3.103 \(\int \frac {x^4 (a+b \tan ^{-1}(c x))^2}{(d+i c d x)^2} \, dx\)

Optimal. Leaf size=433 \[ \frac {4 b \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{c^5 d^2 (-c x+i)}+\frac {11 i \left (a+b \tan ^{-1}(c x)\right )^2}{6 c^5 d^2}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^2 (-c x+i)}-\frac {4 i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^5 d^2}+\frac {20 b \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^5 d^2}+\frac {2 i a b x}{c^4 d^2}+\frac {3 x \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^2}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^2}+\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c^3 d^2}-\frac {x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^2 d^2}+\frac {10 i b^2 \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{3 c^5 d^2}-\frac {2 i b^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )}{c^5 d^2}+\frac {b^2}{2 c^5 d^2 (-c x+i)}-\frac {b^2 \tan ^{-1}(c x)}{6 c^5 d^2}-\frac {b^2 x}{3 c^4 d^2}+\frac {2 i b^2 x \tan ^{-1}(c x)}{c^4 d^2}-\frac {i b^2 \log \left (c^2 x^2+1\right )}{c^5 d^2} \]

[Out]

10/3*I*b^2*polylog(2,1-2/(1+I*c*x))/c^5/d^2-1/3*b^2*x/c^4/d^2+1/2*b^2/c^5/d^2/(I-c*x)-1/6*b^2*arctan(c*x)/c^5/
d^2-I*x^2*(a+b*arctan(c*x))^2/c^3/d^2+1/3*b*x^2*(a+b*arctan(c*x))/c^3/d^2-I*b^2*ln(c^2*x^2+1)/c^5/d^2-4*I*(a+b
*arctan(c*x))^2*ln(2/(1+I*c*x))/c^5/d^2+3*x*(a+b*arctan(c*x))^2/c^4/d^2+I*b*(a+b*arctan(c*x))/c^5/d^2/(I-c*x)-
1/3*x^3*(a+b*arctan(c*x))^2/c^2/d^2-(a+b*arctan(c*x))^2/c^5/d^2/(I-c*x)+20/3*b*(a+b*arctan(c*x))*ln(2/(1+I*c*x
))/c^5/d^2-2*I*b^2*polylog(3,1-2/(1+I*c*x))/c^5/d^2+2*I*b^2*x*arctan(c*x)/c^4/d^2+11/6*I*(a+b*arctan(c*x))^2/c
^5/d^2+4*b*(a+b*arctan(c*x))*polylog(2,1-2/(1+I*c*x))/c^5/d^2+2*I*a*b*x/c^4/d^2

________________________________________________________________________________________

Rubi [A]  time = 0.83, antiderivative size = 433, normalized size of antiderivative = 1.00, number of steps used = 33, number of rules used = 18, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {4876, 4846, 4920, 4854, 2402, 2315, 4852, 4916, 260, 4884, 321, 203, 4864, 4862, 627, 44, 4994, 6610} \[ \frac {4 b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^2}+\frac {10 i b^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{3 c^5 d^2}-\frac {2 i b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{c^5 d^2}-\frac {x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^2 d^2}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^2}+\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c^3 d^2}+\frac {2 i a b x}{c^4 d^2}+\frac {3 x \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{c^5 d^2 (-c x+i)}+\frac {11 i \left (a+b \tan ^{-1}(c x)\right )^2}{6 c^5 d^2}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^2 (-c x+i)}-\frac {4 i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^5 d^2}+\frac {20 b \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^5 d^2}-\frac {i b^2 \log \left (c^2 x^2+1\right )}{c^5 d^2}-\frac {b^2 x}{3 c^4 d^2}+\frac {b^2}{2 c^5 d^2 (-c x+i)}+\frac {2 i b^2 x \tan ^{-1}(c x)}{c^4 d^2}-\frac {b^2 \tan ^{-1}(c x)}{6 c^5 d^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcTan[c*x])^2)/(d + I*c*d*x)^2,x]

[Out]

((2*I)*a*b*x)/(c^4*d^2) - (b^2*x)/(3*c^4*d^2) + b^2/(2*c^5*d^2*(I - c*x)) - (b^2*ArcTan[c*x])/(6*c^5*d^2) + ((
2*I)*b^2*x*ArcTan[c*x])/(c^4*d^2) + (b*x^2*(a + b*ArcTan[c*x]))/(3*c^3*d^2) + (I*b*(a + b*ArcTan[c*x]))/(c^5*d
^2*(I - c*x)) + (((11*I)/6)*(a + b*ArcTan[c*x])^2)/(c^5*d^2) + (3*x*(a + b*ArcTan[c*x])^2)/(c^4*d^2) - (I*x^2*
(a + b*ArcTan[c*x])^2)/(c^3*d^2) - (x^3*(a + b*ArcTan[c*x])^2)/(3*c^2*d^2) - (a + b*ArcTan[c*x])^2/(c^5*d^2*(I
- c*x)) + (20*b*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(3*c^5*d^2) - ((4*I)*(a + b*ArcTan[c*x])^2*Log[2/(1 +
I*c*x)])/(c^5*d^2) - (I*b^2*Log[1 + c^2*x^2])/(c^5*d^2) + (((10*I)/3)*b^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^5
*d^2) + (4*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^5*d^2) - ((2*I)*b^2*PolyLog[3, 1 - 2/(1 + I
*c*x)])/(c^5*d^2)

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 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 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 321

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

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 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 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 4846

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x])^p, x] - Dist[b*c*p, Int[
(x*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 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 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 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 4864

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

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 4884

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

Rule 4916

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(d*f^2)/e, Int[((f*x)^(m - 2)*(a + b*ArcTan[c*x])^p)
/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]

Rule 4920

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

Rule 4994

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*Arc
Tan[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[2, 1 - u
])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[(1 - u)^2 - (1 - (2*
I)/(I - c*x))^2, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int \frac {x^4 \left (a+b \tan ^{-1}(c x)\right )^2}{(d+i c d x)^2} \, dx &=\int \left (\frac {3 \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^2}-\frac {2 i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^2}-\frac {x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^2 (-i+c x)^2}+\frac {4 i \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^2 (-i+c x)}\right ) \, dx\\ &=\frac {(4 i) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{-i+c x} \, dx}{c^4 d^2}-\frac {\int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^2} \, dx}{c^4 d^2}+\frac {3 \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{c^4 d^2}-\frac {(2 i) \int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{c^3 d^2}-\frac {\int x^2 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{c^2 d^2}\\ &=\frac {3 x \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^2}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^2}-\frac {x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^2 d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{c^5 d^2 (i-c x)}-\frac {4 i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^5 d^2}+\frac {(8 i b) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^4 d^2}-\frac {(2 b) \int \left (-\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 (-i+c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{c^4 d^2}-\frac {(6 b) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{c^3 d^2}+\frac {(2 i b) \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{c^2 d^2}+\frac {(2 b) \int \frac {x^3 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{3 c d^2}\\ &=\frac {3 i \left (a+b \tan ^{-1}(c x)\right )^2}{c^5 d^2}+\frac {3 x \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^2}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^2}-\frac {x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^2 d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{c^5 d^2 (i-c x)}-\frac {4 i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^5 d^2}+\frac {4 b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^5 d^2}+\frac {(i b) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{c^4 d^2}-\frac {(i b) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{c^4 d^2}+\frac {(2 i b) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c^4 d^2}-\frac {(2 i b) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{c^4 d^2}+\frac {(6 b) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx}{c^4 d^2}-\frac {\left (4 b^2\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^4 d^2}+\frac {(2 b) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx}{3 c^3 d^2}-\frac {(2 b) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{3 c^3 d^2}\\ &=\frac {2 i a b x}{c^4 d^2}+\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c^3 d^2}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^2 (i-c x)}+\frac {11 i \left (a+b \tan ^{-1}(c x)\right )^2}{6 c^5 d^2}+\frac {3 x \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^2}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^2}-\frac {x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^2 d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{c^5 d^2 (i-c x)}+\frac {6 b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^5 d^2}-\frac {4 i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^5 d^2}+\frac {4 b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^5 d^2}-\frac {2 i b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{c^5 d^2}+\frac {(2 b) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx}{3 c^4 d^2}+\frac {\left (i b^2\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{c^4 d^2}+\frac {\left (2 i b^2\right ) \int \tan ^{-1}(c x) \, dx}{c^4 d^2}-\frac {\left (6 b^2\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^4 d^2}-\frac {b^2 \int \frac {x^2}{1+c^2 x^2} \, dx}{3 c^2 d^2}\\ &=\frac {2 i a b x}{c^4 d^2}-\frac {b^2 x}{3 c^4 d^2}+\frac {2 i b^2 x \tan ^{-1}(c x)}{c^4 d^2}+\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c^3 d^2}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^2 (i-c x)}+\frac {11 i \left (a+b \tan ^{-1}(c x)\right )^2}{6 c^5 d^2}+\frac {3 x \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^2}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^2}-\frac {x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^2 d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{c^5 d^2 (i-c x)}+\frac {20 b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{3 c^5 d^2}-\frac {4 i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^5 d^2}+\frac {4 b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^5 d^2}-\frac {2 i b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{c^5 d^2}+\frac {\left (6 i b^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{c^5 d^2}+\frac {\left (i b^2\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{c^4 d^2}+\frac {b^2 \int \frac {1}{1+c^2 x^2} \, dx}{3 c^4 d^2}-\frac {\left (2 b^2\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{3 c^4 d^2}-\frac {\left (2 i b^2\right ) \int \frac {x}{1+c^2 x^2} \, dx}{c^3 d^2}\\ &=\frac {2 i a b x}{c^4 d^2}-\frac {b^2 x}{3 c^4 d^2}+\frac {b^2 \tan ^{-1}(c x)}{3 c^5 d^2}+\frac {2 i b^2 x \tan ^{-1}(c x)}{c^4 d^2}+\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c^3 d^2}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^2 (i-c x)}+\frac {11 i \left (a+b \tan ^{-1}(c x)\right )^2}{6 c^5 d^2}+\frac {3 x \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^2}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^2}-\frac {x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^2 d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{c^5 d^2 (i-c x)}+\frac {20 b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{3 c^5 d^2}-\frac {4 i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^5 d^2}-\frac {i b^2 \log \left (1+c^2 x^2\right )}{c^5 d^2}+\frac {3 i b^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^5 d^2}+\frac {4 b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^5 d^2}-\frac {2 i b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{c^5 d^2}+\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{3 c^5 d^2}+\frac {\left (i b^2\right ) \int \left (-\frac {i}{2 (-i+c x)^2}+\frac {i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{c^4 d^2}\\ &=\frac {2 i a b x}{c^4 d^2}-\frac {b^2 x}{3 c^4 d^2}+\frac {b^2}{2 c^5 d^2 (i-c x)}+\frac {b^2 \tan ^{-1}(c x)}{3 c^5 d^2}+\frac {2 i b^2 x \tan ^{-1}(c x)}{c^4 d^2}+\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c^3 d^2}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^2 (i-c x)}+\frac {11 i \left (a+b \tan ^{-1}(c x)\right )^2}{6 c^5 d^2}+\frac {3 x \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^2}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^2}-\frac {x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^2 d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{c^5 d^2 (i-c x)}+\frac {20 b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{3 c^5 d^2}-\frac {4 i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^5 d^2}-\frac {i b^2 \log \left (1+c^2 x^2\right )}{c^5 d^2}+\frac {10 i b^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{3 c^5 d^2}+\frac {4 b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^5 d^2}-\frac {2 i b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{c^5 d^2}-\frac {b^2 \int \frac {1}{1+c^2 x^2} \, dx}{2 c^4 d^2}\\ &=\frac {2 i a b x}{c^4 d^2}-\frac {b^2 x}{3 c^4 d^2}+\frac {b^2}{2 c^5 d^2 (i-c x)}-\frac {b^2 \tan ^{-1}(c x)}{6 c^5 d^2}+\frac {2 i b^2 x \tan ^{-1}(c x)}{c^4 d^2}+\frac {b x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c^3 d^2}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^2 (i-c x)}+\frac {11 i \left (a+b \tan ^{-1}(c x)\right )^2}{6 c^5 d^2}+\frac {3 x \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^2}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^2}-\frac {x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^2 d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{c^5 d^2 (i-c x)}+\frac {20 b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{3 c^5 d^2}-\frac {4 i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^5 d^2}-\frac {i b^2 \log \left (1+c^2 x^2\right )}{c^5 d^2}+\frac {10 i b^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{3 c^5 d^2}+\frac {4 b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^5 d^2}-\frac {2 i b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{c^5 d^2}\\ \end {align*}

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Mathematica [A]  time = 2.54, size = 502, normalized size = 1.16 \[ -\frac {4 a^2 c^3 x^3+12 i a^2 c^2 x^2-24 i a^2 \log \left (c^2 x^2+1\right )-36 a^2 c x-\frac {12 a^2}{c x-i}+48 a^2 \tan ^{-1}(c x)+2 a b \left (-2 c^2 x^2+20 \log \left (c^2 x^2+1\right )+2 \tan ^{-1}(c x) \left (2 c^3 x^3+6 i c^2 x^2-18 c x+24 i \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-3 \sin \left (2 \tan ^{-1}(c x)\right )-3 i \cos \left (2 \tan ^{-1}(c x)\right )+6 i\right )+24 \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )-12 i c x+48 \tan ^{-1}(c x)^2+3 i \sin \left (2 \tan ^{-1}(c x)\right )-3 \cos \left (2 \tan ^{-1}(c x)\right )-2\right )+b^2 \left (4 c^3 x^3 \tan ^{-1}(c x)^2+12 i \log \left (c^2 x^2+1\right )+12 i c^2 x^2 \tan ^{-1}(c x)^2-4 c^2 x^2 \tan ^{-1}(c x)+8 \left (6 \tan ^{-1}(c x)+5 i\right ) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+24 i \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )+4 c x-36 c x \tan ^{-1}(c x)^2-24 i c x \tan ^{-1}(c x)+32 \tan ^{-1}(c x)^3+52 i \tan ^{-1}(c x)^2-4 \tan ^{-1}(c x)+48 i \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-80 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-6 \tan ^{-1}(c x)^2 \sin \left (2 \tan ^{-1}(c x)\right )+6 i \tan ^{-1}(c x) \sin \left (2 \tan ^{-1}(c x)\right )+3 \sin \left (2 \tan ^{-1}(c x)\right )-6 i \tan ^{-1}(c x)^2 \cos \left (2 \tan ^{-1}(c x)\right )-6 \tan ^{-1}(c x) \cos \left (2 \tan ^{-1}(c x)\right )+3 i \cos \left (2 \tan ^{-1}(c x)\right )\right )}{12 c^5 d^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*ArcTan[c*x])^2)/(d + I*c*d*x)^2,x]

[Out]

-1/12*(-36*a^2*c*x + (12*I)*a^2*c^2*x^2 + 4*a^2*c^3*x^3 - (12*a^2)/(-I + c*x) + 48*a^2*ArcTan[c*x] - (24*I)*a^
2*Log[1 + c^2*x^2] + 2*a*b*(-2 - (12*I)*c*x - 2*c^2*x^2 + 48*ArcTan[c*x]^2 - 3*Cos[2*ArcTan[c*x]] + 20*Log[1 +
c^2*x^2] + 24*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 2*ArcTan[c*x]*(6*I - 18*c*x + (6*I)*c^2*x^2 + 2*c^3*x^3 -
(3*I)*Cos[2*ArcTan[c*x]] + (24*I)*Log[1 + E^((2*I)*ArcTan[c*x])] - 3*Sin[2*ArcTan[c*x]]) + (3*I)*Sin[2*ArcTan[
c*x]]) + b^2*(4*c*x - 4*ArcTan[c*x] - (24*I)*c*x*ArcTan[c*x] - 4*c^2*x^2*ArcTan[c*x] + (52*I)*ArcTan[c*x]^2 -
36*c*x*ArcTan[c*x]^2 + (12*I)*c^2*x^2*ArcTan[c*x]^2 + 4*c^3*x^3*ArcTan[c*x]^2 + 32*ArcTan[c*x]^3 + (3*I)*Cos[2
*ArcTan[c*x]] - 6*ArcTan[c*x]*Cos[2*ArcTan[c*x]] - (6*I)*ArcTan[c*x]^2*Cos[2*ArcTan[c*x]] - 80*ArcTan[c*x]*Log
[1 + E^((2*I)*ArcTan[c*x])] + (48*I)*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + (12*I)*Log[1 + c^2*x^2] +
8*(5*I + 6*ArcTan[c*x])*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + (24*I)*PolyLog[3, -E^((2*I)*ArcTan[c*x])] + 3*Sin
[2*ArcTan[c*x]] + (6*I)*ArcTan[c*x]*Sin[2*ArcTan[c*x]] - 6*ArcTan[c*x]^2*Sin[2*ArcTan[c*x]]))/(c^5*d^2)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{4} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - 4 i \, a b x^{4} \log \left (-\frac {c x + i}{c x - i}\right ) - 4 \, a^{2} x^{4}}{4 \, {\left (c^{2} d^{2} x^{2} - 2 i \, c d^{2} x - d^{2}\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctan(c*x))^2/(d+I*c*d*x)^2,x, algorithm="fricas")

[Out]

integral(1/4*(b^2*x^4*log(-(c*x + I)/(c*x - I))^2 - 4*I*a*b*x^4*log(-(c*x + I)/(c*x - I)) - 4*a^2*x^4)/(c^2*d^
2*x^2 - 2*I*c*d^2*x - d^2), x)

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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(x^4*(a+b*arctan(c*x))^2/(d+I*c*d*x)^2,x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 7.90, size = 1498, normalized size = 3.46 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctan(c*x))^2/(d+I*c*d*x)^2,x)

[Out]

2/c^5*b^2/d^2*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))
^2*arctan(c*x)^2-2/c^5*b^2/d^2*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*
c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2-1/3*b^2*x/c^4/d^2-29/6*I/c^5*b^2/d^2*arctan(c*x)
^2-2*I/c^5*b^2/d^2*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))-20/3*I/c^5*b^2/d^2*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2
))-20/3*I/c^5*b^2/d^2*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*I/c^5*b^2/d^2*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+2*I
/c^5*a^2/d^2*ln(c^2*x^2+1)-I/c^3*a^2/d^2*x^2+1/3/c^3*b^2/d^2*arctan(c*x)*x^2+3/c^4*b^2/d^2*arctan(c*x)^2*x+4/c
^5*b^2/d^2*Pi*arctan(c*x)^2-2/c^5*a*b/d^2*ln(c*x-I)^2+4/c^5*a*b/d^2*dilog(-1/2*I*(I+c*x))+1/3/c^3*a*b/d^2*x^2+
7/3*b^2*arctan(c*x)/c^5/d^2+6/c^4*a*b/d^2*arctan(c*x)*x-1/2/c^4*b^2/d^2*arctan(c*x)/(c*x-I)*x-2/c^5*b^2/d^2*Pi
*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2-4/c^5*b^2/d^2*Pi*csgn((1+I*c*x)^2/(
c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+4/c^5*a*b/d^2*ln(c*x-I)*ln(-1/2*I*(I+c*x))+2/c^5*a*b/d
^2*arctan(c*x)/(c*x-I)-I/c^3*b^2/d^2*arctan(c*x)^2*x^2+2*I/c^4*b^2/d^2/(8*c*x-8*I)*x+11/12*I/c^5*a*b/d^2*arcta
n(1/6*c^3*x^3+7/6*c*x)-29/6*I/c^5*a*b/d^2*arctan(c*x)+11/6*I/c^5*a*b/d^2*arctan(1/2*c*x-1/2*I)-11/12*I/c^5*a*b
/d^2*arctan(1/2*c*x)+4*I/c^5*b^2/d^2*arctan(c*x)^2*ln(c*x-I)-1/2*I/c^5*b^2/d^2*arctan(c*x)/(c*x-I)-4*I/c^5*b^2
/d^2*arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))-I/c^5*a*b/d^2/(c*x-I)+7/3/c^5*a*b/d^2-1/3*I/c^5*b^2/d^2-1/3
/c^2*a^2/d^2*x^3+3/c^4*a^2/d^2*x-2/c^5*b^2/d^2/(8*c*x-8*I)-8/3/c^5*b^2/d^2*arctan(c*x)^3-4/c^5*a^2/d^2*arctan(
c*x)+1/c^5*a^2/d^2/(c*x-I)-2/c^5*b^2/d^2*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c
*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+8*I/c^5*a*b/d^2*arctan(c*x)*ln(c*x-I)-2*I/c^3*a*b/d^2*arctan(c*x)*x^2+2*
I*a*b*x/d^2/c^4+2*I*b^2*x*arctan(c*x)/d^2/c^4-11/24/c^5*a*b/d^2*ln(c^4*x^4+10*c^2*x^2+9)-29/12/c^5*a*b/d^2*ln(
c^2*x^2+1)+1/c^5*b^2/d^2*arctan(c*x)^2/(c*x-I)+20/3/c^5*b^2/d^2*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2)
)+20/3/c^5*b^2/d^2*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-4/c^5*b^2/d^2*arctan(c*x)*polylog(2,-(1+I*c
*x)^2/(c^2*x^2+1))-1/3/c^2*b^2/d^2*arctan(c*x)^2*x^3-2/3/c^2*a*b/d^2*arctan(c*x)*x^3

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maxima [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(x^4*(a+b*arctan(c*x))^2/(d+I*c*d*x)^2,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*atan(c*x))^2)/(d + c*d*x*1i)^2,x)

[Out]

int((x^4*(a + b*atan(c*x))^2)/(d + c*d*x*1i)^2, 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(x**4*(a+b*atan(c*x))**2/(d+I*c*d*x)**2,x)

[Out]

Timed out

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