\(\int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x} \, dx\) [24]

Optimal result

Integrand size = 23, antiderivative size = 170 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x} \, dx=3 i a c d^3 x+\frac {3}{2} b c d^3 x+\frac {1}{6} i b c^2 d^3 x^2-\frac {3}{2} b d^3 \arctan (c x)+3 i b c d^3 x \arctan (c x)-\frac {3}{2} c^2 d^3 x^2 (a+b \arctan (c x))-\frac {1}{3} i c^3 d^3 x^3 (a+b \arctan (c x))+a d^3 \log (x)-\frac {5}{3} i b d^3 \log \left (1+c^2 x^2\right )+\frac {1}{2} i b d^3 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^3 \operatorname {PolyLog}(2,i c x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4996, 4930, 266, 4940, 2438, 4946, 327, 209, 272, 45} \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x} \, dx=-\frac {1}{3} i c^3 d^3 x^3 (a+b \arctan (c x))-\frac {3}{2} c^2 d^3 x^2 (a+b \arctan (c x))+3 i a c d^3 x+a d^3 \log (x)-\frac {3}{2} b d^3 \arctan (c x)+3 i b c d^3 x \arctan (c x)+\frac {1}{6} i b c^2 d^3 x^2-\frac {5}{3} i b d^3 \log \left (c^2 x^2+1\right )+\frac {1}{2} i b d^3 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b d^3 \operatorname {PolyLog}(2,i c x)+\frac {3}{2} b c d^3 x \]

[In]

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

[Out]

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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 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 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 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 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 4996

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])

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.82 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x} \, dx=-\frac {1}{6} i d^3 \left (-18 a c x+9 i b c x-9 i a c^2 x^2-b c^2 x^2+2 a c^3 x^3-9 i b \arctan (c x)-18 b c x \arctan (c x)-9 i b c^2 x^2 \arctan (c x)+2 b c^3 x^3 \arctan (c x)+6 i a \log (x)+10 b \log \left (1+c^2 x^2\right )-3 b \operatorname {PolyLog}(2,-i c x)+3 b \operatorname {PolyLog}(2,i c x)\right ) \]

[In]

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

[Out]

(-1/6*I)*d^3*(-18*a*c*x + (9*I)*b*c*x - (9*I)*a*c^2*x^2 - b*c^2*x^2 + 2*a*c^3*x^3 - (9*I)*b*ArcTan[c*x] - 18*b
*c*x*ArcTan[c*x] - (9*I)*b*c^2*x^2*ArcTan[c*x] + 2*b*c^3*x^3*ArcTan[c*x] + (6*I)*a*Log[x] + 10*b*Log[1 + c^2*x
^2] - 3*b*PolyLog[2, (-I)*c*x] + 3*b*PolyLog[2, I*c*x])

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.97

method	result	size
parts	\(a \,d^{3} \left (-\frac {i c^{3} x^{3}}{3}-\frac {3 c^{2} x^{2}}{2}+3 i c x +\ln \left (x \right )\right )+b \,d^{3} \left (3 i \arctan \left (c x \right ) c x -\frac {i \arctan \left (c x \right ) c^{3} x^{3}}{3}-\frac {3 c^{2} x^{2} \arctan \left (c x \right )}{2}+\arctan \left (c x \right ) \ln \left (c x \right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {3 c x}{2}+\frac {i c^{2} x^{2}}{6}-\frac {5 i \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {3 \arctan \left (c x \right )}{2}\right )\)	\(165\)
derivativedivides	\(a \,d^{3} \left (3 i c x -\frac {i c^{3} x^{3}}{3}-\frac {3 c^{2} x^{2}}{2}+\ln \left (c x \right )\right )+b \,d^{3} \left (3 i \arctan \left (c x \right ) c x -\frac {i \arctan \left (c x \right ) c^{3} x^{3}}{3}-\frac {3 c^{2} x^{2} \arctan \left (c x \right )}{2}+\arctan \left (c x \right ) \ln \left (c x \right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {3 c x}{2}+\frac {i c^{2} x^{2}}{6}-\frac {5 i \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {3 \arctan \left (c x \right )}{2}\right )\)	\(167\)
default	\(a \,d^{3} \left (3 i c x -\frac {i c^{3} x^{3}}{3}-\frac {3 c^{2} x^{2}}{2}+\ln \left (c x \right )\right )+b \,d^{3} \left (3 i \arctan \left (c x \right ) c x -\frac {i \arctan \left (c x \right ) c^{3} x^{3}}{3}-\frac {3 c^{2} x^{2} \arctan \left (c x \right )}{2}+\arctan \left (c x \right ) \ln \left (c x \right )+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {3 c x}{2}+\frac {i c^{2} x^{2}}{6}-\frac {5 i \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {3 \arctan \left (c x \right )}{2}\right )\)	\(167\)
risch	\(-\frac {b \,d^{3} \ln \left (i c x +1\right ) c^{3} x^{3}}{6}-\frac {i x^{3} a \,c^{3} d^{3}}{3}+\frac {3 b \,d^{3} \ln \left (i c x +1\right ) c x}{2}+\frac {3 i b \,d^{3} \ln \left (i c x +1\right ) c^{2} x^{2}}{4}+\frac {i x^{2} b \,c^{2} d^{3}}{6}+\frac {3 b c \,d^{3} x}{2}+3 i a c \,d^{3} x -\frac {29 i d^{3} \ln \left (-i c x +1\right ) b}{12}-\frac {i d^{3} \operatorname {dilog}\left (-i c x +1\right ) b}{2}-\frac {3 x^{2} d^{3} c^{2} a}{2}+\frac {65 i b \,d^{3}}{18}-\frac {29 a \,d^{3}}{6}+d^{3} \ln \left (-i c x \right ) a +\frac {d^{3} b \,c^{3} x^{3} \ln \left (-i c x +1\right )}{6}-\frac {11 i b \,d^{3} \ln \left (i c x +1\right )}{12}-\frac {3 d^{3} b c x \ln \left (-i c x +1\right )}{2}-\frac {3 i d^{3} x^{2} b \ln \left (-i c x +1\right ) c^{2}}{4}+\frac {i b \,d^{3} \operatorname {dilog}\left (i c x +1\right )}{2}\)	\(255\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(1/2*(-2*I*a*c^3*d^3*x^3 - 6*a*c^2*d^3*x^2 + 6*I*a*c*d^3*x + 2*a*d^3 + (b*c^3*d^3*x^3 - 3*I*b*c^2*d^3*
x^2 - 3*b*c*d^3*x + I*b*d^3)*log(-(c*x + I)/(c*x - I)))/x, x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.08 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x} \, dx=-\frac {1}{3} i \, a c^{3} d^{3} x^{3} - \frac {3}{2} \, a c^{2} d^{3} x^{2} + \frac {1}{6} i \, b c^{2} d^{3} x^{2} + 3 i \, a c d^{3} x + \frac {3}{2} \, b c d^{3} x - \frac {1}{12} \, {\left (3 \, \pi + 2 i\right )} b d^{3} \log \left (c^{2} x^{2} + 1\right ) + b d^{3} \arctan \left (c x\right ) \log \left (c x\right ) + \frac {3}{2} i \, {\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{3} - \frac {1}{2} i \, b d^{3} {\rm Li}_2\left (i \, c x + 1\right ) + \frac {1}{2} i \, b d^{3} {\rm Li}_2\left (-i \, c x + 1\right ) + a d^{3} \log \left (x\right ) + \frac {1}{6} \, {\left (-2 i \, b c^{3} d^{3} x^{3} - 9 \, b c^{2} d^{3} x^{2} - 9 \, b d^{3}\right )} \arctan \left (c x\right ) \]

[In]

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

[Out]

-1/3*I*a*c^3*d^3*x^3 - 3/2*a*c^2*d^3*x^2 + 1/6*I*b*c^2*d^3*x^2 + 3*I*a*c*d^3*x + 3/2*b*c*d^3*x - 1/12*(3*pi +
2*I)*b*d^3*log(c^2*x^2 + 1) + b*d^3*arctan(c*x)*log(c*x) + 3/2*I*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*d^3
- 1/2*I*b*d^3*dilog(I*c*x + 1) + 1/2*I*b*d^3*dilog(-I*c*x + 1) + a*d^3*log(x) + 1/6*(-2*I*b*c^3*d^3*x^3 - 9*b*
c^2*d^3*x^2 - 9*b*d^3)*arctan(c*x)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.89 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.15 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x} \, dx=\left \{\begin {array}{cl} a\,d^3\,\ln \left (x\right ) & \text {\ if\ \ }c=0\\ a\,d^3\,\ln \left (x\right )-\frac {b\,d^3\,\ln \left (c^2\,x^2+1\right )\,3{}\mathrm {i}}{2}-\frac {b\,d^3\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {b\,d^3\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\frac {3\,a\,c^2\,d^3\,x^2}{2}-\frac {a\,c^3\,d^3\,x^3\,1{}\mathrm {i}}{3}+a\,c\,d^3\,x\,3{}\mathrm {i}+\frac {3\,b\,c\,d^3\,x}{2}+\frac {b\,c^2\,d^3\,\left (\frac {x^2}{2}-\frac {\ln \left (c^2\,x^2+1\right )}{2\,c^2}\right )\,1{}\mathrm {i}}{3}-3\,b\,c^2\,d^3\,\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )-\frac {b\,c^3\,d^3\,x^3\,\mathrm {atan}\left (c\,x\right )\,1{}\mathrm {i}}{3}+b\,c\,d^3\,x\,\mathrm {atan}\left (c\,x\right )\,3{}\mathrm {i} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

[In]

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

[Out]

piecewise(c == 0, a*d^3*log(x), c ~= 0, - (b*d^3*log(c^2*x^2 + 1)*3i)/2 + a*d^3*log(x) - (b*d^3*dilog(- c*x*1i
+ 1)*1i)/2 + (b*d^3*dilog(c*x*1i + 1)*1i)/2 - (3*a*c^2*d^3*x^2)/2 - (a*c^3*d^3*x^3*1i)/3 + a*c*d^3*x*3i + (3*
b*c*d^3*x)/2 + (b*c^2*d^3*(x^2/2 - log(c^2*x^2 + 1)/(2*c^2))*1i)/3 - 3*b*c^2*d^3*atan(c*x)*(1/(2*c^2) + x^2/2)
- (b*c^3*d^3*x^3*atan(c*x)*1i)/3 + b*c*d^3*x*atan(c*x)*3i)