3.1.93 \(\int \frac {(d+i c d x)^3 (a+b \text {ArcTan}(c x))^2}{x^6} \, dx\) [93]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.28, 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 = {47, 37, 4994, 4946, 331, 209, 272, 46, 36, 29, 31, 4940, 2438, 4964, 2449, 2352} \begin {gather*} \frac {12}{5} b c^5 d^3 \log \left (\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))+\frac {5 i b c^4 d^3 (a+b \text {ArcTan}(c x))}{2 x}+\frac {6 b c^3 d^3 (a+b \text {ArcTan}(c x))}{5 x^2}-\frac {i b c^2 d^3 (a+b \text {ArcTan}(c x))}{2 x^3}-\frac {d^3 (1+i c x)^4 (a+b \text {ArcTan}(c x))^2}{5 x^5}+\frac {i c d^3 (1+i c x)^4 (a+b \text {ArcTan}(c x))^2}{20 x^4}-\frac {b c d^3 (a+b \text {ArcTan}(c x))}{10 x^4}+\frac {12}{5} a b c^5 d^3 \log (x)+\frac {13}{10} b^2 c^5 d^3 \text {ArcTan}(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(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 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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30*(b^2*c^2*d^3)/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 46

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] && Lt
Q[m + n + 2, 0])

Rule 47

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 209

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

Rule 272

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 331

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 2352

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

Rule 2438

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 2449

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 4940

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 4946

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

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[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 4994

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 = 0.83, size = 363, normalized size = 0.95 \begin {gather*} \frac {d^3 \left (-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) \text {ArcTan}(c x)^2+6 b \text {ArcTan}(c x) \left (b c x \left (-1-5 i c x+12 c^2 x^2+25 i c^3 x^3+13 c^4 x^4\right )+a \left (-4-15 i c x+20 c^2 x^2+10 i c^3 x^3+25 i c^5 x^5\right )+24 b c^5 x^5 \log \left (1-e^{2 i \text {ArcTan}(c x)}\right )\right )+144 a b c^5 x^5 \log (c x)-180 i b^2 c^5 x^5 \log \left (\frac {c x}{\sqrt {1+c^2 x^2}}\right )-72 a b c^5 x^5 \log \left (1+c^2 x^2\right )-72 i b^2 c^5 x^5 \text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c x)}\right )\right )}{60 x^5} \end {gather*}

Antiderivative was successfully verified.

[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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 759 vs. \(2 (337 ) = 674\).
time = 1.04, size = 760, normalized size = 1.98 Too large to display

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)
- 4*(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)*arctan(c*x)^2 + 4*(10*b^2*c^3*d
^3*x^3 - 20*I*b^2*c^2*d^3*x^2 - 15*b^2*c*d^3*x + 4*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 - 20*(2*a*b + I*b^
2)*c^3*d^3*x^3 - 5*(8*I*a*b + 3*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)*l
og(-(c*x + I)/(c*x - I))^2)/x^5

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

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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