\(\int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^3} \, dx\) [37]

Optimal result

Integrand size = 23, antiderivative size = 173 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^3} \, dx=-\frac {b c d^4}{2 x}-4 i a c^3 d^4 x-\frac {1}{2} b c^3 d^4 x-4 i b c^3 d^4 x \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{2 x^2}-\frac {4 i c d^4 (a+b \arctan (c x))}{x}+\frac {1}{2} c^4 d^4 x^2 (a+b \arctan (c x))-6 a c^2 d^4 \log (x)+4 i b c^2 d^4 \log (x)-3 i b c^2 d^4 \operatorname {PolyLog}(2,-i c x)+3 i b c^2 d^4 \operatorname {PolyLog}(2,i c x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {4996, 4930, 266, 4946, 331, 209, 272, 36, 29, 31, 4940, 2438, 327} \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^3} \, dx=\frac {1}{2} c^4 d^4 x^2 (a+b \arctan (c x))-\frac {d^4 (a+b \arctan (c x))}{2 x^2}-\frac {4 i c d^4 (a+b \arctan (c x))}{x}-4 i a c^3 d^4 x-6 a c^2 d^4 \log (x)-4 i b c^3 d^4 x \arctan (c x)-\frac {1}{2} b c^3 d^4 x-3 i b c^2 d^4 \operatorname {PolyLog}(2,-i c x)+3 i b c^2 d^4 \operatorname {PolyLog}(2,i c x)+4 i b c^2 d^4 \log (x)-\frac {b c d^4}{2 x} \]

[In]

Int[((d + I*c*d*x)^4*(a + b*ArcTan[c*x]))/x^3,x]

[Out]

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

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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 (-4 i c^3 d^4 (a+b \arctan (c x))+\frac {d^4 (a+b \arctan (c x))}{x^3}+\frac {4 i c d^4 (a+b \arctan (c x))}{x^2}-\frac {6 c^2 d^4 (a+b \arctan (c x))}{x}+c^4 d^4 x (a+b \arctan (c x))\right ) \, dx \\ & = d^4 \int \frac {a+b \arctan (c x)}{x^3} \, dx+\left (4 i c d^4\right ) \int \frac {a+b \arctan (c x)}{x^2} \, dx-\left (6 c^2 d^4\right ) \int \frac {a+b \arctan (c x)}{x} \, dx-\left (4 i c^3 d^4\right ) \int (a+b \arctan (c x)) \, dx+\left (c^4 d^4\right ) \int x (a+b \arctan (c x)) \, dx \\ & = -4 i a c^3 d^4 x-\frac {d^4 (a+b \arctan (c x))}{2 x^2}-\frac {4 i c d^4 (a+b \arctan (c x))}{x}+\frac {1}{2} c^4 d^4 x^2 (a+b \arctan (c x))-6 a c^2 d^4 \log (x)+\frac {1}{2} \left (b c d^4\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx-\left (3 i b c^2 d^4\right ) \int \frac {\log (1-i c x)}{x} \, dx+\left (3 i b c^2 d^4\right ) \int \frac {\log (1+i c x)}{x} \, dx+\left (4 i b c^2 d^4\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx-\left (4 i b c^3 d^4\right ) \int \arctan (c x) \, dx-\frac {1}{2} \left (b c^5 d^4\right ) \int \frac {x^2}{1+c^2 x^2} \, dx \\ & = -\frac {b c d^4}{2 x}-4 i a c^3 d^4 x-\frac {1}{2} b c^3 d^4 x-4 i b c^3 d^4 x \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{2 x^2}-\frac {4 i c d^4 (a+b \arctan (c x))}{x}+\frac {1}{2} c^4 d^4 x^2 (a+b \arctan (c x))-6 a c^2 d^4 \log (x)-3 i b c^2 d^4 \operatorname {PolyLog}(2,-i c x)+3 i b c^2 d^4 \operatorname {PolyLog}(2,i c x)+\left (2 i b c^2 d^4\right ) \text {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )+\left (4 i b c^4 d^4\right ) \int \frac {x}{1+c^2 x^2} \, dx \\ & = -\frac {b c d^4}{2 x}-4 i a c^3 d^4 x-\frac {1}{2} b c^3 d^4 x-4 i b c^3 d^4 x \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{2 x^2}-\frac {4 i c d^4 (a+b \arctan (c x))}{x}+\frac {1}{2} c^4 d^4 x^2 (a+b \arctan (c x))-6 a c^2 d^4 \log (x)+2 i b c^2 d^4 \log \left (1+c^2 x^2\right )-3 i b c^2 d^4 \operatorname {PolyLog}(2,-i c x)+3 i b c^2 d^4 \operatorname {PolyLog}(2,i c x)+\left (2 i b c^2 d^4\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\left (2 i b c^4 d^4\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right ) \\ & = -\frac {b c d^4}{2 x}-4 i a c^3 d^4 x-\frac {1}{2} b c^3 d^4 x-4 i b c^3 d^4 x \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{2 x^2}-\frac {4 i c d^4 (a+b \arctan (c x))}{x}+\frac {1}{2} c^4 d^4 x^2 (a+b \arctan (c x))-6 a c^2 d^4 \log (x)+4 i b c^2 d^4 \log (x)-3 i b c^2 d^4 \operatorname {PolyLog}(2,-i c x)+3 i b c^2 d^4 \operatorname {PolyLog}(2,i c x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.94 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^3} \, dx=\frac {d^4 \left (-a-8 i a c x-b c x-8 i a c^3 x^3-b c^3 x^3+a c^4 x^4-b \arctan (c x)-8 i b c x \arctan (c x)-8 i b c^3 x^3 \arctan (c x)+b c^4 x^4 \arctan (c x)-12 a c^2 x^2 \log (x)+8 i b c^2 x^2 \log (c x)-6 i b c^2 x^2 \operatorname {PolyLog}(2,-i c x)+6 i b c^2 x^2 \operatorname {PolyLog}(2,i c x)\right )}{2 x^2} \]

[In]

Integrate[((d + I*c*d*x)^4*(a + b*ArcTan[c*x]))/x^3,x]

[Out]

(d^4*(-a - (8*I)*a*c*x - b*c*x - (8*I)*a*c^3*x^3 - b*c^3*x^3 + a*c^4*x^4 - b*ArcTan[c*x] - (8*I)*b*c*x*ArcTan[
c*x] - (8*I)*b*c^3*x^3*ArcTan[c*x] + b*c^4*x^4*ArcTan[c*x] - 12*a*c^2*x^2*Log[x] + (8*I)*b*c^2*x^2*Log[c*x] -
(6*I)*b*c^2*x^2*PolyLog[2, (-I)*c*x] + (6*I)*b*c^2*x^2*PolyLog[2, I*c*x]))/(2*x^2)

Maple [A] (verified)

Time = 1.85 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.03

method	result	size
parts	\(d^{4} a \left (\frac {c^{4} x^{2}}{2}-4 i c^{3} x -\frac {1}{2 x^{2}}-6 c^{2} \ln \left (x \right )-\frac {4 i c}{x}\right )+d^{4} b \,c^{2} \left (-4 i \arctan \left (c x \right ) c x +\frac {c^{2} x^{2} \arctan \left (c x \right )}{2}-6 \arctan \left (c x \right ) \ln \left (c x \right )-\frac {4 i \arctan \left (c x \right )}{c x}-\frac {\arctan \left (c x \right )}{2 c^{2} x^{2}}-\frac {c x}{2}+4 i \ln \left (c x \right )-\frac {1}{2 c x}-3 i \ln \left (c x \right ) \ln \left (i c x +1\right )+3 i \ln \left (c x \right ) \ln \left (-i c x +1\right )-3 i \operatorname {dilog}\left (i c x +1\right )+3 i \operatorname {dilog}\left (-i c x +1\right )\right )\)	\(178\)
derivativedivides	\(c^{2} \left (d^{4} a \left (-4 i c x +\frac {c^{2} x^{2}}{2}-6 \ln \left (c x \right )-\frac {4 i}{c x}-\frac {1}{2 c^{2} x^{2}}\right )+d^{4} b \left (-4 i \arctan \left (c x \right ) c x +\frac {c^{2} x^{2} \arctan \left (c x \right )}{2}-6 \arctan \left (c x \right ) \ln \left (c x \right )-\frac {4 i \arctan \left (c x \right )}{c x}-\frac {\arctan \left (c x \right )}{2 c^{2} x^{2}}-\frac {c x}{2}+4 i \ln \left (c x \right )-\frac {1}{2 c x}-3 i \ln \left (c x \right ) \ln \left (i c x +1\right )+3 i \ln \left (c x \right ) \ln \left (-i c x +1\right )-3 i \operatorname {dilog}\left (i c x +1\right )+3 i \operatorname {dilog}\left (-i c x +1\right )\right )\right )\)	\(181\)
default	\(c^{2} \left (d^{4} a \left (-4 i c x +\frac {c^{2} x^{2}}{2}-6 \ln \left (c x \right )-\frac {4 i}{c x}-\frac {1}{2 c^{2} x^{2}}\right )+d^{4} b \left (-4 i \arctan \left (c x \right ) c x +\frac {c^{2} x^{2} \arctan \left (c x \right )}{2}-6 \arctan \left (c x \right ) \ln \left (c x \right )-\frac {4 i \arctan \left (c x \right )}{c x}-\frac {\arctan \left (c x \right )}{2 c^{2} x^{2}}-\frac {c x}{2}+4 i \ln \left (c x \right )-\frac {1}{2 c x}-3 i \ln \left (c x \right ) \ln \left (i c x +1\right )+3 i \ln \left (c x \right ) \ln \left (-i c x +1\right )-3 i \operatorname {dilog}\left (i c x +1\right )+3 i \operatorname {dilog}\left (-i c x +1\right )\right )\right )\)	\(181\)
risch	\(\frac {9 a \,c^{2} d^{4}}{2}-\frac {b c \,d^{4}}{2 x}-\frac {b \,c^{3} d^{4} x}{2}-4 i a \,c^{3} d^{4} x +\frac {7 i b \,d^{4} c^{2} \ln \left (i c x \right )}{4}+3 i c^{2} d^{4} b \operatorname {dilog}\left (-i c x +1\right )-\frac {i d^{4} b \ln \left (-i c x +1\right )}{4 x^{2}}-\frac {4 i c \,d^{4} a}{x}+\frac {i b \,d^{4} \ln \left (i c x +1\right )}{4 x^{2}}+\frac {9 i c^{2} d^{4} b \ln \left (-i c x \right )}{4}+\frac {i c^{4} d^{4} b \ln \left (-i c x +1\right ) x^{2}}{4}+\frac {c^{4} d^{4} a \,x^{2}}{2}-\frac {i b \,d^{4} c^{4} \ln \left (i c x +1\right ) x^{2}}{4}-4 i b \,d^{4} c^{2}-6 c^{2} d^{4} a \ln \left (-i c x \right )-3 i b \,d^{4} c^{2} \operatorname {dilog}\left (i c x +1\right )-\frac {d^{4} a}{2 x^{2}}-2 b \,d^{4} c^{3} \ln \left (i c x +1\right ) x -\frac {2 b \,d^{4} c \ln \left (i c x +1\right )}{x}+\frac {2 c \,d^{4} b \ln \left (-i c x +1\right )}{x}+2 c^{3} d^{4} b \ln \left (-i c x +1\right ) x\)	\(317\)

[In]

int((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^3} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{3}} \,d x } \]

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^3,x, algorithm="fricas")

[Out]

integral(1/2*(2*a*c^4*d^4*x^4 - 8*I*a*c^3*d^4*x^3 - 12*a*c^2*d^4*x^2 + 8*I*a*c*d^4*x + 2*a*d^4 + (I*b*c^4*d^4*
x^4 + 4*b*c^3*d^4*x^3 - 6*I*b*c^2*d^4*x^2 - 4*b*c*d^4*x + I*b*d^4)*log(-(c*x + I)/(c*x - I)))/x^3, x)

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^3} \, dx=\text {Timed out} \]

[In]

integrate((d+I*c*d*x)**4*(a+b*atan(c*x))/x**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.45 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^3} \, dx=\frac {1}{2} \, a c^{4} d^{4} x^{2} - 4 i \, a c^{3} d^{4} x - \frac {1}{2} \, b c^{3} d^{4} x + \frac {3}{2} \, \pi b c^{2} d^{4} \log \left (c^{2} x^{2} + 1\right ) - 6 \, b c^{2} d^{4} \arctan \left (c x\right ) \log \left (c x\right ) - 2 i \, {\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b c^{2} d^{4} + 3 i \, b c^{2} d^{4} {\rm Li}_2\left (i \, c x + 1\right ) - 3 i \, b c^{2} d^{4} {\rm Li}_2\left (-i \, c x + 1\right ) - 6 \, a c^{2} d^{4} \log \left (x\right ) - 2 i \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b c d^{4} - \frac {1}{2} \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b d^{4} - \frac {4 i \, a c d^{4}}{x} - \frac {a d^{4}}{2 \, x^{2}} + \frac {1}{2} \, {\left (b c^{4} d^{4} x^{2} + b c^{2} d^{4}\right )} \arctan \left (c x\right ) \]

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^3,x, algorithm="maxima")

[Out]

1/2*a*c^4*d^4*x^2 - 4*I*a*c^3*d^4*x - 1/2*b*c^3*d^4*x + 3/2*pi*b*c^2*d^4*log(c^2*x^2 + 1) - 6*b*c^2*d^4*arctan
(c*x)*log(c*x) - 2*I*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*c^2*d^4 + 3*I*b*c^2*d^4*dilog(I*c*x + 1) - 3*I*b
*c^2*d^4*dilog(-I*c*x + 1) - 6*a*c^2*d^4*log(x) - 2*I*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*c*
d^4 - 1/2*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*b*d^4 - 4*I*a*c*d^4/x - 1/2*a*d^4/x^2 + 1/2*(b*c^4*d^4*x
^2 + b*c^2*d^4)*arctan(c*x)

Giac [F]

\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^3} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{3}} \,d x } \]

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^3,x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.94 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.49 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^3} \, dx=\left \{\begin {array}{cl} -\frac {a\,d^4}{2\,x^2} & \text {\ if\ \ }c=0\\ \frac {a\,c^4\,d^4\,x^2}{2}-\frac {\frac {a\,d^4}{2}+a\,c\,d^4\,x\,4{}\mathrm {i}}{x^2}-6\,a\,c^2\,d^4\,\ln \left (x\right )-\frac {b\,d^4\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )}{2\,c}-\frac {b\,c^3\,d^4\,x}{2}-\frac {b\,d^4\,\mathrm {atan}\left (c\,x\right )}{2\,x^2}+b\,c^4\,d^4\,\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )+b\,d^4\,\left (c^2\,\ln \left (x\right )-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )\,4{}\mathrm {i}+b\,c^2\,d^4\,\ln \left (c^2\,x^2+1\right )\,2{}\mathrm {i}+b\,c^2\,d^4\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,3{}\mathrm {i}-b\,c^2\,d^4\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,3{}\mathrm {i}-a\,c^3\,d^4\,x\,4{}\mathrm {i}-\frac {b\,c\,d^4\,\mathrm {atan}\left (c\,x\right )\,4{}\mathrm {i}}{x}-b\,c^3\,d^4\,x\,\mathrm {atan}\left (c\,x\right )\,4{}\mathrm {i} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

[In]

int(((a + b*atan(c*x))*(d + c*d*x*1i)^4)/x^3,x)

[Out]

piecewise(c == 0, -(a*d^4)/(2*x^2), c ~= 0, - ((a*d^4)/2 + a*c*d^4*x*4i)/x^2 + b*d^4*(c^2*log(x) - (c^2*log(c^
2*x^2 + 1))/2)*4i + b*c^2*d^4*log(c^2*x^2 + 1)*2i + (a*c^4*d^4*x^2)/2 - 6*a*c^2*d^4*log(x) + b*c^2*d^4*dilog(-
c*x*1i + 1)*3i - b*c^2*d^4*dilog(c*x*1i + 1)*3i - (b*d^4*(c^3*atan(c*x) + c^2/x))/(2*c) - a*c^3*d^4*x*4i - (b
*c^3*d^4*x)/2 - (b*d^4*atan(c*x))/(2*x^2) - (b*c*d^4*atan(c*x)*4i)/x - b*c^3*d^4*x*atan(c*x)*4i + b*c^4*d^4*at
an(c*x)*(1/(2*c^2) + x^2/2))