\(\int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx\) [36]

Optimal result

Integrand size = 23, antiderivative size = 190 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=-6 a c^2 d^4 x+2 i b c^2 d^4 x-\frac {1}{6} b c^3 d^4 x^2-2 i b c d^4 \arctan (c x)-6 b c^2 d^4 x \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{x}-2 i c^3 d^4 x^2 (a+b \arctan (c x))+\frac {1}{3} c^4 d^4 x^3 (a+b \arctan (c x))+4 i a c d^4 \log (x)+b c d^4 \log (x)+\frac {8}{3} b c d^4 \log \left (1+c^2 x^2\right )-2 b c d^4 \operatorname {PolyLog}(2,-i c x)+2 b c d^4 \operatorname {PolyLog}(2,i c x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {4996, 4930, 266, 4946, 272, 36, 29, 31, 4940, 2438, 327, 209, 45} \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=\frac {1}{3} c^4 d^4 x^3 (a+b \arctan (c x))-2 i c^3 d^4 x^2 (a+b \arctan (c x))-\frac {d^4 (a+b \arctan (c x))}{x}-6 a c^2 d^4 x+4 i a c d^4 \log (x)-6 b c^2 d^4 x \arctan (c x)-2 i b c d^4 \arctan (c x)-\frac {1}{6} b c^3 d^4 x^2+\frac {8}{3} b c d^4 \log \left (c^2 x^2+1\right )+2 i b c^2 d^4 x-2 b c d^4 \operatorname {PolyLog}(2,-i c x)+2 b c d^4 \operatorname {PolyLog}(2,i c x)+b c d^4 \log (x) \]

[In]

Int[((d + I*c*d*x)^4*(a + b*ArcTan[c*x]))/x^2,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.95 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=\frac {d^4 \left (-6 a-36 a c^2 x^2+12 i b c^2 x^2-12 i a c^3 x^3-b c^3 x^3+2 a c^4 x^4-6 b \arctan (c x)-12 i b c x \arctan (c x)-36 b c^2 x^2 \arctan (c x)-12 i b c^3 x^3 \arctan (c x)+2 b c^4 x^4 \arctan (c x)+24 i a c x \log (x)+6 b c x \log (c x)+16 b c x \log \left (1+c^2 x^2\right )-12 b c x \operatorname {PolyLog}(2,-i c x)+12 b c x \operatorname {PolyLog}(2,i c x)\right )}{6 x} \]

[In]

Integrate[((d + I*c*d*x)^4*(a + b*ArcTan[c*x]))/x^2,x]

[Out]

(d^4*(-6*a - 36*a*c^2*x^2 + (12*I)*b*c^2*x^2 - (12*I)*a*c^3*x^3 - b*c^3*x^3 + 2*a*c^4*x^4 - 6*b*ArcTan[c*x] -
(12*I)*b*c*x*ArcTan[c*x] - 36*b*c^2*x^2*ArcTan[c*x] - (12*I)*b*c^3*x^3*ArcTan[c*x] + 2*b*c^4*x^4*ArcTan[c*x] +
(24*I)*a*c*x*Log[x] + 6*b*c*x*Log[c*x] + 16*b*c*x*Log[1 + c^2*x^2] - 12*b*c*x*PolyLog[2, (-I)*c*x] + 12*b*c*x
*PolyLog[2, I*c*x]))/(6*x)

Maple [A] (verified)

Time = 1.29 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.99

method	result	size
parts	\(d^{4} a \left (\frac {c^{4} x^{3}}{3}-2 i c^{3} x^{2}-6 c^{2} x +4 i c \ln \left (x \right )-\frac {1}{x}\right )+d^{4} b c \left (-6 c x \arctan \left (c x \right )+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}-2 i \arctan \left (c x \right ) c^{2} x^{2}-\frac {\arctan \left (c x \right )}{c x}+4 i \arctan \left (c x \right ) \ln \left (c x \right )-2 \ln \left (c x \right ) \ln \left (i c x +1\right )+2 \ln \left (c x \right ) \ln \left (-i c x +1\right )-2 \operatorname {dilog}\left (i c x +1\right )+2 \operatorname {dilog}\left (-i c x +1\right )+2 i c x -\frac {c^{2} x^{2}}{6}+\ln \left (c x \right )+\frac {8 \ln \left (c^{2} x^{2}+1\right )}{3}-2 i \arctan \left (c x \right )\right )\)	\(189\)
derivativedivides	\(c \left (d^{4} a \left (-6 c x +\frac {c^{3} x^{3}}{3}-2 i c^{2} x^{2}-\frac {1}{c x}+4 i \ln \left (c x \right )\right )+d^{4} b \left (-6 c x \arctan \left (c x \right )+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}-2 i \arctan \left (c x \right ) c^{2} x^{2}-\frac {\arctan \left (c x \right )}{c x}+4 i \arctan \left (c x \right ) \ln \left (c x \right )-2 \ln \left (c x \right ) \ln \left (i c x +1\right )+2 \ln \left (c x \right ) \ln \left (-i c x +1\right )-2 \operatorname {dilog}\left (i c x +1\right )+2 \operatorname {dilog}\left (-i c x +1\right )+2 i c x -\frac {c^{2} x^{2}}{6}+\ln \left (c x \right )+\frac {8 \ln \left (c^{2} x^{2}+1\right )}{3}-2 i \arctan \left (c x \right )\right )\right )\)	\(192\)
default	\(c \left (d^{4} a \left (-6 c x +\frac {c^{3} x^{3}}{3}-2 i c^{2} x^{2}-\frac {1}{c x}+4 i \ln \left (c x \right )\right )+d^{4} b \left (-6 c x \arctan \left (c x \right )+\frac {c^{3} x^{3} \arctan \left (c x \right )}{3}-2 i \arctan \left (c x \right ) c^{2} x^{2}-\frac {\arctan \left (c x \right )}{c x}+4 i \arctan \left (c x \right ) \ln \left (c x \right )-2 \ln \left (c x \right ) \ln \left (i c x +1\right )+2 \ln \left (c x \right ) \ln \left (-i c x +1\right )-2 \operatorname {dilog}\left (i c x +1\right )+2 \operatorname {dilog}\left (-i c x +1\right )+2 i c x -\frac {c^{2} x^{2}}{6}+\ln \left (c x \right )+\frac {8 \ln \left (c^{2} x^{2}+1\right )}{3}-2 i \arctan \left (c x \right )\right )\right )\)	\(192\)
risch	\(-\frac {119 b c \,d^{4}}{18}-6 a \,c^{2} d^{4} x -\frac {b \,c^{3} d^{4} x^{2}}{6}-\frac {i d^{4} b \ln \left (-i c x +1\right )}{2 x}+4 i d^{4} c a \ln \left (-i c x \right )+\frac {i b \,d^{4} \ln \left (i c x +1\right )}{2 x}-\frac {i b \,c^{4} d^{4} \ln \left (i c x +1\right ) x^{3}}{6}-3 i d^{4} c^{2} b x \ln \left (-i c x +1\right )+\frac {5 b c \,d^{4} \ln \left (i c x +1\right )}{3}-2 b c \,d^{4} \operatorname {dilog}\left (i c x +1\right )+\frac {b c \,d^{4} \ln \left (i c x \right )}{2}-\frac {d^{4} a}{x}-2 i d^{4} c^{3} x^{2} a +2 i b \,c^{2} d^{4} x +3 i b \,c^{2} d^{4} \ln \left (i c x +1\right ) x -b \,c^{3} d^{4} \ln \left (i c x +1\right ) x^{2}+\frac {i d^{4} c^{4} b \,x^{3} \ln \left (-i c x +1\right )}{6}+d^{4} c^{3} b \ln \left (-i c x +1\right ) x^{2}+\frac {d^{4} c^{4} a \,x^{3}}{3}+2 d^{4} c b \operatorname {dilog}\left (-i c x +1\right )+\frac {d^{4} c b \ln \left (-i c x \right )}{2}+\frac {11 d^{4} c b \ln \left (-i c x +1\right )}{3}-\frac {25 i d^{4} c a}{3}\)	\(339\)

[In]

int((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{2}} \,d x } \]

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^2,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^2, x)

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=\text {Timed out} \]

[In]

integrate((d+I*c*d*x)**4*(a+b*atan(c*x))/x**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.26 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=\frac {1}{3} \, a c^{4} d^{4} x^{3} - 2 i \, a c^{3} d^{4} x^{2} - \frac {1}{6} \, b c^{3} d^{4} x^{2} - 6 \, a c^{2} d^{4} x + 2 i \, b c^{2} d^{4} x - \frac {1}{6} \, {\left (6 i \, \pi - 1\right )} b c d^{4} \log \left (c^{2} x^{2} + 1\right ) + 4 i \, b c d^{4} \arctan \left (c x\right ) \log \left (c x\right ) - 3 \, {\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b c d^{4} + 2 \, b c d^{4} {\rm Li}_2\left (i \, c x + 1\right ) - 2 \, b c d^{4} {\rm Li}_2\left (-i \, c x + 1\right ) + 4 i \, a c d^{4} \log \left (x\right ) - \frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b d^{4} - \frac {a d^{4}}{x} + \frac {1}{3} \, {\left (b c^{4} d^{4} x^{3} - 6 i \, b c^{3} d^{4} x^{2} - 6 i \, b c d^{4}\right )} \arctan \left (c x\right ) \]

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^2,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{2}} \,d x } \]

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^2,x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.88 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.33 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^2} \, dx=\left \{\begin {array}{cl} -\frac {a\,d^4}{x} & \text {\ if\ \ }c=0\\ \frac {a\,c^4\,d^4\,x^3}{3}-\frac {a\,d^4}{x}+\frac {b\,d^4\,\left (c^2\,\ln \left (x\right )-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )}{c}+2\,b\,c\,d^4\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )+3\,b\,c\,d^4\,\ln \left (c^2\,x^2+1\right )-6\,a\,c^2\,d^4\,x-\frac {b\,c^3\,d^4\,\left (\frac {x^2}{2}-\frac {\ln \left (c^2\,x^2+1\right )}{2\,c^2}\right )}{3}-\frac {b\,d^4\,\mathrm {atan}\left (c\,x\right )}{x}-6\,b\,c^2\,d^4\,x\,\mathrm {atan}\left (c\,x\right )+\frac {b\,c^4\,d^4\,x^3\,\mathrm {atan}\left (c\,x\right )}{3}-a\,c^3\,d^4\,x^2\,2{}\mathrm {i}+b\,c^2\,d^4\,x\,2{}\mathrm {i}+a\,c\,d^4\,\ln \left (x\right )\,4{}\mathrm {i}-b\,c^3\,d^4\,\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )\,4{}\mathrm {i} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

[In]

int(((a + b*atan(c*x))*(d + c*d*x*1i)^4)/x^2,x)

[Out]

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