### 3.84 $$\int \frac{e^{\frac{5}{2} \coth ^{-1}(a x)}}{x^3} \, dx$$

Optimal. Leaf size=351 $-\frac{2 a^2 \left (\frac{1}{a x}+1\right )^{9/4}}{\sqrt [4]{1-\frac{1}{a x}}}-\frac{5}{2} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \left (\frac{1}{a x}+1\right )^{5/4}-\frac{25}{4} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{\frac{1}{a x}+1}-\frac{25 a^2 \log \left (\frac{\sqrt{1-\frac{1}{a x}}}{\sqrt{\frac{1}{a x}+1}}-\frac{\sqrt{2} \sqrt [4]{1-\frac{1}{a x}}}{\sqrt [4]{\frac{1}{a x}+1}}+1\right )}{8 \sqrt{2}}+\frac{25 a^2 \log \left (\frac{\sqrt{1-\frac{1}{a x}}}{\sqrt{\frac{1}{a x}+1}}+\frac{\sqrt{2} \sqrt [4]{1-\frac{1}{a x}}}{\sqrt [4]{\frac{1}{a x}+1}}+1\right )}{8 \sqrt{2}}+\frac{25 a^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-\frac{1}{a x}}}{\sqrt [4]{\frac{1}{a x}+1}}\right )}{4 \sqrt{2}}-\frac{25 a^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{1-\frac{1}{a x}}}{\sqrt [4]{\frac{1}{a x}+1}}+1\right )}{4 \sqrt{2}}$

[Out]

(-25*a^2*(1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(1/4))/4 - (5*a^2*(1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(5/4))/2 - (2*a
^2*(1 + 1/(a*x))^(9/4))/(1 - 1/(a*x))^(1/4) + (25*a^2*ArcTan[1 - (Sqrt[2]*(1 - 1/(a*x))^(1/4))/(1 + 1/(a*x))^(
1/4)])/(4*Sqrt[2]) - (25*a^2*ArcTan[1 + (Sqrt[2]*(1 - 1/(a*x))^(1/4))/(1 + 1/(a*x))^(1/4)])/(4*Sqrt[2]) - (25*
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]) + (25*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])

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Rubi [A]  time = 0.284915, antiderivative size = 351, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.786, Rules used = {6171, 78, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} $-\frac{2 a^2 \left (\frac{1}{a x}+1\right )^{9/4}}{\sqrt [4]{1-\frac{1}{a x}}}-\frac{5}{2} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \left (\frac{1}{a x}+1\right )^{5/4}-\frac{25}{4} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{\frac{1}{a x}+1}-\frac{25 a^2 \log \left (\frac{\sqrt{1-\frac{1}{a x}}}{\sqrt{\frac{1}{a x}+1}}-\frac{\sqrt{2} \sqrt [4]{1-\frac{1}{a x}}}{\sqrt [4]{\frac{1}{a x}+1}}+1\right )}{8 \sqrt{2}}+\frac{25 a^2 \log \left (\frac{\sqrt{1-\frac{1}{a x}}}{\sqrt{\frac{1}{a x}+1}}+\frac{\sqrt{2} \sqrt [4]{1-\frac{1}{a x}}}{\sqrt [4]{\frac{1}{a x}+1}}+1\right )}{8 \sqrt{2}}+\frac{25 a^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-\frac{1}{a x}}}{\sqrt [4]{\frac{1}{a x}+1}}\right )}{4 \sqrt{2}}-\frac{25 a^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{1-\frac{1}{a x}}}{\sqrt [4]{\frac{1}{a x}+1}}+1\right )}{4 \sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^((5*ArcCoth[a*x])/2)/x^3,x]

[Out]

(-25*a^2*(1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(1/4))/4 - (5*a^2*(1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(5/4))/2 - (2*a
^2*(1 + 1/(a*x))^(9/4))/(1 - 1/(a*x))^(1/4) + (25*a^2*ArcTan[1 - (Sqrt[2]*(1 - 1/(a*x))^(1/4))/(1 + 1/(a*x))^(
1/4)])/(4*Sqrt[2]) - (25*a^2*ArcTan[1 + (Sqrt[2]*(1 - 1/(a*x))^(1/4))/(1 + 1/(a*x))^(1/4)])/(4*Sqrt[2]) - (25*
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]) + (25*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 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]

Rule 78

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

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 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 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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{e^{\frac{5}{2} \coth ^{-1}(a x)}}{x^3} \, dx &=-\operatorname{Subst}\left (\int \frac{x \left (1+\frac{x}{a}\right )^{5/4}}{\left (1-\frac{x}{a}\right )^{5/4}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{2 a^2 \left (1+\frac{1}{a x}\right )^{9/4}}{\sqrt [4]{1-\frac{1}{a x}}}+(5 a) \operatorname{Subst}\left (\int \frac{\left (1+\frac{x}{a}\right )^{5/4}}{\sqrt [4]{1-\frac{x}{a}}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{5}{2} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \left (1+\frac{1}{a x}\right )^{5/4}-\frac{2 a^2 \left (1+\frac{1}{a x}\right )^{9/4}}{\sqrt [4]{1-\frac{1}{a x}}}+\frac{1}{4} (25 a) \operatorname{Subst}\left (\int \frac{\sqrt [4]{1+\frac{x}{a}}}{\sqrt [4]{1-\frac{x}{a}}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{25}{4} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{1+\frac{1}{a x}}-\frac{5}{2} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \left (1+\frac{1}{a x}\right )^{5/4}-\frac{2 a^2 \left (1+\frac{1}{a x}\right )^{9/4}}{\sqrt [4]{1-\frac{1}{a x}}}+\frac{1}{8} (25 a) \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{25}{4} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{1+\frac{1}{a x}}-\frac{5}{2} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \left (1+\frac{1}{a x}\right )^{5/4}-\frac{2 a^2 \left (1+\frac{1}{a x}\right )^{9/4}}{\sqrt [4]{1-\frac{1}{a x}}}-\frac{1}{2} \left (25 a^2\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{25}{4} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{1+\frac{1}{a x}}-\frac{5}{2} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \left (1+\frac{1}{a x}\right )^{5/4}-\frac{2 a^2 \left (1+\frac{1}{a x}\right )^{9/4}}{\sqrt [4]{1-\frac{1}{a x}}}-\frac{1}{2} \left (25 a^2\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{25}{4} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{1+\frac{1}{a x}}-\frac{5}{2} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \left (1+\frac{1}{a x}\right )^{5/4}-\frac{2 a^2 \left (1+\frac{1}{a x}\right )^{9/4}}{\sqrt [4]{1-\frac{1}{a x}}}+\frac{1}{4} \left (25 a^2\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}{4} \left (25 a^2\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{25}{4} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{1+\frac{1}{a x}}-\frac{5}{2} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \left (1+\frac{1}{a x}\right )^{5/4}-\frac{2 a^2 \left (1+\frac{1}{a x}\right )^{9/4}}{\sqrt [4]{1-\frac{1}{a x}}}-\frac{1}{8} \left (25 a^2\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}{8} \left (25 a^2\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 (25 a^2\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 )}{8 \sqrt{2}}-\frac{\left (25 a^2\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 )}{8 \sqrt{2}}\\ &=-\frac{25}{4} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{1+\frac{1}{a x}}-\frac{5}{2} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \left (1+\frac{1}{a x}\right )^{5/4}-\frac{2 a^2 \left (1+\frac{1}{a x}\right )^{9/4}}{\sqrt [4]{1-\frac{1}{a x}}}-\frac{25 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{25 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{\left (25 a^2\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 )}{4 \sqrt{2}}+\frac{\left (25 a^2\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 )}{4 \sqrt{2}}\\ &=-\frac{25}{4} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{1+\frac{1}{a x}}-\frac{5}{2} a^2 \left (1-\frac{1}{a x}\right )^{3/4} \left (1+\frac{1}{a x}\right )^{5/4}-\frac{2 a^2 \left (1+\frac{1}{a x}\right )^{9/4}}{\sqrt [4]{1-\frac{1}{a x}}}+\frac{25 a^2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-\frac{1}{a x}}}{\sqrt [4]{1+\frac{1}{a x}}}\right )}{4 \sqrt{2}}-\frac{25 a^2 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-\frac{1}{a x}}}{\sqrt [4]{1+\frac{1}{a x}}}\right )}{4 \sqrt{2}}-\frac{25 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{25 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]  time = 0.2621, size = 186, normalized size = 0.53 $\frac{1}{16} a^2 \left (-128 e^{\frac{1}{2} \coth ^{-1}(a x)}-\frac{104 e^{\frac{1}{2} \coth ^{-1}(a x)}}{e^{2 \coth ^{-1}(a x)}+1}+\frac{32 e^{\frac{1}{2} \coth ^{-1}(a x)}}{\left (e^{2 \coth ^{-1}(a x)}+1\right )^2}-25 \sqrt{2} \log \left (-\sqrt{2} e^{\frac{1}{2} \coth ^{-1}(a x)}+e^{\coth ^{-1}(a x)}+1\right )+25 \sqrt{2} \log \left (\sqrt{2} e^{\frac{1}{2} \coth ^{-1}(a x)}+e^{\coth ^{-1}(a x)}+1\right )-50 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} e^{\frac{1}{2} \coth ^{-1}(a x)}\right )+50 \sqrt{2} \tan ^{-1}\left (\sqrt{2} e^{\frac{1}{2} \coth ^{-1}(a x)}+1\right )\right )$

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((5*ArcCoth[a*x])/2)/x^3,x]

[Out]

(a^2*(-128*E^(ArcCoth[a*x]/2) + (32*E^(ArcCoth[a*x]/2))/(1 + E^(2*ArcCoth[a*x]))^2 - (104*E^(ArcCoth[a*x]/2))/
(1 + E^(2*ArcCoth[a*x])) - 50*Sqrt[2]*ArcTan[1 - Sqrt[2]*E^(ArcCoth[a*x]/2)] + 50*Sqrt[2]*ArcTan[1 + Sqrt[2]*E
^(ArcCoth[a*x]/2)] - 25*Sqrt[2]*Log[1 - Sqrt[2]*E^(ArcCoth[a*x]/2) + E^ArcCoth[a*x]] + 25*Sqrt[2]*Log[1 + Sqrt
[2]*E^(ArcCoth[a*x]/2) + E^ArcCoth[a*x]]))/16

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Maple [F]  time = 0.336, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}} \left ({\frac{ax-1}{ax+1}} \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(5/4)/x^3,x)

[Out]

int(1/((a*x-1)/(a*x+1))^(5/4)/x^3,x)

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Maxima [A]  time = 1.52576, size = 329, normalized size = 0.94 \begin{align*} -\frac{1}{16} \,{\left (25 \,{\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 (\frac{45 \,{\left (a x - 1\right )} a}{a x + 1} + \frac{25 \,{\left (a x - 1\right )}^{2} a}{{\left (a x + 1\right )}^{2}} + 16 \, a\right )}}{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{9}{4}} + 2 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{5}{4}} + \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}}\right )} a \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(5/4)/x^3,x, algorithm="maxima")

[Out]

-1/16*(25*(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*sqr
t(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*(45*(a*x - 1)*a/(a*x + 1) + 25*(a*x - 1)^2*a/(a*x + 1)^2 + 16*a)/(((a*x - 1)/(a*x + 1))^(9/4) + 2*((a*x - 1)
/(a*x + 1))^(5/4) + ((a*x - 1)/(a*x + 1))^(1/4)))*a

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Fricas [A]  time = 1.7602, size = 1247, normalized size = 3.55 \begin{align*} \frac{100 \, \sqrt{2}{\left (a^{8}\right )}^{\frac{1}{4}}{\left (a x^{3} - x^{2}\right )} \arctan \left (-\frac{a^{8} + \sqrt{2}{\left (a^{8}\right )}^{\frac{1}{4}} a^{6} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} - \sqrt{2} \sqrt{a^{12} \sqrt{\frac{a x - 1}{a x + 1}} + \sqrt{a^{8}} a^{8} + \sqrt{2}{\left (a^{8}\right )}^{\frac{3}{4}} a^{6} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}}{\left (a^{8}\right )}^{\frac{1}{4}}}{a^{8}}\right ) + 100 \, \sqrt{2}{\left (a^{8}\right )}^{\frac{1}{4}}{\left (a x^{3} - x^{2}\right )} \arctan \left (\frac{a^{8} - \sqrt{2}{\left (a^{8}\right )}^{\frac{1}{4}} a^{6} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} + \sqrt{2} \sqrt{a^{12} \sqrt{\frac{a x - 1}{a x + 1}} + \sqrt{a^{8}} a^{8} - \sqrt{2}{\left (a^{8}\right )}^{\frac{3}{4}} a^{6} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}}{\left (a^{8}\right )}^{\frac{1}{4}}}{a^{8}}\right ) + 25 \, \sqrt{2}{\left (a^{8}\right )}^{\frac{1}{4}}{\left (a x^{3} - x^{2}\right )} \log \left (244140625 \, a^{12} \sqrt{\frac{a x - 1}{a x + 1}} + 244140625 \, \sqrt{a^{8}} a^{8} + 244140625 \, \sqrt{2}{\left (a^{8}\right )}^{\frac{3}{4}} a^{6} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right ) - 25 \, \sqrt{2}{\left (a^{8}\right )}^{\frac{1}{4}}{\left (a x^{3} - x^{2}\right )} \log \left (244140625 \, a^{12} \sqrt{\frac{a x - 1}{a x + 1}} + 244140625 \, \sqrt{a^{8}} a^{8} - 244140625 \, \sqrt{2}{\left (a^{8}\right )}^{\frac{3}{4}} a^{6} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right ) - 4 \,{\left (43 \, a^{3} x^{3} + 34 \, a^{2} x^{2} - 11 \, a x - 2\right )} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{4}}}{16 \,{\left (a x^{3} - x^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(5/4)/x^3,x, algorithm="fricas")

[Out]

1/16*(100*sqrt(2)*(a^8)^(1/4)*(a*x^3 - x^2)*arctan(-(a^8 + sqrt(2)*(a^8)^(1/4)*a^6*((a*x - 1)/(a*x + 1))^(1/4)
- sqrt(2)*sqrt(a^12*sqrt((a*x - 1)/(a*x + 1)) + sqrt(a^8)*a^8 + sqrt(2)*(a^8)^(3/4)*a^6*((a*x - 1)/(a*x + 1))
^(1/4))*(a^8)^(1/4))/a^8) + 100*sqrt(2)*(a^8)^(1/4)*(a*x^3 - x^2)*arctan((a^8 - sqrt(2)*(a^8)^(1/4)*a^6*((a*x
- 1)/(a*x + 1))^(1/4) + sqrt(2)*sqrt(a^12*sqrt((a*x - 1)/(a*x + 1)) + sqrt(a^8)*a^8 - sqrt(2)*(a^8)^(3/4)*a^6*
((a*x - 1)/(a*x + 1))^(1/4))*(a^8)^(1/4))/a^8) + 25*sqrt(2)*(a^8)^(1/4)*(a*x^3 - x^2)*log(244140625*a^12*sqrt(
(a*x - 1)/(a*x + 1)) + 244140625*sqrt(a^8)*a^8 + 244140625*sqrt(2)*(a^8)^(3/4)*a^6*((a*x - 1)/(a*x + 1))^(1/4)
) - 25*sqrt(2)*(a^8)^(1/4)*(a*x^3 - x^2)*log(244140625*a^12*sqrt((a*x - 1)/(a*x + 1)) + 244140625*sqrt(a^8)*a^
8 - 244140625*sqrt(2)*(a^8)^(3/4)*a^6*((a*x - 1)/(a*x + 1))^(1/4)) - 4*(43*a^3*x^3 + 34*a^2*x^2 - 11*a*x - 2)*
((a*x - 1)/(a*x + 1))^(3/4))/(a*x^3 - x^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))**(5/4)/x**3,x)

[Out]

Timed out

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Giac [A]  time = 1.18918, size = 328, normalized size = 0.93 \begin{align*} -\frac{1}{16} \,{\left (50 \, \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 ) + 50 \, \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 ) - 25 \, \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 ) + 25 \, \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{128 \, a}{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}} + \frac{8 \,{\left (\frac{9 \,{\left (a x - 1\right )} a \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{4}}}{a x + 1} + 13 \, 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 \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(5/4)/x^3,x, algorithm="giac")

[Out]

-1/16*(50*sqrt(2)*a*arctan(1/2*sqrt(2)*(sqrt(2) + 2*((a*x - 1)/(a*x + 1))^(1/4))) + 50*sqrt(2)*a*arctan(-1/2*s
qrt(2)*(sqrt(2) - 2*((a*x - 1)/(a*x + 1))^(1/4))) - 25*sqrt(2)*a*log(sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqr
t((a*x - 1)/(a*x + 1)) + 1) + 25*sqrt(2)*a*log(-sqrt(2)*((a*x - 1)/(a*x + 1))^(1/4) + sqrt((a*x - 1)/(a*x + 1)
) + 1) + 128*a/((a*x - 1)/(a*x + 1))^(1/4) + 8*(9*(a*x - 1)*a*((a*x - 1)/(a*x + 1))^(3/4)/(a*x + 1) + 13*a*((a
*x - 1)/(a*x + 1))^(3/4))/((a*x - 1)/(a*x + 1) + 1)^2)*a