3.1.0 ∫ (d+icdx)3(a+b arctan(cx))2 ________________________________________

Integrand size = 25, antiderivative size = 402 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))^2}{x^2} \, dx=i a b c^2 d^3 x+i b^2 c^2 d^3 x \arctan (c x)-\frac {9}{2} i c d^3 (a+b \arctan (c x))^2-\frac {d^3 (a+b \arctan (c x))^2}{x}-3 c^2 d^3 x (a+b \arctan (c x))^2-\frac {1}{2} i c^3 d^3 x^2 (a+b \arctan (c x))^2+6 i c d^3 (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )-6 b c d^3 (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )-\frac {1}{2} i b^2 c d^3 \log \left (1+c^2 x^2\right )+2 b c d^3 (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )-i b^2 c d^3 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )-3 i b^2 c d^3 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+3 b c d^3 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-3 b c d^3 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {3}{2} i b^2 c d^3 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {3}{2} i b^2 c d^3 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \]

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

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

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

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

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

1+2/(1+I*c*x))-9/2*I*c*d^3*(a+b*arctan(c*x))^2+I*a*b*c^2*d^3*x

Rubi [A] (veriﬁed)

Time = 0.53 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {4996, 4930, 5040, 4964, 2449, 2352, 4946, 5044, 4988, 2497, 4942, 5108, 5004, 5114, 6745, 5036, 266} \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))^2}{x^2} \, dx=6 i c d^3 \text {arctanh}\left (1-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2-\frac {1}{2} i c^3 d^3 x^2 (a+b \arctan (c x))^2-3 c^2 d^3 x (a+b \arctan (c x))^2+3 b c d^3 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right ) (a+b \arctan (c x))-3 b c d^3 \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))-\frac {9}{2} i c d^3 (a+b \arctan (c x))^2-\frac {d^3 (a+b \arctan (c x))^2}{x}-6 b c d^3 \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))+2 b c d^3 \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))+i a b c^2 d^3 x+i b^2 c^2 d^3 x \arctan (c x)-\frac {1}{2} i b^2 c d^3 \log \left (c^2 x^2+1\right )-i b^2 c d^3 \operatorname {PolyLog}\left (2,\frac {2}{1-i c x}-1\right )-3 i b^2 c d^3 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )-\frac {3}{2} i b^2 c d^3 \operatorname {PolyLog}\left (3,1-\frac {2}{i c x+1}\right )+\frac {3}{2} i b^2 c d^3 \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right ) \]

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

I*a*b*c^2*d^3*x + I*b^2*c^2*d^3*x*ArcTan[c*x] - ((9*I)/2)*c*d^3*(a + b*ArcTan[c*x])^2 - (d^3*(a + b*ArcTan[c*x

])^2)/x - 3*c^2*d^3*x*(a + b*ArcTan[c*x])^2 - (I/2)*c^3*d^3*x^2*(a + b*ArcTan[c*x])^2 + (6*I)*c*d^3*(a + b*Arc

Tan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)] - 6*b*c*d^3*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)] - (I/2)*b^2*c*d^3*Lo

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

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

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

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

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*

d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[b*c

*n*p, Int[x^n*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTan[c*x])^p*ArcTanh[1 - 2/(1 +

I*c*x)], x] - Dist[2*b*c*p, Int[(a + b*ArcTan[c*x])^(p - 1)*(ArcTanh[1 - 2/(1 + I*c*x)]/(1 + c^2*x^2)), x], x

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]

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(

Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[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]

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]

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])

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]

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/

e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[d*(f^2/e), 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] && GtQ[m, 1]

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*e*(p + 1))), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b

, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

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]

Int[(ArcTanh[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[L

og[1 + u]*((a + b*ArcTan[c*x])^p/(d + e*x^2)), x], x] - Dist[1/2, Int[Log[1 - u]*((a + b*ArcTan[c*x])^p/(d + e

*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[u^2 - (1 - 2*(I/(I - c*x)))^

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar

cTan[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

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

Rubi steps \begin{align*} \text {integral}& = \int \left (-3 c^2 d^3 (a+b \arctan (c x))^2+\frac {d^3 (a+b \arctan (c x))^2}{x^2}+\frac {3 i c d^3 (a+b \arctan (c x))^2}{x}-i c^3 d^3 x (a+b \arctan (c x))^2\right ) \, dx \\ & = d^3 \int \frac {(a+b \arctan (c x))^2}{x^2} \, dx+\left (3 i c d^3\right ) \int \frac {(a+b \arctan (c x))^2}{x} \, dx-\left (3 c^2 d^3\right ) \int (a+b \arctan (c x))^2 \, dx-\left (i c^3 d^3\right ) \int x (a+b \arctan (c x))^2 \, dx \\ & = -\frac {d^3 (a+b \arctan (c x))^2}{x}-3 c^2 d^3 x (a+b \arctan (c x))^2-\frac {1}{2} i c^3 d^3 x^2 (a+b \arctan (c x))^2+6 i c d^3 (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+\left (2 b c d^3\right ) \int \frac {a+b \arctan (c x)}{x \left (1+c^2 x^2\right )} \, dx-\left (12 i b c^2 d^3\right ) \int \frac {(a+b \arctan (c x)) \text {arctanh}\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx+\left (6 b c^3 d^3\right ) \int \frac {x (a+b \arctan (c x))}{1+c^2 x^2} \, dx+\left (i b c^4 d^3\right ) \int \frac {x^2 (a+b \arctan (c x))}{1+c^2 x^2} \, dx \\ & = -4 i c d^3 (a+b \arctan (c x))^2-\frac {d^3 (a+b \arctan (c x))^2}{x}-3 c^2 d^3 x (a+b \arctan (c x))^2-\frac {1}{2} i c^3 d^3 x^2 (a+b \arctan (c x))^2+6 i c d^3 (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )+\left (2 i b c d^3\right ) \int \frac {a+b \arctan (c x)}{x (i+c x)} \, dx+\left (i b c^2 d^3\right ) \int (a+b \arctan (c x)) \, dx-\left (i b c^2 d^3\right ) \int \frac {a+b \arctan (c x)}{1+c^2 x^2} \, dx+\left (6 i b c^2 d^3\right ) \int \frac {(a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (6 i b c^2 d^3\right ) \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (6 b c^2 d^3\right ) \int \frac {a+b \arctan (c x)}{i-c x} \, dx \\ & = i a b c^2 d^3 x-\frac {9}{2} i c d^3 (a+b \arctan (c x))^2-\frac {d^3 (a+b \arctan (c x))^2}{x}-3 c^2 d^3 x (a+b \arctan (c x))^2-\frac {1}{2} i c^3 d^3 x^2 (a+b \arctan (c x))^2+6 i c d^3 (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )-6 b c d^3 (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )+2 b c d^3 (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )+3 b c d^3 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-3 b c d^3 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )+\left (i b^2 c^2 d^3\right ) \int \arctan (c x) \, dx-\left (2 b^2 c^2 d^3\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx-\left (3 b^2 c^2 d^3\right ) \int \frac {\operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx+\left (3 b^2 c^2 d^3\right ) \int \frac {\operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx+\left (6 b^2 c^2 d^3\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx \\ & = i a b c^2 d^3 x+i b^2 c^2 d^3 x \arctan (c x)-\frac {9}{2} i c d^3 (a+b \arctan (c x))^2-\frac {d^3 (a+b \arctan (c x))^2}{x}-3 c^2 d^3 x (a+b \arctan (c x))^2-\frac {1}{2} i c^3 d^3 x^2 (a+b \arctan (c x))^2+6 i c d^3 (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )-6 b c d^3 (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )+2 b c d^3 (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )-i b^2 c d^3 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )+3 b c d^3 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-3 b c d^3 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {3}{2} i b^2 c d^3 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {3}{2} i b^2 c d^3 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )-\left (6 i b^2 c d^3\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )-\left (i b^2 c^3 d^3\right ) \int \frac {x}{1+c^2 x^2} \, dx \\ & = i a b c^2 d^3 x+i b^2 c^2 d^3 x \arctan (c x)-\frac {9}{2} i c d^3 (a+b \arctan (c x))^2-\frac {d^3 (a+b \arctan (c x))^2}{x}-3 c^2 d^3 x (a+b \arctan (c x))^2-\frac {1}{2} i c^3 d^3 x^2 (a+b \arctan (c x))^2+6 i c d^3 (a+b \arctan (c x))^2 \text {arctanh}\left (1-\frac {2}{1+i c x}\right )-6 b c d^3 (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )-\frac {1}{2} i b^2 c d^3 \log \left (1+c^2 x^2\right )+2 b c d^3 (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )-i b^2 c d^3 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )-3 i b^2 c d^3 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+3 b c d^3 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-3 b c d^3 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {3}{2} i b^2 c d^3 \operatorname {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {3}{2} i b^2 c d^3 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \\ \end{align*}

Mathematica [A] (veriﬁed)

Time = 0.55 (sec) , antiderivative size = 512, normalized size of antiderivative = 1.27 \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))^2}{x^2} \, dx=\frac {d^3 \left (-8 a^2+b^2 c \pi ^3 x-24 a^2 c^2 x^2+8 i a b c^2 x^2-4 i a^2 c^3 x^3-16 a b \arctan (c x)-8 i a b c x \arctan (c x)-48 a b c^2 x^2 \arctan (c x)+8 i b^2 c^2 x^2 \arctan (c x)-8 i a b c^3 x^3 \arctan (c x)-8 b^2 \arctan (c x)^2+12 i b^2 c x \arctan (c x)^2-24 b^2 c^2 x^2 \arctan (c x)^2-4 i b^2 c^3 x^3 \arctan (c x)^2-16 b^2 c x \arctan (c x)^3+24 i b^2 c x \arctan (c x)^2 \log \left (1-e^{-2 i \arctan (c x)}\right )+16 b^2 c x \arctan (c x) \log \left (1-e^{2 i \arctan (c x)}\right )-48 b^2 c x \arctan (c x) \log \left (1+e^{2 i \arctan (c x)}\right )-24 i b^2 c x \arctan (c x)^2 \log \left (1+e^{2 i \arctan (c x)}\right )+24 i a^2 c x \log (x)+16 a b c x \log (c x)+16 a b c x \log \left (1+c^2 x^2\right )-4 i b^2 c x \log \left (1+c^2 x^2\right )-24 b^2 c x \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )-24 b^2 c x (-i+\arctan (c x)) \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )-8 i b^2 c x \operatorname {PolyLog}\left (2,e^{2 i \arctan (c x)}\right )-24 a b c x \operatorname {PolyLog}(2,-i c x)+24 a b c x \operatorname {PolyLog}(2,i c x)+12 i b^2 c x \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )-12 i b^2 c x \operatorname {PolyLog}\left (3,-e^{2 i \arctan (c x)}\right )\right )}{8 x} \]

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

(d^3*(-8*a^2 + b^2*c*Pi^3*x - 24*a^2*c^2*x^2 + (8*I)*a*b*c^2*x^2 - (4*I)*a^2*c^3*x^3 - 16*a*b*ArcTan[c*x] - (8

*I)*a*b*c*x*ArcTan[c*x] - 48*a*b*c^2*x^2*ArcTan[c*x] + (8*I)*b^2*c^2*x^2*ArcTan[c*x] - (8*I)*a*b*c^3*x^3*ArcTa

n[c*x] - 8*b^2*ArcTan[c*x]^2 + (12*I)*b^2*c*x*ArcTan[c*x]^2 - 24*b^2*c^2*x^2*ArcTan[c*x]^2 - (4*I)*b^2*c^3*x^3

*ArcTan[c*x]^2 - 16*b^2*c*x*ArcTan[c*x]^3 + (24*I)*b^2*c*x*ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] + 16*

b^2*c*x*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] - 48*b^2*c*x*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] - (

24*I)*b^2*c*x*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + (24*I)*a^2*c*x*Log[x] + 16*a*b*c*x*Log[c*x] + 16*

a*b*c*x*Log[1 + c^2*x^2] - (4*I)*b^2*c*x*Log[1 + c^2*x^2] - 24*b^2*c*x*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan

[c*x])] - 24*b^2*c*x*(-I + ArcTan[c*x])*PolyLog[2, -E^((2*I)*ArcTan[c*x])] - (8*I)*b^2*c*x*PolyLog[2, E^((2*I)

*ArcTan[c*x])] - 24*a*b*c*x*PolyLog[2, (-I)*c*x] + 24*a*b*c*x*PolyLog[2, I*c*x] + (12*I)*b^2*c*x*PolyLog[3, E^

((-2*I)*ArcTan[c*x])] - (12*I)*b^2*c*x*PolyLog[3, -E^((2*I)*ArcTan[c*x])]))/(8*x)

Maple [C] (warning: unable to verify)

Time = 32.12 (sec) , antiderivative size = 1437, normalized size of antiderivative = 3.57


method	result	size

parts	\(\text {Expression too large to display}\)	\(1437\)
derivativedivides	\(\text {Expression too large to display}\)	\(1439\)
default	\(\text {Expression too large to display}\)	\(1439\)

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

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

*c*x)^2/(c^2*x^2+1))-6*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-6*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2

+1)^(1/2))+6*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*arctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2+1)

^(1/2))-3/2*Pi*arctan(c*x)^2-3/2*Pi*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*cs

gn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*arctan(c*x)^2+3/2*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+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2-3/2*Pi*csgn(I*((1+I*c

*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1))

)*arctan(c*x)^2+3/2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(((1+I*c*x)^2/(c^2*

x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2+3/2*Pi*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(I*((1+I

*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2-3/2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/

(1+(1+I*c*x)^2/(c^2*x^2+1)))^3*arctan(c*x)^2+3/2*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1

)))^2*arctan(c*x)^2-3/2*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3*arctan(c*x)^2-arcta

n(c*x)^2/c/x+2*arctan(c*x)*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2*I*arctan(c*x)^2*c^2*x^2+I*ln(1+(1+I*c*x)^2/(c

^2*x^2+1))-3/2*I*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+6*I*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*I*polylog(3

,(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*I*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*I*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(

1/2))+2*I*dilog((1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*dilog(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+3/2*I*arctan(c*x)^2+3*I*

arctan(c*x)^2*ln(c*x)+I*arctan(c*x)*(c*x-I)-3*I*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+3*I*arctan(c*x)^2*

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

tan(c*x)-1/2*I*arctan(c*x)*c^2*x^2+3*I*arctan(c*x)*ln(c*x)-1/c/x*arctan(c*x)-3/2*ln(c*x)*ln(1+I*c*x)+3/2*ln(c*

x)*ln(1-I*c*x)-3/2*dilog(1+I*c*x)+3/2*dilog(1-I*c*x)+1/2*I*c*x+ln(c*x)+ln(c^2*x^2+1)-1/2*I*arctan(c*x))

Fricas [F]

\[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))^2}{x^2} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{2}} \,d x } \]

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

integral(1/4*(-4*I*a^2*c^3*d^3*x^3 - 12*a^2*c^2*d^3*x^2 + 12*I*a^2*c*d^3*x + 4*a^2*d^3 + (I*b^2*c^3*d^3*x^3 +

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

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))^2}{x^2} \, dx=\text {Timed out} \]

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

Maxima [F]

\[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))^2}{x^2} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{3} {\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{2}} \,d x } \]

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

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

log(x) - (c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*a*b*d^3 - a^2*d^3/x + 1/96*(12*(-I*b^2*c^3*d^3*x^

3 - 6*b^2*c^2*d^3*x^2 - 2*b^2*d^3)*arctan(c*x)^2 + 12*(b^2*c^3*d^3*x^3 - 6*I*b^2*c^2*d^3*x^2 - 2*I*b^2*d^3)*ar

ctan(c*x)*log(c^2*x^2 + 1) - 3*(-I*b^2*c^3*d^3*x^3 - 6*b^2*c^2*d^3*x^2 - 2*b^2*d^3)*log(c^2*x^2 + 1)^2 - 2*I*(

576*b^2*c^5*d^3*integrate(1/16*x^5*arctan(c*x)^2/(c^2*x^4 + x^2), x) + 48*b^2*c^5*d^3*integrate(1/16*x^5*log(c

^2*x^2 + 1)^2/(c^2*x^4 + x^2), x) + 1536*a*b*c^5*d^3*integrate(1/16*x^5*arctan(c*x)/(c^2*x^4 + x^2), x) + 96*b

^2*c^5*d^3*integrate(1/16*x^5*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) - 576*b^2*c^4*d^3*integrate(1/16*x^4*arctan

(c*x)*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) - 1344*b^2*c^4*d^3*integrate(1/16*x^4*arctan(c*x)/(c^2*x^4 + x^2),

x) - 1152*b^2*c^3*d^3*integrate(1/16*x^3*arctan(c*x)^2/(c^2*x^4 + x^2), x) - 3072*a*b*c^3*d^3*integrate(1/16*x

^3*arctan(c*x)/(c^2*x^4 + x^2), x) - b^2*c*d^3*log(c^2*x^2 + 1)^3 - 12*b^2*c*d^3*arctan(c*x)^2 - 384*b^2*c^2*d

^3*integrate(1/16*x^2*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) - 9*b^2*c*d^3*log(c^2*x^2 + 1)^2 - 1728

*b^2*c*d^3*integrate(1/16*x*arctan(c*x)^2/(c^2*x^4 + x^2), x) - 144*b^2*c*d^3*integrate(1/16*x*log(c^2*x^2 + 1

)^2/(c^2*x^4 + x^2), x) - 4608*a*b*c*d^3*integrate(1/16*x*arctan(c*x)/(c^2*x^4 + x^2), x) - 192*b^2*c*d^3*inte

grate(1/16*x*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) + 192*b^2*d^3*integrate(1/16*arctan(c*x)*log(c^2*x^2 + 1)/(c

^2*x^4 + x^2), x))*x - 48*(8*b^2*c^5*d^3*integrate(1/16*x^5*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) +

8*b^2*c^5*d^3*integrate(1/16*x^5*arctan(c*x)/(c^2*x^4 + x^2), x) + 72*b^2*c^4*d^3*integrate(1/16*x^4*arctan(c

*x)^2/(c^2*x^4 + x^2), x) + 6*b^2*c^4*d^3*integrate(1/16*x^4*log(c^2*x^2 + 1)^2/(c^2*x^4 + x^2), x) + 28*b^2*c

^4*d^3*integrate(1/16*x^4*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) + b^2*c*d^3*arctan(c*x)^3 - 16*b^2*c^3*d^3*inte

grate(1/16*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) - 48*b^2*c^3*d^3*integrate(1/16*x^3*arctan(c*x

)/(c^2*x^4 + x^2), x) + 4*b^2*c^2*d^3*integrate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^2*x^4 + x^2), x) + 8*b^2*c^2*d^

3*integrate(1/16*x^2*log(c^2*x^2 + 1)/(c^2*x^4 + x^2), x) - 24*b^2*c*d^3*integrate(1/16*x*arctan(c*x)*log(c^2*

x^2 + 1)/(c^2*x^4 + x^2), x) - 16*b^2*c*d^3*integrate(1/16*x*arctan(c*x)/(c^2*x^4 + x^2), x) - 24*b^2*d^3*inte

grate(1/16*arctan(c*x)^2/(c^2*x^4 + x^2), x) - 2*b^2*d^3*integrate(1/16*log(c^2*x^2 + 1)^2/(c^2*x^4 + x^2), x)

Giac [F(-1)]

Timed out. \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))^2}{x^2} \, dx=\text {Timed out} \]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+i c d x)^3 (a+b \arctan (c x))^2}{x^2} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3}{x^2} \,d x \]

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

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

\(\int \frac {(d+i c d x)^3 (a+b \arctan (c x))^2}{x^2} \, dx\) [89]

Optimal result