} \log \left (a^{6} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \left (-a^{8}\right )^{\frac {3}{4}}\right ) - i \, \left (-a^{8}\right )^{\frac {1}{4}} x^{2} \log \left (a^{6} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + i \, \left (-a^{8}\right )^{\frac {3}{4}}\right ) + i \, \left (-a^{8}\right )^{\frac {1}{4}} x^{2} \log \left (a^{6} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - i \, \left (-a^{8}\right )^{\frac {3}{4}}\right ) - \left (-a^{8}\right )^{\frac {1}{4}} x^{2} \log \left (a^{6} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - \left (-a^{8}\right )^{\frac {3}{4}}\right ) + 2 \, {\left (3 \, a^{2} x^{2} + 5 \, a x + 2\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{8 \, x^{2}} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/4)/x^3,x, algorithm="fricas")

[Out]

1/8*((-a^8)^(1/4)*x^2*log(a^6*((a*x - 1)/(a*x + 1))^(1/4) + (-a^8)^(3/4)) - I*(-a^8)^(1/4)*x^2*log(a^6*((a*x -
1)/(a*x + 1))^(1/4) + I*(-a^8)^(3/4)) + I*(-a^8)^(1/4)*x^2*log(a^6*((a*x - 1)/(a*x + 1))^(1/4) - I*(-a^8)^(3/
4)) - (-a^8)^(1/4)*x^2*log(a^6*((a*x - 1)/(a*x + 1))^(1/4) - (-a^8)^(3/4)) + 2*(3*a^2*x^2 + 5*a*x + 2)*((a*x -
1)/(a*x + 1))^(3/4))/x^2

Sympy [F]

\[ \int \frac {e^{\frac {1}{2} \coth ^{-1}(a x)}}{x^3} \, dx=\int \frac {1}{x^{3} \sqrt [4]{\frac {a x - 1}{a x + 1}}}\, dx \]

[In]

integrate(1/((a*x-1)/(a*x+1))**(1/4)/x**3,x)

[Out]

Integral(1/(x**3*((a*x - 1)/(a*x + 1))**(1/4)), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {1}{2} \coth ^{-1}(a x)}}{x^3} \, dx=\frac {1}{16} \, {\left ({\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) - \sqrt {2} \log \left (\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + \sqrt {2} \log \left (-\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right )\right )} a + \frac {8 \, {\left (a \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{4}} + 5 \, a \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{\frac {2 \, {\left (a x - 1\right )}}{a x + 1} + \frac {{\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + 1}\right )} a \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/4)/x^3,x, algorithm="maxima")

[Out]

1/16*((2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*((a*x - 1)/(a*x + 1))^(1/4))) + 2*sqrt(2)*arctan(-1/2*sqrt(2)
*(sqrt(2) - 2*((a*x - 1)/(a*x + 1))^(1/4))) - sqrt(2)*log(sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt((a*x - 1)
/(a*x + 1)) + 1) + sqrt(2)*log(-sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt((a*x - 1)/(a*x + 1)) + 1))*a + 8*(a
*((a*x - 1)/(a*x + 1))^(7/4) + 5*a*((a*x - 1)/(a*x + 1))^(3/4))/(2*(a*x - 1)/(a*x + 1) + (a*x - 1)^2/(a*x + 1)
^2 + 1))*a

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\frac {1}{2} \coth ^{-1}(a x)}}{x^3} \, dx=\frac {1}{16} \, {\left (2 \, \sqrt {2} a \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 2 \, \sqrt {2} a \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) - \sqrt {2} a \log \left (\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + \sqrt {2} a \log \left (-\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + \frac {8 \, {\left (\frac {{\left (a x - 1\right )} a \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{a x + 1} + 5 \, a \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{{\left (\frac {a x - 1}{a x + 1} + 1\right )}^{2}}\right )} a \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/4)/x^3,x, algorithm="giac")

[Out]

1/16*(2*sqrt(2)*a*arctan(1/2*sqrt(2)*(sqrt(2) + 2*((a*x - 1)/(a*x + 1))^(1/4))) + 2*sqrt(2)*a*arctan(-1/2*sqrt
(2)*(sqrt(2) - 2*((a*x - 1)/(a*x + 1))^(1/4))) - sqrt(2)*a*log(sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt((a*x
- 1)/(a*x + 1)) + 1) + sqrt(2)*a*log(-sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt((a*x - 1)/(a*x + 1)) + 1) +
8*((a*x - 1)*a*((a*x - 1)/(a*x + 1))^(3/4)/(a*x + 1) + 5*a*((a*x - 1)/(a*x + 1))^(3/4))/((a*x - 1)/(a*x + 1) +
1)^2)*a

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.41 \[ \int \frac {e^{\frac {1}{2} \coth ^{-1}(a x)}}{x^3} \, dx=\frac {\frac {5\,a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/4}}{2}+\frac {a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/4}}{2}}{\frac {{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}+\frac {2\,\left (a\,x-1\right )}{a\,x+1}+1}+\frac {{\left (-1\right )}^{1/4}\,a^2\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{4}-\frac {{\left (-1\right )}^{1/4}\,a^2\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{4} \]

[In]

int(1/(x^3*((a*x - 1)/(a*x + 1))^(1/4)),x)

[Out]

((5*a^2*((a*x - 1)/(a*x + 1))^(3/4))/2 + (a^2*((a*x - 1)/(a*x + 1))^(7/4))/2)/((a*x - 1)^2/(a*x + 1)^2 + (2*(a
*x - 1))/(a*x + 1) + 1) + ((-1)^(1/4)*a^2*atan((-1)^(1/4)*((a*x - 1)/(a*x + 1))^(1/4)))/4 - ((-1)^(1/4)*a^2*at
anh((-1)^(1/4)*((a*x - 1)/(a*x + 1))^(1/4)))/4