3.1.74 \(\int \frac {(d+i c d x) (a+b \text {ArcTan}(c x))^2}{x^3} \, dx\) [74]

Optimal. Leaf size=159 \[ -\frac {b c d (a+b \text {ArcTan}(c x))}{x}+\frac {1}{2} c^2 d (a+b \text {ArcTan}(c x))^2-\frac {d (a+b \text {ArcTan}(c x))^2}{2 x^2}-\frac {i c d (a+b \text {ArcTan}(c x))^2}{x}+b^2 c^2 d \log (x)-\frac {1}{2} b^2 c^2 d \log \left (1+c^2 x^2\right )+2 i b c^2 d (a+b \text {ArcTan}(c x)) \log \left (2-\frac {2}{1-i c x}\right )+b^2 c^2 d \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right ) \]

[Out]

-b*c*d*(a+b*arctan(c*x))/x+1/2*c^2*d*(a+b*arctan(c*x))^2-1/2*d*(a+b*arctan(c*x))^2/x^2-I*c*d*(a+b*arctan(c*x))
^2/x+b^2*c^2*d*ln(x)-1/2*b^2*c^2*d*ln(c^2*x^2+1)+2*I*b*c^2*d*(a+b*arctan(c*x))*ln(2-2/(1-I*c*x))+b^2*c^2*d*pol
ylog(2,-1+2/(1-I*c*x))

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Rubi [A]
time = 0.25, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {4996, 4946, 5038, 272, 36, 29, 31, 5004, 5044, 4988, 2497} \begin {gather*} \frac {1}{2} c^2 d (a+b \text {ArcTan}(c x))^2+2 i b c^2 d \log \left (2-\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))-\frac {d (a+b \text {ArcTan}(c x))^2}{2 x^2}-\frac {i c d (a+b \text {ArcTan}(c x))^2}{x}-\frac {b c d (a+b \text {ArcTan}(c x))}{x}+b^2 c^2 d \text {Li}_2\left (\frac {2}{1-i c x}-1\right )-\frac {1}{2} b^2 c^2 d \log \left (c^2 x^2+1\right )+b^2 c^2 d \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)*(a + b*ArcTan[c*x])^2)/x^3,x]

[Out]

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

Rule 5004

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

Rule 5038

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

Rule 5044

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

Rubi steps

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

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Mathematica [A]
time = 0.15, size = 190, normalized size = 1.19 \begin {gather*} -\frac {d \left (a^2+2 i a^2 c x+2 a b c x-b^2 (-i+c x)^2 \text {ArcTan}(c x)^2+2 b \text {ArcTan}(c x) \left (a+2 i a c x+b c x+a c^2 x^2-2 i b c^2 x^2 \log \left (1-e^{2 i \text {ArcTan}(c x)}\right )\right )-4 i a b c^2 x^2 \log (c x)-2 b^2 c^2 x^2 \log \left (\frac {c x}{\sqrt {1+c^2 x^2}}\right )+2 i a b c^2 x^2 \log \left (1+c^2 x^2\right )-2 b^2 c^2 x^2 \text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c x)}\right )\right )}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + I*c*d*x)*(a + b*ArcTan[c*x])^2)/x^3,x]

[Out]

-1/2*(d*(a^2 + (2*I)*a^2*c*x + 2*a*b*c*x - b^2*(-I + c*x)^2*ArcTan[c*x]^2 + 2*b*ArcTan[c*x]*(a + (2*I)*a*c*x +
b*c*x + a*c^2*x^2 - (2*I)*b*c^2*x^2*Log[1 - E^((2*I)*ArcTan[c*x])]) - (4*I)*a*b*c^2*x^2*Log[c*x] - 2*b^2*c^2*
x^2*Log[(c*x)/Sqrt[1 + c^2*x^2]] + (2*I)*a*b*c^2*x^2*Log[1 + c^2*x^2] - 2*b^2*c^2*x^2*PolyLog[2, E^((2*I)*ArcT
an[c*x])]))/x^2

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 447 vs. \(2 (149 ) = 298\).
time = 0.55, size = 448, normalized size = 2.82

method	result	size
derivativedivides	\(c^{2} \left (-i d \,b^{2} \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )-i d a b \ln \left (c^{2} x^{2}+1\right )-\frac {i d \,b^{2} \arctan \left (c x \right )^{2}}{c x}+\frac {d \,b^{2} \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{2}-\frac {d \,b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}-\frac {d \,b^{2} \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\frac {d \,b^{2} \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{2}-\frac {d a b \arctan \left (c x \right )}{c^{2} x^{2}}-d a b \arctan \left (c x \right )-\frac {d a b}{c x}-\frac {d \,b^{2} \arctan \left (c x \right )^{2}}{2 c^{2} x^{2}}-\frac {d \,b^{2} \arctan \left (c x \right )}{c x}-\frac {d \,b^{2} \arctan \left (c x \right )^{2}}{2}-\frac {d \,b^{2} \ln \left (c^{2} x^{2}+1\right )}{2}+d \,b^{2} \ln \left (c x \right )+2 i d \,b^{2} \arctan \left (c x \right ) \ln \left (c x \right )+2 i d a b \ln \left (c x \right )-d \,b^{2} \ln \left (c x \right ) \ln \left (i c x +1\right )+d \,b^{2} \ln \left (c x \right ) \ln \left (-i c x +1\right )-\frac {d \,b^{2} \ln \left (c x -i\right )^{2}}{4}+\frac {d \,b^{2} \ln \left (c x +i\right )^{2}}{4}-\frac {d \,b^{2} \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}+\frac {d \,b^{2} \dilog \left (\frac {i \left (c x -i\right )}{2}\right )}{2}-\frac {2 i d a b \arctan \left (c x \right )}{c x}+d \,a^{2} \left (-\frac {1}{2 c^{2} x^{2}}-\frac {i}{c x}\right )-d \,b^{2} \dilog \left (i c x +1\right )+d \,b^{2} \dilog \left (-i c x +1\right )\right )\)	\(448\)
default	\(c^{2} \left (-i d \,b^{2} \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )-i d a b \ln \left (c^{2} x^{2}+1\right )-\frac {i d \,b^{2} \arctan \left (c x \right )^{2}}{c x}+\frac {d \,b^{2} \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{2}-\frac {d \,b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}-\frac {d \,b^{2} \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\frac {d \,b^{2} \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{2}-\frac {d a b \arctan \left (c x \right )}{c^{2} x^{2}}-d a b \arctan \left (c x \right )-\frac {d a b}{c x}-\frac {d \,b^{2} \arctan \left (c x \right )^{2}}{2 c^{2} x^{2}}-\frac {d \,b^{2} \arctan \left (c x \right )}{c x}-\frac {d \,b^{2} \arctan \left (c x \right )^{2}}{2}-\frac {d \,b^{2} \ln \left (c^{2} x^{2}+1\right )}{2}+d \,b^{2} \ln \left (c x \right )+2 i d \,b^{2} \arctan \left (c x \right ) \ln \left (c x \right )+2 i d a b \ln \left (c x \right )-d \,b^{2} \ln \left (c x \right ) \ln \left (i c x +1\right )+d \,b^{2} \ln \left (c x \right ) \ln \left (-i c x +1\right )-\frac {d \,b^{2} \ln \left (c x -i\right )^{2}}{4}+\frac {d \,b^{2} \ln \left (c x +i\right )^{2}}{4}-\frac {d \,b^{2} \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2}+\frac {d \,b^{2} \dilog \left (\frac {i \left (c x -i\right )}{2}\right )}{2}-\frac {2 i d a b \arctan \left (c x \right )}{c x}+d \,a^{2} \left (-\frac {1}{2 c^{2} x^{2}}-\frac {i}{c x}\right )-d \,b^{2} \dilog \left (i c x +1\right )+d \,b^{2} \dilog \left (-i c x +1\right )\right )\)	\(448\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)*(a+b*arctan(c*x))^2/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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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)*(a+b*arctan(c*x))^2/x^3,x, algorithm="maxima")

[Out]

-I*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*a*b*c*d - ((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*a
*b*d + 1/2*((arctan(c*x)^2 - log(c^2*x^2 + 1) + 2*log(x))*c^2 - 2*(c*arctan(c*x) + 1/x)*c*arctan(c*x))*b^2*d +
1/16*I*(4*(c*arctan(c*x)^3 + 4*c^2*integrate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^2*x^4 + x^2), x) - 16*c^2*integra
te(1/16*x^2*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) + 32*c*integrate(1/16*x*arctan(c*x)/(c^2*x^4 + x^2), x) + 48*
integrate(1/16*arctan(c*x)^2/(c^2*x^4 + x^2), x) + 4*integrate(1/16*log(c^2*x^2 + 1)^2/(c^2*x^4 + x^2), x))*x
- 4*arctan(c*x)^2 + log(c^2*x^2 + 1)^2)*b^2*c*d/x - I*a^2*c*d/x - 1/2*b^2*d*arctan(c*x)^2/x^2 - 1/2*a^2*d/x^2

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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)*(a+b*arctan(c*x))^2/x^3,x, algorithm="fricas")

[Out]

1/8*(8*x^2*integral(1/2*(2*I*a^2*c^3*d*x^3 + 2*a^2*c^2*d*x^2 + 2*I*a^2*c*d*x + 2*a^2*d - (2*a*b*c^3*d*x^3 + 2*
(-I*a*b + b^2)*c^2*d*x^2 + (2*a*b - I*b^2)*c*d*x - 2*I*a*b*d)*log(-(c*x + I)/(c*x - I)))/(c^2*x^5 + x^3), x) +
(2*I*b^2*c*d*x + b^2*d)*log(-(c*x + I)/(c*x - I))^2)/x^2

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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)*(a+b*atan(c*x))**2/x**3,x)

[Out]

Timed out

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Giac [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)*(a+b*arctan(c*x))^2/x^3,x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i))/x^3,x)

[Out]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i))/x^3, x)

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