\(\int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx\) [39]

Optimal result

Integrand size = 23, antiderivative size = 227 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=-\frac {b c d^4}{12 x^3}-\frac {2 i b c^2 d^4}{3 x^2}+\frac {13 b c^3 d^4}{4 x}+\frac {13}{4} b c^4 d^4 \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{4 x^4}-\frac {4 i c d^4 (a+b \arctan (c x))}{3 x^3}+\frac {3 c^2 d^4 (a+b \arctan (c x))}{x^2}+\frac {4 i c^3 d^4 (a+b \arctan (c x))}{x}+a c^4 d^4 \log (x)-\frac {16}{3} i b c^4 d^4 \log (x)+\frac {8}{3} i b c^4 d^4 \log \left (1+c^2 x^2\right )+\frac {1}{2} i b c^4 d^4 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b c^4 d^4 \operatorname {PolyLog}(2,i c x) \]

[Out]

-1/12*b*c*d^4/x^3-2/3*I*b*c^2*d^4/x^2+13/4*b*c^3*d^4/x+13/4*b*c^4*d^4*arctan(c*x)-1/4*d^4*(a+b*arctan(c*x))/x^
4-4/3*I*c*d^4*(a+b*arctan(c*x))/x^3+3*c^2*d^4*(a+b*arctan(c*x))/x^2+4*I*c^3*d^4*(a+b*arctan(c*x))/x+a*c^4*d^4*
ln(x)-16/3*I*b*c^4*d^4*ln(x)+8/3*I*b*c^4*d^4*ln(c^2*x^2+1)+1/2*I*b*c^4*d^4*polylog(2,-I*c*x)-1/2*I*b*c^4*d^4*p
olylog(2,I*c*x)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {4996, 4946, 331, 209, 272, 46, 36, 29, 31, 4940, 2438} \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=\frac {4 i c^3 d^4 (a+b \arctan (c x))}{x}+\frac {3 c^2 d^4 (a+b \arctan (c x))}{x^2}-\frac {d^4 (a+b \arctan (c x))}{4 x^4}-\frac {4 i c d^4 (a+b \arctan (c x))}{3 x^3}+a c^4 d^4 \log (x)+\frac {13}{4} b c^4 d^4 \arctan (c x)+\frac {1}{2} i b c^4 d^4 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b c^4 d^4 \operatorname {PolyLog}(2,i c x)-\frac {16}{3} i b c^4 d^4 \log (x)+\frac {13 b c^3 d^4}{4 x}-\frac {2 i b c^2 d^4}{3 x^2}+\frac {8}{3} i b c^4 d^4 \log \left (c^2 x^2+1\right )-\frac {b c d^4}{12 x^3} \]

[In]

Int[((d + I*c*d*x)^4*(a + b*ArcTan[c*x]))/x^5,x]

[Out]

-1/12*(b*c*d^4)/x^3 - (((2*I)/3)*b*c^2*d^4)/x^2 + (13*b*c^3*d^4)/(4*x) + (13*b*c^4*d^4*ArcTan[c*x])/4 - (d^4*(
a + b*ArcTan[c*x]))/(4*x^4) - (((4*I)/3)*c*d^4*(a + b*ArcTan[c*x]))/x^3 + (3*c^2*d^4*(a + b*ArcTan[c*x]))/x^2
+ ((4*I)*c^3*d^4*(a + b*ArcTan[c*x]))/x + a*c^4*d^4*Log[x] - ((16*I)/3)*b*c^4*d^4*Log[x] + ((8*I)/3)*b*c^4*d^4
*Log[1 + c^2*x^2] + (I/2)*b*c^4*d^4*PolyLog[2, (-I)*c*x] - (I/2)*b*c^4*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 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 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 (\frac {d^4 (a+b \arctan (c x))}{x^5}+\frac {4 i c d^4 (a+b \arctan (c x))}{x^4}-\frac {6 c^2 d^4 (a+b \arctan (c x))}{x^3}-\frac {4 i c^3 d^4 (a+b \arctan (c x))}{x^2}+\frac {c^4 d^4 (a+b \arctan (c x))}{x}\right ) \, dx \\ & = d^4 \int \frac {a+b \arctan (c x)}{x^5} \, dx+\left (4 i c d^4\right ) \int \frac {a+b \arctan (c x)}{x^4} \, dx-\left (6 c^2 d^4\right ) \int \frac {a+b \arctan (c x)}{x^3} \, dx-\left (4 i c^3 d^4\right ) \int \frac {a+b \arctan (c x)}{x^2} \, dx+\left (c^4 d^4\right ) \int \frac {a+b \arctan (c x)}{x} \, dx \\ & = -\frac {d^4 (a+b \arctan (c x))}{4 x^4}-\frac {4 i c d^4 (a+b \arctan (c x))}{3 x^3}+\frac {3 c^2 d^4 (a+b \arctan (c x))}{x^2}+\frac {4 i c^3 d^4 (a+b \arctan (c x))}{x}+a c^4 d^4 \log (x)+\frac {1}{4} \left (b c d^4\right ) \int \frac {1}{x^4 \left (1+c^2 x^2\right )} \, dx+\frac {1}{3} \left (4 i b c^2 d^4\right ) \int \frac {1}{x^3 \left (1+c^2 x^2\right )} \, dx-\left (3 b c^3 d^4\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx+\frac {1}{2} \left (i b c^4 d^4\right ) \int \frac {\log (1-i c x)}{x} \, dx-\frac {1}{2} \left (i b c^4 d^4\right ) \int \frac {\log (1+i c x)}{x} \, dx-\left (4 i b c^4 d^4\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx \\ & = -\frac {b c d^4}{12 x^3}+\frac {3 b c^3 d^4}{x}-\frac {d^4 (a+b \arctan (c x))}{4 x^4}-\frac {4 i c d^4 (a+b \arctan (c x))}{3 x^3}+\frac {3 c^2 d^4 (a+b \arctan (c x))}{x^2}+\frac {4 i c^3 d^4 (a+b \arctan (c x))}{x}+a c^4 d^4 \log (x)+\frac {1}{2} i b c^4 d^4 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b c^4 d^4 \operatorname {PolyLog}(2,i c x)+\frac {1}{3} \left (2 i b c^2 d^4\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac {1}{4} \left (b c^3 d^4\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx-\left (2 i b c^4 d^4\right ) \text {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )+\left (3 b c^5 d^4\right ) \int \frac {1}{1+c^2 x^2} \, dx \\ & = -\frac {b c d^4}{12 x^3}+\frac {13 b c^3 d^4}{4 x}+3 b c^4 d^4 \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{4 x^4}-\frac {4 i c d^4 (a+b \arctan (c x))}{3 x^3}+\frac {3 c^2 d^4 (a+b \arctan (c x))}{x^2}+\frac {4 i c^3 d^4 (a+b \arctan (c x))}{x}+a c^4 d^4 \log (x)+\frac {1}{2} i b c^4 d^4 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b c^4 d^4 \operatorname {PolyLog}(2,i c x)+\frac {1}{3} \left (2 i b c^2 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 (2 i b c^4 d^4\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{4} \left (b c^5 d^4\right ) \int \frac {1}{1+c^2 x^2} \, dx+\left (2 i b c^6 d^4\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right ) \\ & = -\frac {b c d^4}{12 x^3}-\frac {2 i b c^2 d^4}{3 x^2}+\frac {13 b c^3 d^4}{4 x}+\frac {13}{4} b c^4 d^4 \arctan (c x)-\frac {d^4 (a+b \arctan (c x))}{4 x^4}-\frac {4 i c d^4 (a+b \arctan (c x))}{3 x^3}+\frac {3 c^2 d^4 (a+b \arctan (c x))}{x^2}+\frac {4 i c^3 d^4 (a+b \arctan (c x))}{x}+a c^4 d^4 \log (x)-\frac {16}{3} i b c^4 d^4 \log (x)+\frac {8}{3} i b c^4 d^4 \log \left (1+c^2 x^2\right )+\frac {1}{2} i b c^4 d^4 \operatorname {PolyLog}(2,-i c x)-\frac {1}{2} i b c^4 d^4 \operatorname {PolyLog}(2,i c x) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.09 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.00 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=\frac {d^4 \left (-3 a-16 i a c x+36 a c^2 x^2-8 i b c^2 x^2+48 i a c^3 x^3-3 b \arctan (c x)-16 i b c x \arctan (c x)+36 b c^2 x^2 \arctan (c x)+48 i b c^3 x^3 \arctan (c x)-b c x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-c^2 x^2\right )+36 b c^3 x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-c^2 x^2\right )+12 a c^4 x^4 \log (x)-64 i b c^4 x^4 \log (x)+32 i b c^4 x^4 \log \left (1+c^2 x^2\right )+6 i b c^4 x^4 \operatorname {PolyLog}(2,-i c x)-6 i b c^4 x^4 \operatorname {PolyLog}(2,i c x)\right )}{12 x^4} \]

[In]

Integrate[((d + I*c*d*x)^4*(a + b*ArcTan[c*x]))/x^5,x]

[Out]

(d^4*(-3*a - (16*I)*a*c*x + 36*a*c^2*x^2 - (8*I)*b*c^2*x^2 + (48*I)*a*c^3*x^3 - 3*b*ArcTan[c*x] - (16*I)*b*c*x
*ArcTan[c*x] + 36*b*c^2*x^2*ArcTan[c*x] + (48*I)*b*c^3*x^3*ArcTan[c*x] - b*c*x*Hypergeometric2F1[-3/2, 1, -1/2
, -(c^2*x^2)] + 36*b*c^3*x^3*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)] + 12*a*c^4*x^4*Log[x] - (64*I)*b*c^4*
x^4*Log[x] + (32*I)*b*c^4*x^4*Log[1 + c^2*x^2] + (6*I)*b*c^4*x^4*PolyLog[2, (-I)*c*x] - (6*I)*b*c^4*x^4*PolyLo
g[2, I*c*x]))/(12*x^4)

Maple [A] (verified)

Time = 1.90 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.94

method	result	size
parts	\(d^{4} a \left (\frac {3 c^{2}}{x^{2}}+c^{4} \ln \left (x \right )-\frac {1}{4 x^{4}}+\frac {4 i c^{3}}{x}-\frac {4 i c}{3 x^{3}}\right )+d^{4} b \,c^{4} \left (-\frac {\arctan \left (c x \right )}{4 c^{4} x^{4}}-\frac {4 i \arctan \left (c x \right )}{3 c^{3} x^{3}}+\arctan \left (c x \right ) \ln \left (c x \right )+\frac {4 i \arctan \left (c x \right )}{c x}+\frac {3 \arctan \left (c x \right )}{c^{2} x^{2}}-\frac {2 i}{3 c^{2} x^{2}}-\frac {16 i \ln \left (c x \right )}{3}-\frac {1}{12 c^{3} x^{3}}+\frac {13}{4 c x}+\frac {8 i \ln \left (c^{2} x^{2}+1\right )}{3}+\frac {13 \arctan \left (c x \right )}{4}+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\)	\(214\)
derivativedivides	\(c^{4} \left (d^{4} a \left (-\frac {1}{4 c^{4} x^{4}}-\frac {4 i}{3 c^{3} x^{3}}+\ln \left (c x \right )+\frac {4 i}{c x}+\frac {3}{c^{2} x^{2}}\right )+d^{4} b \left (-\frac {\arctan \left (c x \right )}{4 c^{4} x^{4}}-\frac {4 i \arctan \left (c x \right )}{3 c^{3} x^{3}}+\arctan \left (c x \right ) \ln \left (c x \right )+\frac {4 i \arctan \left (c x \right )}{c x}+\frac {3 \arctan \left (c x \right )}{c^{2} x^{2}}-\frac {2 i}{3 c^{2} x^{2}}-\frac {16 i \ln \left (c x \right )}{3}-\frac {1}{12 c^{3} x^{3}}+\frac {13}{4 c x}+\frac {8 i \ln \left (c^{2} x^{2}+1\right )}{3}+\frac {13 \arctan \left (c x \right )}{4}+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )\)	\(218\)
default	\(c^{4} \left (d^{4} a \left (-\frac {1}{4 c^{4} x^{4}}-\frac {4 i}{3 c^{3} x^{3}}+\ln \left (c x \right )+\frac {4 i}{c x}+\frac {3}{c^{2} x^{2}}\right )+d^{4} b \left (-\frac {\arctan \left (c x \right )}{4 c^{4} x^{4}}-\frac {4 i \arctan \left (c x \right )}{3 c^{3} x^{3}}+\arctan \left (c x \right ) \ln \left (c x \right )+\frac {4 i \arctan \left (c x \right )}{c x}+\frac {3 \arctan \left (c x \right )}{c^{2} x^{2}}-\frac {2 i}{3 c^{2} x^{2}}-\frac {16 i \ln \left (c x \right )}{3}-\frac {1}{12 c^{3} x^{3}}+\frac {13}{4 c x}+\frac {8 i \ln \left (c^{2} x^{2}+1\right )}{3}+\frac {13 \arctan \left (c x \right )}{4}+\frac {i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {i \operatorname {dilog}\left (i c x +1\right )}{2}-\frac {i \operatorname {dilog}\left (-i c x +1\right )}{2}\right )\right )\)	\(218\)
risch	\(-\frac {b c \,d^{4}}{12 x^{3}}+\frac {13 b \,c^{3} d^{4}}{4 x}-\frac {25 i b \,d^{4} c^{4} \ln \left (i c x \right )}{24}-\frac {3 i b \,d^{4} c^{2} \ln \left (i c x +1\right )}{2 x^{2}}+\frac {8 i b \,c^{4} d^{4} \ln \left (c^{2} x^{2}+1\right )}{3}+\frac {3 i c^{2} d^{4} b \ln \left (-i c x +1\right )}{2 x^{2}}+\frac {i b \,d^{4} c^{4} \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {13 b \,c^{4} d^{4} \arctan \left (c x \right )}{4}+\frac {i b \,d^{4} \ln \left (i c x +1\right )}{8 x^{4}}-\frac {103 i c^{4} d^{4} b \ln \left (-i c x \right )}{24}+\frac {4 i c^{3} d^{4} a}{x}-\frac {i d^{4} b \ln \left (-i c x +1\right )}{8 x^{4}}-\frac {i c^{4} d^{4} b \operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {4 i c \,d^{4} a}{3 x^{3}}+\frac {2 b \,d^{4} c^{3} \ln \left (i c x +1\right )}{x}-\frac {2 b \,d^{4} c \ln \left (i c x +1\right )}{3 x^{3}}+\frac {2 c \,d^{4} b \ln \left (-i c x +1\right )}{3 x^{3}}-\frac {2 c^{3} d^{4} b \ln \left (-i c x +1\right )}{x}+c^{4} d^{4} a \ln \left (-i c x \right )+\frac {3 c^{2} d^{4} a}{x^{2}}-\frac {2 i b \,c^{2} d^{4}}{3 x^{2}}-\frac {d^{4} a}{4 x^{4}}\)	\(351\)

[In]

int((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^5,x,method=_RETURNVERBOSE)

[Out]

d^4*a*(3*c^2/x^2+c^4*ln(x)-1/4/x^4+4*I*c^3/x-4/3*I*c/x^3)+d^4*b*c^4*(-1/4*arctan(c*x)/c^4/x^4-4/3*I*arctan(c*x
)/c^3/x^3+arctan(c*x)*ln(c*x)+4*I*arctan(c*x)/c/x+3/c^2/x^2*arctan(c*x)-2/3*I/c^2/x^2-16/3*I*ln(c*x)-1/12/c^3/
x^3+13/4/c/x+8/3*I*ln(c^2*x^2+1)+13/4*arctan(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*di
log(1+I*c*x)-1/2*I*dilog(1-I*c*x))

Fricas [F]

\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{5}} \,d x } \]

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^5,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^5, x)

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=\text {Timed out} \]

[In]

integrate((d+I*c*d*x)**4*(a+b*atan(c*x))/x**5,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{5}} \,d x } \]

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^5,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{5}} \,d x } \]

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^5,x, algorithm="giac")

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 1.01 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.31 \[ \int \frac {(d+i c d x)^4 (a+b \arctan (c x))}{x^5} \, dx=\left \{\begin {array}{cl} -\frac {a\,d^4}{4\,x^4} & \text {\ if\ \ }c=0\\ 3\,b\,c\,d^4\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )-\frac {b\,d^4\,\left (\frac {\frac {c^2}{3}-c^4\,x^2}{x^3}-c^5\,\mathrm {atan}\left (c\,x\right )\right )}{4\,c}-\frac {b\,c^4\,d^4\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {b\,c^4\,d^4\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-b\,c^2\,d^4\,\left (c^2\,\ln \left (x\right )-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )\,4{}\mathrm {i}+\frac {a\,d^4\,\left (36\,c^2\,x^2+12\,c^4\,x^4\,\ln \left (x\right )-3-c\,x\,16{}\mathrm {i}+c^3\,x^3\,48{}\mathrm {i}\right )}{12\,x^4}-\frac {b\,d^4\,\mathrm {atan}\left (c\,x\right )}{4\,x^4}-\frac {b\,d^4\,\left (c^4\,\ln \left (x\right )-\frac {c^4\,\ln \left (-\frac {c^4\,\left (3\,c^2\,x^2+1\right )}{2}-c^4\right )}{2}+\frac {c^2}{2\,x^2}\right )\,4{}\mathrm {i}}{3}-\frac {b\,c\,d^4\,\mathrm {atan}\left (c\,x\right )\,4{}\mathrm {i}}{3\,x^3}+\frac {3\,b\,c^2\,d^4\,\mathrm {atan}\left (c\,x\right )}{x^2}+\frac {b\,c^3\,d^4\,\mathrm {atan}\left (c\,x\right )\,4{}\mathrm {i}}{x} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

[In]

int(((a + b*atan(c*x))*(d + c*d*x*1i)^4)/x^5,x)

[Out]

piecewise(c == 0, -(a*d^4)/(4*x^4), c ~= 0, - (b*d^4*(c^4*log(x) - (c^4*log(- (c^4*(3*c^2*x^2 + 1))/2 - c^4))/
2 + c^2/(2*x^2))*4i)/3 - (b*d^4*((c^2/3 - c^4*x^2)/x^3 - c^5*atan(c*x)))/(4*c) - (b*c^4*d^4*dilog(- c*x*1i + 1
)*1i)/2 + (b*c^4*d^4*dilog(c*x*1i + 1)*1i)/2 - b*c^2*d^4*(c^2*log(x) - (c^2*log(c^2*x^2 + 1))/2)*4i + (a*d^4*(
- c*x*16i + 36*c^2*x^2 + c^3*x^3*48i + 12*c^4*x^4*log(x) - 3))/(12*x^4) - (b*d^4*atan(c*x))/(4*x^4) + 3*b*c*d^
4*(c^3*atan(c*x) + c^2/x) - (b*c*d^4*atan(c*x)*4i)/(3*x^3) + (3*b*c^2*d^4*atan(c*x))/x^2 + (b*c^3*d^4*atan(c*x
)*4i)/x)