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

Optimal. Leaf size=384 \[ \frac {12}{5} a b c^5 d^3 \log (x)+\frac {12}{5} b c^5 d^3 \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac {5 i b c^4 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}+\frac {6 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^2}-\frac {i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^3}-\frac {d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{5 x^5}+\frac {i c d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{20 x^4}-\frac {b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{10 x^4}+\frac {6}{5} i b^2 c^5 d^3 \text {Li}_2(-i c x)-\frac {6}{5} i b^2 c^5 d^3 \text {Li}_2(i c x)-\frac {6}{5} i b^2 c^5 d^3 \text {Li}_2\left (1-\frac {2}{1-i c x}\right )-3 i b^2 c^5 d^3 \log (x)+\frac {13}{10} b^2 c^5 d^3 \tan ^{-1}(c x)+\frac {13 b^2 c^4 d^3}{10 x}-\frac {i b^2 c^3 d^3}{4 x^2}-\frac {b^2 c^2 d^3}{30 x^3}+\frac {3}{2} i b^2 c^5 d^3 \log \left (c^2 x^2+1\right ) \]

[Out]

-1/30*b^2*c^2*d^3/x^3+5/2*I*b*c^4*d^3*(a+b*arctan(c*x))/x+13/10*b^2*c^4*d^3/x+13/10*b^2*c^5*d^3*arctan(c*x)-1/
10*b*c*d^3*(a+b*arctan(c*x))/x^4-3*I*b^2*c^5*d^3*ln(x)+6/5*b*c^3*d^3*(a+b*arctan(c*x))/x^2-1/2*I*b*c^2*d^3*(a+
b*arctan(c*x))/x^3-1/5*d^3*(1+I*c*x)^4*(a+b*arctan(c*x))^2/x^5-1/4*I*b^2*c^3*d^3/x^2+12/5*a*b*c^5*d^3*ln(x)+6/
5*I*b^2*c^5*d^3*polylog(2,-I*c*x)+12/5*b*c^5*d^3*(a+b*arctan(c*x))*ln(2/(1-I*c*x))-6/5*I*b^2*c^5*d^3*polylog(2
,I*c*x)-6/5*I*b^2*c^5*d^3*polylog(2,1-2/(1-I*c*x))+3/2*I*b^2*c^5*d^3*ln(c^2*x^2+1)+1/20*I*c*d^3*(1+I*c*x)^4*(a
+b*arctan(c*x))^2/x^4

________________________________________________________________________________________

Rubi [A]  time = 0.37, antiderivative size = 384, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 16, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, 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 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 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 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 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 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 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 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 2315

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

Rule 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 1.40, size = 363, normalized size = 0.95 \[ \frac {d^3 \left (30 i a^2 c^3 x^3+60 a^2 c^2 x^2-45 i a^2 c x-12 a^2+144 a b c^5 x^5 \log (c x)+150 i a b c^4 x^4+72 a b c^3 x^3-30 i a b c^2 x^2-72 a b c^5 x^5 \log \left (c^2 x^2+1\right )+6 b \tan ^{-1}(c x) \left (a \left (25 i c^5 x^5+10 i c^3 x^3+20 c^2 x^2-15 i c x-4\right )+24 b c^5 x^5 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+b c x \left (13 c^4 x^4+25 i c^3 x^3+12 c^2 x^2-5 i c x-1\right )\right )-6 a b c x-72 i b^2 c^5 x^5 \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )-15 i b^2 c^5 x^5+78 b^2 c^4 x^4-15 i b^2 c^3 x^3-2 b^2 c^2 x^2-180 i b^2 c^5 x^5 \log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )+3 i b^2 (c x-i)^4 (c x+4 i) \tan ^{-1}(c x)^2\right )}{60 x^5} \]

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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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ \frac {80 \, x^{5} {\rm integral}\left (\frac {-20 i \, a^{2} c^{5} d^{3} x^{5} - 60 \, a^{2} c^{4} d^{3} x^{4} + 40 i \, a^{2} c^{3} d^{3} x^{3} - 40 \, a^{2} c^{2} d^{3} x^{2} + 60 i \, a^{2} c d^{3} x + 20 \, a^{2} d^{3} + {\left (20 \, a b c^{5} d^{3} x^{5} - 10 \, {\left (6 i \, a b - b^{2}\right )} c^{4} d^{3} x^{4} - {\left (40 \, a b + 20 i \, b^{2}\right )} c^{3} d^{3} x^{3} - 5 \, {\left (8 i \, a b + 3 \, b^{2}\right )} c^{2} d^{3} x^{2} - {\left (60 \, a b - 4 i \, b^{2}\right )} c d^{3} x + 20 i \, a b d^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{20 \, {\left (c^{2} x^{8} + x^{6}\right )}}, x\right ) + {\left (-10 i \, b^{2} c^{3} d^{3} x^{3} - 20 \, b^{2} c^{2} d^{3} x^{2} + 15 i \, b^{2} c d^{3} x + 4 \, b^{2} d^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2}}{80 \, x^{5}} \]

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 - 10*(6*I*a*b - b^2)*c^4*d^3*x^4 - (40*a*b + 20*I*b
^2)*c^3*d^3*x^3 - 5*(8*I*a*b + 3*b^2)*c^2*d^3*x^2 - (60*a*b - 4*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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [B]  time = 0.12, size = 816, normalized size = 2.12 \[ -\frac {d^{3} a^{2}}{5 x^{5}}-\frac {3 i c^{5} d^{3} b^{2} \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{5}+\frac {3 i c^{5} d^{3} b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{5}+\frac {6 i c^{5} d^{3} b^{2} \ln \left (c x \right ) \ln \left (i c x +1\right )}{5}-\frac {6 i c^{5} d^{3} b^{2} \ln \left (c x \right ) \ln \left (-i c x +1\right )}{5}+\frac {3 i c^{5} d^{3} b^{2} \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{5}+\frac {i c^{3} d^{3} b^{2} \arctan \left (c x \right )^{2}}{2 x^{2}}-\frac {3 i c \,d^{3} b^{2} \arctan \left (c x \right )^{2}}{4 x^{4}}-\frac {i c^{2} d^{3} b^{2} \arctan \left (c x \right )}{2 x^{3}}+\frac {5 i c^{4} d^{3} b^{2} \arctan \left (c x \right )}{2 x}+\frac {5 i c^{5} d^{3} a b \arctan \left (c x \right )}{2}-\frac {i c^{2} d^{3} a b}{2 x^{3}}+\frac {5 i c^{4} d^{3} a b}{2 x}-\frac {3 i c^{5} d^{3} b^{2} \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{5}+\frac {6 i c^{5} d^{3} b^{2} \dilog \left (i c x +1\right )}{5}-\frac {6 i c^{5} d^{3} b^{2} \dilog \left (-i c x +1\right )}{5}-\frac {3 i c^{5} d^{3} b^{2} \dilog \left (\frac {i \left (c x -i\right )}{2}\right )}{5}-\frac {c \,d^{3} b^{2} \arctan \left (c x \right )}{10 x^{4}}+\frac {6 c^{3} d^{3} b^{2} \arctan \left (c x \right )}{5 x^{2}}+\frac {c^{2} d^{3} b^{2} \arctan \left (c x \right )^{2}}{x^{3}}+\frac {12 c^{5} d^{3} b^{2} \arctan \left (c x \right ) \ln \left (c x \right )}{5}-\frac {6 c^{5} d^{3} b^{2} \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )}{5}-\frac {2 d^{3} a b \arctan \left (c x \right )}{5 x^{5}}-\frac {c \,d^{3} a b}{10 x^{4}}+\frac {12 c^{5} d^{3} a b \ln \left (c x \right )}{5}-\frac {6 c^{5} d^{3} a b \ln \left (c^{2} x^{2}+1\right )}{5}+\frac {6 c^{3} d^{3} a b}{5 x^{2}}+\frac {2 c^{2} d^{3} a b \arctan \left (c x \right )}{x^{3}}+\frac {i c^{3} d^{3} a b \arctan \left (c x \right )}{x^{2}}-\frac {3 i c \,d^{3} a b \arctan \left (c x \right )}{2 x^{4}}-\frac {d^{3} b^{2} \arctan \left (c x \right )^{2}}{5 x^{5}}+\frac {c^{2} d^{3} a^{2}}{x^{3}}+\frac {3 i c^{5} d^{3} b^{2} \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{5}-\frac {3 i c \,d^{3} a^{2}}{4 x^{4}}+\frac {i c^{3} d^{3} a^{2}}{2 x^{2}}-\frac {3 i c^{5} d^{3} b^{2} \ln \left (c x +i\right )^{2}}{10}+\frac {3 i c^{5} d^{3} b^{2} \ln \left (c x -i\right )^{2}}{10}-3 i c^{5} d^{3} b^{2} \ln \left (c x \right )-\frac {i b^{2} c^{3} d^{3}}{4 x^{2}}+\frac {3 i b^{2} c^{5} d^{3} \ln \left (c^{2} x^{2}+1\right )}{2}+\frac {5 i c^{5} d^{3} b^{2} \arctan \left (c x \right )^{2}}{4}-\frac {b^{2} c^{2} d^{3}}{30 x^{3}}+\frac {13 b^{2} c^{4} d^{3}}{10 x}+\frac {13 b^{2} c^{5} d^{3} \arctan \left (c x \right )}{10} \]

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]

-1/5*d^3*a^2/x^5-1/10*c*d^3*b^2*arctan(c*x)/x^4+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+
12/5*c^5*d^3*b^2*arctan(c*x)*ln(c*x)-6/5*c^5*d^3*b^2*arctan(c*x)*ln(c^2*x^2+1)-2/5*d^3*a*b*arctan(c*x)/x^5-1/1
0*c*d^3*a*b/x^4+12/5*c^5*d^3*a*b*ln(c*x)-6/5*c^5*d^3*a*b*ln(c^2*x^2+1)+6/5*c^3*d^3*a*b/x^2-6/5*I*c^5*d^3*b^2*l
n(c*x)*ln(1-I*c*x)+3/5*I*c^5*d^3*b^2*ln(I+c*x)*ln(c^2*x^2+1)+1/2*I*c^3*d^3*b^2*arctan(c*x)^2/x^2-3/4*I*c*d^3*b
^2*arctan(c*x)^2/x^4-1/2*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+5/2*I*c^5*d^3*a*b*arcta
n(c*x)-1/2*I*c^2*d^3*a*b/x^3+5/2*I*c^4*d^3*a*b/x-3/5*I*c^5*d^3*b^2*ln(I+c*x)*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)+3/5*I*c^5*d^3*b^2*ln(c*x-I)*ln(-1/2*I*(I+c*x))+2*c^2*d^3*a*b*arctan(c*x)/x^3+6/5*I*
c^5*d^3*b^2*ln(c*x)*ln(1+I*c*x)-3/4*I*c*d^3*a^2/x^4+1/2*I*c^3*d^3*a^2/x^2+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)-3/10*I*c^5*d^3*b^2*ln(I+c*x)^2-3/5*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*dilog(-1/2*I*(I+c*x))-3*I*c^5*d^3*b^2*ln(c*x)-1/4*I*b^2*c^3*d^3/x^2+I*c
^3*d^3*a*b*arctan(c*x)/x^2-3/2*I*c*d^3*a*b*arctan(c*x)/x^4-1/5*d^3*b^2*arctan(c*x)^2/x^5+c^2*d^3*a^2/x^3+3/2*I
*b^2*c^5*d^3*ln(c^2*x^2+1)+5/4*I*c^5*d^3*b^2*arctan(c*x)^2-1/30*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 = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i)^3)/x^6,x)

[Out]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i)^3)/x^6, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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