\(\int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^2} \, dx\) [25]

Optimal result

Integrand size = 23, antiderivative size = 162 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^2} \, dx=-3 a c^2 d^3 x+\frac {1}{2} i b c^2 d^3 x-\frac {1}{2} i b c d^3 \arctan (c x)-3 b c^2 d^3 x \arctan (c x)-\frac {d^3 (a+b \arctan (c x))}{x}-\frac {1}{2} i c^3 d^3 x^2 (a+b \arctan (c x))+3 i a c d^3 \log (x)+b c d^3 \log (x)+b c d^3 \log \left (1+c^2 x^2\right )-\frac {3}{2} b c d^3 \operatorname {PolyLog}(2,-i c x)+\frac {3}{2} b c d^3 \operatorname {PolyLog}(2,i c x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {4996, 4930, 266, 4946, 272, 36, 29, 31, 4940, 2438, 327, 209} \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^2} \, dx=-\frac {1}{2} i c^3 d^3 x^2 (a+b \arctan (c x))-\frac {d^3 (a+b \arctan (c x))}{x}-3 a c^2 d^3 x+3 i a c d^3 \log (x)-3 b c^2 d^3 x \arctan (c x)-\frac {1}{2} i b c d^3 \arctan (c x)+b c d^3 \log \left (c^2 x^2+1\right )+\frac {1}{2} i b c^2 d^3 x-\frac {3}{2} b c d^3 \operatorname {PolyLog}(2,-i c x)+\frac {3}{2} b c d^3 \operatorname {PolyLog}(2,i c x)+b c d^3 \log (x) \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.93 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^2} \, dx=\frac {d^3 \left (-2 a-6 a c^2 x^2+i b c^2 x^2-i a c^3 x^3-2 b \arctan (c x)-i b c x \arctan (c x)-6 b c^2 x^2 \arctan (c x)-i b c^3 x^3 \arctan (c x)+6 i a c x \log (x)+2 b c x \log (c x)+2 b c x \log \left (1+c^2 x^2\right )-3 b c x \operatorname {PolyLog}(2,-i c x)+3 b c x \operatorname {PolyLog}(2,i c x)\right )}{2 x} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.11 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.98

method	result	size
parts	\(a \,d^{3} \left (-\frac {i c^{3} x^{2}}{2}-3 c^{2} x +3 i c \ln \left (x \right )-\frac {1}{x}\right )+b \,d^{3} c \left (-3 c x \arctan \left (c x \right )-\frac {i \arctan \left (c x \right ) c^{2} x^{2}}{2}+3 i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\arctan \left (c x \right )}{c x}-\frac {3 \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {3 \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {3 \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {3 \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i c x}{2}+\ln \left (c x \right )+\ln \left (c^{2} x^{2}+1\right )-\frac {i \arctan \left (c x \right )}{2}\right )\)	\(159\)
derivativedivides	\(c \left (a \,d^{3} \left (-3 c x -\frac {i c^{2} x^{2}}{2}+3 i \ln \left (c x \right )-\frac {1}{c x}\right )+b \,d^{3} \left (-3 c x \arctan \left (c x \right )-\frac {i \arctan \left (c x \right ) c^{2} x^{2}}{2}+3 i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\arctan \left (c x \right )}{c x}-\frac {3 \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {3 \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {3 \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {3 \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i c x}{2}+\ln \left (c x \right )+\ln \left (c^{2} x^{2}+1\right )-\frac {i \arctan \left (c x \right )}{2}\right )\right )\)	\(162\)
default	\(c \left (a \,d^{3} \left (-3 c x -\frac {i c^{2} x^{2}}{2}+3 i \ln \left (c x \right )-\frac {1}{c x}\right )+b \,d^{3} \left (-3 c x \arctan \left (c x \right )-\frac {i \arctan \left (c x \right ) c^{2} x^{2}}{2}+3 i \arctan \left (c x \right ) \ln \left (c x \right )-\frac {\arctan \left (c x \right )}{c x}-\frac {3 \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {3 \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {3 \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {3 \operatorname {dilog}\left (-i c x +1\right )}{2}+\frac {i c x}{2}+\ln \left (c x \right )+\ln \left (c^{2} x^{2}+1\right )-\frac {i \arctan \left (c x \right )}{2}\right )\right )\)	\(162\)
risch	\(-\frac {b \,c^{3} d^{3} \ln \left (i c x +1\right ) x^{2}}{4}-\frac {7 i d^{3} c a}{2}+\frac {3 b c \,d^{3} \ln \left (i c x +1\right )}{4}+\frac {i b \,d^{3} \ln \left (i c x +1\right )}{2 x}-3 b c \,d^{3}-\frac {3 b c \,d^{3} \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {b c \,d^{3} \ln \left (i c x \right )}{2}-\frac {3 i c^{2} d^{3} b x \ln \left (-i c x +1\right )}{2}+\frac {3 i b \,c^{2} d^{3} \ln \left (i c x +1\right ) x}{2}-3 c^{2} x \,d^{3} a +\frac {i b \,c^{2} d^{3} x}{2}+3 i c \,d^{3} \ln \left (-i c x \right ) a -\frac {d^{3} a}{x}+\frac {c^{3} d^{3} x^{2} b \ln \left (-i c x +1\right )}{4}-\frac {i c^{3} d^{3} a \,x^{2}}{2}+\frac {5 c \,d^{3} \ln \left (-i c x +1\right ) b}{4}-\frac {i d^{3} b \ln \left (-i c x +1\right )}{2 x}+\frac {3 c \,d^{3} \operatorname {dilog}\left (-i c x +1\right ) b}{2}+\frac {c \,d^{3} b \ln \left (-i c x \right )}{2}\)	\(274\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.24 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^2} \, dx=-\frac {1}{2} i \, a c^{3} d^{3} x^{2} - 3 \, a c^{2} d^{3} x + \frac {1}{2} i \, b c^{2} d^{3} x - \frac {3}{4} i \, \pi b c d^{3} \log \left (c^{2} x^{2} + 1\right ) + 3 i \, b c d^{3} \arctan \left (c x\right ) \log \left (c x\right ) - \frac {3}{2} \, {\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b c d^{3} + \frac {3}{2} \, b c d^{3} {\rm Li}_2\left (i \, c x + 1\right ) - \frac {3}{2} \, b c d^{3} {\rm Li}_2\left (-i \, c x + 1\right ) + 3 i \, a c d^{3} \log \left (x\right ) - \frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b d^{3} - \frac {a d^{3}}{x} + \frac {1}{2} \, {\left (-i \, b c^{3} d^{3} x^{2} - i \, b c d^{3}\right )} \arctan \left (c x\right ) \]

[In]

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

[Out]

-1/2*I*a*c^3*d^3*x^2 - 3*a*c^2*d^3*x + 1/2*I*b*c^2*d^3*x - 3/4*I*pi*b*c*d^3*log(c^2*x^2 + 1) + 3*I*b*c*d^3*arc
tan(c*x)*log(c*x) - 3/2*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*c*d^3 + 3/2*b*c*d^3*dilog(I*c*x + 1) - 3/2*b*
c*d^3*dilog(-I*c*x + 1) + 3*I*a*c*d^3*log(x) - 1/2*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*d^3 -
a*d^3/x + 1/2*(-I*b*c^3*d^3*x^2 - I*b*c*d^3)*arctan(c*x)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.20 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))}{x^2} \, dx=\left \{\begin {array}{cl} -\frac {a\,d^3}{x} & \text {\ if\ \ }c=0\\ \frac {b\,d^3\,\left (c^2\,\ln \left (x\right )-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )}{c}-\frac {a\,c^3\,d^3\,x^2\,1{}\mathrm {i}}{2}-\frac {a\,d^3}{x}+\frac {3\,b\,c\,d^3\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )}{2}+\frac {3\,b\,c\,d^3\,\ln \left (c^2\,x^2+1\right )}{2}-3\,a\,c^2\,d^3\,x+\frac {b\,c^2\,d^3\,x\,1{}\mathrm {i}}{2}+a\,c\,d^3\,\ln \left (x\right )\,3{}\mathrm {i}-\frac {b\,d^3\,\mathrm {atan}\left (c\,x\right )}{x}-3\,b\,c^2\,d^3\,x\,\mathrm {atan}\left (c\,x\right )-b\,c^3\,d^3\,\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )\,1{}\mathrm {i} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

[In]

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

[Out]

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