3.93 \(\int \frac{(d+i c d x)^3 (a+b \tan ^{-1}(c x))^2}{x^6} \, dx\)

Optimal. Leaf size=384 \[ \frac{6}{5} i b^2 c^5 d^3 \text{PolyLog}(2,-i c x)-\frac{6}{5} i b^2 c^5 d^3 \text{PolyLog}(2,i c x)-\frac{6}{5} i b^2 c^5 d^3 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )+\frac{6 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^2}-\frac{i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^3}+\frac{12}{5} a b c^5 d^3 \log (x)+\frac{5 i b c^4 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}+\frac{12}{5} b c^5 d^3 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac{i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 x^4}-\frac{b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{10 x^4}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{5 x^5}-\frac{i b^2 c^3 d^3}{4 x^2}-\frac{b^2 c^2 d^3}{30 x^3}+\frac{3}{2} i b^2 c^5 d^3 \log \left (c^2 x^2+1\right )+\frac{13 b^2 c^4 d^3}{10 x}-3 i b^2 c^5 d^3 \log (x)+\frac{13}{10} b^2 c^5 d^3 \tan ^{-1}(c x) \]

[Out]

-(b^2*c^2*d^3)/(30*x^3) - ((I/4)*b^2*c^3*d^3)/x^2 + (13*b^2*c^4*d^3)/(10*x) + (13*b^2*c^5*d^3*ArcTan[c*x])/10
- (b*c*d^3*(a + b*ArcTan[c*x]))/(10*x^4) - ((I/2)*b*c^2*d^3*(a + b*ArcTan[c*x]))/x^3 + (6*b*c^3*d^3*(a + b*Arc
Tan[c*x]))/(5*x^2) + (((5*I)/2)*b*c^4*d^3*(a + b*ArcTan[c*x]))/x - (d^3*(1 + I*c*x)^4*(a + b*ArcTan[c*x])^2)/(
5*x^5) + ((I/20)*c*d^3*(1 + I*c*x)^4*(a + b*ArcTan[c*x])^2)/x^4 + (12*a*b*c^5*d^3*Log[x])/5 - (3*I)*b^2*c^5*d^
3*Log[x] + (12*b*c^5*d^3*(a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/5 + ((3*I)/2)*b^2*c^5*d^3*Log[1 + c^2*x^2] +
((6*I)/5)*b^2*c^5*d^3*PolyLog[2, (-I)*c*x] - ((6*I)/5)*b^2*c^5*d^3*PolyLog[2, I*c*x] - ((6*I)/5)*b^2*c^5*d^3*P
olyLog[2, 1 - 2/(1 - I*c*x)]

________________________________________________________________________________________

Rubi [A]  time = 0.366611, antiderivative size = 384, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 16, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.64, Rules used = {45, 37, 4874, 4852, 325, 203, 266, 44, 36, 29, 31, 4848, 2391, 4854, 2402, 2315} \[ \frac{6}{5} i b^2 c^5 d^3 \text{PolyLog}(2,-i c x)-\frac{6}{5} i b^2 c^5 d^3 \text{PolyLog}(2,i c x)-\frac{6}{5} i b^2 c^5 d^3 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )+\frac{6 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^2}-\frac{i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^3}+\frac{12}{5} a b c^5 d^3 \log (x)+\frac{5 i b c^4 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}+\frac{12}{5} b c^5 d^3 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac{i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 x^4}-\frac{b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{10 x^4}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{5 x^5}-\frac{i b^2 c^3 d^3}{4 x^2}-\frac{b^2 c^2 d^3}{30 x^3}+\frac{3}{2} i b^2 c^5 d^3 \log \left (c^2 x^2+1\right )+\frac{13 b^2 c^4 d^3}{10 x}-3 i b^2 c^5 d^3 \log (x)+\frac{13}{10} b^2 c^5 d^3 \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^3*(a + b*ArcTan[c*x])^2)/x^6,x]

[Out]

-(b^2*c^2*d^3)/(30*x^3) - ((I/4)*b^2*c^3*d^3)/x^2 + (13*b^2*c^4*d^3)/(10*x) + (13*b^2*c^5*d^3*ArcTan[c*x])/10
- (b*c*d^3*(a + b*ArcTan[c*x]))/(10*x^4) - ((I/2)*b*c^2*d^3*(a + b*ArcTan[c*x]))/x^3 + (6*b*c^3*d^3*(a + b*Arc
Tan[c*x]))/(5*x^2) + (((5*I)/2)*b*c^4*d^3*(a + b*ArcTan[c*x]))/x - (d^3*(1 + I*c*x)^4*(a + b*ArcTan[c*x])^2)/(
5*x^5) + ((I/20)*c*d^3*(1 + I*c*x)^4*(a + b*ArcTan[c*x])^2)/x^4 + (12*a*b*c^5*d^3*Log[x])/5 - (3*I)*b^2*c^5*d^
3*Log[x] + (12*b*c^5*d^3*(a + b*ArcTan[c*x])*Log[2/(1 - I*c*x)])/5 + ((3*I)/2)*b^2*c^5*d^3*Log[1 + c^2*x^2] +
((6*I)/5)*b^2*c^5*d^3*PolyLog[2, (-I)*c*x] - ((6*I)/5)*b^2*c^5*d^3*PolyLog[2, I*c*x] - ((6*I)/5)*b^2*c^5*d^3*P
olyLog[2, 1 - 2/(1 - I*c*x)]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

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

Rule 4874

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_))^(q_), x_Symbol] :> With[{u
= IntHide[(f*x)^m*(d + e*x)^q, x]}, Dist[(a + b*ArcTan[c*x])^p, u, x] - Dist[b*c*p, Int[ExpandIntegrand[(a +
b*ArcTan[c*x])^(p - 1), u/(1 + c^2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && IGtQ[p, 1] && EqQ[c
^2*d^2 + e^2, 0] && IntegersQ[m, q] && NeQ[m, -1] && NeQ[q, -1] && ILtQ[m + q + 1, 0] && LtQ[m*q, 0]

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 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 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 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{(d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x^6} \, dx &=-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{5 x^5}+\frac{i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 x^4}-(2 b c) \int \left (-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac{3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}+\frac{6 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^3}+\frac{5 i c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{4 x^2}-\frac{6 c^4 d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x}+\frac{6 c^5 d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 (i+c x)}\right ) \, dx\\ &=-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{5 x^5}+\frac{i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 x^4}+\frac{1}{5} \left (2 b c d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^5} \, dx+\frac{1}{2} \left (3 i b c^2 d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^4} \, dx-\frac{1}{5} \left (12 b c^3 d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^3} \, dx-\frac{1}{2} \left (5 i b c^4 d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^2} \, dx+\frac{1}{5} \left (12 b c^5 d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{x} \, dx-\frac{1}{5} \left (12 b c^6 d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{i+c x} \, dx\\ &=-\frac{b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{10 x^4}-\frac{i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^3}+\frac{6 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^2}+\frac{5 i b c^4 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{5 x^5}+\frac{i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 x^4}+\frac{12}{5} a b c^5 d^3 \log (x)+\frac{12}{5} b c^5 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )+\frac{1}{10} \left (b^2 c^2 d^3\right ) \int \frac{1}{x^4 \left (1+c^2 x^2\right )} \, dx+\frac{1}{2} \left (i b^2 c^3 d^3\right ) \int \frac{1}{x^3 \left (1+c^2 x^2\right )} \, dx-\frac{1}{5} \left (6 b^2 c^4 d^3\right ) \int \frac{1}{x^2 \left (1+c^2 x^2\right )} \, dx+\frac{1}{5} \left (6 i b^2 c^5 d^3\right ) \int \frac{\log (1-i c x)}{x} \, dx-\frac{1}{5} \left (6 i b^2 c^5 d^3\right ) \int \frac{\log (1+i c x)}{x} \, dx-\frac{1}{2} \left (5 i b^2 c^5 d^3\right ) \int \frac{1}{x \left (1+c^2 x^2\right )} \, dx-\frac{1}{5} \left (12 b^2 c^6 d^3\right ) \int \frac{\log \left (\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx\\ &=-\frac{b^2 c^2 d^3}{30 x^3}+\frac{6 b^2 c^4 d^3}{5 x}-\frac{b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{10 x^4}-\frac{i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^3}+\frac{6 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^2}+\frac{5 i b c^4 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{5 x^5}+\frac{i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 x^4}+\frac{12}{5} a b c^5 d^3 \log (x)+\frac{12}{5} b c^5 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )+\frac{6}{5} i b^2 c^5 d^3 \text{Li}_2(-i c x)-\frac{6}{5} i b^2 c^5 d^3 \text{Li}_2(i c x)+\frac{1}{4} \left (i b^2 c^3 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac{1}{10} \left (b^2 c^4 d^3\right ) \int \frac{1}{x^2 \left (1+c^2 x^2\right )} \, dx-\frac{1}{4} \left (5 i b^2 c^5 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac{1}{5} \left (12 i b^2 c^5 d^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-i c x}\right )+\frac{1}{5} \left (6 b^2 c^6 d^3\right ) \int \frac{1}{1+c^2 x^2} \, dx\\ &=-\frac{b^2 c^2 d^3}{30 x^3}+\frac{13 b^2 c^4 d^3}{10 x}+\frac{6}{5} b^2 c^5 d^3 \tan ^{-1}(c x)-\frac{b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{10 x^4}-\frac{i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^3}+\frac{6 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^2}+\frac{5 i b c^4 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{5 x^5}+\frac{i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 x^4}+\frac{12}{5} a b c^5 d^3 \log (x)+\frac{12}{5} b c^5 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )+\frac{6}{5} i b^2 c^5 d^3 \text{Li}_2(-i c x)-\frac{6}{5} i b^2 c^5 d^3 \text{Li}_2(i c x)-\frac{6}{5} i b^2 c^5 d^3 \text{Li}_2\left (1-\frac{2}{1-i c x}\right )+\frac{1}{4} \left (i b^2 c^3 d^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{x^2}-\frac{c^2}{x}+\frac{c^4}{1+c^2 x}\right ) \, dx,x,x^2\right )-\frac{1}{4} \left (5 i b^2 c^5 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{10} \left (b^2 c^6 d^3\right ) \int \frac{1}{1+c^2 x^2} \, dx+\frac{1}{4} \left (5 i b^2 c^7 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac{b^2 c^2 d^3}{30 x^3}-\frac{i b^2 c^3 d^3}{4 x^2}+\frac{13 b^2 c^4 d^3}{10 x}+\frac{13}{10} b^2 c^5 d^3 \tan ^{-1}(c x)-\frac{b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{10 x^4}-\frac{i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^3}+\frac{6 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^2}+\frac{5 i b c^4 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{5 x^5}+\frac{i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 x^4}+\frac{12}{5} a b c^5 d^3 \log (x)-3 i b^2 c^5 d^3 \log (x)+\frac{12}{5} b c^5 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )+\frac{3}{2} i b^2 c^5 d^3 \log \left (1+c^2 x^2\right )+\frac{6}{5} i b^2 c^5 d^3 \text{Li}_2(-i c x)-\frac{6}{5} i b^2 c^5 d^3 \text{Li}_2(i c x)-\frac{6}{5} i b^2 c^5 d^3 \text{Li}_2\left (1-\frac{2}{1-i c x}\right )\\ \end{align*}

Mathematica [A]  time = 1.25055, size = 363, normalized size = 0.95 \[ \frac{d^3 \left (-72 i b^2 c^5 x^5 \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )+30 i a^2 c^3 x^3+60 a^2 c^2 x^2-45 i a^2 c x-12 a^2+150 i a b c^4 x^4+72 a b c^3 x^3-30 i a b c^2 x^2+144 a b c^5 x^5 \log (c x)-72 a b c^5 x^5 \log \left (c^2 x^2+1\right )+6 b \tan ^{-1}(c x) \left (a \left (25 i c^5 x^5+10 i c^3 x^3+20 c^2 x^2-15 i c x-4\right )+b c x \left (13 c^4 x^4+25 i c^3 x^3+12 c^2 x^2-5 i c x-1\right )+24 b c^5 x^5 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )-6 a b c x-15 i b^2 c^5 x^5+78 b^2 c^4 x^4-15 i b^2 c^3 x^3-2 b^2 c^2 x^2-180 i b^2 c^5 x^5 \log \left (\frac{c x}{\sqrt{c^2 x^2+1}}\right )+3 i b^2 (c x-i)^4 (c x+4 i) \tan ^{-1}(c x)^2\right )}{60 x^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + I*c*d*x)^3*(a + b*ArcTan[c*x])^2)/x^6,x]

[Out]

(d^3*(-12*a^2 - (45*I)*a^2*c*x - 6*a*b*c*x + 60*a^2*c^2*x^2 - (30*I)*a*b*c^2*x^2 - 2*b^2*c^2*x^2 + (30*I)*a^2*
c^3*x^3 + 72*a*b*c^3*x^3 - (15*I)*b^2*c^3*x^3 + (150*I)*a*b*c^4*x^4 + 78*b^2*c^4*x^4 - (15*I)*b^2*c^5*x^5 + (3
*I)*b^2*(-I + c*x)^4*(4*I + c*x)*ArcTan[c*x]^2 + 6*b*ArcTan[c*x]*(b*c*x*(-1 - (5*I)*c*x + 12*c^2*x^2 + (25*I)*
c^3*x^3 + 13*c^4*x^4) + a*(-4 - (15*I)*c*x + 20*c^2*x^2 + (10*I)*c^3*x^3 + (25*I)*c^5*x^5) + 24*b*c^5*x^5*Log[
1 - E^((2*I)*ArcTan[c*x])]) + 144*a*b*c^5*x^5*Log[c*x] - (180*I)*b^2*c^5*x^5*Log[(c*x)/Sqrt[1 + c^2*x^2]] - 72
*a*b*c^5*x^5*Log[1 + c^2*x^2] - (72*I)*b^2*c^5*x^5*PolyLog[2, E^((2*I)*ArcTan[c*x])]))/(60*x^5)

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Maple [B]  time = 0.115, size = 816, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)^3*(a+b*arctan(c*x))^2/x^6,x)

[Out]

5/4*I*c^5*d^3*b^2*arctan(c*x)^2+6/5*c^3*d^3*a*b/x^2-1/10*c*d^3*a*b/x^4-6/5*c^5*d^3*b^2*arctan(c*x)*ln(c^2*x^2+
1)+12/5*c^5*d^3*b^2*arctan(c*x)*ln(c*x)+6/5*I*c^5*d^3*b^2*dilog(1+I*c*x)-6/5*I*c^5*d^3*b^2*dilog(1-I*c*x)-3*I*
c^5*d^3*b^2*ln(c*x)-3/5*I*c^5*d^3*b^2*dilog(1/2*I*(c*x-I))+3/10*I*c^5*d^3*b^2*ln(c*x-I)^2+3/5*I*c^5*d^3*b^2*di
log(-1/2*I*(c*x+I))+1/2*I*c^3*d^3*a^2/x^2-3/4*I*c*d^3*a^2/x^4-6/5*c^5*d^3*a*b*ln(c^2*x^2+1)+12/5*c^5*d^3*a*b*l
n(c*x)+6/5*c^3*d^3*b^2*arctan(c*x)/x^2+c^2*d^3*b^2*arctan(c*x)^2/x^3-1/10*c*d^3*b^2*arctan(c*x)/x^4-2/5*d^3*a*
b*arctan(c*x)/x^5+3/5*I*c^5*d^3*b^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))-6/5*I*c^5*d^3*b^2*ln(c*x)*ln(1-I*c*x)+2*c^2*d
^3*a*b*arctan(c*x)/x^3+1/2*I*c^3*d^3*b^2*arctan(c*x)^2/x^2+6/5*I*c^5*d^3*b^2*ln(c*x)*ln(1+I*c*x)+5/2*I*c^5*d^3
*a*b*arctan(c*x)+5/2*I*c^4*d^3*a*b/x-1/2*I*c^2*d^3*a*b/x^3-1/2*I*c^2*d^3*b^2*arctan(c*x)/x^3+5/2*I*c^4*d^3*b^2
*arctan(c*x)/x-3/4*I*c*d^3*b^2*arctan(c*x)^2/x^4-3/5*I*c^5*d^3*b^2*ln(c*x-I)*ln(c^2*x^2+1)-3/5*I*c^5*d^3*b^2*l
n(c*x+I)*ln(1/2*I*(c*x-I))+3/5*I*c^5*d^3*b^2*ln(c*x+I)*ln(c^2*x^2+1)-3/10*I*c^5*d^3*b^2*ln(c*x+I)^2-1/5*d^3*a^
2/x^5-1/4*I*b^2*c^3*d^3/x^2+c^2*d^3*a^2/x^3-1/5*d^3*b^2*arctan(c*x)^2/x^5+I*c^3*d^3*a*b*arctan(c*x)/x^2-3/2*I*
c*d^3*a*b*arctan(c*x)/x^4+3/2*I*b^2*c^5*d^3*ln(c^2*x^2+1)-1/30*b^2*c^2*d^3/x^3+13/10*b^2*c^4*d^3/x+13/10*b^2*c
^5*d^3*arctan(c*x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^3*(a+b*arctan(c*x))^2/x^6,x, algorithm="maxima")

[Out]

I*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*a*b*c^3*d^3 - ((c^2*log(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c -
2*arctan(c*x)/x^3)*a*b*c^2*d^3 + 1/2*I*((3*c^3*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*a*b*
c*d^3 - 1/10*((2*c^4*log(c^2*x^2 + 1) - 2*c^4*log(x^2) - (2*c^2*x^2 - 1)/x^4)*c + 4*arctan(c*x)/x^5)*a*b*d^3 +
1/2*I*a^2*c^3*d^3/x^2 + a^2*c^2*d^3/x^3 - 3/4*I*a^2*c*d^3/x^4 - 1/5*a^2*d^3/x^5 - 1/320*(320*I*x^5*integrate(
1/80*(60*(b^2*c^5*d^3*x^5 - 2*b^2*c^3*d^3*x^3 - 3*b^2*c*d^3*x)*arctan(c*x)^2 + 5*(b^2*c^5*d^3*x^5 - 2*b^2*c^3*
d^3*x^3 - 3*b^2*c*d^3*x)*log(c^2*x^2 + 1)^2 + 2*(30*b^2*c^4*d^3*x^4 - 19*b^2*c^2*d^3*x^2)*arctan(c*x) - (10*b^
2*c^5*d^3*x^5 - 35*b^2*c^3*d^3*x^3 + 4*b^2*c*d^3*x + 20*(3*b^2*c^4*d^3*x^4 + 2*b^2*c^2*d^3*x^2 - b^2*d^3)*arct
an(c*x))*log(c^2*x^2 + 1))/(c^2*x^8 + x^6), x) + 320*x^5*integrate(1/80*(60*(3*b^2*c^4*d^3*x^4 + 2*b^2*c^2*d^3
*x^2 - b^2*d^3)*arctan(c*x)^2 + 5*(3*b^2*c^4*d^3*x^4 + 2*b^2*c^2*d^3*x^2 - b^2*d^3)*log(c^2*x^2 + 1)^2 - 2*(10
*b^2*c^5*d^3*x^5 - 35*b^2*c^3*d^3*x^3 + 4*b^2*c*d^3*x)*arctan(c*x) - (30*b^2*c^4*d^3*x^4 - 19*b^2*c^2*d^3*x^2
- 20*(b^2*c^5*d^3*x^5 - 2*b^2*c^3*d^3*x^3 - 3*b^2*c*d^3*x)*arctan(c*x))*log(c^2*x^2 + 1))/(c^2*x^8 + x^6), x)
+ (-40*I*b^2*c^3*d^3*x^3 - 80*b^2*c^2*d^3*x^2 + 60*I*b^2*c*d^3*x + 16*b^2*d^3)*arctan(c*x)^2 + (40*b^2*c^3*d^3
*x^3 - 80*I*b^2*c^2*d^3*x^2 - 60*b^2*c*d^3*x + 16*I*b^2*d^3)*arctan(c*x)*log(c^2*x^2 + 1) + (10*I*b^2*c^3*d^3*
x^3 + 20*b^2*c^2*d^3*x^2 - 15*I*b^2*c*d^3*x - 4*b^2*d^3)*log(c^2*x^2 + 1)^2)/x^5

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{80 \, x^{5}{\rm integral}\left (\frac{-20 i \, a^{2} c^{5} d^{3} x^{5} - 60 \, a^{2} c^{4} d^{3} x^{4} + 40 i \, a^{2} c^{3} d^{3} x^{3} - 40 \, a^{2} c^{2} d^{3} x^{2} + 60 i \, a^{2} c d^{3} x + 20 \, a^{2} d^{3} +{\left (20 \, a b c^{5} d^{3} x^{5} +{\left (-60 i \, a b + 10 \, b^{2}\right )} c^{4} d^{3} x^{4} - 20 \,{\left (2 \, a b + i \, b^{2}\right )} c^{3} d^{3} x^{3} +{\left (-40 i \, a b - 15 \, b^{2}\right )} c^{2} d^{3} x^{2} - 4 \,{\left (15 \, a b - i \, b^{2}\right )} c d^{3} x + 20 i \, a b d^{3}\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{20 \,{\left (c^{2} x^{8} + x^{6}\right )}}, x\right ) +{\left (-10 i \, b^{2} c^{3} d^{3} x^{3} - 20 \, b^{2} c^{2} d^{3} x^{2} + 15 i \, b^{2} c d^{3} x + 4 \, b^{2} d^{3}\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2}}{80 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^3*(a+b*arctan(c*x))^2/x^6,x, algorithm="fricas")

[Out]

1/80*(80*x^5*integral(1/20*(-20*I*a^2*c^5*d^3*x^5 - 60*a^2*c^4*d^3*x^4 + 40*I*a^2*c^3*d^3*x^3 - 40*a^2*c^2*d^3
*x^2 + 60*I*a^2*c*d^3*x + 20*a^2*d^3 + (20*a*b*c^5*d^3*x^5 + (-60*I*a*b + 10*b^2)*c^4*d^3*x^4 - 20*(2*a*b + I*
b^2)*c^3*d^3*x^3 + (-40*I*a*b - 15*b^2)*c^2*d^3*x^2 - 4*(15*a*b - I*b^2)*c*d^3*x + 20*I*a*b*d^3)*log(-(c*x + I
)/(c*x - I)))/(c^2*x^8 + x^6), x) + (-10*I*b^2*c^3*d^3*x^3 - 20*b^2*c^2*d^3*x^2 + 15*I*b^2*c*d^3*x + 4*b^2*d^3
)*log(-(c*x + I)/(c*x - I))^2)/x^5

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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((d+I*c*d*x)**3*(a+b*atan(c*x))**2/x**6,x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, c d x + d\right )}^{3}{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^3*(a+b*arctan(c*x))^2/x^6,x, algorithm="giac")

[Out]

integrate((I*c*d*x + d)^3*(b*arctan(c*x) + a)^2/x^6, x)