3.1.27 \(\int \frac {(d+i c d x)^3 (a+b \text {ArcTan}(c x))}{x^4} \, dx\) [27]

Optimal. Leaf size=189 \[ -\frac {b c d^3}{6 x^2}-\frac {3 i b c^2 d^3}{2 x}-\frac {3}{2} i b c^3 d^3 \text {ArcTan}(c x)-\frac {d^3 (a+b \text {ArcTan}(c x))}{3 x^3}-\frac {3 i c d^3 (a+b \text {ArcTan}(c x))}{2 x^2}+\frac {3 c^2 d^3 (a+b \text {ArcTan}(c x))}{x}-i a c^3 d^3 \log (x)-\frac {10}{3} b c^3 d^3 \log (x)+\frac {5}{3} b c^3 d^3 \log \left (1+c^2 x^2\right )+\frac {1}{2} b c^3 d^3 \text {PolyLog}(2,-i c x)-\frac {1}{2} b c^3 d^3 \text {PolyLog}(2,i c x) \]

[Out]

-1/6*b*c*d^3/x^2-3/2*I*b*c^2*d^3/x-3/2*I*b*c^3*d^3*arctan(c*x)-1/3*d^3*(a+b*arctan(c*x))/x^3-3/2*I*c*d^3*(a+b*
arctan(c*x))/x^2+3*c^2*d^3*(a+b*arctan(c*x))/x-I*a*c^3*d^3*ln(x)-10/3*b*c^3*d^3*ln(x)+5/3*b*c^3*d^3*ln(c^2*x^2
+1)+1/2*b*c^3*d^3*polylog(2,-I*c*x)-1/2*b*c^3*d^3*polylog(2,I*c*x)

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Rubi [A]
time = 0.15, antiderivative size = 189, 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 = {4996, 4946, 272, 46, 331, 209, 36, 29, 31, 4940, 2438} \begin {gather*} \frac {3 c^2 d^3 (a+b \text {ArcTan}(c x))}{x}-\frac {d^3 (a+b \text {ArcTan}(c x))}{3 x^3}-\frac {3 i c d^3 (a+b \text {ArcTan}(c x))}{2 x^2}-i a c^3 d^3 \log (x)-\frac {3}{2} i b c^3 d^3 \text {ArcTan}(c x)+\frac {1}{2} b c^3 d^3 \text {Li}_2(-i c x)-\frac {1}{2} b c^3 d^3 \text {Li}_2(i c x)-\frac {10}{3} b c^3 d^3 \log (x)-\frac {3 i b c^2 d^3}{2 x}+\frac {5}{3} b c^3 d^3 \log \left (c^2 x^2+1\right )-\frac {b c d^3}{6 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(b*c*d^3)/x^2 - (((3*I)/2)*b*c^2*d^3)/x - ((3*I)/2)*b*c^3*d^3*ArcTan[c*x] - (d^3*(a + b*ArcTan[c*x]))/(3*
x^3) - (((3*I)/2)*c*d^3*(a + b*ArcTan[c*x]))/x^2 + (3*c^2*d^3*(a + b*ArcTan[c*x]))/x - I*a*c^3*d^3*Log[x] - (1
0*b*c^3*d^3*Log[x])/3 + (5*b*c^3*d^3*Log[1 + c^2*x^2])/3 + (b*c^3*d^3*PolyLog[2, (-I)*c*x])/2 - (b*c^3*d^3*Pol
yLog[2, I*c*x])/2

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*} \int \frac {(d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )}{x^4} \, dx &=\int \left (\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^4}+\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-\frac {i c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d^3 \int \frac {a+b \tan ^{-1}(c x)}{x^4} \, dx+\left (3 i c d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^3} \, dx-\left (3 c^2 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx-\left (i c^3 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x} \, dx\\ &=-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-i a c^3 d^3 \log (x)+\frac {1}{3} \left (b c d^3\right ) \int \frac {1}{x^3 \left (1+c^2 x^2\right )} \, dx+\frac {1}{2} \left (3 i b c^2 d^3\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx+\frac {1}{2} \left (b c^3 d^3\right ) \int \frac {\log (1-i c x)}{x} \, dx-\frac {1}{2} \left (b c^3 d^3\right ) \int \frac {\log (1+i c x)}{x} \, dx-\left (3 b c^3 d^3\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {3 i b c^2 d^3}{2 x}-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-i a c^3 d^3 \log (x)+\frac {1}{2} b c^3 d^3 \text {Li}_2(-i c x)-\frac {1}{2} b c^3 d^3 \text {Li}_2(i c x)+\frac {1}{6} \left (b c d^3\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac {1}{2} \left (3 b c^3 d^3\right ) \text {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac {1}{2} \left (3 i b c^4 d^3\right ) \int \frac {1}{1+c^2 x^2} \, dx\\ &=-\frac {3 i b c^2 d^3}{2 x}-\frac {3}{2} i b c^3 d^3 \tan ^{-1}(c x)-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-i a c^3 d^3 \log (x)+\frac {1}{2} b c^3 d^3 \text {Li}_2(-i c x)-\frac {1}{2} b c^3 d^3 \text {Li}_2(i c x)+\frac {1}{6} \left (b c d^3\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 )-\frac {1}{2} \left (3 b c^3 d^3\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (3 b c^5 d^3\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b c d^3}{6 x^2}-\frac {3 i b c^2 d^3}{2 x}-\frac {3}{2} i b c^3 d^3 \tan ^{-1}(c x)-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-i a c^3 d^3 \log (x)-\frac {10}{3} b c^3 d^3 \log (x)+\frac {5}{3} b c^3 d^3 \log \left (1+c^2 x^2\right )+\frac {1}{2} b c^3 d^3 \text {Li}_2(-i c x)-\frac {1}{2} b c^3 d^3 \text {Li}_2(i c x)\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 171, normalized size = 0.90 \begin {gather*} \frac {d^3 \left (-2 a-9 i a c x-b c x+18 a c^2 x^2-9 i b c^2 x^2-2 b \text {ArcTan}(c x)-9 i b c x \text {ArcTan}(c x)+18 b c^2 x^2 \text {ArcTan}(c x)-9 i b c^3 x^3 \text {ArcTan}(c x)-6 i a c^3 x^3 \log (x)-20 b c^3 x^3 \log (c x)+10 b c^3 x^3 \log \left (1+c^2 x^2\right )+3 b c^3 x^3 \text {PolyLog}(2,-i c x)-3 b c^3 x^3 \text {PolyLog}(2,i c x)\right )}{6 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*(-2*a - (9*I)*a*c*x - b*c*x + 18*a*c^2*x^2 - (9*I)*b*c^2*x^2 - 2*b*ArcTan[c*x] - (9*I)*b*c*x*ArcTan[c*x]
+ 18*b*c^2*x^2*ArcTan[c*x] - (9*I)*b*c^3*x^3*ArcTan[c*x] - (6*I)*a*c^3*x^3*Log[x] - 20*b*c^3*x^3*Log[c*x] + 10
*b*c^3*x^3*Log[1 + c^2*x^2] + 3*b*c^3*x^3*PolyLog[2, (-I)*c*x] - 3*b*c^3*x^3*PolyLog[2, I*c*x]))/(6*x^3)

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Maple [A]
time = 0.09, size = 244, normalized size = 1.29

method	result	size
derivativedivides	\(c^{3} \left (-\frac {3 i d^{3} b}{2 c x}-\frac {d^{3} a}{3 c^{3} x^{3}}+\frac {3 d^{3} a}{c x}-\frac {3 i d^{3} b \arctan \left (c x \right )}{2}-i d^{3} a \ln \left (c x \right )-\frac {d^{3} b \arctan \left (c x \right )}{3 c^{3} x^{3}}+\frac {3 d^{3} b \arctan \left (c x \right )}{c x}-i d^{3} b \arctan \left (c x \right ) \ln \left (c x \right )+\frac {d^{3} b \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {d^{3} b \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {d^{3} b \dilog \left (-i c x +1\right )}{2}+\frac {d^{3} b \dilog \left (i c x +1\right )}{2}+\frac {5 b \ln \left (c^{2} x^{2}+1\right ) d^{3}}{3}-\frac {3 i d^{3} a}{2 c^{2} x^{2}}-\frac {3 i d^{3} b \arctan \left (c x \right )}{2 c^{2} x^{2}}-\frac {d^{3} b}{6 c^{2} x^{2}}-\frac {10 d^{3} b \ln \left (c x \right )}{3}\right )\)	\(244\)
default	\(c^{3} \left (-\frac {3 i d^{3} b}{2 c x}-\frac {d^{3} a}{3 c^{3} x^{3}}+\frac {3 d^{3} a}{c x}-\frac {3 i d^{3} b \arctan \left (c x \right )}{2}-i d^{3} a \ln \left (c x \right )-\frac {d^{3} b \arctan \left (c x \right )}{3 c^{3} x^{3}}+\frac {3 d^{3} b \arctan \left (c x \right )}{c x}-i d^{3} b \arctan \left (c x \right ) \ln \left (c x \right )+\frac {d^{3} b \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}-\frac {d^{3} b \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {d^{3} b \dilog \left (-i c x +1\right )}{2}+\frac {d^{3} b \dilog \left (i c x +1\right )}{2}+\frac {5 b \ln \left (c^{2} x^{2}+1\right ) d^{3}}{3}-\frac {3 i d^{3} a}{2 c^{2} x^{2}}-\frac {3 i d^{3} b \arctan \left (c x \right )}{2 c^{2} x^{2}}-\frac {d^{3} b}{6 c^{2} x^{2}}-\frac {10 d^{3} b \ln \left (c x \right )}{3}\right )\)	\(244\)
risch	\(\frac {d^{3} b \,c^{3} \dilog \left (i c x +1\right )}{2}-\frac {11 d^{3} b \,c^{3} \ln \left (i c x \right )}{12}+\frac {5 b \,c^{3} d^{3} \ln \left (c^{2} x^{2}+1\right )}{3}-i d^{3} c^{3} a \ln \left (-i c x \right )-\frac {3 i b \,c^{2} d^{3}}{2 x}-\frac {3 i d^{3} b \,c^{2} \ln \left (i c x +1\right )}{2 x}-\frac {b c \,d^{3}}{6 x^{2}}+\frac {3 i d^{3} c^{2} b \ln \left (-i c x +1\right )}{2 x}-\frac {3 d^{3} b c \ln \left (i c x +1\right )}{4 x^{2}}-\frac {i d^{3} b \ln \left (-i c x +1\right )}{6 x^{3}}+\frac {3 d^{3} c^{2} a}{x}-\frac {d^{3} a}{3 x^{3}}-\frac {3 i d^{3} c a}{2 x^{2}}-\frac {d^{3} c^{3} b \dilog \left (-i c x +1\right )}{2}-\frac {29 d^{3} c^{3} b \ln \left (-i c x \right )}{12}-\frac {3 i b \,c^{3} d^{3} \arctan \left (c x \right )}{2}+\frac {i d^{3} b \ln \left (i c x +1\right )}{6 x^{3}}+\frac {3 d^{3} c b \ln \left (-i c x +1\right )}{4 x^{2}}\)	\(284\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*(-3/2*I*d^3*b/c/x-1/3*d^3*a/c^3/x^3+3*d^3*a/c/x-3/2*I*d^3*b*arctan(c*x)-I*d^3*a*ln(c*x)-1/3*d^3*b*arctan(c
*x)/c^3/x^3+3*d^3*b*arctan(c*x)/c/x-I*d^3*b*arctan(c*x)*ln(c*x)+1/2*d^3*b*ln(c*x)*ln(1+I*c*x)-1/2*d^3*b*ln(c*x
)*ln(1-I*c*x)-1/2*d^3*b*dilog(1-I*c*x)+1/2*d^3*b*dilog(1+I*c*x)+5/3*b*ln(c^2*x^2+1)*d^3-3/2*I*d^3*a/c^2/x^2-3/
2*I*d^3*b*arctan(c*x)/c^2/x^2-1/6*d^3*b/c^2/x^2-10/3*d^3*b*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((d+I*c*d*x)^3*(a+b*arctan(c*x))/x^4,x, algorithm="maxima")

[Out]

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

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

[Out]

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

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

[Out]

sage0*x

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Mupad [B]
time = 0.97, size = 221, normalized size = 1.17 \begin {gather*} \left \{\begin {array}{cl} -\frac {a\,d^3}{3\,x^3} & \text {\ if\ \ }c=0\\ \frac {b\,c^3\,d^3\,\ln \left (-\frac {3\,c^6\,x^2}{2}-\frac {3\,c^4}{2}\right )}{6}-\frac {b\,c^3\,d^3\,\ln \left (x\right )}{3}-\frac {b\,c^3\,d^3\,\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 )}{2}-3\,b\,c\,d^3\,\left (c^2\,\ln \left (x\right )-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )-\frac {b\,c\,d^3}{6\,x^2}-\frac {a\,d^3\,\left (2-18\,c^2\,x^2+c\,x\,9{}\mathrm {i}+c^3\,x^3\,\ln \left (x\right )\,6{}\mathrm {i}\right )}{6\,x^3}-\frac {b\,d^3\,\mathrm {atan}\left (c\,x\right )}{3\,x^3}+\frac {3\,b\,c^2\,d^3\,\mathrm {atan}\left (c\,x\right )}{x}-\frac {b\,d^3\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )\,3{}\mathrm {i}}{2}-\frac {b\,c\,d^3\,\mathrm {atan}\left (c\,x\right )\,3{}\mathrm {i}}{2\,x^2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

piecewise(c == 0, -(a*d^3)/(3*x^3), c ~= 0, - (b*d^3*(c^3*atan(c*x) + c^2/x)*3i)/2 - (b*c^3*d^3*(dilog(- c*x*1
i + 1) - dilog(c*x*1i + 1)))/2 - (b*c^3*d^3*log(x))/3 + (b*c^3*d^3*log(- (3*c^4)/2 - (3*c^6*x^2)/2))/6 - 3*b*c
*d^3*(c^2*log(x) - (c^2*log(c^2*x^2 + 1))/2) - (b*c*d^3)/(6*x^2) - (a*d^3*(c*x*9i - 18*c^2*x^2 + c^3*x^3*log(x
)*6i + 2))/(6*x^3) - (b*d^3*atan(c*x))/(3*x^3) - (b*c*d^3*atan(c*x)*3i)/(2*x^2) + (3*b*c^2*d^3*atan(c*x))/x)

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