\(\int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^4} \, dx\) [38]

Optimal result

Integrand size = 23, antiderivative size = 201 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^4} \, dx=-\frac {b c d^4}{6 x^2}-\frac {2 i b c^2 d^4}{x}+a c^4 d^4 x-2 i b c^3 d^4 \arctan (c x)+b c^4 d^4 x \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{3 x^3}-\frac {2 i c d^4 (a+b \arctan (c x))}{x^2}+\frac {6 c^2 d^4 (a+b \arctan (c x))}{x}-4 i a c^3 d^4 \log (x)-\frac {19}{3} b c^3 d^4 \log (x)+\frac {8}{3} b c^3 d^4 \log \left (1+c^2 x^2\right )+2 b c^3 d^4 \operatorname {PolyLog}(2,-i c x)-2 b c^3 d^4 \operatorname {PolyLog}(2,i c x) \]

[Out]

-1/6*b*c*d^4/x^2-2*I*b*c^2*d^4/x+a*c^4*d^4*x-2*I*b*c^3*d^4*arctan(c*x)+b*c^4*d^4*x*arctan(c*x)-1/3*d^4*(a+b*ar
ctan(c*x))/x^3-2*I*c*d^4*(a+b*arctan(c*x))/x^2+6*c^2*d^4*(a+b*arctan(c*x))/x-4*I*a*c^3*d^4*ln(x)-19/3*b*c^3*d^
4*ln(x)+8/3*b*c^3*d^4*ln(c^2*x^2+1)+2*b*c^3*d^4*polylog(2,-I*c*x)-2*b*c^3*d^4*polylog(2,I*c*x)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 201, 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, 46, 331, 209, 36, 29, 31, 4940, 2438} \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^4} \, dx=\frac {6 c^2 d^4 (a+b \arctan (c x))}{x}-\frac {d^4 (a+b \arctan (c x))}{3 x^3}-\frac {2 i c d^4 (a+b \arctan (c x))}{x^2}+a c^4 d^4 x-4 i a c^3 d^4 \log (x)+b c^4 d^4 x \arctan (c x)-2 i b c^3 d^4 \arctan (c x)+2 b c^3 d^4 \operatorname {PolyLog}(2,-i c x)-2 b c^3 d^4 \operatorname {PolyLog}(2,i c x)-\frac {19}{3} b c^3 d^4 \log (x)-\frac {2 i b c^2 d^4}{x}+\frac {8}{3} b c^3 d^4 \log \left (c^2 x^2+1\right )-\frac {b c d^4}{6 x^2} \]

[In]

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

[Out]

-1/6*(b*c*d^4)/x^2 - ((2*I)*b*c^2*d^4)/x + a*c^4*d^4*x - (2*I)*b*c^3*d^4*ArcTan[c*x] + b*c^4*d^4*x*ArcTan[c*x]
- (d^4*(a + b*ArcTan[c*x]))/(3*x^3) - ((2*I)*c*d^4*(a + b*ArcTan[c*x]))/x^2 + (6*c^2*d^4*(a + b*ArcTan[c*x]))
/x - (4*I)*a*c^3*d^4*Log[x] - (19*b*c^3*d^4*Log[x])/3 + (8*b*c^3*d^4*Log[1 + c^2*x^2])/3 + 2*b*c^3*d^4*PolyLog
[2, (-I)*c*x] - 2*b*c^3*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 46

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] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[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 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 (c^4 d^4 (a+b \arctan (c x))+\frac {d^4 (a+b \arctan (c x))}{x^4}+\frac {4 i c d^4 (a+b \arctan (c x))}{x^3}-\frac {6 c^2 d^4 (a+b \arctan (c x))}{x^2}-\frac {4 i c^3 d^4 (a+b \arctan (c x))}{x}\right ) \, dx \\ & = d^4 \int \frac {a+b \arctan (c x)}{x^4} \, dx+\left (4 i c d^4\right ) \int \frac {a+b \arctan (c x)}{x^3} \, dx-\left (6 c^2 d^4\right ) \int \frac {a+b \arctan (c x)}{x^2} \, dx-\left (4 i c^3 d^4\right ) \int \frac {a+b \arctan (c x)}{x} \, dx+\left (c^4 d^4\right ) \int (a+b \arctan (c x)) \, dx \\ & = a c^4 d^4 x-\frac {d^4 (a+b \arctan (c x))}{3 x^3}-\frac {2 i c d^4 (a+b \arctan (c x))}{x^2}+\frac {6 c^2 d^4 (a+b \arctan (c x))}{x}-4 i a c^3 d^4 \log (x)+\frac {1}{3} \left (b c d^4\right ) \int \frac {1}{x^3 \left (1+c^2 x^2\right )} \, dx+\left (2 i b c^2 d^4\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx+\left (2 b c^3 d^4\right ) \int \frac {\log (1-i c x)}{x} \, dx-\left (2 b c^3 d^4\right ) \int \frac {\log (1+i c x)}{x} \, dx-\left (6 b c^3 d^4\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx+\left (b c^4 d^4\right ) \int \arctan (c x) \, dx \\ & = -\frac {2 i b c^2 d^4}{x}+a c^4 d^4 x+b c^4 d^4 x \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{3 x^3}-\frac {2 i c d^4 (a+b \arctan (c x))}{x^2}+\frac {6 c^2 d^4 (a+b \arctan (c x))}{x}-4 i a c^3 d^4 \log (x)+2 b c^3 d^4 \operatorname {PolyLog}(2,-i c x)-2 b c^3 d^4 \operatorname {PolyLog}(2,i c x)+\frac {1}{6} \left (b c d^4\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )-\left (3 b c^3 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^4 d^4\right ) \int \frac {1}{1+c^2 x^2} \, dx-\left (b c^5 d^4\right ) \int \frac {x}{1+c^2 x^2} \, dx \\ & = -\frac {2 i b c^2 d^4}{x}+a c^4 d^4 x-2 i b c^3 d^4 \arctan (c x)+b c^4 d^4 x \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{3 x^3}-\frac {2 i c d^4 (a+b \arctan (c x))}{x^2}+\frac {6 c^2 d^4 (a+b \arctan (c x))}{x}-4 i a c^3 d^4 \log (x)-\frac {1}{2} b c^3 d^4 \log \left (1+c^2 x^2\right )+2 b c^3 d^4 \operatorname {PolyLog}(2,-i c x)-2 b c^3 d^4 \operatorname {PolyLog}(2,i c x)+\frac {1}{6} \left (b c d^4\right ) \text {Subst}\left (\int \left (\frac {1}{x^2}-\frac {c^2}{x}+\frac {c^4}{1+c^2 x}\right ) \, dx,x,x^2\right )-\left (3 b c^3 d^4\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\left (3 b c^5 d^4\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right ) \\ & = -\frac {b c d^4}{6 x^2}-\frac {2 i b c^2 d^4}{x}+a c^4 d^4 x-2 i b c^3 d^4 \arctan (c x)+b c^4 d^4 x \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{3 x^3}-\frac {2 i c d^4 (a+b \arctan (c x))}{x^2}+\frac {6 c^2 d^4 (a+b \arctan (c x))}{x}-4 i a c^3 d^4 \log (x)-\frac {19}{3} b c^3 d^4 \log (x)+\frac {8}{3} b c^3 d^4 \log \left (1+c^2 x^2\right )+2 b c^3 d^4 \operatorname {PolyLog}(2,-i c x)-2 b c^3 d^4 \operatorname {PolyLog}(2,i c x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.96 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^4} \, dx=\frac {d^4 \left (-2 a-12 i a c x-b c x+36 a c^2 x^2-12 i b c^2 x^2+6 a c^4 x^4-2 b \arctan (c x)-12 i b c x \arctan (c x)+36 b c^2 x^2 \arctan (c x)-12 i b c^3 x^3 \arctan (c x)+6 b c^4 x^4 \arctan (c x)-24 i a c^3 x^3 \log (x)-38 b c^3 x^3 \log (c x)+16 b c^3 x^3 \log \left (1+c^2 x^2\right )+12 b c^3 x^3 \operatorname {PolyLog}(2,-i c x)-12 b c^3 x^3 \operatorname {PolyLog}(2,i c x)\right )}{6 x^3} \]

[In]

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

[Out]

(d^4*(-2*a - (12*I)*a*c*x - b*c*x + 36*a*c^2*x^2 - (12*I)*b*c^2*x^2 + 6*a*c^4*x^4 - 2*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] + 6*b*c^4*x^4*ArcTan[c*x] - (24*I)*
a*c^3*x^3*Log[x] - 38*b*c^3*x^3*Log[c*x] + 16*b*c^3*x^3*Log[1 + c^2*x^2] + 12*b*c^3*x^3*PolyLog[2, (-I)*c*x] -
12*b*c^3*x^3*PolyLog[2, I*c*x]))/(6*x^3)

Maple [A] (verified)

Time = 1.76 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.97

method	result	size
parts	\(d^{4} a \left (c^{4} x -\frac {2 i c}{x^{2}}-4 i c^{3} \ln \left (x \right )+\frac {6 c^{2}}{x}-\frac {1}{3 x^{3}}\right )+d^{4} b \,c^{3} \left (c x \arctan \left (c x \right )-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-4 i \arctan \left (c x \right ) \ln \left (c x \right )+\frac {6 \arctan \left (c x \right )}{c x}-\frac {2 i \arctan \left (c x \right )}{c^{2} x^{2}}+2 \ln \left (c x \right ) \ln \left (i c x +1\right )-2 \ln \left (c x \right ) \ln \left (-i c x +1\right )+2 \operatorname {dilog}\left (i c x +1\right )-2 \operatorname {dilog}\left (-i c x +1\right )-\frac {1}{6 c^{2} x^{2}}-\frac {2 i}{c x}-\frac {19 \ln \left (c x \right )}{3}+\frac {8 \ln \left (c^{2} x^{2}+1\right )}{3}-2 i \arctan \left (c x \right )\right )\)	\(195\)
derivativedivides	\(c^{3} \left (d^{4} a \left (c x -\frac {1}{3 c^{3} x^{3}}-4 i \ln \left (c x \right )+\frac {6}{c x}-\frac {2 i}{c^{2} x^{2}}\right )+d^{4} b \left (c x \arctan \left (c x \right )-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-4 i \arctan \left (c x \right ) \ln \left (c x \right )+\frac {6 \arctan \left (c x \right )}{c x}-\frac {2 i \arctan \left (c x \right )}{c^{2} x^{2}}+2 \ln \left (c x \right ) \ln \left (i c x +1\right )-2 \ln \left (c x \right ) \ln \left (-i c x +1\right )+2 \operatorname {dilog}\left (i c x +1\right )-2 \operatorname {dilog}\left (-i c x +1\right )-\frac {1}{6 c^{2} x^{2}}-\frac {2 i}{c x}-\frac {19 \ln \left (c x \right )}{3}+\frac {8 \ln \left (c^{2} x^{2}+1\right )}{3}-2 i \arctan \left (c x \right )\right )\right )\)	\(198\)
default	\(c^{3} \left (d^{4} a \left (c x -\frac {1}{3 c^{3} x^{3}}-4 i \ln \left (c x \right )+\frac {6}{c x}-\frac {2 i}{c^{2} x^{2}}\right )+d^{4} b \left (c x \arctan \left (c x \right )-\frac {\arctan \left (c x \right )}{3 c^{3} x^{3}}-4 i \arctan \left (c x \right ) \ln \left (c x \right )+\frac {6 \arctan \left (c x \right )}{c x}-\frac {2 i \arctan \left (c x \right )}{c^{2} x^{2}}+2 \ln \left (c x \right ) \ln \left (i c x +1\right )-2 \ln \left (c x \right ) \ln \left (-i c x +1\right )+2 \operatorname {dilog}\left (i c x +1\right )-2 \operatorname {dilog}\left (-i c x +1\right )-\frac {1}{6 c^{2} x^{2}}-\frac {2 i}{c x}-\frac {19 \ln \left (c x \right )}{3}+\frac {8 \ln \left (c^{2} x^{2}+1\right )}{3}-2 i \arctan \left (c x \right )\right )\right )\)	\(198\)
risch	\(b \,c^{3} d^{4}-\frac {b c \,d^{4}}{6 x^{2}}+\frac {8 b \,c^{3} d^{4} \ln \left (c^{2} x^{2}+1\right )}{3}+\frac {i d^{4} c^{4} b \ln \left (-i c x +1\right ) x}{2}+i a \,c^{3} d^{4}+a \,c^{4} d^{4} x -2 i b \,c^{3} d^{4} \arctan \left (c x \right )-\frac {i b \,d^{4} c^{4} \ln \left (i c x +1\right ) x}{2}-4 i d^{4} c^{3} a \ln \left (-i c x \right )+\frac {3 i d^{4} c^{2} b \ln \left (-i c x +1\right )}{x}-\frac {3 i b \,d^{4} c^{2} \ln \left (i c x +1\right )}{x}+\frac {d^{4} c b \ln \left (-i c x +1\right )}{x^{2}}-\frac {d^{4} a}{3 x^{3}}-\frac {b \,d^{4} c \ln \left (i c x +1\right )}{x^{2}}+2 b \,d^{4} c^{3} \operatorname {dilog}\left (i c x +1\right )-\frac {13 b \,d^{4} c^{3} \ln \left (i c x \right )}{6}+\frac {6 d^{4} c^{2} a}{x}-2 d^{4} c^{3} b \operatorname {dilog}\left (-i c x +1\right )-\frac {25 d^{4} c^{3} b \ln \left (-i c x \right )}{6}-\frac {2 i b \,c^{2} d^{4}}{x}+\frac {i b \,d^{4} \ln \left (i c x +1\right )}{6 x^{3}}-\frac {i d^{4} b \ln \left (-i c x +1\right )}{6 x^{3}}-\frac {2 i d^{4} c a}{x^{2}}\)	\(348\)

[In]

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

[Out]

d^4*a*(c^4*x-2*I*c/x^2-4*I*c^3*ln(x)+6*c^2/x-1/3/x^3)+d^4*b*c^3*(c*x*arctan(c*x)-1/3*arctan(c*x)/c^3/x^3-4*I*a
rctan(c*x)*ln(c*x)+6/c/x*arctan(c*x)-2*I*arctan(c*x)/c^2/x^2+2*ln(c*x)*ln(1+I*c*x)-2*ln(c*x)*ln(1-I*c*x)+2*dil
og(1+I*c*x)-2*dilog(1-I*c*x)-1/6/c^2/x^2-2*I/c/x-19/3*ln(c*x)+8/3*ln(c^2*x^2+1)-2*I*arctan(c*x))

Fricas [F]

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

[In]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.89 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.30 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^4} \, dx=\left \{\begin {array}{cl} -\frac {a\,d^4}{3\,x^3} & \text {\ if\ \ }c=0\\ \frac {b\,c^3\,d^4\,\ln \left (-\frac {3\,c^6\,x^2}{2}-\frac {3\,c^4}{2}\right )}{6}-\frac {b\,c^3\,d^4\,\ln \left (c^2\,x^2+1\right )}{2}-\frac {b\,c^3\,d^4\,\ln \left (x\right )}{3}-2\,b\,c^3\,d^4\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )-6\,b\,c\,d^4\,\left (c^2\,\ln \left (x\right )-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )-\frac {b\,c\,d^4}{6\,x^2}-\frac {a\,d^4\,\left (1-3\,c^4\,x^4-18\,c^2\,x^2+c\,x\,6{}\mathrm {i}+c^3\,x^3\,\ln \left (x\right )\,12{}\mathrm {i}\right )}{3\,x^3}-\frac {b\,d^4\,\mathrm {atan}\left (c\,x\right )}{3\,x^3}+b\,c^4\,d^4\,x\,\mathrm {atan}\left (c\,x\right )+\frac {6\,b\,c^2\,d^4\,\mathrm {atan}\left (c\,x\right )}{x}-b\,d^4\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )\,2{}\mathrm {i}-\frac {b\,c\,d^4\,\mathrm {atan}\left (c\,x\right )\,2{}\mathrm {i}}{x^2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

[In]

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

[Out]

piecewise(c == 0, -(a*d^4)/(3*x^3), c ~= 0, - b*d^4*(c^3*atan(c*x) + c^2/x)*2i - 2*b*c^3*d^4*(dilog(- c*x*1i +
1) - dilog(c*x*1i + 1)) - (b*c^3*d^4*log(c^2*x^2 + 1))/2 - (b*c^3*d^4*log(x))/3 + (b*c^3*d^4*log(- (3*c^4)/2
- (3*c^6*x^2)/2))/6 - 6*b*c*d^4*(c^2*log(x) - (c^2*log(c^2*x^2 + 1))/2) - (b*c*d^4)/(6*x^2) - (a*d^4*(c*x*6i -
18*c^2*x^2 - 3*c^4*x^4 + c^3*x^3*log(x)*12i + 1))/(3*x^3) - (b*d^4*atan(c*x))/(3*x^3) - (b*c*d^4*atan(c*x)*2i
)/x^2 + b*c^4*d^4*x*atan(c*x) + (6*b*c^2*d^4*atan(c*x))/x)