3.75 \(\int \frac {(d+i c d x) (a+b \tan ^{-1}(c x))^2}{x^4} \, dx\)

Optimal. Leaf size=224 \[ -\frac {1}{6} i c^3 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {2}{3} b c^3 d \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac {i b c^2 d \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {b c d \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}+\frac {1}{3} i b^2 c^3 d \text {Li}_2\left (\frac {2}{1-i c x}-1\right )+i b^2 c^3 d \log (x)-\frac {1}{3} b^2 c^3 d \tan ^{-1}(c x)-\frac {b^2 c^2 d}{3 x}-\frac {1}{2} i b^2 c^3 d \log \left (c^2 x^2+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.43, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {4876, 4852, 4918, 325, 203, 4924, 4868, 2447, 266, 36, 29, 31, 4884} \[ \frac {1}{3} i b^2 c^3 d \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )-\frac {1}{6} i c^3 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i b c^2 d \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {2}{3} b c^3 d \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {b c d \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac {1}{2} i b^2 c^3 d \log \left (c^2 x^2+1\right )-\frac {b^2 c^2 d}{3 x}+i b^2 c^3 d \log (x)-\frac {1}{3} b^2 c^3 d \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)*(a + b*ArcTan[c*x])^2)/x^4,x]

[Out]

-(b^2*c^2*d)/(3*x) - (b^2*c^3*d*ArcTan[c*x])/3 - (b*c*d*(a + b*ArcTan[c*x]))/(3*x^2) - (I*b*c^2*d*(a + b*ArcTa
n[c*x]))/x - (I/6)*c^3*d*(a + b*ArcTan[c*x])^2 - (d*(a + b*ArcTan[c*x])^2)/(3*x^3) - ((I/2)*c*d*(a + b*ArcTan[
c*x])^2)/x^2 + I*b^2*c^3*d*Log[x] - (I/2)*b^2*c^3*d*Log[1 + c^2*x^2] - (2*b*c^3*d*(a + b*ArcTan[c*x])*Log[2 -
2/(1 - I*c*x)])/3 + (I/3)*b^2*c^3*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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

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 325

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 2447

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 4852

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

Rule 4868

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 4876

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 4884

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 4918

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 4924

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[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^4} \, dx &=\int \left (\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{x^4}+\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{x^3}\right ) \, dx\\ &=d \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^4} \, dx+(i c d) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^3} \, dx\\ &=-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+\frac {1}{3} (2 b c d) \int \frac {a+b \tan ^{-1}(c x)}{x^3 \left (1+c^2 x^2\right )} \, dx+\left (i b c^2 d\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+\frac {1}{3} (2 b c d) \int \frac {a+b \tan ^{-1}(c x)}{x^3} \, dx+\left (i b c^2 d\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx-\frac {1}{3} \left (2 b c^3 d\right ) \int \frac {a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx-\left (i b c^4 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 )}{3 x^2}-\frac {i b c^2 d \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {1}{6} i c^3 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+\frac {1}{3} \left (b^2 c^2 d\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx-\frac {1}{3} \left (2 i b c^3 d\right ) \int \frac {a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx+\left (i b^2 c^3 d\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {b^2 c^2 d}{3 x}-\frac {b c d \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac {i b c^2 d \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {1}{6} i c^3 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {2}{3} b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+\frac {1}{2} \left (i b^2 c^3 d\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac {1}{3} \left (b^2 c^4 d\right ) \int \frac {1}{1+c^2 x^2} \, dx+\frac {1}{3} \left (2 b^2 c^4 d\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx\\ &=-\frac {b^2 c^2 d}{3 x}-\frac {1}{3} b^2 c^3 d \tan ^{-1}(c x)-\frac {b c d \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac {i b c^2 d \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {1}{6} i c^3 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {2}{3} b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+\frac {1}{3} i b^2 c^3 d \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )+\frac {1}{2} \left (i b^2 c^3 d\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (i b^2 c^5 d\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b^2 c^2 d}{3 x}-\frac {1}{3} b^2 c^3 d \tan ^{-1}(c x)-\frac {b c d \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac {i b c^2 d \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {1}{6} i c^3 d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac {i c d \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+i b^2 c^3 d \log (x)-\frac {1}{2} i b^2 c^3 d \log \left (1+c^2 x^2\right )-\frac {2}{3} b c^3 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+\frac {1}{3} i b^2 c^3 d \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 240, normalized size = 1.07 \[ \frac {d \left (-3 i a^2 c x-2 a^2-4 a b c^3 x^3 \log (c x)-6 i a b c^2 x^2+2 a b c^3 x^3 \log \left (c^2 x^2+1\right )-2 b \tan ^{-1}(c x) \left (a \left (3 i c^3 x^3+3 i c x+2\right )+2 b c^3 x^3 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+b c x \left (c^2 x^2+3 i c x+1\right )\right )-2 a b c x+2 i b^2 c^3 x^3 \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )-i b^2 \left (c^3 x^3+3 c x-2 i\right ) \tan ^{-1}(c x)^2-2 b^2 c^2 x^2+6 i b^2 c^3 x^3 \log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )\right )}{6 x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + I*c*d*x)*(a + b*ArcTan[c*x])^2)/x^4,x]

[Out]

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

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ \frac {24 \, x^{3} {\rm integral}\left (\frac {6 i \, a^{2} c^{3} d x^{3} + 6 \, a^{2} c^{2} d x^{2} + 6 i \, a^{2} c d x + 6 \, a^{2} d - {\left (6 \, a b c^{3} d x^{3} + 3 \, {\left (-2 i \, a b + b^{2}\right )} c^{2} d x^{2} + {\left (6 \, a b - 2 i \, b^{2}\right )} c d x - 6 i \, a b d\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{6 \, {\left (c^{2} x^{6} + x^{4}\right )}}, x\right ) + {\left (3 i \, b^{2} c d x + 2 \, b^{2} d\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2}}{24 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)*(a+b*arctan(c*x))^2/x^4,x, algorithm="fricas")

[Out]

1/24*(24*x^3*integral(1/6*(6*I*a^2*c^3*d*x^3 + 6*a^2*c^2*d*x^2 + 6*I*a^2*c*d*x + 6*a^2*d - (6*a*b*c^3*d*x^3 +
3*(-2*I*a*b + b^2)*c^2*d*x^2 + (6*a*b - 2*I*b^2)*c*d*x - 6*I*a*b*d)*log(-(c*x + I)/(c*x - I)))/(c^2*x^6 + x^4)
, x) + (3*I*b^2*c*d*x + 2*b^2*d)*log(-(c*x + I)/(c*x - I))^2)/x^3

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)*(a+b*arctan(c*x))^2/x^4,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.12, size = 556, normalized size = 2.48 \[ -\frac {i c^{2} d \,b^{2} \arctan \left (c x \right )}{x}-\frac {d \,a^{2}}{3 x^{3}}-\frac {i c d a b \arctan \left (c x \right )}{x^{2}}-\frac {c d a b}{3 x^{2}}-\frac {2 d a b \arctan \left (c x \right )}{3 x^{3}}+\frac {c^{3} d a b \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {c d \,b^{2} \arctan \left (c x \right )}{3 x^{2}}-\frac {2 c^{3} d a b \ln \left (c x \right )}{3}-\frac {2 c^{3} d \,b^{2} \ln \left (c x \right ) \arctan \left (c x \right )}{3}+\frac {c^{3} d \,b^{2} \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )}{3}+i c^{3} d \,b^{2} \ln \left (c x \right )+\frac {i c^{3} d \,b^{2} \ln \left (c x +i\right )^{2}}{12}-\frac {i c d \,a^{2}}{2 x^{2}}-\frac {i c^{3} d \,b^{2} \arctan \left (c x \right )^{2}}{2}+\frac {i c^{3} d \,b^{2} \dilog \left (\frac {i \left (c x -i\right )}{2}\right )}{6}-\frac {i c^{3} d \,b^{2} \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{6}-\frac {i c^{3} d \,b^{2} \dilog \left (i c x +1\right )}{3}+\frac {i c^{3} d \,b^{2} \dilog \left (-i c x +1\right )}{3}-\frac {i b^{2} c^{3} d \ln \left (c^{2} x^{2}+1\right )}{2}-\frac {i c^{3} d \,b^{2} \ln \left (c x -i\right )^{2}}{12}-\frac {b^{2} c^{2} d}{3 x}-\frac {b^{2} c^{3} d \arctan \left (c x \right )}{3}-\frac {i c^{3} d \,b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{6}-\frac {i c d \,b^{2} \arctan \left (c x \right )^{2}}{2 x^{2}}-i c^{3} d a b \arctan \left (c x \right )+\frac {i c^{3} d \,b^{2} \ln \left (c x \right ) \ln \left (-i c x +1\right )}{3}-\frac {i c^{3} d \,b^{2} \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{6}-\frac {d \,b^{2} \arctan \left (c x \right )^{2}}{3 x^{3}}-\frac {i c^{2} d a b}{x}+\frac {i c^{3} d \,b^{2} \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{6}+\frac {i c^{3} d \,b^{2} \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{6}-\frac {i c^{3} d \,b^{2} \ln \left (c x \right ) \ln \left (i c x +1\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)*(a+b*arctan(c*x))^2/x^4,x)

[Out]

-1/3*d*a^2/x^3-I*c*d*a*b*arctan(c*x)/x^2-1/3*c*d*a*b/x^2-2/3*d*a*b*arctan(c*x)/x^3+1/3*c^3*d*a*b*ln(c^2*x^2+1)
-1/3*c*d*b^2*arctan(c*x)/x^2-2/3*c^3*d*a*b*ln(c*x)-2/3*c^3*d*b^2*ln(c*x)*arctan(c*x)+1/3*c^3*d*b^2*arctan(c*x)
*ln(c^2*x^2+1)+I*c^3*d*b^2*ln(c*x)+1/6*I*c^3*d*b^2*dilog(1/2*I*(c*x-I))-1/6*I*c^3*d*b^2*dilog(-1/2*I*(I+c*x))-
1/3*I*c^3*d*b^2*dilog(1+I*c*x)+1/12*I*c^3*d*b^2*ln(I+c*x)^2-1/2*I*c*d*a^2/x^2-1/2*I*c^3*d*b^2*arctan(c*x)^2+1/
3*I*c^3*d*b^2*dilog(1-I*c*x)-1/3*b^2*c^2*d/x-1/2*I*b^2*c^3*d*ln(c^2*x^2+1)-1/3*b^2*c^3*d*arctan(c*x)-1/3*d*b^2
*arctan(c*x)^2/x^3-I*c^2*d*a*b/x+1/6*I*c^3*d*b^2*ln(c*x-I)*ln(c^2*x^2+1)+1/6*I*c^3*d*b^2*ln(I+c*x)*ln(1/2*I*(c
*x-I))-I*c^3*d*a*b*arctan(c*x)+1/3*I*c^3*d*b^2*ln(c*x)*ln(1-I*c*x)-1/6*I*c^3*d*b^2*ln(I+c*x)*ln(c^2*x^2+1)-1/6
*I*c^3*d*b^2*ln(c*x-I)*ln(-1/2*I*(I+c*x))-1/2*I*c*d*b^2*arctan(c*x)^2/x^2-I*c^2*d*b^2*arctan(c*x)/x-1/3*I*c^3*
d*b^2*ln(c*x)*ln(1+I*c*x)-1/12*I*c^3*d*b^2*ln(c*x-I)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -i \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} a b c d + \frac {1}{3} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} a b d - \frac {i \, a^{2} c d}{2 \, x^{2}} - \frac {a^{2} d}{3 \, x^{3}} + \frac {-12 \, b^{2} c d x \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right ) - 6 \, b^{2} d \arctan \left (c x\right )^{2} - \frac {1}{2} \, b^{2} d \log \left (c^{2} x^{2} + 1\right )^{2} + i \, {\left (20 \, {\left (\arctan \left (c x\right )^{2} - \log \left (c^{2} x^{2} + 1\right ) + 2 \, \log \relax (x)\right )} b^{2} c^{3} d - 40 \, {\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} b^{2} c^{2} d \arctan \left (c x\right ) + \frac {56 \, b^{2} c^{3} d x^{3} \log \relax (x) - 56 \, b^{2} c^{2} d x^{2} \arctan \left (c x\right ) - 3 \, b^{2} c d x \log \left (c^{2} x^{2} + 1\right )^{2} - 4 \, {\left (7 \, b^{2} c^{3} d x^{3} + 9 \, b^{2} c d x\right )} \arctan \left (c x\right )^{2} - 4 \, {\left (7 \, b^{2} c^{3} d x^{3} - 2 \, b^{2} d \arctan \left (c x\right )\right )} \log \left (c^{2} x^{2} + 1\right )}{x^{3}}\right )} x^{3} + 2 \, x^{3} \int -\frac {12 \, b^{2} c^{2} d x^{2} \log \left (c^{2} x^{2} + 1\right ) - 56 \, b^{2} c d x \arctan \left (c x\right ) - 108 \, {\left (b^{2} c^{2} d x^{2} + b^{2} d\right )} \arctan \left (c x\right )^{2} - 9 \, {\left (b^{2} c^{2} d x^{2} + b^{2} d\right )} \log \left (c^{2} x^{2} + 1\right )^{2}}{4 \, {\left (c^{2} x^{6} + x^{4}\right )}}\,{d x} + {\left (-12 i \, b^{2} c d x - 8 \, b^{2} d\right )} \arctan \left (c x\right )^{2} + 4 \, {\left (3 \, b^{2} c d x - 2 i \, b^{2} d\right )} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right ) + {\left (3 i \, b^{2} c d x + 2 \, b^{2} d\right )} \log \left (c^{2} x^{2} + 1\right )^{2}}{96 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)*(a+b*arctan(c*x))^2/x^4,x, algorithm="maxima")

[Out]

-I*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*a*b*c*d + 1/3*((c^2*log(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c
- 2*arctan(c*x)/x^3)*a*b*d - 1/2*I*a^2*c*d/x^2 - 1/3*a^2*d/x^3 + 1/96*(96*I*x^3*integrate(1/48*(20*b^2*c^2*d*x
^2*arctan(c*x) + 36*(b^2*c^3*d*x^3 + b^2*c*d*x)*arctan(c*x)^2 + 3*(b^2*c^3*d*x^3 + b^2*c*d*x)*log(c^2*x^2 + 1)
^2 - 2*(3*b^2*c^3*d*x^3 - 2*b^2*c*d*x + 6*(b^2*c^2*d*x^2 + b^2*d)*arctan(c*x))*log(c^2*x^2 + 1))/(c^2*x^6 + x^
4), x) + 96*x^3*integrate(1/48*(36*(b^2*c^2*d*x^2 + b^2*d)*arctan(c*x)^2 + 3*(b^2*c^2*d*x^2 + b^2*d)*log(c^2*x
^2 + 1)^2 - 4*(3*b^2*c^3*d*x^3 - 2*b^2*c*d*x)*arctan(c*x) - 2*(5*b^2*c^2*d*x^2 - 6*(b^2*c^3*d*x^3 + b^2*c*d*x)
*arctan(c*x))*log(c^2*x^2 + 1))/(c^2*x^6 + x^4), x) + (-12*I*b^2*c*d*x - 8*b^2*d)*arctan(c*x)^2 + 4*(3*b^2*c*d
*x - 2*I*b^2*d)*arctan(c*x)*log(c^2*x^2 + 1) + (3*I*b^2*c*d*x + 2*b^2*d)*log(c^2*x^2 + 1)^2)/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i))/x^4,x)

[Out]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i))/x^4, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)*(a+b*atan(c*x))**2/x**4,x)

[Out]

Timed out

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