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

Optimal. Leaf size=365 \[ -\frac {b c^3 \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac {11 i c^3 \left (a+b \tan ^{-1}(c x)\right )^2}{6 d}-\frac {8 b c^3 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 d}+\frac {i c^3 \log \left (2-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d x}+\frac {i b c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{3 d x^2}+\frac {4 i b^2 c^3 \text {Li}_2\left (\frac {2}{1-i c x}-1\right )}{3 d}+\frac {i b^2 c^3 \text {Li}_3\left (\frac {2}{i c x+1}-1\right )}{2 d}-\frac {i b^2 c^3 \log (x)}{d}-\frac {b^2 c^3 \tan ^{-1}(c x)}{3 d}-\frac {b^2 c^2}{3 d x}+\frac {i b^2 c^3 \log \left (c^2 x^2+1\right )}{2 d} \]

[Out]

-1/3*b^2*c^2/d/x-1/3*b^2*c^3*arctan(c*x)/d-1/3*b*c*(a+b*arctan(c*x))/d/x^2+I*b*c^2*(a+b*arctan(c*x))/d/x+11/6*
I*c^3*(a+b*arctan(c*x))^2/d-1/3*(a+b*arctan(c*x))^2/d/x^3+1/2*I*c*(a+b*arctan(c*x))^2/d/x^2+c^2*(a+b*arctan(c*
x))^2/d/x-I*b^2*c^3*ln(x)/d+1/2*I*b^2*c^3*ln(c^2*x^2+1)/d-8/3*b*c^3*(a+b*arctan(c*x))*ln(2-2/(1-I*c*x))/d+I*c^
3*(a+b*arctan(c*x))^2*ln(2-2/(1+I*c*x))/d+4/3*I*b^2*c^3*polylog(2,-1+2/(1-I*c*x))/d-b*c^3*(a+b*arctan(c*x))*po
lylog(2,-1+2/(1+I*c*x))/d+1/2*I*b^2*c^3*polylog(3,-1+2/(1+I*c*x))/d

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Rubi [A]  time = 0.97, antiderivative size = 365, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 15, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4870, 4852, 4918, 325, 203, 4924, 4868, 2447, 266, 36, 29, 31, 4884, 4994, 6610} \[ -\frac {b c^3 \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac {4 i b^2 c^3 \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )}{3 d}+\frac {i b^2 c^3 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d}+\frac {11 i c^3 \left (a+b \tan ^{-1}(c x)\right )^2}{6 d}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d x}+\frac {i b c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {8 b c^3 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 d}+\frac {i c^3 \log \left (2-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{3 d x^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{3 d x^3}+\frac {i b^2 c^3 \log \left (c^2 x^2+1\right )}{2 d}-\frac {b^2 c^2}{3 d x}-\frac {i b^2 c^3 \log (x)}{d}-\frac {b^2 c^3 \tan ^{-1}(c x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*c^2)/(3*d*x) - (b^2*c^3*ArcTan[c*x])/(3*d) - (b*c*(a + b*ArcTan[c*x]))/(3*d*x^2) + (I*b*c^2*(a + b*ArcTa
n[c*x]))/(d*x) + (((11*I)/6)*c^3*(a + b*ArcTan[c*x])^2)/d - (a + b*ArcTan[c*x])^2/(3*d*x^3) + ((I/2)*c*(a + b*
ArcTan[c*x])^2)/(d*x^2) + (c^2*(a + b*ArcTan[c*x])^2)/(d*x) - (I*b^2*c^3*Log[x])/d + ((I/2)*b^2*c^3*Log[1 + c^
2*x^2])/d - (8*b*c^3*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)])/(3*d) + (I*c^3*(a + b*ArcTan[c*x])^2*Log[2 -
2/(1 + I*c*x)])/d + (((4*I)/3)*b^2*c^3*PolyLog[2, -1 + 2/(1 - I*c*x)])/d - (b*c^3*(a + b*ArcTan[c*x])*PolyLog[
2, -1 + 2/(1 + I*c*x)])/d + ((I/2)*b^2*c^3*PolyLog[3, -1 + 2/(1 + I*c*x)])/d

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 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 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]

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^4 (d+i c d x)} \, dx &=-\left ((i c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^3 (d+i c d x)} \, dx\right )+\frac {\int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^4} \, dx}{d}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{3 d x^3}-c^2 \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^2 (d+i c d x)} \, dx-\frac {(i c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^3} \, dx}{d}+\frac {(2 b c) \int \frac {a+b \tan ^{-1}(c x)}{x^3 \left (1+c^2 x^2\right )} \, dx}{3 d}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\left (i c^3\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x (d+i c d x)} \, dx+\frac {(2 b c) \int \frac {a+b \tan ^{-1}(c x)}{x^3} \, dx}{3 d}-\frac {c^2 \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx}{d}-\frac {\left (i b c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )} \, dx}{d}-\frac {\left (2 b c^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx}{3 d}\\ &=-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{3 d x^2}+\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d x}+\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {\left (i b c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx}{d}+\frac {\left (b^2 c^2\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx}{3 d}-\frac {\left (2 i b c^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx}{3 d}-\frac {\left (2 b c^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx}{d}+\frac {\left (i b c^4\right ) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{d}-\frac {\left (2 i b c^4\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^2 c^2}{3 d x}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{3 d x^2}+\frac {i b c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}+\frac {11 i c^3 \left (a+b \tan ^{-1}(c x)\right )^2}{6 d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac {2 b c^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{3 d}+\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {b c^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d}-\frac {\left (2 i b c^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx}{d}-\frac {\left (i b^2 c^3\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx}{d}-\frac {\left (b^2 c^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{3 d}+\frac {\left (2 b^2 c^4\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{3 d}+\frac {\left (b^2 c^4\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac {b^2 c^2}{3 d x}-\frac {b^2 c^3 \tan ^{-1}(c x)}{3 d}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{3 d x^2}+\frac {i b c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}+\frac {11 i c^3 \left (a+b \tan ^{-1}(c x)\right )^2}{6 d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac {8 b c^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{3 d}+\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {i b^2 c^3 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{3 d}-\frac {b c^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d}+\frac {i b^2 c^3 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {\left (i b^2 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{2 d}+\frac {\left (2 b^2 c^4\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac {b^2 c^2}{3 d x}-\frac {b^2 c^3 \tan ^{-1}(c x)}{3 d}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{3 d x^2}+\frac {i b c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}+\frac {11 i c^3 \left (a+b \tan ^{-1}(c x)\right )^2}{6 d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac {8 b c^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{3 d}+\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {4 i b^2 c^3 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{3 d}-\frac {b c^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d}+\frac {i b^2 c^3 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {\left (i b^2 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d}+\frac {\left (i b^2 c^5\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )}{2 d}\\ &=-\frac {b^2 c^2}{3 d x}-\frac {b^2 c^3 \tan ^{-1}(c x)}{3 d}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{3 d x^2}+\frac {i b c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}+\frac {11 i c^3 \left (a+b \tan ^{-1}(c x)\right )^2}{6 d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d x^2}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d x}-\frac {i b^2 c^3 \log (x)}{d}+\frac {i b^2 c^3 \log \left (1+c^2 x^2\right )}{2 d}-\frac {8 b c^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{3 d}+\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {4 i b^2 c^3 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{3 d}-\frac {b c^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d}+\frac {i b^2 c^3 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d}\\ \end {align*}

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Mathematica [A]  time = 1.21, size = 535, normalized size = 1.47 \[ \frac {i a^2 c^3 \log (x)}{d}+\frac {a^2 c^3 \tan ^{-1}(c x)}{d}+\frac {a^2 c^2}{d x}-\frac {i a^2 c^3 \log \left (c^2 x^2+1\right )}{2 d}+\frac {i a^2 c}{2 d x^2}-\frac {a^2}{3 d x^3}-\frac {2 i a b c^3 \left (-\frac {i \left (c^2 x^2+1\right )}{6 c^2 x^2}-\frac {4}{3} i \log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )-\frac {\left (c^2 x^2+1\right ) \tan ^{-1}(c x)}{2 c^2 x^2}-\frac {i \left (c^2 x^2+1\right ) \tan ^{-1}(c x)}{3 c^3 x^3}+\frac {1}{2} i \left (\tan ^{-1}(c x)^2+\text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )\right )-\frac {1}{2 c x}+\frac {1}{2} i \tan ^{-1}(c x)^2+\frac {4 i \tan ^{-1}(c x)}{3 c x}-\tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )}{d}+\frac {b^2 c^3 \left (-24 i \log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )+\frac {12 i \left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2}{c^2 x^2}-\frac {8 \left (c^2 x^2+1\right ) \tan ^{-1}(c x)}{c^2 x^2}-\frac {8 \left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2}{c^3 x^3}-24 \tan ^{-1}(c x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c x)}\right )+32 i \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )+12 i \text {Li}_3\left (e^{-2 i \tan ^{-1}(c x)}\right )-\frac {8}{c x}+\frac {32 \tan ^{-1}(c x)^2}{c x}+32 i \tan ^{-1}(c x)^2+\frac {24 i \tan ^{-1}(c x)}{c x}+24 i \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-64 \tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+\pi ^3\right )}{24 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/3*a^2/(d*x^3) + ((I/2)*a^2*c)/(d*x^2) + (a^2*c^2)/(d*x) + (a^2*c^3*ArcTan[c*x])/d + (I*a^2*c^3*Log[x])/d -
((I/2)*a^2*c^3*Log[1 + c^2*x^2])/d - ((2*I)*a*b*c^3*(-1/2*1/(c*x) - ((I/6)*(1 + c^2*x^2))/(c^2*x^2) + (((4*I)/
3)*ArcTan[c*x])/(c*x) - ((I/3)*(1 + c^2*x^2)*ArcTan[c*x])/(c^3*x^3) - ((1 + c^2*x^2)*ArcTan[c*x])/(2*c^2*x^2)
+ (I/2)*ArcTan[c*x]^2 - ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] - ((4*I)/3)*Log[(c*x)/Sqrt[1 + c^2*x^2]] +
(I/2)*(ArcTan[c*x]^2 + PolyLog[2, E^((2*I)*ArcTan[c*x])])))/d + (b^2*c^3*(Pi^3 - 8/(c*x) + ((24*I)*ArcTan[c*x]
)/(c*x) - (8*(1 + c^2*x^2)*ArcTan[c*x])/(c^2*x^2) + (32*I)*ArcTan[c*x]^2 + (32*ArcTan[c*x]^2)/(c*x) - (8*(1 +
c^2*x^2)*ArcTan[c*x]^2)/(c^3*x^3) + ((12*I)*(1 + c^2*x^2)*ArcTan[c*x]^2)/(c^2*x^2) + (24*I)*ArcTan[c*x]^2*Log[
1 - E^((-2*I)*ArcTan[c*x])] - 64*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] - (24*I)*Log[(c*x)/Sqrt[1 + c^2*x^
2]] - 24*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + (32*I)*PolyLog[2, E^((2*I)*ArcTan[c*x])] + (12*I)*Po
lyLog[3, E^((-2*I)*ArcTan[c*x])]))/(24*d)

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fricas [F]  time = 1.04, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {i \, b^{2} \log \left (-\frac {c x + i}{c x - i}\right )^{2} + 4 \, a b \log \left (-\frac {c x + i}{c x - i}\right ) - 4 i \, a^{2}}{4 \, c d x^{5} - 4 i \, d x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((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)/(4*c*d*x^5 - 4*I*d*x^
4), x)

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

[Out]

Timed out

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maple [C]  time = 9.98, size = 2380, normalized size = 6.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*c*a*b/d*arctan(c*x)/x^2+1/2*c^3*b^2/d*Pi*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))^2*arctan(c*x)^2-1/2*c^3*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+2*c^3*b^2/d*arct
an(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))+c^2*b^2/d*arctan(c*x)^2/x+1/2*c^3*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*c^3*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*c^3*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)^2-2*I*c^3*a*b/d*arctan(c*x)*ln
(c*x-I)+2*I*c^3*a*b/d*arctan(c*x)*ln(c*x)+1/2*c^3*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+c^2*a^2/d/x-1/3*b^2/d*a
rctan(c*x)^2/x^3+2/3*c^3*b^2/d*arctan(c*x)^3+c^3*a^2/d*arctan(c*x)-4/3*b^2*c^3*arctan(c*x)/d-3/2*c^3*b^2/d*Pi*
arctan(c*x)^2+I*c^3*a^2/d*ln(c*x)+2*c^3*b^2/d*arctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-8/3*c^3*b^2/
d*arctan(c*x)*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*c^2*a*b/d/x+2*c^2*a*b/d*arctan(c*x)/x+I*c^2*b^2/d*arctan(c*x
)/x-c^3*a*b/d*ln(c*x-I)*ln(-1/2*I*(I+c*x))-1/2*c^3*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+I*c^3*b^2/d*arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))+1/2*c^3*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+I*c^3*b^2/d*arctan(c*x)^2*ln(1+(1+I*c
*x)/(c^2*x^2+1)^(1/2))+I*c^3*b^2/d*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2*c^3*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+I*c^3*b^2/d*arctan(c*x)^2*ln(c*x)+c^3*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*arctan(c*x)^2+1/2*c^3*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-c^3*a*b/d*ln(c*x)*ln(1+I*c*x)+c^3*a*b/d*ln(c*x)*ln
(1-I*c*x)+I*c^3*a*b/d*arctan(c*x)+1/2*I*c*b^2/d*arctan(c*x)^2/x^2-1/3*I*c^3*b^2/d/(I*c*x-(c^2*x^2+1)^(1/2)+1)*
(c^2*x^2+1)^(1/2)-I*c^3*b^2/d*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+1/3*I*c^3*b^2/d/(I*c*x+(c^2*x^2+1)^(
1/2)+1)*(c^2*x^2+1)^(1/2)-I*c^3*b^2/d*arctan(c*x)^2*ln(c*x-I)-1/3*a^2/d/x^3-1/3*c*b^2/d*arctan(c*x)/x^2-2/3*a*
b/d*arctan(c*x)/x^3+c^3*a*b/d*dilog(1-I*c*x)-8/3*c^3*a*b/d*ln(c*x)+8/3*I*c^3*b^2/d*dilog(1+(1+I*c*x)/(c^2*x^2+
1)^(1/2))+2*I*c^3*b^2/d*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*I*c^3*b^2/d*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^
(1/2))-1/2*I*c^3*a^2/d*ln(c^2*x^2+1)-I*c^3*b^2/d*ln((1+I*c*x)/(c^2*x^2+1)^(1/2)-1)+1/2*I*c*a^2/d/x^2-I*c^3*b^2
/d*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-8/3*I*c^3*b^2/d*dilog((1+I*c*x)/(c^2*x^2+1)^(1/2))+11/6*I*c^3*b^2/d*arcta
n(c*x)^2-1/3*c*a*b/d/x^2+4/3*c^3*a*b/d*ln(c^2*x^2+1)-c^3*a*b/d*dilog(-1/2*I*(I+c*x))+1/2*c^3*a*b/d*ln(c*x-I)^2
-c^3*a*b/d*dilog(1+I*c*x)-1/2*c^3*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*c^3*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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \left (\int \frac {a^{2}}{c x^{5} - i x^{4}}\, dx + \int \frac {b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{c x^{5} - i x^{4}}\, dx + \int \frac {2 a b \operatorname {atan}{\left (c x \right )}}{c x^{5} - i x^{4}}\, dx\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*(Integral(a**2/(c*x**5 - I*x**4), x) + Integral(b**2*atan(c*x)**2/(c*x**5 - I*x**4), x) + Integral(2*a*b*at
an(c*x)/(c*x**5 - I*x**4), x))/d

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