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

Optimal. Leaf size=84 \[ -\frac{\left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )^2}{2 c^2}-\frac{\left (a+b \cot ^{-1}\left (\frac{x}{c}\right )\right )^2}{2 x^2}+\frac{a b}{c x}-\frac{b^2 \log \left (\frac{c^2}{x^2}+1\right )}{2 c^2}+\frac{b^2 \cot ^{-1}\left (\frac{x}{c}\right )}{c x} \]

[Out]

(a*b)/(c*x) + (b^2*ArcCot[x/c])/(c*x) - (a + b*ArcCot[x/c])^2/(2*c^2) - (a + b*ArcCot[x/c])^2/(2*x^2) - (b^2*L
og[1 + c^2/x^2])/(2*c^2)

________________________________________________________________________________________

Rubi [C]  time = 1.29363, antiderivative size = 836, normalized size of antiderivative = 9.95, number of steps used = 66, number of rules used = 23, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.438, Rules used = {5035, 2454, 2401, 2389, 2296, 2295, 2390, 2305, 2304, 2395, 43, 6742, 30, 2557, 12, 2466, 2462, 260, 2416, 2394, 2393, 2391, 2315} \[ -\frac{\left (1-\frac{i c}{x}\right )^2 b^2}{16 c^2}-\frac{\left (\frac{i c}{x}+1\right )^2 b^2}{16 c^2}-\frac{\left (\frac{i c}{x}+1\right )^2 \log ^2\left (\frac{i c}{x}+1\right ) b^2}{8 c^2}+\frac{\left (\frac{i c}{x}+1\right ) \log ^2\left (\frac{i c}{x}+1\right ) b^2}{4 c^2}+\frac{\log \left (i-\frac{c}{x}\right ) b^2}{8 c^2}-\frac{3 \left (1-\frac{i c}{x}\right ) \log \left (1-\frac{i c}{x}\right ) b^2}{4 c^2}+\frac{\log \left (1-\frac{i c}{x}\right ) b^2}{8 x^2}+\frac{\left (\frac{i c}{x}+1\right )^2 \log \left (\frac{i c}{x}+1\right ) b^2}{8 c^2}-\frac{3 \left (\frac{i c}{x}+1\right ) \log \left (\frac{i c}{x}+1\right ) b^2}{4 c^2}-\frac{\log \left (1-\frac{i c}{x}\right ) \log \left (\frac{i c}{x}+1\right ) b^2}{4 x^2}+\frac{\log \left (\frac{i c}{x}+1\right ) b^2}{8 x^2}+\frac{\log \left (\frac{c}{x}+i\right ) b^2}{8 c^2}-\frac{\log \left (1-\frac{i c}{x}\right ) \log (c-i x) b^2}{4 c^2}-\frac{\log \left (\frac{i c}{x}+1\right ) \log (c+i x) b^2}{4 c^2}+\frac{\log \left (\frac{c-i x}{2 c}\right ) \log (c+i x) b^2}{4 c^2}+\frac{\log (c-i x) \log \left (\frac{c+i x}{2 c}\right ) b^2}{4 c^2}-\frac{\log (c+i x) \log \left (-\frac{i x}{c}\right ) b^2}{4 c^2}-\frac{\log (c-i x) \log \left (\frac{i x}{c}\right ) b^2}{4 c^2}+\frac{\text{PolyLog}\left (2,\frac{c-i x}{2 c}\right ) b^2}{4 c^2}+\frac{\text{PolyLog}\left (2,\frac{c+i x}{2 c}\right ) b^2}{4 c^2}+\frac{\text{PolyLog}\left (2,-\frac{i c}{x}\right ) b^2}{4 c^2}+\frac{\text{PolyLog}\left (2,\frac{i c}{x}\right ) b^2}{4 c^2}-\frac{\text{PolyLog}\left (2,1-\frac{i x}{c}\right ) b^2}{4 c^2}-\frac{\text{PolyLog}\left (2,\frac{i x}{c}+1\right ) b^2}{4 c^2}-\frac{b^2}{8 x^2}+\frac{i a \log \left (i-\frac{c}{x}\right ) b}{2 c^2}-\frac{i \left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right ) b}{8 c^2}+\frac{i a \log \left (\frac{i c}{x}+1\right ) b}{2 x^2}+\frac{3 a b}{2 c x}-\frac{i a b}{4 x^2}+\frac{\left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c^2}-\frac{\left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{4 c^2} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*ArcTan[c/x])^2/x^3,x]

[Out]

-(b^2*(1 - (I*c)/x)^2)/(16*c^2) - (b^2*(1 + (I*c)/x)^2)/(16*c^2) - ((I/4)*a*b)/x^2 - b^2/(8*x^2) + (3*a*b)/(2*
c*x) + ((I/2)*a*b*Log[I - c/x])/c^2 + (b^2*Log[I - c/x])/(8*c^2) - (3*b^2*(1 - (I*c)/x)*Log[1 - (I*c)/x])/(4*c
^2) + (b^2*Log[1 - (I*c)/x])/(8*x^2) - ((I/8)*b*(1 - (I*c)/x)^2*(2*a + I*b*Log[1 - (I*c)/x]))/c^2 - ((1 - (I*c
)/x)*(2*a + I*b*Log[1 - (I*c)/x])^2)/(4*c^2) + ((1 - (I*c)/x)^2*(2*a + I*b*Log[1 - (I*c)/x])^2)/(8*c^2) - (3*b
^2*(1 + (I*c)/x)*Log[1 + (I*c)/x])/(4*c^2) + (b^2*(1 + (I*c)/x)^2*Log[1 + (I*c)/x])/(8*c^2) + ((I/2)*a*b*Log[1
+ (I*c)/x])/x^2 + (b^2*Log[1 + (I*c)/x])/(8*x^2) - (b^2*Log[1 - (I*c)/x]*Log[1 + (I*c)/x])/(4*x^2) + (b^2*(1
+ (I*c)/x)*Log[1 + (I*c)/x]^2)/(4*c^2) - (b^2*(1 + (I*c)/x)^2*Log[1 + (I*c)/x]^2)/(8*c^2) + (b^2*Log[I + c/x])
/(8*c^2) - (b^2*Log[1 - (I*c)/x]*Log[c - I*x])/(4*c^2) - (b^2*Log[1 + (I*c)/x]*Log[c + I*x])/(4*c^2) + (b^2*Lo
g[(c - I*x)/(2*c)]*Log[c + I*x])/(4*c^2) + (b^2*Log[c - I*x]*Log[(c + I*x)/(2*c)])/(4*c^2) - (b^2*Log[c + I*x]
*Log[((-I)*x)/c])/(4*c^2) - (b^2*Log[c - I*x]*Log[(I*x)/c])/(4*c^2) + (b^2*PolyLog[2, (c - I*x)/(2*c)])/(4*c^2
) + (b^2*PolyLog[2, (c + I*x)/(2*c)])/(4*c^2) + (b^2*PolyLog[2, ((-I)*c)/x])/(4*c^2) + (b^2*PolyLog[2, (I*c)/x
])/(4*c^2) - (b^2*PolyLog[2, 1 - (I*x)/c])/(4*c^2) - (b^2*PolyLog[2, 1 + (I*x)/c])/(4*c^2)

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 2401

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Int[Exp
andIntegrand[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[
e*f - d*g, 0] && IGtQ[q, 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 2390

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

Rule 2305

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

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2395

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

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 6742

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2557

Int[Log[v_]*Log[w_]*(u_), x_Symbol] :> With[{z = IntHide[u, x]}, Dist[Log[v]*Log[w], z, x] + (-Int[SimplifyInt
egrand[(z*Log[w]*D[v, x])/v, x], x] - Int[SimplifyIntegrand[(z*Log[v]*D[w, x])/w, x], x]) /; InverseFunctionFr
eeQ[z, x]] /; InverseFunctionFreeQ[v, x] && InverseFunctionFreeQ[w, x]

Rule 12

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

Rule 2466

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

Rule 2462

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

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}\left (\frac{c}{x}\right )\right )^2}{x^3} \, dx &=\int \left (\frac{\left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{4 x^3}+\frac{b \left (-2 i a+b \log \left (1-\frac{i c}{x}\right )\right ) \log \left (1+\frac{i c}{x}\right )}{2 x^3}-\frac{b^2 \log ^2\left (1+\frac{i c}{x}\right )}{4 x^3}\right ) \, dx\\ &=\frac{1}{4} \int \frac{\left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{x^3} \, dx+\frac{1}{2} b \int \frac{\left (-2 i a+b \log \left (1-\frac{i c}{x}\right )\right ) \log \left (1+\frac{i c}{x}\right )}{x^3} \, dx-\frac{1}{4} b^2 \int \frac{\log ^2\left (1+\frac{i c}{x}\right )}{x^3} \, dx\\ &=-\left (\frac{1}{4} \operatorname{Subst}\left (\int x (2 a+i b \log (1-i c x))^2 \, dx,x,\frac{1}{x}\right )\right )+\frac{1}{2} b \int \left (-\frac{2 i a \log \left (1+\frac{i c}{x}\right )}{x^3}+\frac{b \log \left (1-\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{x^3}\right ) \, dx+\frac{1}{4} b^2 \operatorname{Subst}\left (\int x \log ^2(1+i c x) \, dx,x,\frac{1}{x}\right )\\ &=-\left (\frac{1}{4} \operatorname{Subst}\left (\int \left (-\frac{i (2 a+i b \log (1-i c x))^2}{c}+\frac{i (1-i c x) (2 a+i b \log (1-i c x))^2}{c}\right ) \, dx,x,\frac{1}{x}\right )\right )-(i a b) \int \frac{\log \left (1+\frac{i c}{x}\right )}{x^3} \, dx+\frac{1}{4} b^2 \operatorname{Subst}\left (\int \left (\frac{i \log ^2(1+i c x)}{c}-\frac{i (1+i c x) \log ^2(1+i c x)}{c}\right ) \, dx,x,\frac{1}{x}\right )+\frac{1}{2} b^2 \int \frac{\log \left (1-\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{x^3} \, dx\\ &=-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 x^2}+(i a b) \operatorname{Subst}\left (\int x \log (1+i c x) \, dx,x,\frac{1}{x}\right )-\frac{1}{2} b^2 \int \frac{c \log \left (1-\frac{i c}{x}\right )}{2 (c-i x) x^3} \, dx-\frac{1}{2} b^2 \int \frac{c \log \left (1+\frac{i c}{x}\right )}{2 (c+i x) x^3} \, dx+\frac{i \operatorname{Subst}\left (\int (2 a+i b \log (1-i c x))^2 \, dx,x,\frac{1}{x}\right )}{4 c}-\frac{i \operatorname{Subst}\left (\int (1-i c x) (2 a+i b \log (1-i c x))^2 \, dx,x,\frac{1}{x}\right )}{4 c}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log ^2(1+i c x) \, dx,x,\frac{1}{x}\right )}{4 c}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int (1+i c x) \log ^2(1+i c x) \, dx,x,\frac{1}{x}\right )}{4 c}\\ &=\frac{i a b \log \left (1+\frac{i c}{x}\right )}{2 x^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 x^2}-\frac{\operatorname{Subst}\left (\int (2 a+i b \log (x))^2 \, dx,x,1-\frac{i c}{x}\right )}{4 c^2}+\frac{\operatorname{Subst}\left (\int x (2 a+i b \log (x))^2 \, dx,x,1-\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \operatorname{Subst}\left (\int \log ^2(x) \, dx,x,1+\frac{i c}{x}\right )}{4 c^2}-\frac{b^2 \operatorname{Subst}\left (\int x \log ^2(x) \, dx,x,1+\frac{i c}{x}\right )}{4 c^2}+\frac{1}{2} (a b c) \operatorname{Subst}\left (\int \frac{x^2}{1+i c x} \, dx,x,\frac{1}{x}\right )-\frac{1}{4} \left (b^2 c\right ) \int \frac{\log \left (1-\frac{i c}{x}\right )}{(c-i x) x^3} \, dx-\frac{1}{4} \left (b^2 c\right ) \int \frac{\log \left (1+\frac{i c}{x}\right )}{(c+i x) x^3} \, dx\\ &=-\frac{\left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{4 c^2}+\frac{\left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c^2}+\frac{i a b \log \left (1+\frac{i c}{x}\right )}{2 x^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 x^2}+\frac{b^2 \left (1+\frac{i c}{x}\right ) \log ^2\left (1+\frac{i c}{x}\right )}{4 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log ^2\left (1+\frac{i c}{x}\right )}{8 c^2}-\frac{(i b) \operatorname{Subst}\left (\int x (2 a+i b \log (x)) \, dx,x,1-\frac{i c}{x}\right )}{4 c^2}+\frac{(i b) \operatorname{Subst}\left (\int (2 a+i b \log (x)) \, dx,x,1-\frac{i c}{x}\right )}{2 c^2}+\frac{b^2 \operatorname{Subst}\left (\int x \log (x) \, dx,x,1+\frac{i c}{x}\right )}{4 c^2}-\frac{b^2 \operatorname{Subst}\left (\int \log (x) \, dx,x,1+\frac{i c}{x}\right )}{2 c^2}+\frac{1}{2} (a b c) \operatorname{Subst}\left (\int \left (\frac{1}{c^2}-\frac{i x}{c}+\frac{i}{c^2 (-i+c x)}\right ) \, dx,x,\frac{1}{x}\right )-\frac{1}{4} \left (b^2 c\right ) \int \left (-\frac{i \log \left (1-\frac{i c}{x}\right )}{c^3 (c-i x)}+\frac{\log \left (1-\frac{i c}{x}\right )}{c x^3}+\frac{i \log \left (1-\frac{i c}{x}\right )}{c^2 x^2}-\frac{\log \left (1-\frac{i c}{x}\right )}{c^3 x}\right ) \, dx-\frac{1}{4} \left (b^2 c\right ) \int \left (\frac{i \log \left (1+\frac{i c}{x}\right )}{c^3 (c+i x)}+\frac{\log \left (1+\frac{i c}{x}\right )}{c x^3}-\frac{i \log \left (1+\frac{i c}{x}\right )}{c^2 x^2}-\frac{\log \left (1+\frac{i c}{x}\right )}{c^3 x}\right ) \, dx\\ &=-\frac{b^2 \left (1-\frac{i c}{x}\right )^2}{16 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2}{16 c^2}-\frac{i a b}{4 x^2}+\frac{3 a b}{2 c x}+\frac{i b^2}{2 c x}+\frac{i a b \log \left (i-\frac{c}{x}\right )}{2 c^2}-\frac{i b \left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )}{8 c^2}-\frac{\left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{4 c^2}+\frac{\left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{2 c^2}+\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 c^2}+\frac{i a b \log \left (1+\frac{i c}{x}\right )}{2 x^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 x^2}+\frac{b^2 \left (1+\frac{i c}{x}\right ) \log ^2\left (1+\frac{i c}{x}\right )}{4 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log ^2\left (1+\frac{i c}{x}\right )}{8 c^2}-\frac{1}{4} b^2 \int \frac{\log \left (1-\frac{i c}{x}\right )}{x^3} \, dx-\frac{1}{4} b^2 \int \frac{\log \left (1+\frac{i c}{x}\right )}{x^3} \, dx+\frac{\left (i b^2\right ) \int \frac{\log \left (1-\frac{i c}{x}\right )}{c-i x} \, dx}{4 c^2}-\frac{\left (i b^2\right ) \int \frac{\log \left (1+\frac{i c}{x}\right )}{c+i x} \, dx}{4 c^2}+\frac{b^2 \int \frac{\log \left (1-\frac{i c}{x}\right )}{x} \, dx}{4 c^2}+\frac{b^2 \int \frac{\log \left (1+\frac{i c}{x}\right )}{x} \, dx}{4 c^2}-\frac{b^2 \operatorname{Subst}\left (\int \log (x) \, dx,x,1-\frac{i c}{x}\right )}{2 c^2}-\frac{\left (i b^2\right ) \int \frac{\log \left (1-\frac{i c}{x}\right )}{x^2} \, dx}{4 c}+\frac{\left (i b^2\right ) \int \frac{\log \left (1+\frac{i c}{x}\right )}{x^2} \, dx}{4 c}\\ &=-\frac{b^2 \left (1-\frac{i c}{x}\right )^2}{16 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2}{16 c^2}-\frac{i a b}{4 x^2}+\frac{3 a b}{2 c x}+\frac{i a b \log \left (i-\frac{c}{x}\right )}{2 c^2}-\frac{b^2 \left (1-\frac{i c}{x}\right ) \log \left (1-\frac{i c}{x}\right )}{2 c^2}-\frac{i b \left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )}{8 c^2}-\frac{\left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{4 c^2}+\frac{\left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{2 c^2}+\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 c^2}+\frac{i a b \log \left (1+\frac{i c}{x}\right )}{2 x^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 x^2}+\frac{b^2 \left (1+\frac{i c}{x}\right ) \log ^2\left (1+\frac{i c}{x}\right )}{4 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log ^2\left (1+\frac{i c}{x}\right )}{8 c^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log (c-i x)}{4 c^2}-\frac{b^2 \log \left (1+\frac{i c}{x}\right ) \log (c+i x)}{4 c^2}+\frac{b^2 \text{Li}_2\left (-\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \text{Li}_2\left (\frac{i c}{x}\right )}{4 c^2}+\frac{1}{4} b^2 \operatorname{Subst}\left (\int x \log (1-i c x) \, dx,x,\frac{1}{x}\right )+\frac{1}{4} b^2 \operatorname{Subst}\left (\int x \log (1+i c x) \, dx,x,\frac{1}{x}\right )+\frac{\left (i b^2\right ) \int \frac{\log (c-i x)}{\left (1-\frac{i c}{x}\right ) x^2} \, dx}{4 c}-\frac{\left (i b^2\right ) \int \frac{\log (c+i x)}{\left (1+\frac{i c}{x}\right ) x^2} \, dx}{4 c}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log (1-i c x) \, dx,x,\frac{1}{x}\right )}{4 c}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log (1+i c x) \, dx,x,\frac{1}{x}\right )}{4 c}\\ &=-\frac{b^2 \left (1-\frac{i c}{x}\right )^2}{16 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2}{16 c^2}-\frac{i a b}{4 x^2}+\frac{3 a b}{2 c x}+\frac{i a b \log \left (i-\frac{c}{x}\right )}{2 c^2}-\frac{b^2 \left (1-\frac{i c}{x}\right ) \log \left (1-\frac{i c}{x}\right )}{2 c^2}+\frac{b^2 \log \left (1-\frac{i c}{x}\right )}{8 x^2}-\frac{i b \left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )}{8 c^2}-\frac{\left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{4 c^2}+\frac{\left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{2 c^2}+\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 c^2}+\frac{i a b \log \left (1+\frac{i c}{x}\right )}{2 x^2}+\frac{b^2 \log \left (1+\frac{i c}{x}\right )}{8 x^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 x^2}+\frac{b^2 \left (1+\frac{i c}{x}\right ) \log ^2\left (1+\frac{i c}{x}\right )}{4 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log ^2\left (1+\frac{i c}{x}\right )}{8 c^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log (c-i x)}{4 c^2}-\frac{b^2 \log \left (1+\frac{i c}{x}\right ) \log (c+i x)}{4 c^2}+\frac{b^2 \text{Li}_2\left (-\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \text{Li}_2\left (\frac{i c}{x}\right )}{4 c^2}-\frac{b^2 \operatorname{Subst}\left (\int \log (x) \, dx,x,1-\frac{i c}{x}\right )}{4 c^2}-\frac{b^2 \operatorname{Subst}\left (\int \log (x) \, dx,x,1+\frac{i c}{x}\right )}{4 c^2}+\frac{\left (i b^2\right ) \int \left (\frac{\log (c-i x)}{c (c+i x)}+\frac{i \log (c-i x)}{c x}\right ) \, dx}{4 c}-\frac{\left (i b^2\right ) \int \left (\frac{\log (c+i x)}{c (c-i x)}-\frac{i \log (c+i x)}{c x}\right ) \, dx}{4 c}+\frac{1}{8} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-i c x} \, dx,x,\frac{1}{x}\right )-\frac{1}{8} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{x^2}{1+i c x} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{b^2 \left (1-\frac{i c}{x}\right )^2}{16 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2}{16 c^2}-\frac{i a b}{4 x^2}+\frac{3 a b}{2 c x}+\frac{i a b \log \left (i-\frac{c}{x}\right )}{2 c^2}-\frac{3 b^2 \left (1-\frac{i c}{x}\right ) \log \left (1-\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \log \left (1-\frac{i c}{x}\right )}{8 x^2}-\frac{i b \left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )}{8 c^2}-\frac{\left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{4 c^2}+\frac{\left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c^2}-\frac{3 b^2 \left (1+\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 c^2}+\frac{i a b \log \left (1+\frac{i c}{x}\right )}{2 x^2}+\frac{b^2 \log \left (1+\frac{i c}{x}\right )}{8 x^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 x^2}+\frac{b^2 \left (1+\frac{i c}{x}\right ) \log ^2\left (1+\frac{i c}{x}\right )}{4 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log ^2\left (1+\frac{i c}{x}\right )}{8 c^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log (c-i x)}{4 c^2}-\frac{b^2 \log \left (1+\frac{i c}{x}\right ) \log (c+i x)}{4 c^2}+\frac{b^2 \text{Li}_2\left (-\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \text{Li}_2\left (\frac{i c}{x}\right )}{4 c^2}+\frac{\left (i b^2\right ) \int \frac{\log (c-i x)}{c+i x} \, dx}{4 c^2}-\frac{\left (i b^2\right ) \int \frac{\log (c+i x)}{c-i x} \, dx}{4 c^2}-\frac{b^2 \int \frac{\log (c-i x)}{x} \, dx}{4 c^2}-\frac{b^2 \int \frac{\log (c+i x)}{x} \, dx}{4 c^2}-\frac{1}{8} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \left (\frac{1}{c^2}-\frac{i x}{c}+\frac{i}{c^2 (-i+c x)}\right ) \, dx,x,\frac{1}{x}\right )+\frac{1}{8} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \left (\frac{1}{c^2}+\frac{i x}{c}-\frac{i}{c^2 (i+c x)}\right ) \, dx,x,\frac{1}{x}\right )\\ &=-\frac{b^2 \left (1-\frac{i c}{x}\right )^2}{16 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2}{16 c^2}-\frac{i a b}{4 x^2}-\frac{b^2}{8 x^2}+\frac{3 a b}{2 c x}+\frac{i a b \log \left (i-\frac{c}{x}\right )}{2 c^2}+\frac{b^2 \log \left (i-\frac{c}{x}\right )}{8 c^2}-\frac{3 b^2 \left (1-\frac{i c}{x}\right ) \log \left (1-\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \log \left (1-\frac{i c}{x}\right )}{8 x^2}-\frac{i b \left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )}{8 c^2}-\frac{\left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{4 c^2}+\frac{\left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c^2}-\frac{3 b^2 \left (1+\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 c^2}+\frac{i a b \log \left (1+\frac{i c}{x}\right )}{2 x^2}+\frac{b^2 \log \left (1+\frac{i c}{x}\right )}{8 x^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 x^2}+\frac{b^2 \left (1+\frac{i c}{x}\right ) \log ^2\left (1+\frac{i c}{x}\right )}{4 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log ^2\left (1+\frac{i c}{x}\right )}{8 c^2}+\frac{b^2 \log \left (i+\frac{c}{x}\right )}{8 c^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log (c-i x)}{4 c^2}-\frac{b^2 \log \left (1+\frac{i c}{x}\right ) \log (c+i x)}{4 c^2}+\frac{b^2 \log \left (\frac{c-i x}{2 c}\right ) \log (c+i x)}{4 c^2}+\frac{b^2 \log (c-i x) \log \left (\frac{c+i x}{2 c}\right )}{4 c^2}-\frac{b^2 \log (c+i x) \log \left (-\frac{i x}{c}\right )}{4 c^2}-\frac{b^2 \log (c-i x) \log \left (\frac{i x}{c}\right )}{4 c^2}+\frac{b^2 \text{Li}_2\left (-\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \text{Li}_2\left (\frac{i c}{x}\right )}{4 c^2}-\frac{\left (i b^2\right ) \int \frac{\log \left (\frac{c-i x}{2 c}\right )}{c+i x} \, dx}{4 c^2}+\frac{\left (i b^2\right ) \int \frac{\log \left (\frac{c+i x}{2 c}\right )}{c-i x} \, dx}{4 c^2}+\frac{\left (i b^2\right ) \int \frac{\log \left (-\frac{i x}{c}\right )}{c+i x} \, dx}{4 c^2}-\frac{\left (i b^2\right ) \int \frac{\log \left (\frac{i x}{c}\right )}{c-i x} \, dx}{4 c^2}\\ &=-\frac{b^2 \left (1-\frac{i c}{x}\right )^2}{16 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2}{16 c^2}-\frac{i a b}{4 x^2}-\frac{b^2}{8 x^2}+\frac{3 a b}{2 c x}+\frac{i a b \log \left (i-\frac{c}{x}\right )}{2 c^2}+\frac{b^2 \log \left (i-\frac{c}{x}\right )}{8 c^2}-\frac{3 b^2 \left (1-\frac{i c}{x}\right ) \log \left (1-\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \log \left (1-\frac{i c}{x}\right )}{8 x^2}-\frac{i b \left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )}{8 c^2}-\frac{\left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{4 c^2}+\frac{\left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c^2}-\frac{3 b^2 \left (1+\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 c^2}+\frac{i a b \log \left (1+\frac{i c}{x}\right )}{2 x^2}+\frac{b^2 \log \left (1+\frac{i c}{x}\right )}{8 x^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 x^2}+\frac{b^2 \left (1+\frac{i c}{x}\right ) \log ^2\left (1+\frac{i c}{x}\right )}{4 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log ^2\left (1+\frac{i c}{x}\right )}{8 c^2}+\frac{b^2 \log \left (i+\frac{c}{x}\right )}{8 c^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log (c-i x)}{4 c^2}-\frac{b^2 \log \left (1+\frac{i c}{x}\right ) \log (c+i x)}{4 c^2}+\frac{b^2 \log \left (\frac{c-i x}{2 c}\right ) \log (c+i x)}{4 c^2}+\frac{b^2 \log (c-i x) \log \left (\frac{c+i x}{2 c}\right )}{4 c^2}-\frac{b^2 \log (c+i x) \log \left (-\frac{i x}{c}\right )}{4 c^2}-\frac{b^2 \log (c-i x) \log \left (\frac{i x}{c}\right )}{4 c^2}+\frac{b^2 \text{Li}_2\left (-\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \text{Li}_2\left (\frac{i c}{x}\right )}{4 c^2}-\frac{b^2 \text{Li}_2\left (1-\frac{i x}{c}\right )}{4 c^2}-\frac{b^2 \text{Li}_2\left (1+\frac{i x}{c}\right )}{4 c^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2 c}\right )}{x} \, dx,x,c-i x\right )}{4 c^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2 c}\right )}{x} \, dx,x,c+i x\right )}{4 c^2}\\ &=-\frac{b^2 \left (1-\frac{i c}{x}\right )^2}{16 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2}{16 c^2}-\frac{i a b}{4 x^2}-\frac{b^2}{8 x^2}+\frac{3 a b}{2 c x}+\frac{i a b \log \left (i-\frac{c}{x}\right )}{2 c^2}+\frac{b^2 \log \left (i-\frac{c}{x}\right )}{8 c^2}-\frac{3 b^2 \left (1-\frac{i c}{x}\right ) \log \left (1-\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \log \left (1-\frac{i c}{x}\right )}{8 x^2}-\frac{i b \left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )}{8 c^2}-\frac{\left (1-\frac{i c}{x}\right ) \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{4 c^2}+\frac{\left (1-\frac{i c}{x}\right )^2 \left (2 a+i b \log \left (1-\frac{i c}{x}\right )\right )^2}{8 c^2}-\frac{3 b^2 \left (1+\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log \left (1+\frac{i c}{x}\right )}{8 c^2}+\frac{i a b \log \left (1+\frac{i c}{x}\right )}{2 x^2}+\frac{b^2 \log \left (1+\frac{i c}{x}\right )}{8 x^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log \left (1+\frac{i c}{x}\right )}{4 x^2}+\frac{b^2 \left (1+\frac{i c}{x}\right ) \log ^2\left (1+\frac{i c}{x}\right )}{4 c^2}-\frac{b^2 \left (1+\frac{i c}{x}\right )^2 \log ^2\left (1+\frac{i c}{x}\right )}{8 c^2}+\frac{b^2 \log \left (i+\frac{c}{x}\right )}{8 c^2}-\frac{b^2 \log \left (1-\frac{i c}{x}\right ) \log (c-i x)}{4 c^2}-\frac{b^2 \log \left (1+\frac{i c}{x}\right ) \log (c+i x)}{4 c^2}+\frac{b^2 \log \left (\frac{c-i x}{2 c}\right ) \log (c+i x)}{4 c^2}+\frac{b^2 \log (c-i x) \log \left (\frac{c+i x}{2 c}\right )}{4 c^2}-\frac{b^2 \log (c+i x) \log \left (-\frac{i x}{c}\right )}{4 c^2}-\frac{b^2 \log (c-i x) \log \left (\frac{i x}{c}\right )}{4 c^2}+\frac{b^2 \text{Li}_2\left (\frac{c-i x}{2 c}\right )}{4 c^2}+\frac{b^2 \text{Li}_2\left (\frac{c+i x}{2 c}\right )}{4 c^2}+\frac{b^2 \text{Li}_2\left (-\frac{i c}{x}\right )}{4 c^2}+\frac{b^2 \text{Li}_2\left (\frac{i c}{x}\right )}{4 c^2}-\frac{b^2 \text{Li}_2\left (1-\frac{i x}{c}\right )}{4 c^2}-\frac{b^2 \text{Li}_2\left (1+\frac{i x}{c}\right )}{4 c^2}\\ \end{align*}

Mathematica [A]  time = 0.073959, size = 99, normalized size = 1.18 \[ -\frac{a^2 c^2-2 a b x^2 \tan ^{-1}\left (\frac{x}{c}\right )-2 a b c x+2 b c \tan ^{-1}\left (\frac{c}{x}\right ) (a c-b x)+b^2 x^2 \log \left (c^2+x^2\right )+b^2 \left (c^2+x^2\right ) \tan ^{-1}\left (\frac{c}{x}\right )^2-2 b^2 x^2 \log (x)}{2 c^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*c^2 - 2*a*b*c*x + 2*b*c*(a*c - b*x)*ArcTan[c/x] + b^2*(c^2 + x^2)*ArcTan[c/x]^2 - 2*a*b*x^2*ArcTan[x/c]
- 2*b^2*x^2*Log[x] + b^2*x^2*Log[c^2 + x^2])/(2*c^2*x^2)

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Maple [A]  time = 0.034, size = 110, normalized size = 1.3 \begin{align*} -{\frac{{a}^{2}}{2\,{x}^{2}}}-{\frac{{b}^{2}}{2\,{x}^{2}} \left ( \arctan \left ({\frac{c}{x}} \right ) \right ) ^{2}}-{\frac{{b}^{2}}{2\,{c}^{2}} \left ( \arctan \left ({\frac{c}{x}} \right ) \right ) ^{2}}+{\frac{{b}^{2}}{cx}\arctan \left ({\frac{c}{x}} \right ) }-{\frac{{b}^{2}}{2\,{c}^{2}}\ln \left ( 1+{\frac{{c}^{2}}{{x}^{2}}} \right ) }-{\frac{ab}{{x}^{2}}\arctan \left ({\frac{c}{x}} \right ) }+{\frac{ab}{{c}^{2}}\arctan \left ({\frac{x}{c}} \right ) }+{\frac{ab}{cx}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a^2/x^2-1/2/x^2*b^2*arctan(c/x)^2-1/2/c^2*b^2*arctan(c/x)^2+1/c*b^2*arctan(c/x)/x-1/2*b^2*ln(1+c^2/x^2)/c
^2-b/x^2*a*arctan(c/x)+1/c^2*a*b*arctan(x/c)+a*b/c/x

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Maxima [A]  time = 1.62002, size = 158, normalized size = 1.88 \begin{align*}{\left (c{\left (\frac{\arctan \left (\frac{x}{c}\right )}{c^{3}} + \frac{1}{c^{2} x}\right )} - \frac{\arctan \left (\frac{c}{x}\right )}{x^{2}}\right )} a b + \frac{1}{2} \,{\left (2 \, c{\left (\frac{\arctan \left (\frac{x}{c}\right )}{c^{3}} + \frac{1}{c^{2} x}\right )} \arctan \left (\frac{c}{x}\right ) + \frac{\arctan \left (x, c\right )^{2} - \log \left (c^{2} + x^{2}\right ) + 2 \, \log \left (x\right )}{c^{2}}\right )} b^{2} - \frac{b^{2} \arctan \left (\frac{c}{x}\right )^{2}}{2 \, x^{2}} - \frac{a^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*(arctan(x/c)/c^3 + 1/(c^2*x)) - arctan(c/x)/x^2)*a*b + 1/2*(2*c*(arctan(x/c)/c^3 + 1/(c^2*x))*arctan(c/x) +
(arctan2(x, c)^2 - log(c^2 + x^2) + 2*log(x))/c^2)*b^2 - 1/2*b^2*arctan(c/x)^2/x^2 - 1/2*a^2/x^2

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Fricas [A]  time = 2.31232, size = 239, normalized size = 2.85 \begin{align*} \frac{2 \, a b x^{2} \arctan \left (\frac{x}{c}\right ) - b^{2} x^{2} \log \left (c^{2} + x^{2}\right ) + 2 \, b^{2} x^{2} \log \left (x\right ) - a^{2} c^{2} + 2 \, a b c x -{\left (b^{2} c^{2} + b^{2} x^{2}\right )} \arctan \left (\frac{c}{x}\right )^{2} - 2 \,{\left (a b c^{2} - b^{2} c x\right )} \arctan \left (\frac{c}{x}\right )}{2 \, c^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a*b*x^2*arctan(x/c) - b^2*x^2*log(c^2 + x^2) + 2*b^2*x^2*log(x) - a^2*c^2 + 2*a*b*c*x - (b^2*c^2 + b^2*
x^2)*arctan(c/x)^2 - 2*(a*b*c^2 - b^2*c*x)*arctan(c/x))/(c^2*x^2)

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Sympy [A]  time = 1.43797, size = 117, normalized size = 1.39 \begin{align*} \begin{cases} - \frac{a^{2}}{2 x^{2}} - \frac{a b \operatorname{atan}{\left (\frac{c}{x} \right )}}{x^{2}} + \frac{a b}{c x} - \frac{a b \operatorname{atan}{\left (\frac{c}{x} \right )}}{c^{2}} - \frac{b^{2} \operatorname{atan}^{2}{\left (\frac{c}{x} \right )}}{2 x^{2}} + \frac{b^{2} \operatorname{atan}{\left (\frac{c}{x} \right )}}{c x} + \frac{b^{2} \log{\left (x \right )}}{c^{2}} - \frac{b^{2} \log{\left (c^{2} + x^{2} \right )}}{2 c^{2}} - \frac{b^{2} \operatorname{atan}^{2}{\left (\frac{c}{x} \right )}}{2 c^{2}} & \text{for}\: c \neq 0 \\- \frac{a^{2}}{2 x^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2/(2*x**2) - a*b*atan(c/x)/x**2 + a*b/(c*x) - a*b*atan(c/x)/c**2 - b**2*atan(c/x)**2/(2*x**2) +
b**2*atan(c/x)/(c*x) + b**2*log(x)/c**2 - b**2*log(c**2 + x**2)/(2*c**2) - b**2*atan(c/x)**2/(2*c**2), Ne(c,
0)), (-a**2/(2*x**2), True))

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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 )}^{2}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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