3.67 \(\int \frac {e^{\frac {1}{2} \coth ^{-1}(a x)}}{x^4} \, dx\)

Optimal. Leaf size=356 \[ \frac {1}{12} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \left (\frac {1}{a x}+1\right )^{5/4}+\frac {3}{8} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}+\frac {3 a^3 \log \left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {\frac {1}{a x}+1}}-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{16 \sqrt {2}}-\frac {3 a^3 \log \left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {\frac {1}{a x}+1}}+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{16 \sqrt {2}}-\frac {3 a^3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{8 \sqrt {2}}+\frac {3 a^3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{8 \sqrt {2}}+\frac {a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (\frac {1}{a x}+1\right )^{5/4}}{3 x} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {6171, 90, 80, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} \[ \frac {1}{12} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \left (\frac {1}{a x}+1\right )^{5/4}+\frac {3}{8} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}+\frac {a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (\frac {1}{a x}+1\right )^{5/4}}{3 x}+\frac {3 a^3 \log \left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {\frac {1}{a x}+1}}-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{16 \sqrt {2}}-\frac {3 a^3 \log \left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {\frac {1}{a x}+1}}+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{16 \sqrt {2}}-\frac {3 a^3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}\right )}{8 \sqrt {2}}+\frac {3 a^3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{\frac {1}{a x}+1}}+1\right )}{8 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^3*(1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(1/4))/8 + (a^3*(1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(5/4))/12 + (a^2*(1
- 1/(a*x))^(3/4)*(1 + 1/(a*x))^(5/4))/(3*x) - (3*a^3*ArcTan[1 - (Sqrt[2]*(1 - 1/(a*x))^(1/4))/(1 + 1/(a*x))^(
1/4)])/(8*Sqrt[2]) + (3*a^3*ArcTan[1 + (Sqrt[2]*(1 - 1/(a*x))^(1/4))/(1 + 1/(a*x))^(1/4)])/(8*Sqrt[2]) + (3*a^
3*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)])/(16*Sqrt[2
]) - (3*a^3*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)])/
(16*Sqrt[2])

Rule 50

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 63

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 80

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 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[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 204

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

Rule 297

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 331

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 617

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 628

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 1162

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 1165

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 6171

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*} \int \frac {e^{\frac {1}{2} \coth ^{-1}(a x)}}{x^4} \, dx &=-\operatorname {Subst}\left (\int \frac {x^2 \sqrt [4]{1+\frac {x}{a}}}{\sqrt [4]{1-\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}{3 x}+\frac {1}{3} a^2 \operatorname {Subst}\left (\int \frac {\left (-1-\frac {x}{2 a}\right ) \sqrt [4]{1+\frac {x}{a}}}{\sqrt [4]{1-\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{12} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}+\frac {a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}{3 x}-\frac {1}{8} \left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt [4]{1+\frac {x}{a}}}{\sqrt [4]{1-\frac {x}{a}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {3}{8} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{12} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}+\frac {a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}{3 x}-\frac {1}{16} \left (3 a^2\right ) \operatorname {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 {3}{8} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{12} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}+\frac {a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}{3 x}+\frac {1}{4} \left (3 a^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-\frac {1}{a x}}\right )\\ &=\frac {3}{8} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{12} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}+\frac {a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}{3 x}+\frac {1}{4} \left (3 a^3\right ) \operatorname {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 {3}{8} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{12} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}+\frac {a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}{3 x}-\frac {1}{8} \left (3 a^3\right ) \operatorname {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}{8} \left (3 a^3\right ) \operatorname {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 {3}{8} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{12} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}+\frac {a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}{3 x}+\frac {1}{16} \left (3 a^3\right ) \operatorname {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}{16} \left (3 a^3\right ) \operatorname {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 {\left (3 a^3\right ) \operatorname {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 )}{16 \sqrt {2}}+\frac {\left (3 a^3\right ) \operatorname {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 )}{16 \sqrt {2}}\\ &=\frac {3}{8} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{12} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}+\frac {a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}{3 x}+\frac {3 a^3 \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 )}{16 \sqrt {2}}-\frac {3 a^3 \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 )}{16 \sqrt {2}}+\frac {\left (3 a^3\right ) \operatorname {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 )}{8 \sqrt {2}}-\frac {\left (3 a^3\right ) \operatorname {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 )}{8 \sqrt {2}}\\ &=\frac {3}{8} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}}+\frac {1}{12} a^3 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}+\frac {a^2 \left (1-\frac {1}{a x}\right )^{3/4} \left (1+\frac {1}{a x}\right )^{5/4}}{3 x}-\frac {3 a^3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}}+\frac {3 a^3 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-\frac {1}{a x}}}{\sqrt [4]{1+\frac {1}{a x}}}\right )}{8 \sqrt {2}}+\frac {3 a^3 \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 )}{16 \sqrt {2}}-\frac {3 a^3 \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 )}{16 \sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 93, normalized size = 0.26 \[ \frac {1}{96} a^3 \left (9 \text {RootSum}\left [\text {$\#$1}^4+1\& ,\frac {\coth ^{-1}(a x)-2 \log \left (e^{\frac {1}{2} \coth ^{-1}(a x)}-\text {$\#$1}\right )}{\text {$\#$1}^3}\& \right ]+\frac {8 e^{\frac {1}{2} \coth ^{-1}(a x)} \left (6 e^{2 \coth ^{-1}(a x)}+29 e^{4 \coth ^{-1}(a x)}+9\right )}{\left (e^{2 \coth ^{-1}(a x)}+1\right )^3}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^3*((8*E^(ArcCoth[a*x]/2)*(9 + 6*E^(2*ArcCoth[a*x]) + 29*E^(4*ArcCoth[a*x])))/(1 + E^(2*ArcCoth[a*x]))^3 + 9
*RootSum[1 + #1^4 & , (ArcCoth[a*x] - 2*Log[E^(ArcCoth[a*x]/2) - #1])/#1^3 & ]))/96

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fricas [A]  time = 0.42, size = 427, normalized size = 1.20 \[ -\frac {36 \, \sqrt {2} {\left (a^{12}\right )}^{\frac {1}{4}} x^{3} \arctan \left (-\frac {a^{12} + \sqrt {2} {\left (a^{12}\right )}^{\frac {1}{4}} a^{9} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - \sqrt {2} \sqrt {a^{18} \sqrt {\frac {a x - 1}{a x + 1}} + \sqrt {a^{12}} a^{12} + \sqrt {2} {\left (a^{12}\right )}^{\frac {3}{4}} a^{9} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}} {\left (a^{12}\right )}^{\frac {1}{4}}}{a^{12}}\right ) + 36 \, \sqrt {2} {\left (a^{12}\right )}^{\frac {1}{4}} x^{3} \arctan \left (\frac {a^{12} - \sqrt {2} {\left (a^{12}\right )}^{\frac {1}{4}} a^{9} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {2} \sqrt {a^{18} \sqrt {\frac {a x - 1}{a x + 1}} + \sqrt {a^{12}} a^{12} - \sqrt {2} {\left (a^{12}\right )}^{\frac {3}{4}} a^{9} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}} {\left (a^{12}\right )}^{\frac {1}{4}}}{a^{12}}\right ) + 9 \, \sqrt {2} {\left (a^{12}\right )}^{\frac {1}{4}} x^{3} \log \left (729 \, a^{18} \sqrt {\frac {a x - 1}{a x + 1}} + 729 \, \sqrt {a^{12}} a^{12} + 729 \, \sqrt {2} {\left (a^{12}\right )}^{\frac {3}{4}} a^{9} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 9 \, \sqrt {2} {\left (a^{12}\right )}^{\frac {1}{4}} x^{3} \log \left (729 \, a^{18} \sqrt {\frac {a x - 1}{a x + 1}} + 729 \, \sqrt {a^{12}} a^{12} - 729 \, \sqrt {2} {\left (a^{12}\right )}^{\frac {3}{4}} a^{9} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 4 \, {\left (11 \, a^{3} x^{3} + 21 \, a^{2} x^{2} + 18 \, a x + 8\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{96 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(36*sqrt(2)*(a^12)^(1/4)*x^3*arctan(-(a^12 + sqrt(2)*(a^12)^(1/4)*a^9*((a*x - 1)/(a*x + 1))^(1/4) - sqrt
(2)*sqrt(a^18*sqrt((a*x - 1)/(a*x + 1)) + sqrt(a^12)*a^12 + sqrt(2)*(a^12)^(3/4)*a^9*((a*x - 1)/(a*x + 1))^(1/
4))*(a^12)^(1/4))/a^12) + 36*sqrt(2)*(a^12)^(1/4)*x^3*arctan((a^12 - sqrt(2)*(a^12)^(1/4)*a^9*((a*x - 1)/(a*x
+ 1))^(1/4) + sqrt(2)*sqrt(a^18*sqrt((a*x - 1)/(a*x + 1)) + sqrt(a^12)*a^12 - sqrt(2)*(a^12)^(3/4)*a^9*((a*x -
1)/(a*x + 1))^(1/4))*(a^12)^(1/4))/a^12) + 9*sqrt(2)*(a^12)^(1/4)*x^3*log(729*a^18*sqrt((a*x - 1)/(a*x + 1))
+ 729*sqrt(a^12)*a^12 + 729*sqrt(2)*(a^12)^(3/4)*a^9*((a*x - 1)/(a*x + 1))^(1/4)) - 9*sqrt(2)*(a^12)^(1/4)*x^3
*log(729*a^18*sqrt((a*x - 1)/(a*x + 1)) + 729*sqrt(a^12)*a^12 - 729*sqrt(2)*(a^12)^(3/4)*a^9*((a*x - 1)/(a*x +
1))^(1/4)) - 4*(11*a^3*x^3 + 21*a^2*x^2 + 18*a*x + 8)*((a*x - 1)/(a*x + 1))^(3/4))/x^3

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giac [A]  time = 0.22, size = 271, normalized size = 0.76 \[ \frac {1}{96} \, {\left (18 \, \sqrt {2} a^{2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 18 \, \sqrt {2} a^{2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) - 9 \, \sqrt {2} a^{2} \log \left (\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + 9 \, \sqrt {2} a^{2} \log \left (-\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + \frac {8 \, {\left (\frac {6 \, {\left (a x - 1\right )} a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{a x + 1} + \frac {9 \, {\left (a x - 1\right )}^{2} a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{{\left (a x + 1\right )}^{2}} + 29 \, a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{{\left (\frac {a x - 1}{a x + 1} + 1\right )}^{3}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(18*sqrt(2)*a^2*arctan(1/2*sqrt(2)*(sqrt(2) + 2*((a*x - 1)/(a*x + 1))^(1/4))) + 18*sqrt(2)*a^2*arctan(-1/
2*sqrt(2)*(sqrt(2) - 2*((a*x - 1)/(a*x + 1))^(1/4))) - 9*sqrt(2)*a^2*log(sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) +
sqrt((a*x - 1)/(a*x + 1)) + 1) + 9*sqrt(2)*a^2*log(-sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt((a*x - 1)/(a*x
+ 1)) + 1) + 8*(6*(a*x - 1)*a^2*((a*x - 1)/(a*x + 1))^(3/4)/(a*x + 1) + 9*(a*x - 1)^2*a^2*((a*x - 1)/(a*x + 1
))^(3/4)/(a*x + 1)^2 + 29*a^2*((a*x - 1)/(a*x + 1))^(3/4))/((a*x - 1)/(a*x + 1) + 1)^3)*a

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (\frac {a x -1}{a x +1}\right )^{\frac {1}{4}} x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 270, normalized size = 0.76 \[ \frac {1}{96} \, {\left (9 \, {\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}\right ) - \sqrt {2} \log \left (\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + \sqrt {2} \log \left (-\sqrt {2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + \sqrt {\frac {a x - 1}{a x + 1}} + 1\right )\right )} a^{2} + \frac {8 \, {\left (9 \, a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {11}{4}} + 6 \, a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{4}} + 29 \, a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{\frac {3 \, {\left (a x - 1\right )}}{a x + 1} + \frac {3 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {{\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} + 1}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(9*(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^2 +
8*(9*a^2*((a*x - 1)/(a*x + 1))^(11/4) + 6*a^2*((a*x - 1)/(a*x + 1))^(7/4) + 29*a^2*((a*x - 1)/(a*x + 1))^(3/4)
)/(3*(a*x - 1)/(a*x + 1) + 3*(a*x - 1)^2/(a*x + 1)^2 + (a*x - 1)^3/(a*x + 1)^3 + 1))*a

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mupad [B]  time = 1.21, size = 168, normalized size = 0.47 \[ \frac {\frac {29\,a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/4}}{12}+\frac {a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/4}}{2}+\frac {3\,a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{11/4}}{4}}{\frac {3\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}+\frac {{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}+\frac {3\,\left (a\,x-1\right )}{a\,x+1}+1}+\frac {3\,{\left (-1\right )}^{1/4}\,a^3\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8}-\frac {3\,{\left (-1\right )}^{1/4}\,a^3\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((29*a^3*((a*x - 1)/(a*x + 1))^(3/4))/12 + (a^3*((a*x - 1)/(a*x + 1))^(7/4))/2 + (3*a^3*((a*x - 1)/(a*x + 1))^
(11/4))/4)/((3*(a*x - 1)^2)/(a*x + 1)^2 + (a*x - 1)^3/(a*x + 1)^3 + (3*(a*x - 1))/(a*x + 1) + 1) + (3*(-1)^(1/
4)*a^3*atan((-1)^(1/4)*((a*x - 1)/(a*x + 1))^(1/4)))/8 - (3*(-1)^(1/4)*a^3*atanh((-1)^(1/4)*((a*x - 1)/(a*x +
1))^(1/4)))/8

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{4} \sqrt [4]{\frac {a x - 1}{a x + 1}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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