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

Optimal. Leaf size=225 \[ \frac{3 i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c^4 d^3}-\frac{3 i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3 (-c x+i)}-\frac{a+b \tan ^{-1}(c x)}{2 c^4 d^3 (-c x+i)^2}+\frac{3 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3}+\frac{i a x}{c^3 d^3}-\frac{i b \log \left (c^2 x^2+1\right )}{2 c^4 d^3}-\frac{11 b}{8 c^4 d^3 (-c x+i)}+\frac{i b}{8 c^4 d^3 (-c x+i)^2}+\frac{i b x \tan ^{-1}(c x)}{c^3 d^3}+\frac{11 b \tan ^{-1}(c x)}{8 c^4 d^3} \]

[Out]

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

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Rubi [A]  time = 0.24278, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {4876, 4846, 260, 4862, 627, 44, 203, 4854, 2402, 2315} \[ \frac{3 i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c^4 d^3}-\frac{3 i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3 (-c x+i)}-\frac{a+b \tan ^{-1}(c x)}{2 c^4 d^3 (-c x+i)^2}+\frac{3 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3}+\frac{i a x}{c^3 d^3}-\frac{i b \log \left (c^2 x^2+1\right )}{2 c^4 d^3}-\frac{11 b}{8 c^4 d^3 (-c x+i)}+\frac{i b}{8 c^4 d^3 (-c x+i)^2}+\frac{i b x \tan ^{-1}(c x)}{c^3 d^3}+\frac{11 b \tan ^{-1}(c x)}{8 c^4 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4846

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x])^p, x] - Dist[b*c*p, Int[
(x*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 4862

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b*
ArcTan[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x] /; FreeQ[{
a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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 4854

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

Rule 2402

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]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rubi steps

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

Mathematica [A]  time = 0.788684, size = 216, normalized size = 0.96 \[ \frac{i b \left (-48 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )-16 \log \left (c^2 x^2+1\right )-96 \tan ^{-1}(c x)^2-20 i \sin \left (2 \tan ^{-1}(c x)\right )+i \sin \left (4 \tan ^{-1}(c x)\right )+20 \cos \left (2 \tan ^{-1}(c x)\right )-\cos \left (4 \tan ^{-1}(c x)\right )+4 \tan ^{-1}(c x) \left (8 c x-24 i \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+10 \sin \left (2 \tan ^{-1}(c x)\right )-\sin \left (4 \tan ^{-1}(c x)\right )+10 i \cos \left (2 \tan ^{-1}(c x)\right )-i \cos \left (4 \tan ^{-1}(c x)\right )\right )\right )-48 a \log \left (c^2 x^2+1\right )+32 i a c x+\frac{96 i a}{c x-i}-\frac{16 a}{(c x-i)^2}-96 i a \tan ^{-1}(c x)}{32 c^4 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((32*I)*a*c*x - (16*a)/(-I + c*x)^2 + ((96*I)*a)/(-I + c*x) - (96*I)*a*ArcTan[c*x] - 48*a*Log[1 + c^2*x^2] + I
*b*(-96*ArcTan[c*x]^2 + 20*Cos[2*ArcTan[c*x]] - Cos[4*ArcTan[c*x]] - 16*Log[1 + c^2*x^2] - 48*PolyLog[2, -E^((
2*I)*ArcTan[c*x])] - (20*I)*Sin[2*ArcTan[c*x]] + 4*ArcTan[c*x]*(8*c*x + (10*I)*Cos[2*ArcTan[c*x]] - I*Cos[4*Ar
cTan[c*x]] - (24*I)*Log[1 + E^((2*I)*ArcTan[c*x])] + 10*Sin[2*ArcTan[c*x]] - Sin[4*ArcTan[c*x]]) + I*Sin[4*Arc
Tan[c*x]]))/(32*c^4*d^3)

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Maple [A]  time = 0.056, size = 375, normalized size = 1.7 \begin{align*}{\frac{iax}{{c}^{3}{d}^{3}}}-{\frac{3\,a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,{c}^{4}{d}^{3}}}+{\frac{3\,ib\arctan \left ( cx \right ) }{{c}^{4}{d}^{3} \left ( cx-i \right ) }}-{\frac{a}{2\,{c}^{4}{d}^{3} \left ( cx-i \right ) ^{2}}}+{\frac{{\frac{3\,i}{2}}b\ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) \ln \left ( cx-i \right ) }{{c}^{4}{d}^{3}}}-{\frac{{\frac{19\,i}{32}}b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{{c}^{4}{d}^{3}}}-3\,{\frac{b\arctan \left ( cx \right ) \ln \left ( cx-i \right ) }{{c}^{4}{d}^{3}}}-{\frac{b\arctan \left ( cx \right ) }{2\,{c}^{4}{d}^{3} \left ( cx-i \right ) ^{2}}}+{\frac{{\frac{i}{8}}b}{{c}^{4}{d}^{3} \left ( cx-i \right ) ^{2}}}+{\frac{{\frac{3\,i}{64}}b\ln \left ({c}^{4}{x}^{4}+10\,{c}^{2}{x}^{2}+9 \right ) }{{c}^{4}{d}^{3}}}+{\frac{3\,b}{32\,{c}^{4}{d}^{3}}\arctan \left ({\frac{{c}^{3}{x}^{3}}{6}}+{\frac{7\,cx}{6}} \right ) }-{\frac{3\,b}{32\,{c}^{4}{d}^{3}}\arctan \left ({\frac{cx}{2}} \right ) }+{\frac{3\,b}{16\,{c}^{4}{d}^{3}}\arctan \left ({\frac{cx}{2}}-{\frac{i}{2}} \right ) }+{\frac{3\,ia}{{c}^{4}{d}^{3} \left ( cx-i \right ) }}+{\frac{ibx\arctan \left ( cx \right ) }{{c}^{3}{d}^{3}}}+{\frac{19\,b\arctan \left ( cx \right ) }{16\,{c}^{4}{d}^{3}}}+{\frac{11\,b}{8\,{c}^{4}{d}^{3} \left ( cx-i \right ) }}+{\frac{{\frac{3\,i}{2}}b{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{c}^{4}{d}^{3}}}-{\frac{{\frac{3\,i}{4}}b \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{{c}^{4}{d}^{3}}}-{\frac{3\,ia\arctan \left ( cx \right ) }{{c}^{4}{d}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*a*x/c^3/d^3-3/2/c^4*a/d^3*ln(c^2*x^2+1)+3*I/c^4*b/d^3*arctan(c*x)/(c*x-I)-1/2/c^4*a/d^3/(c*x-I)^2+3/2*I/c^4*
b/d^3*ln(-1/2*I*(c*x+I))*ln(c*x-I)-19/32*I/c^4*b/d^3*ln(c^2*x^2+1)-3/c^4*b/d^3*arctan(c*x)*ln(c*x-I)-1/2/c^4*b
/d^3*arctan(c*x)/(c*x-I)^2+1/8*I/c^4*b/d^3/(c*x-I)^2+3/64*I/c^4*b/d^3*ln(c^4*x^4+10*c^2*x^2+9)+3/32/c^4*b/d^3*
arctan(1/6*c^3*x^3+7/6*c*x)-3/32/c^4*b/d^3*arctan(1/2*c*x)+3/16/c^4*b/d^3*arctan(1/2*c*x-1/2*I)+3*I/c^4*a/d^3/
(c*x-I)+I*b*x*arctan(c*x)/c^3/d^3+19/16*b*arctan(c*x)/c^4/d^3+11/8/c^4*b/d^3/(c*x-I)+3/2*I/c^4*b/d^3*dilog(-1/
2*I*(c*x+I))-3/4*I/c^4*b/d^3*ln(c*x-I)^2-3*I/c^4*a/d^3*arctan(c*x)

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Maxima [A]  time = 1.27976, size = 441, normalized size = 1.96 \begin{align*} -\frac{-16 i \, a c^{3} x^{3} - 32 \, a c^{2} x^{2} +{\left (-32 i \, a - 22 \, b\right )} c x +{\left (12 i \, b c^{2} x^{2} + 24 \, b c x - 12 i \, b\right )} \arctan \left (c x\right )^{2} +{\left (3 i \, b c^{2} x^{2} + 6 \, b c x - 3 i \, b\right )} \log \left (c^{2} x^{2} + 1\right )^{2} + 12 \,{\left (b c^{2} x^{2} - 2 i \, b c x - b\right )} \arctan \left (c x\right ) \log \left (\frac{1}{4} \, c^{2} x^{2} + \frac{1}{4}\right ) +{\left (-16 i \, b c^{3} x^{3} +{\left (48 i \, a - 51 \, b\right )} c^{2} x^{2} + 6 \,{\left (16 \, a + i \, b\right )} c x - 48 i \, a - 21 \, b\right )} \arctan \left (c x\right ) + 3 \,{\left (b c^{2} x^{2} - 2 i \, b c x - b\right )} \arctan \left (c x, -1\right ) +{\left (-24 i \, b c^{2} x^{2} - 48 \, b c x + 24 i \, b\right )}{\rm Li}_2\left (\frac{1}{2} i \, c x + \frac{1}{2}\right ) +{\left (8 \,{\left (3 \, a + i \, b\right )} c^{2} x^{2} +{\left (-48 i \, a + 16 \, b\right )} c x +{\left (-6 i \, b c^{2} x^{2} - 12 \, b c x + 6 i \, b\right )} \log \left (\frac{1}{4} \, c^{2} x^{2} + \frac{1}{4}\right ) - 24 \, a - 8 i \, b\right )} \log \left (c^{2} x^{2} + 1\right ) - 40 \, a + 20 i \, b}{16 \, c^{6} d^{3} x^{2} - 32 i \, c^{5} d^{3} x - 16 \, c^{4} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-16*I*a*c^3*x^3 - 32*a*c^2*x^2 + (-32*I*a - 22*b)*c*x + (12*I*b*c^2*x^2 + 24*b*c*x - 12*I*b)*arctan(c*x)^2 +
(3*I*b*c^2*x^2 + 6*b*c*x - 3*I*b)*log(c^2*x^2 + 1)^2 + 12*(b*c^2*x^2 - 2*I*b*c*x - b)*arctan(c*x)*log(1/4*c^2
*x^2 + 1/4) + (-16*I*b*c^3*x^3 + (48*I*a - 51*b)*c^2*x^2 + 6*(16*a + I*b)*c*x - 48*I*a - 21*b)*arctan(c*x) + 3
*(b*c^2*x^2 - 2*I*b*c*x - b)*arctan2(c*x, -1) + (-24*I*b*c^2*x^2 - 48*b*c*x + 24*I*b)*dilog(1/2*I*c*x + 1/2) +
(8*(3*a + I*b)*c^2*x^2 + (-48*I*a + 16*b)*c*x + (-6*I*b*c^2*x^2 - 12*b*c*x + 6*I*b)*log(1/4*c^2*x^2 + 1/4) -
24*a - 8*I*b)*log(c^2*x^2 + 1) - 40*a + 20*I*b)/(16*c^6*d^3*x^2 - 32*I*c^5*d^3*x - 16*c^4*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*x^3*log(-(c*x + I)/(c*x - I)) - 2*I*a*x^3)/(2*c^3*d^3*x^3 - 6*I*c^2*d^3*x^2 - 6*c*d^3*x + 2*I*d^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(x**3*(a+b*atan(c*x))/(d+I*c*d*x)**3,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 )} x^{3}}{{\left (i \, c d x + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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