3.65 \(\int \frac{a+b \tan ^{-1}(c x)}{x^3 (d+i c d x)^3} \, dx\)

Optimal. Leaf size=306 \[ -\frac{3 i b c^2 \text{PolyLog}(2,-i c x)}{d^3}+\frac{3 i b c^2 \text{PolyLog}(2,i c x)}{d^3}-\frac{3 i b c^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{d^3}-\frac{3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-c x+i)}+\frac{c^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (-c x+i)^2}-\frac{6 c^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}-\frac{a+b \tan ^{-1}(c x)}{2 d^3 x^2}+\frac{3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}-\frac{6 a c^2 \log (x)}{d^3}+\frac{3 i b c^2 \log \left (c^2 x^2+1\right )}{2 d^3}-\frac{13 b c^2}{8 d^3 (-c x+i)}-\frac{i b c^2}{8 d^3 (-c x+i)^2}-\frac{3 i b c^2 \log (x)}{d^3}+\frac{9 b c^2 \tan ^{-1}(c x)}{8 d^3}-\frac{b c}{2 d^3 x} \]

[Out]

-(b*c)/(2*d^3*x) - ((I/8)*b*c^2)/(d^3*(I - c*x)^2) - (13*b*c^2)/(8*d^3*(I - c*x)) + (9*b*c^2*ArcTan[c*x])/(8*d
^3) - (a + b*ArcTan[c*x])/(2*d^3*x^2) + ((3*I)*c*(a + b*ArcTan[c*x]))/(d^3*x) + (c^2*(a + b*ArcTan[c*x]))/(2*d
^3*(I - c*x)^2) - ((3*I)*c^2*(a + b*ArcTan[c*x]))/(d^3*(I - c*x)) - (6*a*c^2*Log[x])/d^3 - ((3*I)*b*c^2*Log[x]
)/d^3 - (6*c^2*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/d^3 + (((3*I)/2)*b*c^2*Log[1 + c^2*x^2])/d^3 - ((3*I)*b
*c^2*PolyLog[2, (-I)*c*x])/d^3 + ((3*I)*b*c^2*PolyLog[2, I*c*x])/d^3 - ((3*I)*b*c^2*PolyLog[2, 1 - 2/(1 + I*c*
x)])/d^3

________________________________________________________________________________________

Rubi [A]  time = 0.320179, antiderivative size = 306, normalized size of antiderivative = 1., number of steps used = 26, number of rules used = 16, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.696, Rules used = {4876, 4852, 325, 203, 266, 36, 29, 31, 4848, 2391, 4862, 627, 44, 4854, 2402, 2315} \[ -\frac{3 i b c^2 \text{PolyLog}(2,-i c x)}{d^3}+\frac{3 i b c^2 \text{PolyLog}(2,i c x)}{d^3}-\frac{3 i b c^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{d^3}-\frac{3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-c x+i)}+\frac{c^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (-c x+i)^2}-\frac{6 c^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}-\frac{a+b \tan ^{-1}(c x)}{2 d^3 x^2}+\frac{3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}-\frac{6 a c^2 \log (x)}{d^3}+\frac{3 i b c^2 \log \left (c^2 x^2+1\right )}{2 d^3}-\frac{13 b c^2}{8 d^3 (-c x+i)}-\frac{i b c^2}{8 d^3 (-c x+i)^2}-\frac{3 i b c^2 \log (x)}{d^3}+\frac{9 b c^2 \tan ^{-1}(c x)}{8 d^3}-\frac{b c}{2 d^3 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])/(x^3*(d + I*c*d*x)^3),x]

[Out]

-(b*c)/(2*d^3*x) - ((I/8)*b*c^2)/(d^3*(I - c*x)^2) - (13*b*c^2)/(8*d^3*(I - c*x)) + (9*b*c^2*ArcTan[c*x])/(8*d
^3) - (a + b*ArcTan[c*x])/(2*d^3*x^2) + ((3*I)*c*(a + b*ArcTan[c*x]))/(d^3*x) + (c^2*(a + b*ArcTan[c*x]))/(2*d
^3*(I - c*x)^2) - ((3*I)*c^2*(a + b*ArcTan[c*x]))/(d^3*(I - c*x)) - (6*a*c^2*Log[x])/d^3 - ((3*I)*b*c^2*Log[x]
)/d^3 - (6*c^2*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/d^3 + (((3*I)/2)*b*c^2*Log[1 + c^2*x^2])/d^3 - ((3*I)*b
*c^2*PolyLog[2, (-I)*c*x])/d^3 + ((3*I)*b*c^2*PolyLog[2, I*c*x])/d^3 - ((3*I)*b*c^2*PolyLog[2, 1 - 2/(1 + I*c*
x)])/d^3

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 4852

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 203

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 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 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 4862

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

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 + c^2*x^
2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[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 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{a+b \tan ^{-1}(c x)}{x^3 (d+i c d x)^3} \, dx &=\int \left (\frac{a+b \tan ^{-1}(c x)}{d^3 x^3}-\frac{3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x^2}-\frac{6 c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}-\frac{c^3 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-i+c x)^3}-\frac{3 i c^3 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-i+c x)^2}+\frac{6 c^3 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-i+c x)}\right ) \, dx\\ &=\frac{\int \frac{a+b \tan ^{-1}(c x)}{x^3} \, dx}{d^3}-\frac{(3 i c) \int \frac{a+b \tan ^{-1}(c x)}{x^2} \, dx}{d^3}-\frac{\left (6 c^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{x} \, dx}{d^3}-\frac{\left (3 i c^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{d^3}-\frac{c^3 \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{d^3}+\frac{\left (6 c^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{-i+c x} \, dx}{d^3}\\ &=-\frac{a+b \tan ^{-1}(c x)}{2 d^3 x^2}+\frac{3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}+\frac{c^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}-\frac{3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac{6 a c^2 \log (x)}{d^3}-\frac{6 c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{d^3}+\frac{(b c) \int \frac{1}{x^2 \left (1+c^2 x^2\right )} \, dx}{2 d^3}-\frac{\left (3 i b c^2\right ) \int \frac{1}{x \left (1+c^2 x^2\right )} \, dx}{d^3}-\frac{\left (3 i b c^2\right ) \int \frac{\log (1-i c x)}{x} \, dx}{d^3}+\frac{\left (3 i b c^2\right ) \int \frac{\log (1+i c x)}{x} \, dx}{d^3}-\frac{\left (3 i b c^3\right ) \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{d^3}-\frac{\left (b c^3\right ) \int \frac{1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{2 d^3}+\frac{\left (6 b c^3\right ) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d^3}\\ &=-\frac{b c}{2 d^3 x}-\frac{a+b \tan ^{-1}(c x)}{2 d^3 x^2}+\frac{3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}+\frac{c^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}-\frac{3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac{6 a c^2 \log (x)}{d^3}-\frac{6 c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{d^3}-\frac{3 i b c^2 \text{Li}_2(-i c x)}{d^3}+\frac{3 i b c^2 \text{Li}_2(i c x)}{d^3}-\frac{\left (3 i b c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{2 d^3}-\frac{\left (6 i b c^2\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{d^3}-\frac{\left (3 i b c^3\right ) \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{d^3}-\frac{\left (b c^3\right ) \int \frac{1}{(-i+c x)^3 (i+c x)} \, dx}{2 d^3}-\frac{\left (b c^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{2 d^3}\\ &=-\frac{b c}{2 d^3 x}-\frac{b c^2 \tan ^{-1}(c x)}{2 d^3}-\frac{a+b \tan ^{-1}(c x)}{2 d^3 x^2}+\frac{3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}+\frac{c^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}-\frac{3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac{6 a c^2 \log (x)}{d^3}-\frac{6 c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{d^3}-\frac{3 i b c^2 \text{Li}_2(-i c x)}{d^3}+\frac{3 i b c^2 \text{Li}_2(i c x)}{d^3}-\frac{3 i b c^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{d^3}-\frac{\left (3 i b c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )}{2 d^3}-\frac{\left (3 i b c^3\right ) \int \left (-\frac{i}{2 (-i+c x)^2}+\frac{i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{d^3}-\frac{\left (b c^3\right ) \int \left (-\frac{i}{2 (-i+c x)^3}+\frac{1}{4 (-i+c x)^2}-\frac{1}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{2 d^3}+\frac{\left (3 i b c^4\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x} \, dx,x,x^2\right )}{2 d^3}\\ &=-\frac{b c}{2 d^3 x}-\frac{i b c^2}{8 d^3 (i-c x)^2}-\frac{13 b c^2}{8 d^3 (i-c x)}-\frac{b c^2 \tan ^{-1}(c x)}{2 d^3}-\frac{a+b \tan ^{-1}(c x)}{2 d^3 x^2}+\frac{3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}+\frac{c^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}-\frac{3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac{6 a c^2 \log (x)}{d^3}-\frac{3 i b c^2 \log (x)}{d^3}-\frac{6 c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{d^3}+\frac{3 i b c^2 \log \left (1+c^2 x^2\right )}{2 d^3}-\frac{3 i b c^2 \text{Li}_2(-i c x)}{d^3}+\frac{3 i b c^2 \text{Li}_2(i c x)}{d^3}-\frac{3 i b c^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{d^3}+\frac{\left (b c^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{8 d^3}+\frac{\left (3 b c^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{2 d^3}\\ &=-\frac{b c}{2 d^3 x}-\frac{i b c^2}{8 d^3 (i-c x)^2}-\frac{13 b c^2}{8 d^3 (i-c x)}+\frac{9 b c^2 \tan ^{-1}(c x)}{8 d^3}-\frac{a+b \tan ^{-1}(c x)}{2 d^3 x^2}+\frac{3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}+\frac{c^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}-\frac{3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac{6 a c^2 \log (x)}{d^3}-\frac{3 i b c^2 \log (x)}{d^3}-\frac{6 c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{d^3}+\frac{3 i b c^2 \log \left (1+c^2 x^2\right )}{2 d^3}-\frac{3 i b c^2 \text{Li}_2(-i c x)}{d^3}+\frac{3 i b c^2 \text{Li}_2(i c x)}{d^3}-\frac{3 i b c^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{d^3}\\ \end{align*}

Mathematica [C]  time = 0.54173, size = 285, normalized size = 0.93 \[ -\frac{\frac{4 b c \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-c^2 x^2\right )}{x}+24 i b c^2 \text{PolyLog}(2,-i c x)-24 i b c^2 \text{PolyLog}(2,i c x)+24 i b c^2 \text{PolyLog}\left (2,\frac{c x+i}{c x-i}\right )-\frac{24 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{c x-i}-\frac{4 c^2 \left (a+b \tan ^{-1}(c x)\right )}{(c x-i)^2}+48 c^2 \log \left (\frac{2 i}{-c x+i}\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac{4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-\frac{24 i c \left (a+b \tan ^{-1}(c x)\right )}{x}+48 a c^2 \log (x)+12 i b c^2 \left (2 \log (x)-\log \left (c^2 x^2+1\right )\right )+12 b c^2 \left (-\tan ^{-1}(c x)+\frac{1}{-c x+i}\right )-\frac{b c^2 \left (c x+(c x-i)^2 \tan ^{-1}(c x)-2 i\right )}{(c x-i)^2}}{8 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(12*b*c^2*((I - c*x)^(-1) - ArcTan[c*x]) + (4*(a + b*ArcTan[c*x]))/x^2 - ((24*I)*c*(a + b*ArcTan[c*x]))/x - (
4*c^2*(a + b*ArcTan[c*x]))/(-I + c*x)^2 - ((24*I)*c^2*(a + b*ArcTan[c*x]))/(-I + c*x) - (b*c^2*(-2*I + c*x + (
-I + c*x)^2*ArcTan[c*x]))/(-I + c*x)^2 + (4*b*c*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)])/x + 48*a*c^2*Log[
x] + 48*c^2*(a + b*ArcTan[c*x])*Log[(2*I)/(I - c*x)] + (12*I)*b*c^2*(2*Log[x] - Log[1 + c^2*x^2]) + (24*I)*b*c
^2*PolyLog[2, (-I)*c*x] - (24*I)*b*c^2*PolyLog[2, I*c*x] + (24*I)*b*c^2*PolyLog[2, (I + c*x)/(-I + c*x)])/(8*d
^3)

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Maple [A]  time = 0.075, size = 481, normalized size = 1.6 \begin{align*}{\frac{-3\,i{c}^{2}b\ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) \ln \left ( cx-i \right ) }{{d}^{3}}}+{\frac{3\,icb\arctan \left ( cx \right ) }{{d}^{3}x}}+{\frac{3\,i{c}^{2}b\ln \left ( -i \left ( cx+i \right ) \right ) \ln \left ( cx \right ) }{{d}^{3}}}+6\,{\frac{{c}^{2}b\arctan \left ( cx \right ) \ln \left ( cx-i \right ) }{{d}^{3}}}-{\frac{bc}{2\,{d}^{3}x}}+{\frac{9\,{c}^{2}b\arctan \left ( cx \right ) }{8\,{d}^{3}}}-{\frac{b\arctan \left ( cx \right ) }{2\,{d}^{3}{x}^{2}}}+3\,{\frac{{c}^{2}a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{{d}^{3}}}+{\frac{13\,{c}^{2}b}{8\,{d}^{3} \left ( cx-i \right ) }}-6\,{\frac{{c}^{2}a\ln \left ( cx \right ) }{{d}^{3}}}-{\frac{a}{2\,{d}^{3}{x}^{2}}}+{\frac{{\frac{3\,i}{2}}{c}^{2}b \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{{d}^{3}}}-{\frac{3\,i{c}^{2}b{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{d}^{3}}}+{\frac{3\,i{c}^{2}b{\it dilog} \left ( -icx \right ) }{{d}^{3}}}+{\frac{6\,i{c}^{2}a\arctan \left ( cx \right ) }{{d}^{3}}}+{\frac{3\,i{c}^{2}b{\it dilog} \left ( -i \left ( cx+i \right ) \right ) }{{d}^{3}}}+{\frac{{c}^{2}a}{2\,{d}^{3} \left ( cx-i \right ) ^{2}}}+{\frac{{c}^{2}b\arctan \left ( cx \right ) }{2\,{d}^{3} \left ( cx-i \right ) ^{2}}}-6\,{\frac{{c}^{2}b\arctan \left ( cx \right ) \ln \left ( cx \right ) }{{d}^{3}}}+{\frac{3\,ica}{{d}^{3}x}}+{\frac{3\,i{c}^{2}a}{{d}^{3} \left ( cx-i \right ) }}-{\frac{3\,i{c}^{2}b\ln \left ( cx \right ) }{{d}^{3}}}+{\frac{3\,i{c}^{2}b\ln \left ( -icx \right ) \ln \left ( -i \left ( -cx+i \right ) \right ) }{{d}^{3}}}+{\frac{{\frac{3\,i}{2}}{c}^{2}b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{{d}^{3}}}-{\frac{3\,i{c}^{2}b\ln \left ( -i \left ( -cx+i \right ) \right ) \ln \left ( cx \right ) }{{d}^{3}}}-{\frac{{\frac{i}{8}}{c}^{2}b}{{d}^{3} \left ( cx-i \right ) ^{2}}}+{\frac{3\,i{c}^{2}b\arctan \left ( cx \right ) }{{d}^{3} \left ( cx-i \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*I*c^2*b/d^3*ln(-1/2*I*(c*x+I))*ln(c*x-I)+3*I*c*b/d^3*arctan(c*x)/x+3*I*c^2*b/d^3*ln(-I*(c*x+I))*ln(c*x)+6*c
^2*b/d^3*arctan(c*x)*ln(c*x-I)-1/2*b*c/d^3/x+9/8*b*c^2*arctan(c*x)/d^3-1/2*b/d^3*arctan(c*x)/x^2+3*c^2*a/d^3*l
n(c^2*x^2+1)+13/8*c^2*b/d^3/(c*x-I)-6*c^2*a/d^3*ln(c*x)-1/2*a/d^3/x^2+3/2*I*c^2*b/d^3*ln(c*x-I)^2-3*I*c^2*b/d^
3*dilog(-1/2*I*(c*x+I))+3*I*c^2*b/d^3*dilog(-I*c*x)+6*I*c^2*a/d^3*arctan(c*x)+3*I*c^2*b/d^3*dilog(-I*(c*x+I))+
1/2*c^2*a/d^3/(c*x-I)^2+1/2*c^2*b/d^3*arctan(c*x)/(c*x-I)^2-6*c^2*b/d^3*arctan(c*x)*ln(c*x)+3*I*c*a/d^3/x+3*I*
c^2*a/d^3/(c*x-I)-3*I*c^2*b/d^3*ln(c*x)+3*I*c^2*b/d^3*ln(-I*c*x)*ln(-I*(-c*x+I))+3/2*I*b*c^2*ln(c^2*x^2+1)/d^3
-3*I*c^2*b/d^3*ln(-I*(-c*x+I))*ln(c*x)-1/8*I*c^2*b/d^3/(c*x-I)^2+3*I*c^2*b/d^3*arctan(c*x)/(c*x-I)

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Maxima [B]  time = 1.46372, size = 829, normalized size = 2.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(33*b*c^4*x^4*arctan2(1, c*x) + (6*b*(-11*I*arctan2(1, c*x) - 3) - 96*I*a)*c^3*x^3 - (b*(33*arctan2(1, c*x) -
 12*I) + 144*a)*c^2*x^2 - (-32*I*a + 8*b)*c*x - (24*I*b*c^4*x^4 + 48*b*c^3*x^3 - 24*I*b*c^2*x^2)*arctan(c*x)^2
 - (6*I*b*c^4*x^4 + 12*b*c^3*x^3 - 6*I*b*c^2*x^2)*log(c^2*x^2 + 1)^2 - 24*(b*c^4*x^4 - 2*I*b*c^3*x^3 - b*c^2*x
^2)*arctan(c*x)*log(1/4*c^2*x^2 + 1/4) + 96*(b*c^4*x^4 - 2*I*b*c^3*x^3 - b*c^2*x^2)*arctan(c*x)*log(x*abs(c))
+ ((3*b*(32*I*arctan2(0, c) + 5) - 96*I*a)*c^4*x^4 + (b*(192*arctan2(0, c) - 126*I) - 192*a)*c^3*x^3 + (3*b*(-
32*I*arctan2(0, c) - 53) + 96*I*a)*c^2*x^2 + 32*I*b*c*x - 8*b)*arctan(c*x) - (48*I*b*c^4*x^4 + 96*b*c^3*x^3 -
48*I*b*c^2*x^2)*dilog(I*c*x + 1) - (-48*I*b*c^4*x^4 - 96*b*c^3*x^3 + 48*I*b*c^2*x^2)*dilog(1/2*I*c*x + 1/2) -
(-48*I*b*c^4*x^4 - 96*b*c^3*x^3 + 48*I*b*c^2*x^2)*dilog(-I*c*x + 1) - (((24*pi + 24*I)*b + 48*a)*c^4*x^4 - (48
*(I*pi - 1)*b + 96*I*a)*c^3*x^3 - ((24*pi + 24*I)*b + 48*a)*c^2*x^2 + (-12*I*b*c^4*x^4 - 24*b*c^3*x^3 + 12*I*b
*c^2*x^2)*log(1/4*c^2*x^2 + 1/4))*log(c^2*x^2 + 1) + (48*(2*a + I*b)*c^4*x^4 - (192*I*a - 96*b)*c^3*x^3 - 48*(
2*a + I*b)*c^2*x^2)*log(x) - 8*a)/(16*c^2*d^3*x^4 - 32*I*c*d^3*x^3 - 16*d^3*x^2)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b \log \left (-\frac{c x + i}{c x - i}\right ) - 2 i \, a}{2 \, c^{3} d^{3} x^{6} - 6 i \, c^{2} d^{3} x^{5} - 6 \, c d^{3} x^{4} + 2 i \, d^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*log(-(c*x + I)/(c*x - I)) - 2*I*a)/(2*c^3*d^3*x^6 - 6*I*c^2*d^3*x^5 - 6*c*d^3*x^4 + 2*I*d^3*x^3),
 x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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