\(\int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx\) [95]

Optimal result

Integrand size = 25, antiderivative size = 356 \[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx=-\frac {a b x}{c^3 d}-\frac {i b^2 x}{3 c^3 d}+\frac {i b^2 \arctan (c x)}{3 c^4 d}-\frac {b^2 x \arctan (c x)}{c^3 d}+\frac {i b x^2 (a+b \arctan (c x))}{3 c^2 d}-\frac {5 (a+b \arctan (c x))^2}{6 c^4 d}+\frac {i x (a+b \arctan (c x))^2}{c^3 d}+\frac {x^2 (a+b \arctan (c x))^2}{2 c^2 d}-\frac {i x^3 (a+b \arctan (c x))^2}{3 c d}+\frac {8 i b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{3 c^4 d}+\frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^4 d}+\frac {b^2 \log \left (1+c^2 x^2\right )}{2 c^4 d}-\frac {4 b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{3 c^4 d}+\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^4 d}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^4 d} \]

[Out]

-a*b*x/c^3/d-1/3*I*b^2*x/c^3/d+1/3*I*b^2*arctan(c*x)/c^4/d-b^2*x*arctan(c*x)/c^3/d+1/3*I*b*x^2*(a+b*arctan(c*x
))/c^2/d-5/6*(a+b*arctan(c*x))^2/c^4/d+I*x*(a+b*arctan(c*x))^2/c^3/d+1/2*x^2*(a+b*arctan(c*x))^2/c^2/d-1/3*I*x
^3*(a+b*arctan(c*x))^2/c/d+8/3*I*b*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c^4/d+(a+b*arctan(c*x))^2*ln(2/(1+I*c*x))
/c^4/d+1/2*b^2*ln(c^2*x^2+1)/c^4/d-4/3*b^2*polylog(2,1-2/(1+I*c*x))/c^4/d+I*b*(a+b*arctan(c*x))*polylog(2,1-2/
(1+I*c*x))/c^4/d+1/2*b^2*polylog(3,1-2/(1+I*c*x))/c^4/d

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {4986, 4946, 5036, 327, 209, 5040, 4964, 2449, 2352, 4930, 266, 5004, 5114, 6745} \[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx=\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))}{c^4 d}-\frac {5 (a+b \arctan (c x))^2}{6 c^4 d}+\frac {\log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{c^4 d}+\frac {8 i b \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{3 c^4 d}+\frac {i x (a+b \arctan (c x))^2}{c^3 d}+\frac {x^2 (a+b \arctan (c x))^2}{2 c^2 d}+\frac {i b x^2 (a+b \arctan (c x))}{3 c^2 d}-\frac {i x^3 (a+b \arctan (c x))^2}{3 c d}-\frac {a b x}{c^3 d}+\frac {i b^2 \arctan (c x)}{3 c^4 d}-\frac {b^2 x \arctan (c x)}{c^3 d}-\frac {4 b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{3 c^4 d}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )}{2 c^4 d}-\frac {i b^2 x}{3 c^3 d}+\frac {b^2 \log \left (c^2 x^2+1\right )}{2 c^4 d} \]

[In]

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

[Out]

-((a*b*x)/(c^3*d)) - ((I/3)*b^2*x)/(c^3*d) + ((I/3)*b^2*ArcTan[c*x])/(c^4*d) - (b^2*x*ArcTan[c*x])/(c^3*d) + (
(I/3)*b*x^2*(a + b*ArcTan[c*x]))/(c^2*d) - (5*(a + b*ArcTan[c*x])^2)/(6*c^4*d) + (I*x*(a + b*ArcTan[c*x])^2)/(
c^3*d) + (x^2*(a + b*ArcTan[c*x])^2)/(2*c^2*d) - ((I/3)*x^3*(a + b*ArcTan[c*x])^2)/(c*d) + (((8*I)/3)*b*(a + b
*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(c^4*d) + ((a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/(c^4*d) + (b^2*Log[1 +
c^2*x^2])/(2*c^4*d) - (4*b^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/(3*c^4*d) + (I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1
- 2/(1 + I*c*x)])/(c^4*d) + (b^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/(2*c^4*d)

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 266

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 327

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 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 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 4930

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

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 4986

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

Rule 5004

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 5036

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 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[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 5114

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar
cTan[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 6745

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*} \text {integral}& = \frac {i \int \frac {x^2 (a+b \arctan (c x))^2}{d+i c d x} \, dx}{c}-\frac {i \int x^2 (a+b \arctan (c x))^2 \, dx}{c d} \\ & = -\frac {i x^3 (a+b \arctan (c x))^2}{3 c d}-\frac {\int \frac {x (a+b \arctan (c x))^2}{d+i c d x} \, dx}{c^2}+\frac {(2 i b) \int \frac {x^3 (a+b \arctan (c x))}{1+c^2 x^2} \, dx}{3 d}+\frac {\int x (a+b \arctan (c x))^2 \, dx}{c^2 d} \\ & = \frac {x^2 (a+b \arctan (c x))^2}{2 c^2 d}-\frac {i x^3 (a+b \arctan (c x))^2}{3 c d}-\frac {i \int \frac {(a+b \arctan (c x))^2}{d+i c d x} \, dx}{c^3}+\frac {i \int (a+b \arctan (c x))^2 \, dx}{c^3 d}+\frac {(2 i b) \int x (a+b \arctan (c x)) \, dx}{3 c^2 d}-\frac {(2 i b) \int \frac {x (a+b \arctan (c x))}{1+c^2 x^2} \, dx}{3 c^2 d}-\frac {b \int \frac {x^2 (a+b \arctan (c x))}{1+c^2 x^2} \, dx}{c d} \\ & = \frac {i b x^2 (a+b \arctan (c x))}{3 c^2 d}-\frac {(a+b \arctan (c x))^2}{3 c^4 d}+\frac {i x (a+b \arctan (c x))^2}{c^3 d}+\frac {x^2 (a+b \arctan (c x))^2}{2 c^2 d}-\frac {i x^3 (a+b \arctan (c x))^2}{3 c d}+\frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^4 d}+\frac {(2 i b) \int \frac {a+b \arctan (c x)}{i-c x} \, dx}{3 c^3 d}-\frac {b \int (a+b \arctan (c x)) \, dx}{c^3 d}+\frac {b \int \frac {a+b \arctan (c x)}{1+c^2 x^2} \, dx}{c^3 d}-\frac {(2 b) \int \frac {(a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^3 d}-\frac {(2 i b) \int \frac {x (a+b \arctan (c x))}{1+c^2 x^2} \, dx}{c^2 d}-\frac {\left (i b^2\right ) \int \frac {x^2}{1+c^2 x^2} \, dx}{3 c d} \\ & = -\frac {a b x}{c^3 d}-\frac {i b^2 x}{3 c^3 d}+\frac {i b x^2 (a+b \arctan (c x))}{3 c^2 d}-\frac {5 (a+b \arctan (c x))^2}{6 c^4 d}+\frac {i x (a+b \arctan (c x))^2}{c^3 d}+\frac {x^2 (a+b \arctan (c x))^2}{2 c^2 d}-\frac {i x^3 (a+b \arctan (c x))^2}{3 c d}+\frac {2 i b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{3 c^4 d}+\frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^4 d}+\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^4 d}+\frac {(2 i b) \int \frac {a+b \arctan (c x)}{i-c x} \, dx}{c^3 d}+\frac {\left (i b^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{3 c^3 d}-\frac {\left (2 i b^2\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{3 c^3 d}-\frac {\left (i b^2\right ) \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^3 d}-\frac {b^2 \int \arctan (c x) \, dx}{c^3 d} \\ & = -\frac {a b x}{c^3 d}-\frac {i b^2 x}{3 c^3 d}+\frac {i b^2 \arctan (c x)}{3 c^4 d}-\frac {b^2 x \arctan (c x)}{c^3 d}+\frac {i b x^2 (a+b \arctan (c x))}{3 c^2 d}-\frac {5 (a+b \arctan (c x))^2}{6 c^4 d}+\frac {i x (a+b \arctan (c x))^2}{c^3 d}+\frac {x^2 (a+b \arctan (c x))^2}{2 c^2 d}-\frac {i x^3 (a+b \arctan (c x))^2}{3 c d}+\frac {8 i b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{3 c^4 d}+\frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^4 d}+\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^4 d}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^4 d}-\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{3 c^4 d}-\frac {\left (2 i b^2\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^3 d}+\frac {b^2 \int \frac {x}{1+c^2 x^2} \, dx}{c^2 d} \\ & = -\frac {a b x}{c^3 d}-\frac {i b^2 x}{3 c^3 d}+\frac {i b^2 \arctan (c x)}{3 c^4 d}-\frac {b^2 x \arctan (c x)}{c^3 d}+\frac {i b x^2 (a+b \arctan (c x))}{3 c^2 d}-\frac {5 (a+b \arctan (c x))^2}{6 c^4 d}+\frac {i x (a+b \arctan (c x))^2}{c^3 d}+\frac {x^2 (a+b \arctan (c x))^2}{2 c^2 d}-\frac {i x^3 (a+b \arctan (c x))^2}{3 c d}+\frac {8 i b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{3 c^4 d}+\frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^4 d}+\frac {b^2 \log \left (1+c^2 x^2\right )}{2 c^4 d}-\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{3 c^4 d}+\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^4 d}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^4 d}-\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{c^4 d} \\ & = -\frac {a b x}{c^3 d}-\frac {i b^2 x}{3 c^3 d}+\frac {i b^2 \arctan (c x)}{3 c^4 d}-\frac {b^2 x \arctan (c x)}{c^3 d}+\frac {i b x^2 (a+b \arctan (c x))}{3 c^2 d}-\frac {5 (a+b \arctan (c x))^2}{6 c^4 d}+\frac {i x (a+b \arctan (c x))^2}{c^3 d}+\frac {x^2 (a+b \arctan (c x))^2}{2 c^2 d}-\frac {i x^3 (a+b \arctan (c x))^2}{3 c d}+\frac {8 i b (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{3 c^4 d}+\frac {(a+b \arctan (c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c^4 d}+\frac {b^2 \log \left (1+c^2 x^2\right )}{2 c^4 d}-\frac {4 b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{3 c^4 d}+\frac {i b (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^4 d}+\frac {b^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.18 \[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx=\frac {i a^2 x}{c^3 d}+\frac {a^2 x^2}{2 c^2 d}-\frac {i a^2 x^3}{3 c d}-\frac {i a^2 \arctan (c x)}{c^4 d}-\frac {a^2 \log \left (1+c^2 x^2\right )}{2 c^4 d}-\frac {i a b \left (-3 i c x-8 c x \arctan (c x)+6 \arctan (c x)^2+\left (1+c^2 x^2\right ) (-1+3 i \arctan (c x)+2 c x \arctan (c x))+6 i \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )-8 \log \left (\frac {1}{\sqrt {1+c^2 x^2}}\right )+3 \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )\right )}{3 c^4 d}-\frac {i b^2 \left (2 c x-6 i c x \arctan (c x)-2 \left (1+c^2 x^2\right ) \arctan (c x)+8 i \arctan (c x)^2-8 c x \arctan (c x)^2+3 i \left (1+c^2 x^2\right ) \arctan (c x)^2+2 c x \left (1+c^2 x^2\right ) \arctan (c x)^2+4 \arctan (c x)^3-16 \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )+6 i \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )-6 i \log \left (\frac {1}{\sqrt {1+c^2 x^2}}\right )+(8 i+6 \arctan (c x)) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+3 i \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right )}{6 c^4 d} \]

[In]

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

[Out]

(I*a^2*x)/(c^3*d) + (a^2*x^2)/(2*c^2*d) - ((I/3)*a^2*x^3)/(c*d) - (I*a^2*ArcTan[c*x])/(c^4*d) - (a^2*Log[1 + c
^2*x^2])/(2*c^4*d) - ((I/3)*a*b*((-3*I)*c*x - 8*c*x*ArcTan[c*x] + 6*ArcTan[c*x]^2 + (1 + c^2*x^2)*(-1 + (3*I)*
ArcTan[c*x] + 2*c*x*ArcTan[c*x]) + (6*I)*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] - 8*Log[1/Sqrt[1 + c^2*x^2
]] + 3*PolyLog[2, -E^((2*I)*ArcTan[c*x])]))/(c^4*d) - ((I/6)*b^2*(2*c*x - (6*I)*c*x*ArcTan[c*x] - 2*(1 + c^2*x
^2)*ArcTan[c*x] + (8*I)*ArcTan[c*x]^2 - 8*c*x*ArcTan[c*x]^2 + (3*I)*(1 + c^2*x^2)*ArcTan[c*x]^2 + 2*c*x*(1 + c
^2*x^2)*ArcTan[c*x]^2 + 4*ArcTan[c*x]^3 - 16*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + (6*I)*ArcTan[c*x]^2*
Log[1 + E^((2*I)*ArcTan[c*x])] - (6*I)*Log[1/Sqrt[1 + c^2*x^2]] + (8*I + 6*ArcTan[c*x])*PolyLog[2, -E^((2*I)*A
rcTan[c*x])] + (3*I)*PolyLog[3, -E^((2*I)*ArcTan[c*x])]))/(c^4*d)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 47.77 (sec) , antiderivative size = 1105, normalized size of antiderivative = 3.10

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1105\)
default	\(\text {Expression too large to display}\)	\(1105\)
parts	\(\text {Expression too large to display}\)	\(1149\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{i \, c d x + d} \,d x } \]

[In]

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

[Out]

integral(1/4*(I*b^2*x^3*log(-(c*x + I)/(c*x - I))^2 + 4*a*b*x^3*log(-(c*x + I)/(c*x - I)) - 4*I*a^2*x^3)/(c*d*
x - I*d), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{i \, c d x + d} \,d x } \]

[In]

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

[Out]

-1/6*a^2*(I*(2*c^2*x^3 + 3*I*c*x^2 - 6*x)/(c^3*d) + 6*log(I*c*x + 1)/(c^4*d)) - 1/96*(16*I*(216*b^2*c^4*integr
ate(1/48*x^4*arctan(c*x)^2/(c^5*d*x^2 + c^3*d), x) + 18*b^2*c^4*integrate(1/48*x^4*log(c^2*x^2 + 1)^2/(c^5*d*x
^2 + c^3*d), x) + 576*a*b*c^4*integrate(1/48*x^4*arctan(c*x)/(c^5*d*x^2 + c^3*d), x) + 24*b^2*c^4*integrate(1/
48*x^4*log(c^2*x^2 + 1)/(c^5*d*x^2 + c^3*d), x) + 72*b^2*c^3*integrate(1/48*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(
c^5*d*x^2 + c^3*d), x) + 24*b^2*c^3*integrate(1/48*x^3*arctan(c*x)/(c^5*d*x^2 + c^3*d), x) - 36*b^2*c^2*integr
ate(1/48*x^2*log(c^2*x^2 + 1)/(c^5*d*x^2 + c^3*d), x) + 144*b^2*c*integrate(1/48*x*arctan(c*x)/(c^5*d*x^2 + c^
3*d), x) - 36*b^2*integrate(1/48*log(c^2*x^2 + 1)^2/(c^5*d*x^2 + c^3*d), x) - b^2*arctan(c*x)^3/(c^4*d))*c^4*d
- 96*c^4*d*integrate(1/48*(12*(3*b^2*c^2*x^3 - 2*b^2*x)*arctan(c*x)^2 + 3*(b^2*c^2*x^3 - 2*b^2*x)*log(c^2*x^2
+ 1)^2 - 4*(2*b^2*c^3*x^4 - 24*a*b*c^2*x^3 - 3*b^2*c*x^2)*arctan(c*x) - 2*(6*b^2*c^3*x^4*arctan(c*x) - b^2*c^
2*x^3 - 6*b^2*x)*log(c^2*x^2 + 1))/(c^4*d*x^2 + c^2*d), x) + 24*I*b^2*arctan(c*x)^3 - 3*b^2*log(c^2*x^2 + 1)^3
- 4*(-2*I*b^2*c^3*x^3 + 3*b^2*c^2*x^2 + 6*I*b^2*c*x)*arctan(c*x)^2 + (-2*I*b^2*c^3*x^3 + 3*b^2*c^2*x^2 + 6*I*
b^2*c*x + 6*I*b^2*arctan(c*x))*log(c^2*x^2 + 1)^2 - 4*(3*b^2*arctan(c*x)^2 + (2*b^2*c^3*x^3 + 3*I*b^2*c^2*x^2
- 6*b^2*c*x)*arctan(c*x))*log(c^2*x^2 + 1))/(c^4*d)

Giac [F]

\[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{i \, c d x + d} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 (a+b \arctan (c x))^2}{d+i c d x} \, dx=\int \frac {x^3\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{d+c\,d\,x\,1{}\mathrm {i}} \,d x \]

[In]

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

[Out]

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