\(\int \frac {x^3 (a+b \arctan (c x))}{d+i c d x} \, dx\) [43]

Optimal result

Integrand size = 23, antiderivative size = 196 \[ \int \frac {x^3 (a+b \arctan (c x))}{d+i c d x} \, dx=\frac {i a x}{c^3 d}-\frac {b x}{2 c^3 d}+\frac {i b x^2}{6 c^2 d}+\frac {b \arctan (c x)}{2 c^4 d}+\frac {i b x \arctan (c x)}{c^3 d}+\frac {x^2 (a+b \arctan (c x))}{2 c^2 d}-\frac {i x^3 (a+b \arctan (c x))}{3 c d}+\frac {(a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c^4 d}-\frac {2 i b \log \left (1+c^2 x^2\right )}{3 c^4 d}+\frac {i b \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c^4 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.85 \[ \int \frac {x^3 (a+b \arctan (c x))}{d+i c d x} \, dx=-\frac {i \left (-b-6 a c x-3 i b c x+3 i a c^2 x^2-b c^2 x^2+2 a c^3 x^3+6 b \arctan (c x)^2+\arctan (c x) \left (6 a+b \left (3 i-6 c x+3 i c^2 x^2+2 c^3 x^3\right )+6 i b \log \left (1+e^{2 i \arctan (c x)}\right )\right )-3 i a \log \left (1+c^2 x^2\right )+4 b \log \left (1+c^2 x^2\right )+3 b \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )\right )}{6 c^4 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.10, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {5401, 27, 5361, 243, 49, 2009, 5401, 5361, 262, 216, 5401, 2009, 5379, 2849, 2752}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^3 (a+b \arctan (c x))}{d+i c d x} \, dx\)

\(\Big \downarrow \) 5401

\(\displaystyle \frac {i \int \frac {x^2 (a+b \arctan (c x))}{d (i c x+1)}dx}{c}-\frac {i \int x^2 (a+b \arctan (c x))dx}{c d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {i \int \frac {x^2 (a+b \arctan (c x))}{i c x+1}dx}{c d}-\frac {i \int x^2 (a+b \arctan (c x))dx}{c d}\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {i \int \frac {x^2 (a+b \arctan (c x))}{i c x+1}dx}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{3} b c \int \frac {x^3}{c^2 x^2+1}dx\right )}{c d}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {i \int \frac {x^2 (a+b \arctan (c x))}{i c x+1}dx}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \int \frac {x^2}{c^2 x^2+1}dx^2\right )}{c d}\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {i \int \frac {x^2 (a+b \arctan (c x))}{i c x+1}dx}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \int \left (\frac {1}{c^2}-\frac {1}{c^2 \left (c^2 x^2+1\right )}\right )dx^2\right )}{c d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {i \int \frac {x^2 (a+b \arctan (c x))}{i c x+1}dx}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )\right )}{c d}\)

\(\Big \downarrow \) 5401

\(\displaystyle \frac {i \left (\frac {i \int \frac {x (a+b \arctan (c x))}{i c x+1}dx}{c}-\frac {i \int x (a+b \arctan (c x))dx}{c}\right )}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )\right )}{c d}\)

\(\Big \downarrow \) 5361

\(\displaystyle \frac {i \left (\frac {i \int \frac {x (a+b \arctan (c x))}{i c x+1}dx}{c}-\frac {i \left (\frac {1}{2} x^2 (a+b \arctan (c x))-\frac {1}{2} b c \int \frac {x^2}{c^2 x^2+1}dx\right )}{c}\right )}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )\right )}{c d}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {i \left (\frac {i \int \frac {x (a+b \arctan (c x))}{i c x+1}dx}{c}-\frac {i \left (\frac {1}{2} x^2 (a+b \arctan (c x))-\frac {1}{2} b c \left (\frac {x}{c^2}-\frac {\int \frac {1}{c^2 x^2+1}dx}{c^2}\right )\right )}{c}\right )}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )\right )}{c d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {i \left (\frac {i \int \frac {x (a+b \arctan (c x))}{i c x+1}dx}{c}-\frac {i \left (\frac {1}{2} x^2 (a+b \arctan (c x))-\frac {1}{2} b c \left (\frac {x}{c^2}-\frac {\arctan (c x)}{c^3}\right )\right )}{c}\right )}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )\right )}{c d}\)

\(\Big \downarrow \) 5401

\(\displaystyle \frac {i \left (\frac {i \left (\frac {i \int \frac {a+b \arctan (c x)}{i c x+1}dx}{c}-\frac {i \int (a+b \arctan (c x))dx}{c}\right )}{c}-\frac {i \left (\frac {1}{2} x^2 (a+b \arctan (c x))-\frac {1}{2} b c \left (\frac {x}{c^2}-\frac {\arctan (c x)}{c^3}\right )\right )}{c}\right )}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )\right )}{c d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {i \left (\frac {i \left (\frac {i \int \frac {a+b \arctan (c x)}{i c x+1}dx}{c}-\frac {i \left (a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}\right )}{c}\right )}{c}-\frac {i \left (\frac {1}{2} x^2 (a+b \arctan (c x))-\frac {1}{2} b c \left (\frac {x}{c^2}-\frac {\arctan (c x)}{c^3}\right )\right )}{c}\right )}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )\right )}{c d}\)

\(\Big \downarrow \) 5379

\(\displaystyle \frac {i \left (\frac {i \left (\frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c}-i b \int \frac {\log \left (\frac {2}{i c x+1}\right )}{c^2 x^2+1}dx\right )}{c}-\frac {i \left (a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}\right )}{c}\right )}{c}-\frac {i \left (\frac {1}{2} x^2 (a+b \arctan (c x))-\frac {1}{2} b c \left (\frac {x}{c^2}-\frac {\arctan (c x)}{c^3}\right )\right )}{c}\right )}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )\right )}{c d}\)

\(\Big \downarrow \) 2849

\(\displaystyle \frac {i \left (\frac {i \left (\frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c}-\frac {b \int \frac {\log \left (\frac {2}{i c x+1}\right )}{1-\frac {2}{i c x+1}}d\frac {1}{i c x+1}}{c}\right )}{c}-\frac {i \left (a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}\right )}{c}\right )}{c}-\frac {i \left (\frac {1}{2} x^2 (a+b \arctan (c x))-\frac {1}{2} b c \left (\frac {x}{c^2}-\frac {\arctan (c x)}{c^3}\right )\right )}{c}\right )}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )\right )}{c d}\)

\(\Big \downarrow \) 2752

\(\displaystyle \frac {i \left (\frac {i \left (\frac {i \left (\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{2 c}\right )}{c}-\frac {i \left (a x+b x \arctan (c x)-\frac {b \log \left (c^2 x^2+1\right )}{2 c}\right )}{c}\right )}{c}-\frac {i \left (\frac {1}{2} x^2 (a+b \arctan (c x))-\frac {1}{2} b c \left (\frac {x}{c^2}-\frac {\arctan (c x)}{c^3}\right )\right )}{c}\right )}{c d}-\frac {i \left (\frac {1}{3} x^3 (a+b \arctan (c x))-\frac {1}{6} b c \left (\frac {x^2}{c^2}-\frac {\log \left (c^2 x^2+1\right )}{c^4}\right )\right )}{c d}\)

Input:

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

Output:

((-I)*((x^3*(a + b*ArcTan[c*x]))/3 - (b*c*(x^2/c^2 - Log[1 + c^2*x^2]/c^4) 
)/6))/(c*d) + (I*(((-I)*((x^2*(a + b*ArcTan[c*x]))/2 - (b*c*(x/c^2 - ArcTa 
n[c*x]/c^3))/2))/c + (I*(((-I)*(a*x + b*x*ArcTan[c*x] - (b*Log[1 + c^2*x^2 
])/(2*c)))/c + (I*((I*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/c - (b*PolyL 
og[2, 1 - 2/(1 + I*c*x)])/(2*c)))/c))/c))/(c*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 49

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] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 216

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2752

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

rule 2849

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp 
[-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 5361

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] - Simp[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 5379

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] + Simp[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 5401

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + ( 
e_.)*(x_)), x_Symbol] :> Simp[f/e   Int[(f*x)^(m - 1)*(a + b*ArcTan[c*x])^p 
, x], x] - Simp[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]
 

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.59

method	result	size
derivativedivides	\(\frac {\frac {2 i b}{3 d}+\frac {i b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d}+\frac {a \,c^{2} x^{2}}{2 d}-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d}+\frac {i b \operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d}+\frac {i a c x}{d}+\frac {i b \,c^{2} x^{2}}{6 d}+\frac {b \arctan \left (c x \right ) c^{2} x^{2}}{2 d}-\frac {b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d}-\frac {i a \,c^{3} x^{3}}{3 d}-\frac {i a \arctan \left (c x \right )}{d}-\frac {11 i b \ln \left (c^{2} x^{2}+1\right )}{24 d}-\frac {b c x}{2 d}+\frac {i b \arctan \left (c x \right ) c x}{d}-\frac {i b \arctan \left (c x \right ) c^{3} x^{3}}{3 d}-\frac {i b \ln \left (c x -i\right )^{2}}{4 d}+\frac {5 b \arctan \left (\frac {c x}{2}\right )}{24 d}-\frac {5 b \arctan \left (\frac {1}{6} c^{3} x^{3}+\frac {7}{6} c x \right )}{24 d}-\frac {5 b \arctan \left (\frac {c x}{2}-\frac {i}{2}\right )}{12 d}-\frac {5 i b \ln \left (c^{4} x^{4}+10 c^{2} x^{2}+9\right )}{48 d}+\frac {11 b \arctan \left (c x \right )}{12 d}}{c^{4}}\)	\(312\)
default	\(\frac {\frac {2 i b}{3 d}+\frac {i b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d}+\frac {a \,c^{2} x^{2}}{2 d}-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d}+\frac {i b \operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d}+\frac {i a c x}{d}+\frac {i b \,c^{2} x^{2}}{6 d}+\frac {b \arctan \left (c x \right ) c^{2} x^{2}}{2 d}-\frac {b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d}-\frac {i a \,c^{3} x^{3}}{3 d}-\frac {i a \arctan \left (c x \right )}{d}-\frac {11 i b \ln \left (c^{2} x^{2}+1\right )}{24 d}-\frac {b c x}{2 d}+\frac {i b \arctan \left (c x \right ) c x}{d}-\frac {i b \arctan \left (c x \right ) c^{3} x^{3}}{3 d}-\frac {i b \ln \left (c x -i\right )^{2}}{4 d}+\frac {5 b \arctan \left (\frac {c x}{2}\right )}{24 d}-\frac {5 b \arctan \left (\frac {1}{6} c^{3} x^{3}+\frac {7}{6} c x \right )}{24 d}-\frac {5 b \arctan \left (\frac {c x}{2}-\frac {i}{2}\right )}{12 d}-\frac {5 i b \ln \left (c^{4} x^{4}+10 c^{2} x^{2}+9\right )}{48 d}+\frac {11 b \arctan \left (c x \right )}{12 d}}{c^{4}}\)	\(312\)
risch	\(\frac {i b \operatorname {dilog}\left (\frac {1}{2}-\frac {i c x}{2}\right )}{2 d \,c^{4}}-\frac {b \left (\frac {1}{3} x^{3} c^{2}+\frac {1}{2} i c \,x^{2}-x \right ) \ln \left (i c x +1\right )}{2 c^{3} d}-\frac {i a \arctan \left (c x \right )}{d \,c^{4}}-\frac {b x}{2 c^{3} d}-\frac {5 a}{6 d \,c^{4}}+\frac {i a x}{c^{3} d}+\frac {b \ln \left (-i c x +1\right ) x^{3}}{6 d c}-\frac {i x^{3} a}{3 d c}-\frac {b \ln \left (-i c x +1\right ) x}{2 d \,c^{3}}+\frac {i b \,x^{2}}{6 c^{2} d}+\frac {i b \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (\frac {1}{2}-\frac {i c x}{2}\right )}{2 d \,c^{4}}-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d \,c^{4}}-\frac {5 i b \ln \left (-i c x +1\right )}{12 d \,c^{4}}+\frac {a \,x^{2}}{2 d \,c^{2}}+\frac {31 i b}{72 d \,c^{4}}+\frac {i x^{2} b \ln \left (-i c x +1\right )}{4 d \,c^{2}}-\frac {i b \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (-i c x +1\right )}{2 d \,c^{4}}+\frac {i b \ln \left (i c x +1\right )^{2}}{4 c^{4} d}-\frac {11 i b \ln \left (c^{2} x^{2}+1\right )}{24 c^{4} d}+\frac {11 b \arctan \left (c x \right )}{12 c^{4} d}\)	\(349\)
parts	\(-\frac {i a \arctan \left (c x \right )}{d \,c^{4}}+\frac {a \,x^{2}}{2 d \,c^{2}}-\frac {i b \ln \left (c x -i\right )^{2}}{4 d \,c^{4}}-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d \,c^{4}}-\frac {5 i b \ln \left (c^{4} x^{4}+10 c^{2} x^{2}+9\right )}{48 d \,c^{4}}+\frac {i a x}{c^{3} d}+\frac {i b \operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d \,c^{4}}+\frac {b \arctan \left (c x \right ) x^{2}}{2 d \,c^{2}}-\frac {b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d \,c^{4}}-\frac {i x^{3} a}{3 d c}+\frac {i b x \arctan \left (c x \right )}{c^{3} d}+\frac {i b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d \,c^{4}}-\frac {b x}{2 c^{3} d}+\frac {i b \,x^{2}}{6 c^{2} d}-\frac {i b \arctan \left (c x \right ) x^{3}}{3 d c}+\frac {2 i b}{3 d \,c^{4}}+\frac {5 b \arctan \left (\frac {c x}{2}\right )}{24 d \,c^{4}}-\frac {5 b \arctan \left (\frac {1}{6} c^{3} x^{3}+\frac {7}{6} c x \right )}{24 d \,c^{4}}-\frac {5 b \arctan \left (\frac {c x}{2}-\frac {i}{2}\right )}{12 d \,c^{4}}-\frac {11 i b \ln \left (c^{2} x^{2}+1\right )}{24 c^{4} d}+\frac {11 b \arctan \left (c x \right )}{12 c^{4} d}\)	\(353\)

Input:

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

Output:

1/c^4*(2/3*I*b/d+1/2*I*b/d*ln(c*x-I)*ln(-1/2*I*(c*x+I))+1/2*a/d*c^2*x^2-1/ 
2*a/d*ln(c^2*x^2+1)+1/2*I*b/d*dilog(-1/2*I*(c*x+I))+I*a/d*c*x+1/6*I*b/d*c^ 
2*x^2+1/2*b/d*arctan(c*x)*c^2*x^2-b/d*arctan(c*x)*ln(c*x-I)-1/3*I*a/d*c^3* 
x^3-I*a/d*arctan(c*x)-11/24*I*b/d*ln(c^2*x^2+1)-1/2*b/d*c*x+I*b/d*arctan(c 
*x)*c*x-1/3*I*b/d*arctan(c*x)*c^3*x^3-1/4*I*b/d*ln(c*x-I)^2+5/24*b/d*arcta 
n(1/2*c*x)-5/24*b/d*arctan(1/6*c^3*x^3+7/6*c*x)-5/12*b/d*arctan(1/2*c*x-1/ 
2*I)-5/48*I*b/d*ln(c^4*x^4+10*c^2*x^2+9)+11/12*b/d*arctan(c*x))
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

-1/6*a*(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) 
) - 1/72*(432*I*c^8*d*integrate(1/12*x^4*arctan(c*x)/(c^5*d*x^2 + c^3*d), 
x) + 216*c^8*d*integrate(1/12*x^4*log(c^2*x^2 + 1)/(c^5*d*x^2 + c^3*d), x) 
 - 432*c^7*d*integrate(1/12*x^3*arctan(c*x)/(c^5*d*x^2 + c^3*d), x) + 216* 
I*c^7*d*integrate(1/12*x^3*log(c^2*x^2 + 1)/(c^5*d*x^2 + c^3*d), x) + 432* 
c^5*d*integrate(1/12*x*arctan(c*x)/(c^5*d*x^2 + c^3*d), x) - 216*I*c^5*d*i 
ntegrate(1/12*x*log(c^2*x^2 + 1)/(c^5*d*x^2 + c^3*d), x) + 4*c^3*x^3 - 216 
*c^4*d*integrate(1/12*log(c^2*x^2 + 1)/(c^5*d*x^2 + c^3*d), x) + 3*I*c^2*x 
^2 - 30*c*x - 6*(-2*I*c^3*x^3 + 3*c^2*x^2 + 6*I*c*x - 5)*arctan(c*x) + 18* 
I*arctan(c*x)^2 - 3*(2*c^3*x^3 + 3*I*c^2*x^2 - 6*c*x + I)*log(c^2*x^2 + 1) 
 + 9*I*log(c^2*x^2 + 1)^2 + 18*I*log(12*c^5*d*x^2 + 12*c^3*d))*b/(c^4*d)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^3 (a+b \arctan (c x))}{d+i c d x} \, dx=\frac {6 \left (\int \frac {\mathit {atan} \left (c x \right ) x^{3}}{c i x +1}d x \right ) b \,c^{4}-6 \,\mathrm {log}\left (c i x +1\right ) a -2 a \,c^{3} i \,x^{3}+3 a \,c^{2} x^{2}+6 a c i x}{6 c^{4} d} \] Input:

int(x^3*(a+b*atan(c*x))/(d+I*c*d*x),x)
                                                                                    
                                                                                    
 

Output:

(6*int((atan(c*x)*x**3)/(c*i*x + 1),x)*b*c**4 - 6*log(c*i*x + 1)*a - 2*a*c 
**3*i*x**3 + 3*a*c**2*x**2 + 6*a*c*i*x)/(6*c**4*d)