3.152 \(\int \frac{(a+b \tan ^{-1}(\frac{c}{x}))^3}{x^2} \, dx\)

Optimal. Leaf size=136 \[ -\frac{3 i b^2 \text{PolyLog}\left (2,1-\frac{2}{1+\frac{i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )}{c}-\frac{3 b^3 \text{PolyLog}\left (3,1-\frac{2}{1+\frac{i c}{x}}\right )}{2 c}-\frac{i \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )^3}{c}-\frac{\left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )^3}{x}-\frac{3 b \log \left (\frac{2}{1+\frac{i c}{x}}\right ) \left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )^2}{c} \]

[Out]

((-I)*(a + b*ArcCot[x/c])^3)/c - (a + b*ArcCot[x/c])^3/x - (3*b*(a + b*ArcCot[x/c])^2*Log[2/(1 + (I*c)/x)])/c
- ((3*I)*b^2*(a + b*ArcCot[x/c])*PolyLog[2, 1 - 2/(1 + (I*c)/x)])/c - (3*b^3*PolyLog[3, 1 - 2/(1 + (I*c)/x)])/
(2*c)

________________________________________________________________________________________

Rubi [B]  time = 2.35871, antiderivative size = 551, normalized size of antiderivative = 4.05, number of steps used = 82, number of rules used = 23, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.438, Rules used = {5035, 2454, 2389, 2296, 2295, 6715, 2430, 2416, 2396, 2433, 2374, 6589, 2411, 2346, 2301, 6742, 43, 2394, 2393, 2391, 2375, 2317, 2425} \[ \frac{3 b^2 \text{PolyLog}\left (2,-\frac{-x+i c}{2 x}\right ) \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )}{2 c}+\frac{3 b^3 \text{PolyLog}\left (3,-\frac{-x+i c}{2 x}\right )}{2 c}+\frac{3 b^3 \text{PolyLog}\left (3,\frac{x+i c}{2 x}\right )}{2 c}-\frac{3 b^3 \log \left (1+\frac{i c}{x}\right ) \text{PolyLog}\left (2,\frac{x+i c}{2 x}\right )}{2 c}-\frac{3 b^2 \log ^2\left (1+\frac{i c}{x}\right ) \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )}{8 c}-\frac{3 i b^2 \log ^2\left (1+\frac{i c}{x}\right ) \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )}{8 x}-\frac{3 b \left (1-\frac{i c}{x}\right ) \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c}-\frac{3 b \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c}+\frac{3 b \log \left (1+\frac{i c}{x}\right ) \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c}-\frac{3 i b \log \left (1+\frac{i c}{x}\right ) \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 x}-\frac{3 b \log \left (\frac{x+i c}{2 x}\right ) \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2}{4 c}-\frac{i \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^3}{8 c}-\frac{b^3 \left (1+\frac{i c}{x}\right ) \log ^3\left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 b^3 \log ^2\left (1+\frac{i c}{x}\right ) \log \left (-\frac{-x+i c}{2 x}\right )}{4 c} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*ArcTan[c/x])^3/x^2,x]

[Out]

(-3*b*(1 - (I*c)/x)*((2*I)*a - b*Log[1 - (I*c)/x])^2)/(8*c) - (3*b*(1 - (I*c)/x)*(2*a + I*b*Log[1 - (I*c)/x])^
2)/(8*c) - ((I/8)*(1 - (I*c)/x)*(2*a + I*b*Log[1 - (I*c)/x])^3)/c + (3*b*((2*I)*a - b*Log[1 - (I*c)/x])^2*Log[
1 + (I*c)/x])/(8*c) - (((3*I)/8)*b*((2*I)*a - b*Log[1 - (I*c)/x])^2*Log[1 + (I*c)/x])/x - (3*b^2*((2*I)*a - b*
Log[1 - (I*c)/x])*Log[1 + (I*c)/x]^2)/(8*c) - (((3*I)/8)*b^2*((2*I)*a - b*Log[1 - (I*c)/x])*Log[1 + (I*c)/x]^2
)/x - (b^3*(1 + (I*c)/x)*Log[1 + (I*c)/x]^3)/(8*c) - (3*b^3*Log[1 + (I*c)/x]^2*Log[-(I*c - x)/(2*x)])/(4*c) -
(3*b*((2*I)*a - b*Log[1 - (I*c)/x])^2*Log[(I*c + x)/(2*x)])/(4*c) + (3*b^2*((2*I)*a - b*Log[1 - (I*c)/x])*Poly
Log[2, -(I*c - x)/(2*x)])/(2*c) - (3*b^3*Log[1 + (I*c)/x]*PolyLog[2, (I*c + x)/(2*x)])/(2*c) + (3*b^3*PolyLog[
3, -(I*c - x)/(2*x)])/(2*c) + (3*b^3*PolyLog[3, (I*c + x)/(2*x)])/(2*c)

Rule 5035

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(d*x)^
m*(a + (I*b*Log[1 - I*c*x^n])/2 - (I*b*Log[1 + I*c*x^n])/2)^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IGtQ[
p, 0] && IntegerQ[m] && IntegerQ[n]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2389

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

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 2430

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.)), x_Symbol] :> Simp[x*(a + b*Log[c*(d + e*x)^n])^p*(f + g*Log[h*(i + j*x)^m]), x] + (-Dist[g*j*m, Int[(x
*(a + b*Log[c*(d + e*x)^n])^p)/(i + j*x), x], x] - Dist[b*e*n*p, Int[(x*(a + b*Log[c*(d + e*x)^n])^(p - 1)*(f
+ g*Log[h*(i + j*x)^m]))/(d + e*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, i, j, m, n}, x] && IGtQ[p, 0]

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 2396

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

Rule 2433

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((k*x)/d)^r*(a + b*Log[c*x^n])^p*(f + g*Lo
g[h*((e*i - d*j)/e + (j*x)/e)^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r},
x] && EqQ[e*k - d*l, 0]

Rule 2374

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

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 6742

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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 2375

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :
> Simp[(Log[d*(e + f*x^m)^r]*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1)), x] - Dist[(f*m*r)/(b*n*(p + 1)), Int[(
x^(m - 1)*(a + b*Log[c*x^n])^(p + 1))/(e + f*x^m), x], x] /; FreeQ[{a, b, c, d, e, f, r, m, n}, x] && IGtQ[p,
0] && NeQ[d*e, 1]

Rule 2317

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

Rule 2425

Int[(Log[(f_.)*(x_)^(m_.)]*((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.)))/(x_), x_Symbol] :> Simp[(Log[
f*x^m]^2*(a + b*Log[c*(d + e*x)^n]))/(2*m), x] - Dist[(b*e*n)/(2*m), Int[Log[f*x^m]^2/(d + e*x), x], x] /; Fre
eQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )^3}{x^2} \, dx &=\int \left (\frac{\left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^3}{8 x^2}+\frac{3 i b \left (-2 i a+b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 x^2}-\frac{3 i b^2 \left (-2 i a+b \log \left (1-\frac{i c}{x}\right )\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 x^2}+\frac{i b^3 \log ^3\left (1+\frac{i c}{x}\right )}{8 x^2}\right ) \, dx\\ &=\frac{1}{8} \int \frac{\left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^3}{x^2} \, dx+\frac{1}{8} (3 i b) \int \frac{\left (-2 i a+b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (1+\frac{i c}{x}\right )}{x^2} \, dx-\frac{1}{8} \left (3 i b^2\right ) \int \frac{\left (-2 i a+b \log \left (1-\frac{i c}{x}\right )\right ) \log ^2\left (1+\frac{i c}{x}\right )}{x^2} \, dx+\frac{1}{8} \left (i b^3\right ) \int \frac{\log ^3\left (1+\frac{i c}{x}\right )}{x^2} \, dx\\ &=-\left (\frac{1}{8} \operatorname{Subst}\left (\int (2 a+i b \log (1-i c x))^3 \, dx,x,\frac{1}{x}\right )\right )-\frac{1}{8} (3 i b) \operatorname{Subst}\left (\int (-2 i a+b \log (1-i c x))^2 \log (1+i c x) \, dx,x,\frac{1}{x}\right )+\frac{1}{8} \left (3 i b^2\right ) \operatorname{Subst}\left (\int (-2 i a+b \log (1-i c x)) \log ^2(1+i c x) \, dx,x,\frac{1}{x}\right )-\frac{1}{8} \left (i b^3\right ) \operatorname{Subst}\left (\int \log ^3(1+i c x) \, dx,x,\frac{1}{x}\right )\\ &=-\frac{3 i b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 x}-\frac{3 i b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 x}-\frac{i \operatorname{Subst}\left (\int (2 a+i b \log (x))^3 \, dx,x,1-\frac{i c}{x}\right )}{8 c}-\frac{b^3 \operatorname{Subst}\left (\int \log ^3(x) \, dx,x,1+\frac{i c}{x}\right )}{8 c}-\frac{1}{8} (3 b c) \operatorname{Subst}\left (\int \frac{x (-2 i a+b \log (1-i c x))^2}{1+i c x} \, dx,x,\frac{1}{x}\right )+\frac{1}{4} \left (3 b^2 c\right ) \operatorname{Subst}\left (\int \frac{x (-2 i a+b \log (1-i c x)) \log (1+i c x)}{1-i c x} \, dx,x,\frac{1}{x}\right )+\frac{1}{4} \left (3 b^2 c\right ) \operatorname{Subst}\left (\int \frac{x (-2 i a+b \log (1-i c x)) \log (1+i c x)}{1+i c x} \, dx,x,\frac{1}{x}\right )-\frac{1}{8} \left (3 b^3 c\right ) \operatorname{Subst}\left (\int \frac{x \log ^2(1+i c x)}{1-i c x} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{i \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^3}{8 c}-\frac{3 i b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 x}-\frac{3 i b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 x}-\frac{b^3 \left (1+\frac{i c}{x}\right ) \log ^3\left (1+\frac{i c}{x}\right )}{8 c}-\frac{(3 b) \operatorname{Subst}\left (\int (2 a+i b \log (x))^2 \, dx,x,1-\frac{i c}{x}\right )}{8 c}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \log ^2(x) \, dx,x,1+\frac{i c}{x}\right )}{8 c}-\frac{1}{8} (3 b c) \operatorname{Subst}\left (\int \left (-\frac{i (-2 i a+b \log (1-i c x))^2}{c}+\frac{(-2 i a+b \log (1-i c x))^2}{c (-i+c x)}\right ) \, dx,x,\frac{1}{x}\right )+\frac{1}{4} \left (3 b^2 c\right ) \operatorname{Subst}\left (\int \left (-\frac{(2 a+i b \log (1-i c x)) \log (1+i c x)}{c}+\frac{(-2 i a+b \log (1-i c x)) \log (1+i c x)}{c (-i+c x)}\right ) \, dx,x,\frac{1}{x}\right )+\frac{1}{4} \left (3 b^2 c\right ) \operatorname{Subst}\left (\int \left (\frac{(2 a+i b \log (1-i c x)) \log (1+i c x)}{c}+\frac{(-2 i a+b \log (1-i c x)) \log (1+i c x)}{c (i+c x)}\right ) \, dx,x,\frac{1}{x}\right )-\frac{1}{8} \left (3 b^3 c\right ) \operatorname{Subst}\left (\int \left (\frac{i \log ^2(1+i c x)}{c}+\frac{\log ^2(1+i c x)}{c (i+c x)}\right ) \, dx,x,\frac{1}{x}\right )\\ &=-\frac{3 b \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c}-\frac{i \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^3}{8 c}-\frac{3 i b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 x}+\frac{3 b^3 \left (1+\frac{i c}{x}\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 i b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 x}-\frac{b^3 \left (1+\frac{i c}{x}\right ) \log ^3\left (1+\frac{i c}{x}\right )}{8 c}+\frac{1}{8} (3 i b) \operatorname{Subst}\left (\int (-2 i a+b \log (1-i c x))^2 \, dx,x,\frac{1}{x}\right )-\frac{1}{8} (3 b) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x))^2}{-i+c x} \, dx,x,\frac{1}{x}\right )+\frac{1}{4} \left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x)) \log (1+i c x)}{-i+c x} \, dx,x,\frac{1}{x}\right )+\frac{1}{4} \left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x)) \log (1+i c x)}{i+c x} \, dx,x,\frac{1}{x}\right )-\frac{1}{8} \left (3 i b^3\right ) \operatorname{Subst}\left (\int \log ^2(1+i c x) \, dx,x,\frac{1}{x}\right )-\frac{1}{8} \left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\log ^2(1+i c x)}{i+c x} \, dx,x,\frac{1}{x}\right )+\frac{\left (3 i b^2\right ) \operatorname{Subst}\left (\int (2 a+i b \log (x)) \, dx,x,1-\frac{i c}{x}\right )}{4 c}-\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \log (x) \, dx,x,1+\frac{i c}{x}\right )}{4 c}\\ &=\frac{3 a b^2}{2 x}+\frac{3 i b^3}{4 x}-\frac{3 b \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c}-\frac{i \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^3}{8 c}-\frac{3 b^3 \left (1+\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 c}-\frac{3 i b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 x}+\frac{3 b^3 \left (1+\frac{i c}{x}\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 i b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 x}-\frac{b^3 \left (1+\frac{i c}{x}\right ) \log ^3\left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 b^3 \log ^2\left (1+\frac{i c}{x}\right ) \log \left (-\frac{i c-x}{2 x}\right )}{8 c}-\frac{3 b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (\frac{i c+x}{2 x}\right )}{8 c}-\frac{1}{4} \left (3 i b^2\right ) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (1-i c x)) \log \left (\frac{1}{2} i (-i+c x)\right )}{1-i c x} \, dx,x,\frac{1}{x}\right )+\frac{1}{4} \left (3 i b^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+i c x) \log \left (-\frac{1}{2} i (i+c x)\right )}{1+i c x} \, dx,x,\frac{1}{x}\right )-\frac{(3 b) \operatorname{Subst}\left (\int (-2 i a+b \log (x))^2 \, dx,x,1-\frac{i c}{x}\right )}{8 c}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (2-x)) \log (x)}{x} \, dx,x,1+\frac{i c}{x}\right )}{4 c}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\log (2-x) (-2 i a+b \log (x))}{x} \, dx,x,1-\frac{i c}{x}\right )}{4 c}-\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \log ^2(x) \, dx,x,1+\frac{i c}{x}\right )}{8 c}-\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \log (x) \, dx,x,1-\frac{i c}{x}\right )}{4 c}\\ &=\frac{3 a b^2}{2 x}-\frac{3 b^3 \left (1-\frac{i c}{x}\right ) \log \left (1-\frac{i c}{x}\right )}{4 c}-\frac{3 b \left (1-\frac{i c}{x}\right ) \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c}-\frac{3 b \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c}-\frac{i \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^3}{8 c}-\frac{3 b^3 \left (1+\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 c}+\frac{3 b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 i b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 x}-\frac{3 b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 i b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 x}-\frac{b^3 \left (1+\frac{i c}{x}\right ) \log ^3\left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 b^3 \log ^2\left (1+\frac{i c}{x}\right ) \log \left (-\frac{i c-x}{2 x}\right )}{8 c}-\frac{3 b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (\frac{i c+x}{2 x}\right )}{8 c}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{(-2 i a+b \log (x))^2}{2-x} \, dx,x,1-\frac{i c}{x}\right )}{8 c}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int (-2 i a+b \log (x)) \, dx,x,1-\frac{i c}{x}\right )}{4 c}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{1}{2} i (-2 i+i x)\right ) (-2 i a+b \log (x))}{x} \, dx,x,1-\frac{i c}{x}\right )}{4 c}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\log ^2(x)}{2-x} \, dx,x,1+\frac{i c}{x}\right )}{8 c}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \log (x) \, dx,x,1+\frac{i c}{x}\right )}{4 c}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\log \left (-\frac{1}{2} i (2 i-i x)\right ) \log (x)}{x} \, dx,x,1+\frac{i c}{x}\right )}{4 c}\\ &=-\frac{3 i b^3}{4 x}-\frac{3 b^3 \left (1-\frac{i c}{x}\right ) \log \left (1-\frac{i c}{x}\right )}{4 c}-\frac{3 b \left (1-\frac{i c}{x}\right ) \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c}-\frac{3 b \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c}-\frac{i \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^3}{8 c}+\frac{3 b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 i b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 x}-\frac{3 b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 i b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 x}-\frac{b^3 \left (1+\frac{i c}{x}\right ) \log ^3\left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 b^3 \log ^2\left (1+\frac{i c}{x}\right ) \log \left (-\frac{i c-x}{2 x}\right )}{4 c}-\frac{3 b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (\frac{i c+x}{2 x}\right )}{4 c}+\frac{3 b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \text{Li}_2\left (-\frac{i c-x}{2 x}\right )}{4 c}-\frac{3 b^3 \log \left (1+\frac{i c}{x}\right ) \text{Li}_2\left (\frac{i c+x}{2 x}\right )}{4 c}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right ) (-2 i a+b \log (x))}{x} \, dx,x,1-\frac{i c}{x}\right )}{4 c}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \log (x) \, dx,x,1-\frac{i c}{x}\right )}{4 c}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right ) \log (x)}{x} \, dx,x,1+\frac{i c}{x}\right )}{4 c}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{x}{2}\right )}{x} \, dx,x,1-\frac{i c}{x}\right )}{4 c}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{x}{2}\right )}{x} \, dx,x,1+\frac{i c}{x}\right )}{4 c}\\ &=-\frac{3 b \left (1-\frac{i c}{x}\right ) \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c}-\frac{3 b \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c}-\frac{i \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^3}{8 c}+\frac{3 b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 i b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 x}-\frac{3 b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 i b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 x}-\frac{b^3 \left (1+\frac{i c}{x}\right ) \log ^3\left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 b^3 \log ^2\left (1+\frac{i c}{x}\right ) \log \left (-\frac{i c-x}{2 x}\right )}{4 c}-\frac{3 b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (\frac{i c+x}{2 x}\right )}{4 c}+\frac{3 b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \text{Li}_2\left (-\frac{i c-x}{2 x}\right )}{2 c}-\frac{3 b^3 \log \left (1+\frac{i c}{x}\right ) \text{Li}_2\left (\frac{i c+x}{2 x}\right )}{2 c}+\frac{3 b^3 \text{Li}_3\left (-\frac{i c-x}{2 x}\right )}{4 c}+\frac{3 b^3 \text{Li}_3\left (\frac{i c+x}{2 x}\right )}{4 c}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{x}{2}\right )}{x} \, dx,x,1-\frac{i c}{x}\right )}{4 c}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{x}{2}\right )}{x} \, dx,x,1+\frac{i c}{x}\right )}{4 c}\\ &=-\frac{3 b \left (1-\frac{i c}{x}\right ) \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c}-\frac{3 b \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c}-\frac{i \left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^3}{8 c}+\frac{3 b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 i b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 x}-\frac{3 b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 i b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \log ^2\left (1+\frac{i c}{x}\right )}{8 x}-\frac{b^3 \left (1+\frac{i c}{x}\right ) \log ^3\left (1+\frac{i c}{x}\right )}{8 c}-\frac{3 b^3 \log ^2\left (1+\frac{i c}{x}\right ) \log \left (-\frac{i c-x}{2 x}\right )}{4 c}-\frac{3 b \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right )^2 \log \left (\frac{i c+x}{2 x}\right )}{4 c}+\frac{3 b^2 \left (2 i a-b \log \left (1-\frac{i c}{x}\right )\right ) \text{Li}_2\left (-\frac{i c-x}{2 x}\right )}{2 c}-\frac{3 b^3 \log \left (1+\frac{i c}{x}\right ) \text{Li}_2\left (\frac{i c+x}{2 x}\right )}{2 c}+\frac{3 b^3 \text{Li}_3\left (-\frac{i c-x}{2 x}\right )}{2 c}+\frac{3 b^3 \text{Li}_3\left (\frac{i c+x}{2 x}\right )}{2 c}\\ \end{align*}

Mathematica [A]  time = 0.118877, size = 222, normalized size = 1.63 \[ -\frac{-6 i b^2 x \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}\left (\frac{c}{x}\right )}\right ) \left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )+3 b^3 x \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}\left (\frac{c}{x}\right )}\right )-3 a^2 b x \log \left (\frac{c^2}{x^2}+1\right )+6 a^2 b c \tan ^{-1}\left (\frac{c}{x}\right )+2 a^3 c+6 a b^2 c \tan ^{-1}\left (\frac{c}{x}\right )^2-6 i a b^2 x \tan ^{-1}\left (\frac{c}{x}\right )^2+12 a b^2 x \tan ^{-1}\left (\frac{c}{x}\right ) \log \left (1+e^{2 i \tan ^{-1}\left (\frac{c}{x}\right )}\right )+2 b^3 c \tan ^{-1}\left (\frac{c}{x}\right )^3-2 i b^3 x \tan ^{-1}\left (\frac{c}{x}\right )^3+6 b^3 x \tan ^{-1}\left (\frac{c}{x}\right )^2 \log \left (1+e^{2 i \tan ^{-1}\left (\frac{c}{x}\right )}\right )}{2 c x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTan[c/x])^3/x^2,x]

[Out]

-(2*a^3*c + 6*a^2*b*c*ArcTan[c/x] + 6*a*b^2*c*ArcTan[c/x]^2 - (6*I)*a*b^2*x*ArcTan[c/x]^2 + 2*b^3*c*ArcTan[c/x
]^3 - (2*I)*b^3*x*ArcTan[c/x]^3 + 12*a*b^2*x*ArcTan[c/x]*Log[1 + E^((2*I)*ArcTan[c/x])] + 6*b^3*x*ArcTan[c/x]^
2*Log[1 + E^((2*I)*ArcTan[c/x])] - 3*a^2*b*x*Log[1 + c^2/x^2] - (6*I)*b^2*x*(a + b*ArcTan[c/x])*PolyLog[2, -E^
((2*I)*ArcTan[c/x])] + 3*b^3*x*PolyLog[3, -E^((2*I)*ArcTan[c/x])])/(2*c*x)

________________________________________________________________________________________

Maple [B]  time = 0.125, size = 306, normalized size = 2.3 \begin{align*} -{\frac{{a}^{3}}{x}}+{\frac{i{b}^{3}}{c} \left ( \arctan \left ({\frac{c}{x}} \right ) \right ) ^{3}}-{\frac{{b}^{3}}{x} \left ( \arctan \left ({\frac{c}{x}} \right ) \right ) ^{3}}-3\,{\frac{{b}^{3}}{c} \left ( \arctan \left ({\frac{c}{x}} \right ) \right ) ^{2}\ln \left ({ \left ( 1+{\frac{ic}{x}} \right ) ^{2} \left ( 1+{\frac{{c}^{2}}{{x}^{2}}} \right ) ^{-1}}+1 \right ) }+{\frac{3\,i{b}^{3}}{c}\arctan \left ({\frac{c}{x}} \right ){\it polylog} \left ( 2,-{ \left ( 1+{\frac{ic}{x}} \right ) ^{2} \left ( 1+{\frac{{c}^{2}}{{x}^{2}}} \right ) ^{-1}} \right ) }-{\frac{3\,{b}^{3}}{2\,c}{\it polylog} \left ( 3,-{ \left ( 1+{\frac{ic}{x}} \right ) ^{2} \left ( 1+{\frac{{c}^{2}}{{x}^{2}}} \right ) ^{-1}} \right ) }+{\frac{3\,ia{b}^{2}}{c} \left ( \arctan \left ({\frac{c}{x}} \right ) \right ) ^{2}}-3\,{\frac{a{b}^{2}}{x} \left ( \arctan \left ({\frac{c}{x}} \right ) \right ) ^{2}}-6\,{\frac{a{b}^{2}}{c}\arctan \left ({\frac{c}{x}} \right ) \ln \left ({ \left ( 1+{\frac{ic}{x}} \right ) ^{2} \left ( 1+{\frac{{c}^{2}}{{x}^{2}}} \right ) ^{-1}}+1 \right ) }+{\frac{3\,ia{b}^{2}}{c}{\it polylog} \left ( 2,-{ \left ( 1+{\frac{ic}{x}} \right ) ^{2} \left ( 1+{\frac{{c}^{2}}{{x}^{2}}} \right ) ^{-1}} \right ) }-3\,{\frac{{a}^{2}b}{x}\arctan \left ({\frac{c}{x}} \right ) }+{\frac{3\,{a}^{2}b}{2\,c}\ln \left ( 1+{\frac{{c}^{2}}{{x}^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c/x))^3/x^2,x)

[Out]

-a^3/x+I/c*b^3*arctan(c/x)^3-b^3*arctan(c/x)^3/x-3/c*b^3*arctan(c/x)^2*ln((1+I*c/x)^2/(1+c^2/x^2)+1)+3*I/c*b^3
*arctan(c/x)*polylog(2,-(1+I*c/x)^2/(1+c^2/x^2))-3/2/c*b^3*polylog(3,-(1+I*c/x)^2/(1+c^2/x^2))+3*I/c*arctan(c/
x)^2*a*b^2-3*arctan(c/x)^2/x*a*b^2-6/c*arctan(c/x)*ln((1+I*c/x)^2/(1+c^2/x^2)+1)*a*b^2+3*I/c*polylog(2,-(1+I*c
/x)^2/(1+c^2/x^2))*a*b^2-3*a^2*b/x*arctan(c/x)+3/2/c*a^2*b*ln(1+c^2/x^2)

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Maxima [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*arctan(c/x))^3/x^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \arctan \left (\frac{c}{x}\right )^{3} + 3 \, a b^{2} \arctan \left (\frac{c}{x}\right )^{2} + 3 \, a^{2} b \arctan \left (\frac{c}{x}\right ) + a^{3}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c/x))^3/x^2,x, algorithm="fricas")

[Out]

integral((b^3*arctan(c/x)^3 + 3*a*b^2*arctan(c/x)^2 + 3*a^2*b*arctan(c/x) + a^3)/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (\frac{c}{x} \right )}\right )^{3}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c/x))**3/x**2,x)

[Out]

Integral((a + b*atan(c/x))**3/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (\frac{c}{x}\right ) + a\right )}^{3}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c/x))^3/x^2,x, algorithm="giac")

[Out]

integrate((b*arctan(c/x) + a)^3/x^2, x)