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

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

[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)]

________________________________________________________________________________________

Rubi [A]  time = 0.432427, antiderivative size = 224, normalized size of antiderivative = 1., 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 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 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 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 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 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 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]

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 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 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 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 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 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]

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*}

Mathematica [A]  time = 0.495674, size = 240, normalized size = 1.07 \[ \frac{d \left (2 i b^2 c^3 x^3 \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )-3 i a^2 c x-2 a^2-6 i a b c^2 x^2-4 a b c^3 x^3 \log (c x)+2 a b c^3 x^3 \log \left (c^2 x^2+1\right )-2 b \tan ^{-1}(c x) \left (a \left (3 i c^3 x^3+3 i c x+2\right )+b c x \left (c^2 x^2+3 i c x+1\right )+2 b c^3 x^3 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )-2 a b c x-2 b^2 c^2 x^2+6 i b^2 c^3 x^3 \log \left (\frac{c x}{\sqrt{c^2 x^2+1}}\right )-i b^2 \left (c^3 x^3+3 c x-2 i\right ) \tan ^{-1}(c x)^2\right )}{6 x^3} \]

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)

________________________________________________________________________________________

Maple [B]  time = 0.105, size = 556, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -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 \left (x\right )\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 \left (x\right ) - 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}} \end{align*}

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} -{\left (6 i \, a b - 3 \, b^{2}\right )} c^{2} d x^{2} + 2 \,{\left (3 \, a b - 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}} \end{align*}

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 -
(6*I*a*b - 3*b^2)*c^2*d*x^2 + 2*(3*a*b - 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

________________________________________________________________________________________

Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

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]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, c d x + d\right )}{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{4}}\,{d x} \end{align*}

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]

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