3.281 \(\int \frac{(a+b \coth ^{-1}(c x)) (d+e \log (f+g x^2))}{x^2} \, dx\)

Optimal. Leaf size=560 \[ -\frac{1}{2} b c e \text{PolyLog}\left (2,\frac{c^2 \left (f+g x^2\right )}{c^2 f+g}\right )+\frac{1}{2} b c e \text{PolyLog}\left (2,\frac{g x^2}{f}+1\right )-\frac{i b e \sqrt{g} \text{PolyLog}\left (2,1+\frac{2 \sqrt{f} \sqrt{g} (1-c x)}{\left (-\sqrt{g}+i c \sqrt{f}\right ) \left (\sqrt{f}-i \sqrt{g} x\right )}\right )}{2 \sqrt{f}}+\frac{i b e \sqrt{g} \text{PolyLog}\left (2,1-\frac{2 \sqrt{f} \sqrt{g} (c x+1)}{\left (\sqrt{g}+i c \sqrt{f}\right ) \left (\sqrt{f}-i \sqrt{g} x\right )}\right )}{2 \sqrt{f}}-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x}+\frac{2 a e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}-\frac{1}{2} b c \log \left (\frac{g \left (1-c^2 x^2\right )}{c^2 f+g}\right ) \left (d+e \log \left (f+g x^2\right )\right )+\frac{1}{2} b c \log \left (-\frac{g x^2}{f}\right ) \left (d+e \log \left (f+g x^2\right )\right )-\frac{b e \sqrt{g} \log \left (1-\frac{1}{c x}\right ) \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}+\frac{b e \sqrt{g} \log \left (\frac{1}{c x}+1\right ) \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}+\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (-\frac{2 \sqrt{f} \sqrt{g} (1-c x)}{\left (-\sqrt{g}+i c \sqrt{f}\right ) \left (\sqrt{f}-i \sqrt{g} x\right )}\right )}{\sqrt{f}}-\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (\frac{2 \sqrt{f} \sqrt{g} (c x+1)}{\left (\sqrt{g}+i c \sqrt{f}\right ) \left (\sqrt{f}-i \sqrt{g} x\right )}\right )}{\sqrt{f}} \]

[Out]

(2*a*e*Sqrt[g]*ArcTan[(Sqrt[g]*x)/Sqrt[f]])/Sqrt[f] - (b*e*Sqrt[g]*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[1 - 1/(c*x)
])/Sqrt[f] + (b*e*Sqrt[g]*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[1 + 1/(c*x)])/Sqrt[f] + (b*e*Sqrt[g]*ArcTan[(Sqrt[g]
*x)/Sqrt[f]]*Log[(-2*Sqrt[f]*Sqrt[g]*(1 - c*x))/((I*c*Sqrt[f] - Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x))])/Sqrt[f] -
(b*e*Sqrt[g]*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[(2*Sqrt[f]*Sqrt[g]*(1 + c*x))/((I*c*Sqrt[f] + Sqrt[g])*(Sqrt[f] -
I*Sqrt[g]*x))])/Sqrt[f] - ((a + b*ArcCoth[c*x])*(d + e*Log[f + g*x^2]))/x + (b*c*Log[-((g*x^2)/f)]*(d + e*Log
[f + g*x^2]))/2 - (b*c*Log[(g*(1 - c^2*x^2))/(c^2*f + g)]*(d + e*Log[f + g*x^2]))/2 - (b*c*e*PolyLog[2, (c^2*(
f + g*x^2))/(c^2*f + g)])/2 + (b*c*e*PolyLog[2, 1 + (g*x^2)/f])/2 - ((I/2)*b*e*Sqrt[g]*PolyLog[2, 1 + (2*Sqrt[
f]*Sqrt[g]*(1 - c*x))/((I*c*Sqrt[f] - Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x))])/Sqrt[f] + ((I/2)*b*e*Sqrt[g]*PolyLog
[2, 1 - (2*Sqrt[f]*Sqrt[g]*(1 + c*x))/((I*c*Sqrt[f] + Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x))])/Sqrt[f]

________________________________________________________________________________________

Rubi [A]  time = 1.26144, antiderivative size = 560, normalized size of antiderivative = 1., number of steps used = 38, number of rules used = 22, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.917, Rules used = {6082, 2475, 36, 29, 31, 2416, 2394, 2315, 2393, 2391, 5975, 205, 5973, 2470, 12, 260, 6688, 4876, 4848, 4856, 2402, 2447} \[ -\frac{1}{2} b c e \text{PolyLog}\left (2,\frac{c^2 \left (f+g x^2\right )}{c^2 f+g}\right )+\frac{1}{2} b c e \text{PolyLog}\left (2,\frac{g x^2}{f}+1\right )-\frac{i b e \sqrt{g} \text{PolyLog}\left (2,1+\frac{2 \sqrt{f} \sqrt{g} (1-c x)}{\left (-\sqrt{g}+i c \sqrt{f}\right ) \left (\sqrt{f}-i \sqrt{g} x\right )}\right )}{2 \sqrt{f}}+\frac{i b e \sqrt{g} \text{PolyLog}\left (2,1-\frac{2 \sqrt{f} \sqrt{g} (c x+1)}{\left (\sqrt{g}+i c \sqrt{f}\right ) \left (\sqrt{f}-i \sqrt{g} x\right )}\right )}{2 \sqrt{f}}-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x}+\frac{2 a e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}-\frac{1}{2} b c \log \left (\frac{g \left (1-c^2 x^2\right )}{c^2 f+g}\right ) \left (d+e \log \left (f+g x^2\right )\right )+\frac{1}{2} b c \log \left (-\frac{g x^2}{f}\right ) \left (d+e \log \left (f+g x^2\right )\right )-\frac{b e \sqrt{g} \log \left (1-\frac{1}{c x}\right ) \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}+\frac{b e \sqrt{g} \log \left (\frac{1}{c x}+1\right ) \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}+\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (-\frac{2 \sqrt{f} \sqrt{g} (1-c x)}{\left (-\sqrt{g}+i c \sqrt{f}\right ) \left (\sqrt{f}-i \sqrt{g} x\right )}\right )}{\sqrt{f}}-\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (\frac{2 \sqrt{f} \sqrt{g} (c x+1)}{\left (\sqrt{g}+i c \sqrt{f}\right ) \left (\sqrt{f}-i \sqrt{g} x\right )}\right )}{\sqrt{f}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*ArcCoth[c*x])*(d + e*Log[f + g*x^2]))/x^2,x]

[Out]

(2*a*e*Sqrt[g]*ArcTan[(Sqrt[g]*x)/Sqrt[f]])/Sqrt[f] - (b*e*Sqrt[g]*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[1 - 1/(c*x)
])/Sqrt[f] + (b*e*Sqrt[g]*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[1 + 1/(c*x)])/Sqrt[f] + (b*e*Sqrt[g]*ArcTan[(Sqrt[g]
*x)/Sqrt[f]]*Log[(-2*Sqrt[f]*Sqrt[g]*(1 - c*x))/((I*c*Sqrt[f] - Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x))])/Sqrt[f] -
(b*e*Sqrt[g]*ArcTan[(Sqrt[g]*x)/Sqrt[f]]*Log[(2*Sqrt[f]*Sqrt[g]*(1 + c*x))/((I*c*Sqrt[f] + Sqrt[g])*(Sqrt[f] -
I*Sqrt[g]*x))])/Sqrt[f] - ((a + b*ArcCoth[c*x])*(d + e*Log[f + g*x^2]))/x + (b*c*Log[-((g*x^2)/f)]*(d + e*Log
[f + g*x^2]))/2 - (b*c*Log[(g*(1 - c^2*x^2))/(c^2*f + g)]*(d + e*Log[f + g*x^2]))/2 - (b*c*e*PolyLog[2, (c^2*(
f + g*x^2))/(c^2*f + g)])/2 + (b*c*e*PolyLog[2, 1 + (g*x^2)/f])/2 - ((I/2)*b*e*Sqrt[g]*PolyLog[2, 1 + (2*Sqrt[
f]*Sqrt[g]*(1 - c*x))/((I*c*Sqrt[f] - Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x))])/Sqrt[f] + ((I/2)*b*e*Sqrt[g]*PolyLog
[2, 1 - (2*Sqrt[f]*Sqrt[g]*(1 + c*x))/((I*c*Sqrt[f] + Sqrt[g])*(Sqrt[f] - I*Sqrt[g]*x))])/Sqrt[f]

Rule 6082

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> Sim
p[(x^(m + 1)*(d + e*Log[f + g*x^2])*(a + b*ArcCoth[c*x]))/(m + 1), x] + (-Dist[(b*c)/(m + 1), Int[(x^(m + 1)*(
d + e*Log[f + g*x^2]))/(1 - c^2*x^2), x], x] - Dist[(2*e*g)/(m + 1), Int[(x^(m + 2)*(a + b*ArcCoth[c*x]))/(f +
g*x^2), x], x]) /; FreeQ[{a, b, c, d, e, f, g}, x] && ILtQ[m/2, 0]

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 5975

Int[(ArcCoth[(c_.)*(x_)]*(b_.) + (a_))/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Dist[a, Int[1/(d + e*x^2), x], x]
+ Dist[b, Int[ArcCoth[c*x]/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]

Rule 205

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

Rule 5973

Int[ArcCoth[(c_.)*(x_)]/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[Log[1 + 1/(c*x)]/(d + e*x^2), x], x
] - Dist[1/2, Int[Log[1 - 1/(c*x)]/(d + e*x^2), x], x] /; FreeQ[{c, d, e}, x]

Rule 2470

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_) + (g_.)*(x_)^2), x_Symbol] :> With[{u = In
tHide[1/(f + g*x^2), x]}, Simp[u*(a + b*Log[c*(d + e*x^n)^p]), x] - Dist[b*e*n*p, Int[(u*x^(n - 1))/(d + e*x^n
), x], x]] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IntegerQ[n]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 4876

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[Ex
pandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p,
0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 4856

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])*Log[2/(1 -
I*c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d
+ e*x))/((c*d + I*e)*(1 - I*c*x))]/(1 + c^2*x^2), x], x] + Simp[((a + b*ArcTan[c*x])*Log[(2*c*(d + e*x))/((c*d
+ I*e)*(1 - I*c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

\begin{align*} \int \frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x^2} \, dx &=-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x}+(b c) \int \frac{d+e \log \left (f+g x^2\right )}{x \left (1-c^2 x^2\right )} \, dx+(2 e g) \int \frac{a+b \coth ^{-1}(c x)}{f+g x^2} \, dx\\ &=-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x}+\frac{1}{2} (b c) \operatorname{Subst}\left (\int \frac{d+e \log (f+g x)}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )+(2 a e g) \int \frac{1}{f+g x^2} \, dx+(2 b e g) \int \frac{\coth ^{-1}(c x)}{f+g x^2} \, dx\\ &=\frac{2 a e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x}+\frac{1}{2} (b c) \operatorname{Subst}\left (\int \left (\frac{d+e \log (f+g x)}{x}-\frac{c^2 (d+e \log (f+g x))}{-1+c^2 x}\right ) \, dx,x,x^2\right )-(b e g) \int \frac{\log \left (1-\frac{1}{c x}\right )}{f+g x^2} \, dx+(b e g) \int \frac{\log \left (1+\frac{1}{c x}\right )}{f+g x^2} \, dx\\ &=\frac{2 a e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}-\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1-\frac{1}{c x}\right )}{\sqrt{f}}+\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1+\frac{1}{c x}\right )}{\sqrt{f}}-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x}+\frac{1}{2} (b c) \operatorname{Subst}\left (\int \frac{d+e \log (f+g x)}{x} \, dx,x,x^2\right )-\frac{1}{2} \left (b c^3\right ) \operatorname{Subst}\left (\int \frac{d+e \log (f+g x)}{-1+c^2 x} \, dx,x,x^2\right )+\frac{(b e g) \int \frac{\tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f} \sqrt{g} \left (1-\frac{1}{c x}\right ) x^2} \, dx}{c}+\frac{(b e g) \int \frac{\tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f} \sqrt{g} \left (1+\frac{1}{c x}\right ) x^2} \, dx}{c}\\ &=\frac{2 a e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}-\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1-\frac{1}{c x}\right )}{\sqrt{f}}+\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1+\frac{1}{c x}\right )}{\sqrt{f}}-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x}+\frac{1}{2} b c \log \left (-\frac{g x^2}{f}\right ) \left (d+e \log \left (f+g x^2\right )\right )-\frac{1}{2} b c \log \left (\frac{g \left (1-c^2 x^2\right )}{c^2 f+g}\right ) \left (d+e \log \left (f+g x^2\right )\right )+\frac{\left (b e \sqrt{g}\right ) \int \frac{\tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\left (1-\frac{1}{c x}\right ) x^2} \, dx}{c \sqrt{f}}+\frac{\left (b e \sqrt{g}\right ) \int \frac{\tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\left (1+\frac{1}{c x}\right ) x^2} \, dx}{c \sqrt{f}}-\frac{1}{2} (b c e g) \operatorname{Subst}\left (\int \frac{\log \left (-\frac{g x}{f}\right )}{f+g x} \, dx,x,x^2\right )+\frac{1}{2} (b c e g) \operatorname{Subst}\left (\int \frac{\log \left (\frac{g \left (-1+c^2 x\right )}{-c^2 f-g}\right )}{f+g x} \, dx,x,x^2\right )\\ &=\frac{2 a e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}-\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1-\frac{1}{c x}\right )}{\sqrt{f}}+\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1+\frac{1}{c x}\right )}{\sqrt{f}}-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x}+\frac{1}{2} b c \log \left (-\frac{g x^2}{f}\right ) \left (d+e \log \left (f+g x^2\right )\right )-\frac{1}{2} b c \log \left (\frac{g \left (1-c^2 x^2\right )}{c^2 f+g}\right ) \left (d+e \log \left (f+g x^2\right )\right )+\frac{1}{2} b c e \text{Li}_2\left (1+\frac{g x^2}{f}\right )+\frac{1}{2} (b c e) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{c^2 x}{-c^2 f-g}\right )}{x} \, dx,x,f+g x^2\right )+\frac{\left (b e \sqrt{g}\right ) \int \frac{c \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{x (-1+c x)} \, dx}{c \sqrt{f}}+\frac{\left (b e \sqrt{g}\right ) \int \frac{c \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{x (1+c x)} \, dx}{c \sqrt{f}}\\ &=\frac{2 a e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}-\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1-\frac{1}{c x}\right )}{\sqrt{f}}+\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1+\frac{1}{c x}\right )}{\sqrt{f}}-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x}+\frac{1}{2} b c \log \left (-\frac{g x^2}{f}\right ) \left (d+e \log \left (f+g x^2\right )\right )-\frac{1}{2} b c \log \left (\frac{g \left (1-c^2 x^2\right )}{c^2 f+g}\right ) \left (d+e \log \left (f+g x^2\right )\right )-\frac{1}{2} b c e \text{Li}_2\left (\frac{c^2 \left (f+g x^2\right )}{c^2 f+g}\right )+\frac{1}{2} b c e \text{Li}_2\left (1+\frac{g x^2}{f}\right )+\frac{\left (b e \sqrt{g}\right ) \int \frac{\tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{x (-1+c x)} \, dx}{\sqrt{f}}+\frac{\left (b e \sqrt{g}\right ) \int \frac{\tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{x (1+c x)} \, dx}{\sqrt{f}}\\ &=\frac{2 a e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}-\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1-\frac{1}{c x}\right )}{\sqrt{f}}+\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1+\frac{1}{c x}\right )}{\sqrt{f}}-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x}+\frac{1}{2} b c \log \left (-\frac{g x^2}{f}\right ) \left (d+e \log \left (f+g x^2\right )\right )-\frac{1}{2} b c \log \left (\frac{g \left (1-c^2 x^2\right )}{c^2 f+g}\right ) \left (d+e \log \left (f+g x^2\right )\right )-\frac{1}{2} b c e \text{Li}_2\left (\frac{c^2 \left (f+g x^2\right )}{c^2 f+g}\right )+\frac{1}{2} b c e \text{Li}_2\left (1+\frac{g x^2}{f}\right )+\frac{\left (b e \sqrt{g}\right ) \int \left (-\frac{\tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{x}+\frac{c \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{-1+c x}\right ) \, dx}{\sqrt{f}}+\frac{\left (b e \sqrt{g}\right ) \int \left (\frac{\tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{x}-\frac{c \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{1+c x}\right ) \, dx}{\sqrt{f}}\\ &=\frac{2 a e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}-\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1-\frac{1}{c x}\right )}{\sqrt{f}}+\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1+\frac{1}{c x}\right )}{\sqrt{f}}-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x}+\frac{1}{2} b c \log \left (-\frac{g x^2}{f}\right ) \left (d+e \log \left (f+g x^2\right )\right )-\frac{1}{2} b c \log \left (\frac{g \left (1-c^2 x^2\right )}{c^2 f+g}\right ) \left (d+e \log \left (f+g x^2\right )\right )-\frac{1}{2} b c e \text{Li}_2\left (\frac{c^2 \left (f+g x^2\right )}{c^2 f+g}\right )+\frac{1}{2} b c e \text{Li}_2\left (1+\frac{g x^2}{f}\right )+\frac{\left (b c e \sqrt{g}\right ) \int \frac{\tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{-1+c x} \, dx}{\sqrt{f}}-\frac{\left (b c e \sqrt{g}\right ) \int \frac{\tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{1+c x} \, dx}{\sqrt{f}}\\ &=\frac{2 a e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}-\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1-\frac{1}{c x}\right )}{\sqrt{f}}+\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1+\frac{1}{c x}\right )}{\sqrt{f}}+\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (-\frac{2 \sqrt{f} \sqrt{g} (1-c x)}{\left (i c \sqrt{f}-\sqrt{g}\right ) \left (\sqrt{f}-i \sqrt{g} x\right )}\right )}{\sqrt{f}}-\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (\frac{2 \sqrt{f} \sqrt{g} (1+c x)}{\left (i c \sqrt{f}+\sqrt{g}\right ) \left (\sqrt{f}-i \sqrt{g} x\right )}\right )}{\sqrt{f}}-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x}+\frac{1}{2} b c \log \left (-\frac{g x^2}{f}\right ) \left (d+e \log \left (f+g x^2\right )\right )-\frac{1}{2} b c \log \left (\frac{g \left (1-c^2 x^2\right )}{c^2 f+g}\right ) \left (d+e \log \left (f+g x^2\right )\right )-\frac{1}{2} b c e \text{Li}_2\left (\frac{c^2 \left (f+g x^2\right )}{c^2 f+g}\right )+\frac{1}{2} b c e \text{Li}_2\left (1+\frac{g x^2}{f}\right )-\frac{(b e g) \int \frac{\log \left (\frac{2 \sqrt{g} (-1+c x)}{\sqrt{f} \left (i c-\frac{\sqrt{g}}{\sqrt{f}}\right ) \left (1-\frac{i \sqrt{g} x}{\sqrt{f}}\right )}\right )}{1+\frac{g x^2}{f}} \, dx}{f}+\frac{(b e g) \int \frac{\log \left (\frac{2 \sqrt{g} (1+c x)}{\sqrt{f} \left (i c+\frac{\sqrt{g}}{\sqrt{f}}\right ) \left (1-\frac{i \sqrt{g} x}{\sqrt{f}}\right )}\right )}{1+\frac{g x^2}{f}} \, dx}{f}\\ &=\frac{2 a e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right )}{\sqrt{f}}-\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1-\frac{1}{c x}\right )}{\sqrt{f}}+\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (1+\frac{1}{c x}\right )}{\sqrt{f}}+\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (-\frac{2 \sqrt{f} \sqrt{g} (1-c x)}{\left (i c \sqrt{f}-\sqrt{g}\right ) \left (\sqrt{f}-i \sqrt{g} x\right )}\right )}{\sqrt{f}}-\frac{b e \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{f}}\right ) \log \left (\frac{2 \sqrt{f} \sqrt{g} (1+c x)}{\left (i c \sqrt{f}+\sqrt{g}\right ) \left (\sqrt{f}-i \sqrt{g} x\right )}\right )}{\sqrt{f}}-\frac{\left (a+b \coth ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x}+\frac{1}{2} b c \log \left (-\frac{g x^2}{f}\right ) \left (d+e \log \left (f+g x^2\right )\right )-\frac{1}{2} b c \log \left (\frac{g \left (1-c^2 x^2\right )}{c^2 f+g}\right ) \left (d+e \log \left (f+g x^2\right )\right )-\frac{1}{2} b c e \text{Li}_2\left (\frac{c^2 \left (f+g x^2\right )}{c^2 f+g}\right )+\frac{1}{2} b c e \text{Li}_2\left (1+\frac{g x^2}{f}\right )-\frac{i b e \sqrt{g} \text{Li}_2\left (1+\frac{2 \sqrt{f} \sqrt{g} (1-c x)}{\left (i c \sqrt{f}-\sqrt{g}\right ) \left (\sqrt{f}-i \sqrt{g} x\right )}\right )}{2 \sqrt{f}}+\frac{i b e \sqrt{g} \text{Li}_2\left (1-\frac{2 \sqrt{f} \sqrt{g} (1+c x)}{\left (i c \sqrt{f}+\sqrt{g}\right ) \left (\sqrt{f}-i \sqrt{g} x\right )}\right )}{2 \sqrt{f}}\\ \end{align*}

Mathematica [B]  time = 3.56189, size = 1236, normalized size = 2.21 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + b*ArcCoth[c*x])*(d + e*Log[f + g*x^2]))/x^2,x]

[Out]

-((a*d)/x) - (b*d*ArcCoth[c*x])/x + b*c*d*Log[x] - (b*c*d*Log[1 - c^2*x^2])/2 + a*e*((2*Sqrt[g]*ArcTan[(Sqrt[g
]*x)/Sqrt[f]])/Sqrt[f] - Log[f + g*x^2]/x) + (b*e*(-(((2*ArcCoth[c*x] + c*x*(-2*Log[x] + Log[1 - c^2*x^2]))*Lo
g[f + g*x^2])/x) - 2*c*(Log[x]*(Log[1 - (I*Sqrt[g]*x)/Sqrt[f]] + Log[1 + (I*Sqrt[g]*x)/Sqrt[f]]) + PolyLog[2,
((-I)*Sqrt[g]*x)/Sqrt[f]] + PolyLog[2, (I*Sqrt[g]*x)/Sqrt[f]]) + c*(Log[-c^(-1) + x]*Log[(c*(Sqrt[f] - I*Sqrt[
g]*x))/(c*Sqrt[f] - I*Sqrt[g])] + Log[c^(-1) + x]*Log[(c*(Sqrt[f] - I*Sqrt[g]*x))/(c*Sqrt[f] + I*Sqrt[g])] + L
og[-c^(-1) + x]*Log[(c*(Sqrt[f] + I*Sqrt[g]*x))/(c*Sqrt[f] + I*Sqrt[g])] - (Log[-c^(-1) + x] + Log[c^(-1) + x]
- Log[1 - c^2*x^2])*Log[f + g*x^2] + Log[c^(-1) + x]*Log[1 - (Sqrt[g]*(1 + c*x))/(I*c*Sqrt[f] + Sqrt[g])] + P
olyLog[2, (c*Sqrt[g]*(c^(-1) + x))/(I*c*Sqrt[f] + Sqrt[g])] + PolyLog[2, (I*Sqrt[g]*(-1 + c*x))/(c*Sqrt[f] - I
*Sqrt[g])] + PolyLog[2, ((-I)*Sqrt[g]*(-1 + c*x))/(c*Sqrt[f] + I*Sqrt[g])] + PolyLog[2, (I*Sqrt[g]*(1 + c*x))/
(c*Sqrt[f] + I*Sqrt[g])]) - (c*g*((2*I)*ArcCos[(c^2*f - g)/(c^2*f + g)]*ArcTan[(c*f)/(Sqrt[c^2*f*g]*x)] - 4*Ar
cCoth[c*x]*ArcTan[(c*g*x)/Sqrt[c^2*f*g]] + (ArcCos[(c^2*f - g)/(c^2*f + g)] + 2*ArcTan[(c*f)/(Sqrt[c^2*f*g]*x)
])*Log[(2*g*(c^2*f - I*Sqrt[c^2*f*g])*(-1 + c*x))/((c^2*f + g)*(I*Sqrt[c^2*f*g] + c*g*x))] + (ArcCos[(c^2*f -
g)/(c^2*f + g)] - 2*ArcTan[(c*f)/(Sqrt[c^2*f*g]*x)])*Log[(2*g*(c^2*f + I*Sqrt[c^2*f*g])*(1 + c*x))/((c^2*f + g
)*(I*Sqrt[c^2*f*g] + c*g*x))] - (ArcCos[(c^2*f - g)/(c^2*f + g)] + 2*(ArcTan[(c*f)/(Sqrt[c^2*f*g]*x)] + ArcTan
[(c*g*x)/Sqrt[c^2*f*g]]))*Log[(Sqrt[2]*Sqrt[c^2*f*g])/(E^ArcCoth[c*x]*Sqrt[c^2*f + g]*Sqrt[-(c^2*f) + g + (c^2
*f + g)*Cosh[2*ArcCoth[c*x]]])] - (ArcCos[(c^2*f - g)/(c^2*f + g)] - 2*(ArcTan[(c*f)/(Sqrt[c^2*f*g]*x)] + ArcT
an[(c*g*x)/Sqrt[c^2*f*g]]))*Log[(Sqrt[2]*E^ArcCoth[c*x]*Sqrt[c^2*f*g])/(Sqrt[c^2*f + g]*Sqrt[-(c^2*f) + g + (c
^2*f + g)*Cosh[2*ArcCoth[c*x]]])] + I*(PolyLog[2, ((c^2*f - g - (2*I)*Sqrt[c^2*f*g])*(Sqrt[c^2*f*g] + I*c*g*x)
)/((c^2*f + g)*(Sqrt[c^2*f*g] - I*c*g*x))] - PolyLog[2, ((c^2*f - g + (2*I)*Sqrt[c^2*f*g])*(Sqrt[c^2*f*g] + I*
c*g*x))/((c^2*f + g)*(Sqrt[c^2*f*g] - I*c*g*x))])))/Sqrt[c^2*f*g]))/2

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Maple [F]  time = 0.832, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{\rm arccoth} \left (cx\right ) \right ) \left ( d+e\ln \left ( g{x}^{2}+f \right ) \right ) }{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccoth(c*x))*(d+e*ln(g*x^2+f))/x^2,x)

[Out]

int((a+b*arccoth(c*x))*(d+e*ln(g*x^2+f))/x^2,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccoth(c*x))*(d+e*log(g*x^2+f))/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b d \operatorname{arcoth}\left (c x\right ) + a d +{\left (b e \operatorname{arcoth}\left (c x\right ) + a e\right )} \log \left (g x^{2} + f\right )}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccoth(c*x))*(d+e*log(g*x^2+f))/x^2,x, algorithm="fricas")

[Out]

integral((b*d*arccoth(c*x) + a*d + (b*e*arccoth(c*x) + a*e)*log(g*x^2 + f))/x^2, x)

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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((a+b*acoth(c*x))*(d+e*ln(g*x**2+f))/x**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arcoth}\left (c x\right ) + a\right )}{\left (e \log \left (g x^{2} + f\right ) + d\right )}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccoth(c*x))*(d+e*log(g*x^2+f))/x^2,x, algorithm="giac")

[Out]

integrate((b*arccoth(c*x) + a)*(e*log(g*x^2 + f) + d)/x^2, x)