3.101 \(\int \frac{(a+b \tan ^{-1}(c x))^2}{x^3 (d+i c d x)} \, dx\)

Optimal. Leaf size=273 \[ -\frac{i b c^2 \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac{b^2 c^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )}{d}-\frac{b^2 c^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )}{2 d}-\frac{3 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}-\frac{2 i b c^2 \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac{c^2 \log \left (2-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac{b^2 c^2 \log \left (c^2 x^2+1\right )}{2 d}+\frac{b^2 c^2 \log (x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.624345, antiderivative size = 273, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 13, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.52, Rules used = {4870, 4852, 4918, 266, 36, 29, 31, 4884, 4924, 4868, 2447, 4994, 6610} \[ -\frac{i b c^2 \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac{b^2 c^2 \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )}{d}-\frac{b^2 c^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )}{2 d}-\frac{3 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}-\frac{2 i b c^2 \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac{c^2 \log \left (2-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac{b^2 c^2 \log \left (c^2 x^2+1\right )}{2 d}+\frac{b^2 c^2 \log (x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4870

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]

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 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]

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 4994

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

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x^3 (d+i c d x)} \, dx &=-\left ((i c) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x^2 (d+i c d x)} \, dx\right )+\frac{\int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x^3} \, dx}{d}\\ &=-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}-c^2 \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x (d+i c d x)} \, dx-\frac{(i c) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx}{d}+\frac{(b c) \int \frac{a+b \tan ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )} \, dx}{d}\\ &=-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac{c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+i c x}\right )}{d}+\frac{(b c) \int \frac{a+b \tan ^{-1}(c x)}{x^2} \, dx}{d}-\frac{\left (2 i b c^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx}{d}-\frac{\left (b c^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{d}+\frac{\left (2 b c^3\right ) \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac{3 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac{c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+i c x}\right )}{d}-\frac{i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{d}+\frac{\left (2 b c^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx}{d}+\frac{\left (b^2 c^2\right ) \int \frac{1}{x \left (1+c^2 x^2\right )} \, dx}{d}+\frac{\left (i b^2 c^3\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac{3 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac{2 i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1-i c x}\right )}{d}-\frac{c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+i c x}\right )}{d}-\frac{i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{d}-\frac{b^2 c^2 \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )}{2 d}+\frac{\left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{2 d}+\frac{\left (2 i b^2 c^3\right ) \int \frac{\log \left (2-\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac{3 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac{2 i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1-i c x}\right )}{d}-\frac{c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+i c x}\right )}{d}-\frac{b^2 c^2 \text{Li}_2\left (-1+\frac{2}{1-i c x}\right )}{d}-\frac{i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{d}-\frac{b^2 c^2 \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )}{2 d}+\frac{\left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )}{2 d}-\frac{\left (b^2 c^4\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x} \, dx,x,x^2\right )}{2 d}\\ &=-\frac{b c \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac{3 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac{i c \left (a+b \tan ^{-1}(c x)\right )^2}{d x}+\frac{b^2 c^2 \log (x)}{d}-\frac{b^2 c^2 \log \left (1+c^2 x^2\right )}{2 d}-\frac{2 i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1-i c x}\right )}{d}-\frac{c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+i c x}\right )}{d}-\frac{b^2 c^2 \text{Li}_2\left (-1+\frac{2}{1-i c x}\right )}{d}-\frac{i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{d}-\frac{b^2 c^2 \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 1.12109, size = 372, normalized size = 1.36 \[ \frac{\frac{2 i a b \left (c^2 x^2 \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )+c x \left (-2 c x \log \left (\frac{c x}{\sqrt{c^2 x^2+1}}\right )+i\right )+2 c^2 x^2 \tan ^{-1}(c x)^2+\tan ^{-1}(c x) \left (i c^2 x^2+2 i c^2 x^2 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+2 c x+i\right )\right )}{x^2}+2 b^2 c^2 \left (-i \tan ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(c x)}\right )-\text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(c x)}\right )+\log \left (\frac{c x}{\sqrt{c^2 x^2+1}}\right )-\frac{\tan ^{-1}(c x)^2}{2 c^2 x^2}+\frac{i \tan ^{-1}(c x)^2}{c x}-\frac{3}{2} \tan ^{-1}(c x)^2-\frac{\tan ^{-1}(c x)}{c x}+\tan ^{-1}(c x)^2 \left (-\log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )\right )-2 i \tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+\frac{i \pi ^3}{24}\right )+a^2 c^2 \log \left (c^2 x^2+1\right )-2 a^2 c^2 \log (x)+2 i a^2 c^2 \tan ^{-1}(c x)+\frac{2 i a^2 c}{x}-\frac{a^2}{x^2}}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(a^2/x^2) + ((2*I)*a^2*c)/x + (2*I)*a^2*c^2*ArcTan[c*x] - 2*a^2*c^2*Log[x] + a^2*c^2*Log[1 + c^2*x^2] + ((2*
I)*a*b*(2*c^2*x^2*ArcTan[c*x]^2 + ArcTan[c*x]*(I + 2*c*x + I*c^2*x^2 + (2*I)*c^2*x^2*Log[1 - E^((2*I)*ArcTan[c
*x])]) + c*x*(I - 2*c*x*Log[(c*x)/Sqrt[1 + c^2*x^2]]) + c^2*x^2*PolyLog[2, E^((2*I)*ArcTan[c*x])]))/x^2 + 2*b^
2*c^2*((I/24)*Pi^3 - ArcTan[c*x]/(c*x) - (3*ArcTan[c*x]^2)/2 - ArcTan[c*x]^2/(2*c^2*x^2) + (I*ArcTan[c*x]^2)/(
c*x) - ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - (2*I)*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] + Log[
(c*x)/Sqrt[1 + c^2*x^2]] - I*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] - PolyLog[2, E^((2*I)*ArcTan[c*x])
] - PolyLog[3, E^((-2*I)*ArcTan[c*x])]/2))/(2*d)

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Maple [C]  time = 2.36, size = 2221, normalized size = 8.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c^2*b^2/d*dilog((1+I*c*x)/(c^2*x^2+1)^(1/2))+c^2*b^2/d*ln((1+I*c*x)/(c^2*x^2+1)^(1/2)-1)-1/2*b^2/d*arctan(c*
x)^2/x^2-c^2*a^2/d*ln(c*x)+1/2*c^2*a^2/d*ln(c^2*x^2+1)-2*c^2*b^2/d*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*c^
2*b^2/d*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/2*c^2*b^2/d*arctan(c*x)^2-2*c^2*b^2/d*dilog(1+(1+I*c*x)/(c^2
*x^2+1)^(1/2))+c^2*b^2/d*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2*I*c^2*b^2/d*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-
1))*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan
(c*x)^2+1/2*I*c^2*b^2/d*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/
(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2-1/2*a^2/d/x^2-1/2*I*c^2*b^2/d*Pi*csgn(I/((1+I*c*x)^2/(c
^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+1/2*I*c^2*b^2/d*Pi*csg
n((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+1/2*I*c^2
*b^2/d*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^
2*arctan(c*x)^2-1/2*I*c^2*b^2/d*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(((1+I*
c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2+1/2*I*c^2*b^2/d*Pi*csgn(I*((1+I*c*x)^2/(c^2*x
^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c
*x)^2+1/2*I*c^2*b^2/d*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(
c^2*x^2+1)+1))^2*arctan(c*x)^2+I*c^2*b^2/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arct
an(c*x)^2+I*c^2*a*b/d*ln(c*x)*ln(1-I*c*x)+2*I*c*a*b/d*arctan(c*x)/x-I*c^2*a*b/d*ln(c*x-I)*ln(-1/2*I*(c*x+I))+1
/2*I*c^2*b^2/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2-I*c^2*a*b/d*ln(c*x
)*ln(1+I*c*x)-1/2*I*c^2*b^2/d*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2
+1/2*I*c^2*b^2/d*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2-1/2*I*c^2*b^
2/d*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2-c*a*b/d/x+c^2*b^2/d*arc
tan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)-c^2*b^2/d*arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-c^2*b^2/d*a
rctan(c*x)^2*ln(c*x)-c^2*b^2/d*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))-c^2*b^2/d*arctan(c*x)^2*ln(2*I*
(1+I*c*x)^2/(c^2*x^2+1))+c^2*b^2/d*arctan(c*x)^2*ln(c*x-I)+I*c^2*a^2/d*arctan(c*x)-c^2*a*b/d*arctan(c*x)-c*b^2
/d*arctan(c*x)/x+I*c*a^2/d/x-a*b/d*arctan(c*x)/x^2+2/3*I*c^2*b^2/d*arctan(c*x)^3-I*c^2*b^2/d*arctan(c*x)+I*c^2
*a*b/d*dilog(1-I*c*x)+2*c^2*a*b/d*arctan(c*x)*ln(c*x-I)-2*c^2*a*b/d*arctan(c*x)*ln(c*x)+I*c*b^2/d*arctan(c*x)^
2/x+I*c^2*a*b/d*ln(c^2*x^2+1)+2*I*c^2*b^2/d*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*I*c^2*b^2/d*a
rctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*c^2*b^2/d*arctan(c*x)*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))
-2*I*c^2*a*b/d*ln(c*x)-I*c^2*a*b/d*dilog(1+I*c*x)-I*c^2*a*b/d*dilog(-1/2*I*(c*x+I))+1/2*I*c^2*a*b/d*ln(c*x-I)^
2-3/2*I*c^2*b^2/d*Pi*arctan(c*x)^2

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{i \, b^{2} \log \left (-\frac{c x + i}{c x - i}\right )^{2} + 4 \, a b \log \left (-\frac{c x + i}{c x - i}\right ) - 4 i \, a^{2}}{4 \,{\left (c d x^{4} - i \, d x^{3}\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: AttributeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^2/x^3/(d+I*c*d*x),x, algorithm="giac")

[Out]

integrate((b*arctan(c*x) + a)^2/((I*c*d*x + d)*x^3), x)