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

Optimal. Leaf size=203 \[ -\frac{3 i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c^4 d^2}-\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^2 (-c x+i)}-\frac{3 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^2}-\frac{2 i a x}{c^3 d^2}+\frac{i b \log \left (c^2 x^2+1\right )}{c^4 d^2}+\frac{b x}{2 c^3 d^2}+\frac{b}{2 c^4 d^2 (-c x+i)}-\frac{2 i b x \tan ^{-1}(c x)}{c^3 d^2}-\frac{b \tan ^{-1}(c x)}{c^4 d^2} \]

[Out]

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

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Rubi [A]  time = 0.221916, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 12, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.522, Rules used = {4876, 4846, 260, 4852, 321, 203, 4862, 627, 44, 4854, 2402, 2315} \[ -\frac{3 i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c^4 d^2}-\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^2 (-c x+i)}-\frac{3 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^2}-\frac{2 i a x}{c^3 d^2}+\frac{i b \log \left (c^2 x^2+1\right )}{c^4 d^2}+\frac{b x}{2 c^3 d^2}+\frac{b}{2 c^4 d^2 (-c x+i)}-\frac{2 i b x \tan ^{-1}(c x)}{c^3 d^2}-\frac{b \tan ^{-1}(c x)}{c^4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 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 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)^2} \, dx &=\int \left (-\frac{2 i \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^2}-\frac{x \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^2 (-i+c x)^2}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^2 (-i+c x)}\right ) \, dx\\ &=\frac{i \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{c^3 d^2}-\frac{(2 i) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c^3 d^2}+\frac{3 \int \frac{a+b \tan ^{-1}(c x)}{-i+c x} \, dx}{c^3 d^2}-\frac{\int x \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c^2 d^2}\\ &=-\frac{2 i a x}{c^3 d^2}-\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^2 (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^2}+\frac{(i b) \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{c^3 d^2}-\frac{(2 i b) \int \tan ^{-1}(c x) \, dx}{c^3 d^2}+\frac{(3 b) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^3 d^2}+\frac{b \int \frac{x^2}{1+c^2 x^2} \, dx}{2 c d^2}\\ &=-\frac{2 i a x}{c^3 d^2}+\frac{b x}{2 c^3 d^2}-\frac{2 i b x \tan ^{-1}(c x)}{c^3 d^2}-\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^2 (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^2}-\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^2}+\frac{(i b) \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{c^3 d^2}-\frac{b \int \frac{1}{1+c^2 x^2} \, dx}{2 c^3 d^2}+\frac{(2 i b) \int \frac{x}{1+c^2 x^2} \, dx}{c^2 d^2}\\ &=-\frac{2 i a x}{c^3 d^2}+\frac{b x}{2 c^3 d^2}-\frac{b \tan ^{-1}(c x)}{2 c^4 d^2}-\frac{2 i b x \tan ^{-1}(c x)}{c^3 d^2}-\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^2 (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^2}+\frac{i b \log \left (1+c^2 x^2\right )}{c^4 d^2}-\frac{3 i b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 c^4 d^2}+\frac{(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^2}\\ &=-\frac{2 i a x}{c^3 d^2}+\frac{b x}{2 c^3 d^2}+\frac{b}{2 c^4 d^2 (i-c x)}-\frac{b \tan ^{-1}(c x)}{2 c^4 d^2}-\frac{2 i b x \tan ^{-1}(c x)}{c^3 d^2}-\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^2 (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^2}+\frac{i b \log \left (1+c^2 x^2\right )}{c^4 d^2}-\frac{3 i b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 c^4 d^2}-\frac{b \int \frac{1}{1+c^2 x^2} \, dx}{2 c^3 d^2}\\ &=-\frac{2 i a x}{c^3 d^2}+\frac{b x}{2 c^3 d^2}+\frac{b}{2 c^4 d^2 (i-c x)}-\frac{b \tan ^{-1}(c x)}{c^4 d^2}-\frac{2 i b x \tan ^{-1}(c x)}{c^3 d^2}-\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^2 (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^2}+\frac{i b \log \left (1+c^2 x^2\right )}{c^4 d^2}-\frac{3 i b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 c^4 d^2}\\ \end{align*}

Mathematica [A]  time = 0.96972, size = 186, normalized size = 0.92 \[ -\frac{b \left (-6 i \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )-4 i \log \left (c^2 x^2+1\right )+2 \tan ^{-1}(c x) \left (c^2 x^2+4 i c x+6 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+i \sin \left (2 \tan ^{-1}(c x)\right )-\cos \left (2 \tan ^{-1}(c x)\right )+1\right )-2 c x-12 i \tan ^{-1}(c x)^2+\sin \left (2 \tan ^{-1}(c x)\right )+i \cos \left (2 \tan ^{-1}(c x)\right )\right )+2 a c^2 x^2-6 a \log \left (c^2 x^2+1\right )+8 i a c x+\frac{4 i a}{c x-i}-12 i a \tan ^{-1}(c x)}{4 c^4 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((8*I)*a*c*x + 2*a*c^2*x^2 + ((4*I)*a)/(-I + c*x) - (12*I)*a*ArcTan[c*x] - 6*a*Log[1 + c^2*x^2] + b*(-2*c*x -
(12*I)*ArcTan[c*x]^2 + I*Cos[2*ArcTan[c*x]] - (4*I)*Log[1 + c^2*x^2] - (6*I)*PolyLog[2, -E^((2*I)*ArcTan[c*x]
)] + 2*ArcTan[c*x]*(1 + (4*I)*c*x + c^2*x^2 - Cos[2*ArcTan[c*x]] + 6*Log[1 + E^((2*I)*ArcTan[c*x])] + I*Sin[2*
ArcTan[c*x]]) + Sin[2*ArcTan[c*x]]))/(4*c^4*d^2)

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Maple [A]  time = 0.06, size = 367, normalized size = 1.8 \begin{align*}{\frac{-2\,iax}{{c}^{3}{d}^{2}}}-{\frac{a{x}^{2}}{2\,{c}^{2}{d}^{2}}}+{\frac{3\,a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,{d}^{2}{c}^{4}}}-{\frac{2\,ibx\arctan \left ( cx \right ) }{{c}^{3}{d}^{2}}}-{\frac{ia}{{d}^{2}{c}^{4} \left ( cx-i \right ) }}-{\frac{ib\arctan \left ( cx \right ) }{{d}^{2}{c}^{4} \left ( cx-i \right ) }}-{\frac{b{x}^{2}\arctan \left ( cx \right ) }{2\,{c}^{2}{d}^{2}}}+3\,{\frac{b\arctan \left ( cx \right ) \ln \left ( cx-i \right ) }{{d}^{2}{c}^{4}}}+{\frac{{\frac{i}{8}}b\ln \left ({c}^{4}{x}^{4}+10\,{c}^{2}{x}^{2}+9 \right ) }{{d}^{2}{c}^{4}}}-{\frac{{\frac{i}{2}}b}{{d}^{2}{c}^{4}}}-{\frac{{\frac{3\,i}{2}}b\ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) \ln \left ( cx-i \right ) }{{d}^{2}{c}^{4}}}-{\frac{{\frac{3\,i}{2}}b{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{d}^{2}{c}^{4}}}+{\frac{bx}{2\,{c}^{3}{d}^{2}}}+{\frac{3\,ia\arctan \left ( cx \right ) }{{d}^{2}{c}^{4}}}+{\frac{{\frac{3\,i}{4}}b \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{{d}^{2}{c}^{4}}}-{\frac{b}{4\,{d}^{2}{c}^{4}}\arctan \left ({\frac{cx}{2}} \right ) }+{\frac{b}{4\,{d}^{2}{c}^{4}}\arctan \left ({\frac{{c}^{3}{x}^{3}}{6}}+{\frac{7\,cx}{6}} \right ) }+{\frac{b}{2\,{d}^{2}{c}^{4}}\arctan \left ({\frac{cx}{2}}-{\frac{i}{2}} \right ) }-{\frac{b}{2\,{d}^{2}{c}^{4} \left ( cx-i \right ) }}+{\frac{{\frac{3\,i}{4}}b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{{d}^{2}{c}^{4}}}-{\frac{3\,b\arctan \left ( cx \right ) }{2\,{d}^{2}{c}^{4}}} \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)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \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)^2,x, algorithm="maxima")

[Out]

-1/2*a*(2*I/(c^5*d^2*x - I*c^4*d^2) + (c*x^2 + 4*I*x)/(c^3*d^2) - 6*log(c*x - I)/(c^4*d^2)) - 1/16*(-16*I*c^3*
x^3 + 80*c^2*x^2 + 16*c*x*(2*arctan2(1, c*x) - 6*I) + 192*(-I*c*x - 1)*arctan(c*x)^2 + 48*(-I*c*x - 1)*log(c^2
*x^2 + 1)^2 + (48*c^5*d^2*x - 48*I*c^4*d^2)*((c*(x/(c^7*d^2*x^2 + c^5*d^2) + arctan(c*x)/(c^6*d^2)) - 2*arctan
(c*x)/(c^7*d^2*x^2 + c^5*d^2))*c + 16*integrate(1/8*log(c^2*x^2 + 1)/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2),
x)) - 48*(-I*c^5*d^2*x - c^4*d^2)*(c*(c^2/(c^9*d^2*x^2 + c^7*d^2) + log(c^2*x^2 + 1)/(c^7*d^2*x^2 + c^5*d^2))
+ 32*integrate(1/8*arctan(c*x)/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2), x)) - (96*c^6*d^2*x - 96*I*c^5*d^2)*(c
*(x/(c^7*d^2*x^2 + c^5*d^2) + arctan(c*x)/(c^6*d^2)) - 16*c*integrate(1/8*x^2*log(c^2*x^2 + 1)/(c^7*d^2*x^4 +
2*c^5*d^2*x^2 + c^3*d^2), x) - 2*arctan(c*x)/(c^7*d^2*x^2 + c^5*d^2)) + 96*(I*c^6*d^2*x + c^5*d^2)*(32*c*integ
rate(1/8*x^2*arctan(c*x)/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2), x) - c^2/(c^9*d^2*x^2 + c^7*d^2) - log(c^2*x
^2 + 1)/(c^7*d^2*x^2 + c^5*d^2)) + 16*(2*c^3*x^3 + 6*I*c^2*x^2 + 8*c*x + 4*I)*arctan(c*x) + 16*(16*c^9*d^2*x -
16*I*c^8*d^2)*integrate(1/8*(2*c*x^5*arctan(c*x) + x^4*log(c^2*x^2 + 1))/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d
^2), x) + 256*(-I*c^9*d^2*x - c^8*d^2)*integrate(1/8*(c*x^5*log(c^2*x^2 + 1) - 2*x^4*arctan(c*x))/(c^7*d^2*x^4
+ 2*c^5*d^2*x^2 + c^3*d^2), x) + 256*(I*c^8*d^2*x + c^7*d^2)*integrate(1/8*(2*c*x^4*arctan(c*x) + x^3*log(c^2
*x^2 + 1))/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2), x) + 16*(16*c^8*d^2*x - 16*I*c^7*d^2)*integrate(1/8*(c*x^4
*log(c^2*x^2 + 1) - 2*x^3*arctan(c*x))/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2), x) - 16*(48*c^7*d^2*x - 48*I*c
^6*d^2)*integrate(1/8*(2*c*x^3*arctan(c*x) + x^2*log(c^2*x^2 + 1))/(c^7*d^2*x^4 + 2*c^5*d^2*x^2 + c^3*d^2), x)
+ 768*(I*c^7*d^2*x + c^6*d^2)*integrate(1/8*(c*x^3*log(c^2*x^2 + 1) - 2*x^2*arctan(c*x))/(c^7*d^2*x^4 + 2*c^5
*d^2*x^2 + c^3*d^2), x) - 16*(-I*c^3*x^3 + 3*c^2*x^2 - I*c*x + 5)*log(c^2*x^2 + 1) - 32*I*arctan2(1, c*x))*b/(
8*c^5*d^2*x - 8*I*c^4*d^2)

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

[Out]

integral((-I*b*x^3*log(-(c*x + I)/(c*x - I)) - 2*a*x^3)/(2*c^2*d^2*x^2 - 4*I*c*d^2*x - 2*d^2), 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)**2,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 )}^{2}}\,{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)^2,x, algorithm="giac")

[Out]

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