3.1.50 \(\int \frac {a+b \text {ArcTan}(c x)}{x^4 (d+i c d x)} \, dx\) [50]

Optimal. Leaf size=197 \[ -\frac {b c}{6 d x^2}+\frac {i b c^2}{2 d x}+\frac {i b c^3 \text {ArcTan}(c x)}{2 d}-\frac {a+b \text {ArcTan}(c x)}{3 d x^3}+\frac {i c (a+b \text {ArcTan}(c x))}{2 d x^2}+\frac {c^2 (a+b \text {ArcTan}(c x))}{d x}-\frac {4 b c^3 \log (x)}{3 d}+\frac {2 b c^3 \log \left (1+c^2 x^2\right )}{3 d}+\frac {i c^3 (a+b \text {ArcTan}(c x)) \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {b c^3 \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.25, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {4990, 4946, 272, 46, 331, 209, 36, 29, 31, 4988, 2497} \begin {gather*} \frac {i c^3 \log \left (2-\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))}{d}+\frac {c^2 (a+b \text {ArcTan}(c x))}{d x}-\frac {a+b \text {ArcTan}(c x)}{3 d x^3}+\frac {i c (a+b \text {ArcTan}(c x))}{2 d x^2}+\frac {i b c^3 \text {ArcTan}(c x)}{2 d}-\frac {b c^3 \text {Li}_2\left (\frac {2}{i c x+1}-1\right )}{2 d}-\frac {4 b c^3 \log (x)}{3 d}+\frac {i b c^2}{2 d x}+\frac {2 b c^3 \log \left (c^2 x^2+1\right )}{3 d}-\frac {b c}{6 d x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, 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 4988

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTan[c*x])
^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/d))
]/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 4990

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[1/d, I
nt[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f), Int[(f*x)^(m + 1)*((a + b*ArcTan[c*x])^p/(d + e*x)),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b \tan ^{-1}(c x)}{x^4 (d+i c d x)} \, dx &=-\left ((i c) \int \frac {a+b \tan ^{-1}(c x)}{x^3 (d+i c d x)} \, dx\right )+\frac {\int \frac {a+b \tan ^{-1}(c x)}{x^4} \, dx}{d}\\ &=-\frac {a+b \tan ^{-1}(c x)}{3 d x^3}-c^2 \int \frac {a+b \tan ^{-1}(c x)}{x^2 (d+i c d x)} \, dx-\frac {(i c) \int \frac {a+b \tan ^{-1}(c x)}{x^3} \, dx}{d}+\frac {(b c) \int \frac {1}{x^3 \left (1+c^2 x^2\right )} \, dx}{3 d}\\ &=-\frac {a+b \tan ^{-1}(c x)}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d x^2}+\left (i c^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x (d+i c d x)} \, dx+\frac {(b c) \text {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )}{6 d}-\frac {c^2 \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx}{d}-\frac {\left (i b c^2\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx}{2 d}\\ &=\frac {i b c^2}{2 d x}-\frac {a+b \tan ^{-1}(c x)}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d x^2}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}+\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {(b c) \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 )}{6 d}-\frac {\left (b c^3\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx}{d}+\frac {\left (i b c^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d}-\frac {\left (i b c^4\right ) \int \frac {\log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac {b c}{6 d x^2}+\frac {i b c^2}{2 d x}+\frac {i b c^3 \tan ^{-1}(c x)}{2 d}-\frac {a+b \tan ^{-1}(c x)}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d x^2}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {b c^3 \log (x)}{3 d}+\frac {b c^3 \log \left (1+c^2 x^2\right )}{6 d}+\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {b c^3 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {\left (b c^3\right ) \text {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{2 d}\\ &=-\frac {b c}{6 d x^2}+\frac {i b c^2}{2 d x}+\frac {i b c^3 \tan ^{-1}(c x)}{2 d}-\frac {a+b \tan ^{-1}(c x)}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d x^2}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {b c^3 \log (x)}{3 d}+\frac {b c^3 \log \left (1+c^2 x^2\right )}{6 d}+\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {b c^3 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {\left (b c^3\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d}+\frac {\left (b c^5\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )}{2 d}\\ &=-\frac {b c}{6 d x^2}+\frac {i b c^2}{2 d x}+\frac {i b c^3 \tan ^{-1}(c x)}{2 d}-\frac {a+b \tan ^{-1}(c x)}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d x^2}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {4 b c^3 \log (x)}{3 d}+\frac {2 b c^3 \log \left (1+c^2 x^2\right )}{3 d}+\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {b c^3 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{2 d}\\ \end {align*}

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Mathematica [A]
time = 0.33, size = 220, normalized size = 1.12 \begin {gather*} \frac {-2 a+3 i a c x-b c x+6 a c^2 x^2+3 i b c^2 x^2-b c^3 x^3+6 b c^3 x^3 \text {ArcTan}(c x)^2+\text {ArcTan}(c x) \left (6 a c^3 x^3+b \left (-2+3 i c x+6 c^2 x^2+3 i c^3 x^3\right )+6 i b c^3 x^3 \log \left (1-e^{2 i \text {ArcTan}(c x)}\right )\right )+6 i a c^3 x^3 \log (x)-8 b c^3 x^3 \log \left (\frac {c x}{\sqrt {1+c^2 x^2}}\right )-3 i a c^3 x^3 \log \left (1+c^2 x^2\right )+3 b c^3 x^3 \text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c x)}\right )}{6 d x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 340, normalized size = 1.73

method	result	size
derivativedivides	\(c^{3} \left (-\frac {i b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d}+\frac {a \arctan \left (c x \right )}{d}-\frac {a}{3 d \,c^{3} x^{3}}+\frac {i b}{2 d c x}+\frac {i b \arctan \left (c x \right ) \ln \left (c x \right )}{d}+\frac {a}{d c x}-\frac {i a \ln \left (c^{2} x^{2}+1\right )}{2 d}-\frac {b \arctan \left (c x \right )}{3 d \,c^{3} x^{3}}+\frac {i a \ln \left (c x \right )}{d}+\frac {i a}{2 d \,c^{2} x^{2}}+\frac {b \arctan \left (c x \right )}{d c x}+\frac {b \ln \left (c x -i\right )^{2}}{4 d}-\frac {b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d}-\frac {b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d}-\frac {b \ln \left (c x \right ) \ln \left (i c x +1\right )}{2 d}+\frac {b \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2 d}-\frac {b \dilog \left (i c x +1\right )}{2 d}+\frac {b \dilog \left (-i c x +1\right )}{2 d}+\frac {2 b \ln \left (c^{2} x^{2}+1\right )}{3 d}+\frac {i b \arctan \left (c x \right )}{2 d \,c^{2} x^{2}}+\frac {i b \arctan \left (c x \right )}{2 d}-\frac {b}{6 d \,c^{2} x^{2}}-\frac {4 b \ln \left (c x \right )}{3 d}\right )\)	\(340\)
default	\(c^{3} \left (-\frac {i b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d}+\frac {a \arctan \left (c x \right )}{d}-\frac {a}{3 d \,c^{3} x^{3}}+\frac {i b}{2 d c x}+\frac {i b \arctan \left (c x \right ) \ln \left (c x \right )}{d}+\frac {a}{d c x}-\frac {i a \ln \left (c^{2} x^{2}+1\right )}{2 d}-\frac {b \arctan \left (c x \right )}{3 d \,c^{3} x^{3}}+\frac {i a \ln \left (c x \right )}{d}+\frac {i a}{2 d \,c^{2} x^{2}}+\frac {b \arctan \left (c x \right )}{d c x}+\frac {b \ln \left (c x -i\right )^{2}}{4 d}-\frac {b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d}-\frac {b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d}-\frac {b \ln \left (c x \right ) \ln \left (i c x +1\right )}{2 d}+\frac {b \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2 d}-\frac {b \dilog \left (i c x +1\right )}{2 d}+\frac {b \dilog \left (-i c x +1\right )}{2 d}+\frac {2 b \ln \left (c^{2} x^{2}+1\right )}{3 d}+\frac {i b \arctan \left (c x \right )}{2 d \,c^{2} x^{2}}+\frac {i b \arctan \left (c x \right )}{2 d}-\frac {b}{6 d \,c^{2} x^{2}}-\frac {4 b \ln \left (c x \right )}{3 d}\right )\)	\(340\)
risch	\(\frac {i c a}{2 d \,x^{2}}+\frac {b c \ln \left (i c x +1\right )}{4 d \,x^{2}}+\frac {c^{2} a}{d x}+\frac {i b \,c^{3} \arctan \left (c x \right )}{2 d}-\frac {c b \ln \left (-i c x +1\right )}{4 d \,x^{2}}-\frac {c^{3} b \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (\frac {1}{2}-\frac {i c x}{2}\right )}{2 d}+\frac {c^{3} b \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (-i c x +1\right )}{2 d}+\frac {2 b \,c^{3} \ln \left (c^{2} x^{2}+1\right )}{3 d}+\frac {i c^{3} a \ln \left (-i c x \right )}{d}+\frac {c^{3} a \arctan \left (c x \right )}{d}-\frac {c^{3} b \dilog \left (\frac {1}{2}-\frac {i c x}{2}\right )}{2 d}+\frac {c^{3} \dilog \left (-i c x +1\right ) b}{2 d}-\frac {5 c^{3} b \ln \left (-i c x \right )}{12 d}-\frac {11 b \,c^{3} \ln \left (i c x \right )}{12 d}-\frac {b \,c^{3} \ln \left (i c x +1\right )^{2}}{4 d}-\frac {b \,c^{3} \dilog \left (i c x +1\right )}{2 d}-\frac {i b \ln \left (-i c x +1\right )}{6 d \,x^{3}}-\frac {i b \,c^{2} \ln \left (i c x +1\right )}{2 d x}+\frac {i b \,c^{2}}{2 d x}+\frac {i c^{2} b \ln \left (-i c x +1\right )}{2 d x}-\frac {b c}{6 d \,x^{2}}-\frac {a}{3 d \,x^{3}}+\frac {i b \ln \left (i c x +1\right )}{6 d \,x^{3}}-\frac {i c^{3} a \ln \left (c^{2} x^{2}+1\right )}{2 d}\)	\(401\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*I*c^3*log(I*c*x + 1)/d - 6*I*c^3*log(x)/d - (6*c^2*x^2 + 3*I*c*x - 2)/(d*x^3))*a + (-I*c*integrate(arc
tan(c*x)/(c^2*d*x^5 + d*x^3), x) + integrate(arctan(c*x)/(c^2*d*x^6 + d*x^4), x))*b

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Fricas [A]
time = 1.11, size = 155, normalized size = 0.79 \begin {gather*} \frac {6 \, b c^{3} x^{3} {\rm Li}_2\left (\frac {c x + i}{c x - i} + 1\right ) - 4 \, {\left (-3 i \, a + 4 \, b\right )} c^{3} x^{3} \log \left (x\right ) + 5 \, b c^{3} x^{3} \log \left (\frac {c x + i}{c}\right ) + {\left (-12 i \, a + 11 \, b\right )} c^{3} x^{3} \log \left (\frac {c x - i}{c}\right ) + 6 \, {\left (2 \, a + i \, b\right )} c^{2} x^{2} - 2 \, {\left (-3 i \, a + b\right )} c x + {\left (6 i \, b c^{2} x^{2} - 3 \, b c x - 2 i \, b\right )} \log \left (-\frac {c x + i}{c x - i}\right ) - 4 \, a}{12 \, d x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*b*c^3*x^3*dilog((c*x + I)/(c*x - I) + 1) - 4*(-3*I*a + 4*b)*c^3*x^3*log(x) + 5*b*c^3*x^3*log((c*x + I)
/c) + (-12*I*a + 11*b)*c^3*x^3*log((c*x - I)/c) + 6*(2*a + I*b)*c^2*x^2 - 2*(-3*I*a + b)*c*x + (6*I*b*c^2*x^2
- 3*b*c*x - 2*I*b)*log(-(c*x + I)/(c*x - I)) - 4*a)/(d*x^3)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x^4\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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