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

Optimal. Leaf size=195 \[ \frac{i b \text{PolyLog}(2,-i c x)}{2 d^3}-\frac{i b \text{PolyLog}(2,i c x)}{2 d^3}+\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-c x+i)}-\frac{a+b \tan ^{-1}(c x)}{2 d^3 (-c x+i)^2}+\frac{\log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac{a \log (x)}{d^3}+\frac{5 b}{8 d^3 (-c x+i)}+\frac{i b}{8 d^3 (-c x+i)^2}-\frac{5 b \tan ^{-1}(c x)}{8 d^3} \]

[Out]

((I/8)*b)/(d^3*(I - c*x)^2) + (5*b)/(8*d^3*(I - c*x)) - (5*b*ArcTan[c*x])/(8*d^3) - (a + b*ArcTan[c*x])/(2*d^3
*(I - c*x)^2) + (I*(a + b*ArcTan[c*x]))/(d^3*(I - c*x)) + (a*Log[x])/d^3 + ((a + b*ArcTan[c*x])*Log[2/(1 + I*c
*x)])/d^3 + ((I/2)*b*PolyLog[2, (-I)*c*x])/d^3 - ((I/2)*b*PolyLog[2, I*c*x])/d^3 + ((I/2)*b*PolyLog[2, 1 - 2/(
1 + I*c*x)])/d^3

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Rubi [A]  time = 0.241664, antiderivative size = 195, 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, 4848, 2391, 4862, 627, 44, 203, 4854, 2402, 2315} \[ \frac{i b \text{PolyLog}(2,-i c x)}{2 d^3}-\frac{i b \text{PolyLog}(2,i c x)}{2 d^3}+\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-c x+i)}-\frac{a+b \tan ^{-1}(c x)}{2 d^3 (-c x+i)^2}+\frac{\log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac{a \log (x)}{d^3}+\frac{5 b}{8 d^3 (-c x+i)}+\frac{i b}{8 d^3 (-c x+i)^2}-\frac{5 b \tan ^{-1}(c x)}{8 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/8)*b)/(d^3*(I - c*x)^2) + (5*b)/(8*d^3*(I - c*x)) - (5*b*ArcTan[c*x])/(8*d^3) - (a + b*ArcTan[c*x])/(2*d^3
*(I - c*x)^2) + (I*(a + b*ArcTan[c*x]))/(d^3*(I - c*x)) + (a*Log[x])/d^3 + ((a + b*ArcTan[c*x])*Log[2/(1 + I*c
*x)])/d^3 + ((I/2)*b*PolyLog[2, (-I)*c*x])/d^3 - ((I/2)*b*PolyLog[2, I*c*x])/d^3 + ((I/2)*b*PolyLog[2, 1 - 2/(
1 + I*c*x)])/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 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 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{a+b \tan ^{-1}(c x)}{x (d+i c d x)^3} \, dx &=\int \left (\frac{a+b \tan ^{-1}(c x)}{d^3 x}+\frac{c \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-i+c x)^3}+\frac{i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-i+c x)^2}-\frac{c \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-i+c x)}\right ) \, dx\\ &=\frac{\int \frac{a+b \tan ^{-1}(c x)}{x} \, dx}{d^3}+\frac{(i c) \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{d^3}+\frac{c \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{d^3}-\frac{c \int \frac{a+b \tan ^{-1}(c x)}{-i+c x} \, dx}{d^3}\\ &=-\frac{a+b \tan ^{-1}(c x)}{2 d^3 (i-c x)^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}+\frac{a \log (x)}{d^3}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{d^3}+\frac{(i b) \int \frac{\log (1-i c x)}{x} \, dx}{2 d^3}-\frac{(i b) \int \frac{\log (1+i c x)}{x} \, dx}{2 d^3}+\frac{(i b c) \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{d^3}+\frac{(b c) \int \frac{1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{2 d^3}-\frac{(b c) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d^3}\\ &=-\frac{a+b \tan ^{-1}(c x)}{2 d^3 (i-c x)^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}+\frac{a \log (x)}{d^3}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{d^3}+\frac{i b \text{Li}_2(-i c x)}{2 d^3}-\frac{i b \text{Li}_2(i c x)}{2 d^3}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{d^3}+\frac{(i b c) \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{d^3}+\frac{(b c) \int \frac{1}{(-i+c x)^3 (i+c x)} \, dx}{2 d^3}\\ &=-\frac{a+b \tan ^{-1}(c x)}{2 d^3 (i-c x)^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}+\frac{a \log (x)}{d^3}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{d^3}+\frac{i b \text{Li}_2(-i c x)}{2 d^3}-\frac{i b \text{Li}_2(i c x)}{2 d^3}+\frac{i b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 d^3}+\frac{(i b c) \int \left (-\frac{i}{2 (-i+c x)^2}+\frac{i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{d^3}+\frac{(b c) \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 d^3}\\ &=\frac{i b}{8 d^3 (i-c x)^2}+\frac{5 b}{8 d^3 (i-c x)}-\frac{a+b \tan ^{-1}(c x)}{2 d^3 (i-c x)^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}+\frac{a \log (x)}{d^3}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{d^3}+\frac{i b \text{Li}_2(-i c x)}{2 d^3}-\frac{i b \text{Li}_2(i c x)}{2 d^3}+\frac{i b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 d^3}-\frac{(b c) \int \frac{1}{1+c^2 x^2} \, dx}{8 d^3}-\frac{(b c) \int \frac{1}{1+c^2 x^2} \, dx}{2 d^3}\\ &=\frac{i b}{8 d^3 (i-c x)^2}+\frac{5 b}{8 d^3 (i-c x)}-\frac{5 b \tan ^{-1}(c x)}{8 d^3}-\frac{a+b \tan ^{-1}(c x)}{2 d^3 (i-c x)^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}+\frac{a \log (x)}{d^3}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{d^3}+\frac{i b \text{Li}_2(-i c x)}{2 d^3}-\frac{i b \text{Li}_2(i c x)}{2 d^3}+\frac{i b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 d^3}\\ \end{align*}

Mathematica [A]  time = 0.209265, size = 162, normalized size = 0.83 \[ \frac{4 i b \text{PolyLog}(2,-i c x)-4 i b \text{PolyLog}(2,i c x)+4 i b \text{PolyLog}\left (2,\frac{c x+i}{c x-i}\right )-\frac{8 i \left (a+b \tan ^{-1}(c x)\right )}{c x-i}-\frac{4 \left (a+b \tan ^{-1}(c x)\right )}{(c x-i)^2}+8 \log \left (\frac{2 i}{-c x+i}\right ) \left (a+b \tan ^{-1}(c x)\right )+8 a \log (x)+\frac{5 b}{-c x+i}+\frac{i b}{(c x-i)^2}-5 b \tan ^{-1}(c x)}{8 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*b)/(I - c*x) + (I*b)/(-I + c*x)^2 - 5*b*ArcTan[c*x] - (4*(a + b*ArcTan[c*x]))/(-I + c*x)^2 - ((8*I)*(a + b
*ArcTan[c*x]))/(-I + c*x) + 8*a*Log[x] + 8*(a + b*ArcTan[c*x])*Log[(2*I)/(I - c*x)] + (4*I)*b*PolyLog[2, (-I)*
c*x] - (4*I)*b*PolyLog[2, I*c*x] + (4*I)*b*PolyLog[2, (I + c*x)/(-I + c*x)])/(8*d^3)

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Maple [A]  time = 0.066, size = 327, normalized size = 1.7 \begin{align*} -{\frac{a}{2\,{d}^{3} \left ( cx-i \right ) ^{2}}}-{\frac{{\frac{i}{2}}b{\it dilog} \left ( -i \left ( cx+i \right ) \right ) }{{d}^{3}}}-{\frac{a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,{d}^{3}}}-{\frac{ia\arctan \left ( cx \right ) }{{d}^{3}}}+{\frac{a\ln \left ( cx \right ) }{{d}^{3}}}-{\frac{b\arctan \left ( cx \right ) }{2\,{d}^{3} \left ( cx-i \right ) ^{2}}}+{\frac{{\frac{i}{8}}b}{{d}^{3} \left ( cx-i \right ) ^{2}}}-{\frac{b\arctan \left ( cx \right ) \ln \left ( cx-i \right ) }{{d}^{3}}}+{\frac{b\arctan \left ( cx \right ) \ln \left ( cx \right ) }{{d}^{3}}}-{\frac{{\frac{i}{4}}b \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{{d}^{3}}}-{\frac{5\,b\arctan \left ( cx \right ) }{8\,{d}^{3}}}-{\frac{5\,b}{8\,{d}^{3} \left ( cx-i \right ) }}-{\frac{ib\arctan \left ( cx \right ) }{{d}^{3} \left ( cx-i \right ) }}+{\frac{{\frac{i}{2}}b\ln \left ( -i \left ( -cx+i \right ) \right ) \ln \left ( cx \right ) }{{d}^{3}}}-{\frac{{\frac{i}{2}}b\ln \left ( -i \left ( cx+i \right ) \right ) \ln \left ( cx \right ) }{{d}^{3}}}-{\frac{{\frac{i}{2}}b\ln \left ( -icx \right ) \ln \left ( -i \left ( -cx+i \right ) \right ) }{{d}^{3}}}+{\frac{{\frac{i}{2}}b\ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) \ln \left ( cx-i \right ) }{{d}^{3}}}-{\frac{ia}{{d}^{3} \left ( cx-i \right ) }}-{\frac{{\frac{i}{2}}b{\it dilog} \left ( -icx \right ) }{{d}^{3}}}+{\frac{{\frac{i}{2}}b{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{d}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a/d^3/(c*x-I)^2-1/2*I*b/d^3*dilog(-I*(c*x+I))-1/2*a/d^3*ln(c^2*x^2+1)-I*a/d^3*arctan(c*x)+a/d^3*ln(c*x)-1
/2*b/d^3*arctan(c*x)/(c*x-I)^2+1/8*I*b/d^3/(c*x-I)^2-b/d^3*arctan(c*x)*ln(c*x-I)+b/d^3*arctan(c*x)*ln(c*x)-1/4
*I*b/d^3*ln(c*x-I)^2-5/8*b*arctan(c*x)/d^3-5/8*b/d^3/(c*x-I)-I*b/d^3*arctan(c*x)/(c*x-I)+1/2*I*b/d^3*ln(-I*(-c
*x+I))*ln(c*x)-1/2*I*b/d^3*ln(-I*(c*x+I))*ln(c*x)-1/2*I*b/d^3*ln(-I*c*x)*ln(-I*(-c*x+I))+1/2*I*b/d^3*ln(-1/2*I
*(c*x+I))*ln(c*x-I)-I*a/d^3/(c*x-I)-1/2*I*b/d^3*dilog(-I*c*x)+1/2*I*b/d^3*dilog(-1/2*I*(c*x+I))

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Maxima [B]  time = 1.33817, size = 563, normalized size = 2.89 \begin{align*} -\frac{{\left (16 i \, a + 10 \, b\right )} c x +{\left (4 i \, b c^{2} x^{2} + 8 \, b c x - 4 i \, b\right )} \arctan \left (c x\right )^{2} +{\left (i \, b c^{2} x^{2} + 2 \, b c x - i \, b\right )} \log \left (c^{2} x^{2} + 1\right )^{2} + 4 \,{\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 ) - 16 \,{\left (b c^{2} x^{2} - 2 i \, b c x - b\right )} \arctan \left (c x\right ) \log \left (x{\left | c \right |}\right ) +{\left ({\left (b{\left (-16 i \, \arctan \left (0, c\right ) + 5\right )} + 16 i \, a\right )} c^{2} x^{2} -{\left (b{\left (32 \, \arctan \left (0, c\right ) - 6 i\right )} - 32 \, a\right )} c x + b{\left (16 i \, \arctan \left (0, c\right ) + 19\right )} - 16 i \, a\right )} \arctan \left (c x\right ) - 5 \,{\left (b c^{2} x^{2} - 2 i \, b c x - b\right )} \arctan \left (c x, -1\right ) +{\left (8 i \, b c^{2} x^{2} + 16 \, b c x - 8 i \, b\right )}{\rm Li}_2\left (i \, c x + 1\right ) +{\left (-8 i \, b c^{2} x^{2} - 16 \, b c x + 8 i \, b\right )}{\rm Li}_2\left (\frac{1}{2} i \, c x + \frac{1}{2}\right ) +{\left (-8 i \, b c^{2} x^{2} - 16 \, b c x + 8 i \, b\right )}{\rm Li}_2\left (-i \, c x + 1\right ) +{\left (4 \,{\left (\pi b + 2 \, a\right )} c^{2} x^{2} +{\left (-8 i \, \pi b - 16 i \, a\right )} c x - 4 \, \pi b +{\left (-2 i \, b c^{2} x^{2} - 4 \, b c x + 2 i \, b\right )} \log \left (\frac{1}{4} \, c^{2} x^{2} + \frac{1}{4}\right ) - 8 \, a\right )} \log \left (c^{2} x^{2} + 1\right ) -{\left (16 \, a c^{2} x^{2} - 32 i \, a c x - 16 \, a\right )} \log \left (x\right ) + 24 \, a - 12 i \, b}{16 \, c^{2} d^{3} x^{2} - 32 i \, c d^{3} x - 16 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((16*I*a + 10*b)*c*x + (4*I*b*c^2*x^2 + 8*b*c*x - 4*I*b)*arctan(c*x)^2 + (I*b*c^2*x^2 + 2*b*c*x - I*b)*log(c^
2*x^2 + 1)^2 + 4*(b*c^2*x^2 - 2*I*b*c*x - b)*arctan(c*x)*log(1/4*c^2*x^2 + 1/4) - 16*(b*c^2*x^2 - 2*I*b*c*x -
b)*arctan(c*x)*log(x*abs(c)) + ((b*(-16*I*arctan2(0, c) + 5) + 16*I*a)*c^2*x^2 - (b*(32*arctan2(0, c) - 6*I) -
32*a)*c*x + b*(16*I*arctan2(0, c) + 19) - 16*I*a)*arctan(c*x) - 5*(b*c^2*x^2 - 2*I*b*c*x - b)*arctan2(c*x, -1
) + (8*I*b*c^2*x^2 + 16*b*c*x - 8*I*b)*dilog(I*c*x + 1) + (-8*I*b*c^2*x^2 - 16*b*c*x + 8*I*b)*dilog(1/2*I*c*x
+ 1/2) + (-8*I*b*c^2*x^2 - 16*b*c*x + 8*I*b)*dilog(-I*c*x + 1) + (4*(pi*b + 2*a)*c^2*x^2 + (-8*I*pi*b - 16*I*a
)*c*x - 4*pi*b + (-2*I*b*c^2*x^2 - 4*b*c*x + 2*I*b)*log(1/4*c^2*x^2 + 1/4) - 8*a)*log(c^2*x^2 + 1) - (16*a*c^2
*x^2 - 32*I*a*c*x - 16*a)*log(x) + 24*a - 12*I*b)/(16*c^2*d^3*x^2 - 32*I*c*d^3*x - 16*d^3)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b \log \left (-\frac{c x + i}{c x - i}\right ) - 2 i \, a}{2 \, c^{3} d^{3} x^{4} - 6 i \, c^{2} d^{3} x^{3} - 6 \, c d^{3} x^{2} + 2 i \, d^{3} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*log(-(c*x + I)/(c*x - I)) - 2*I*a)/(2*c^3*d^3*x^4 - 6*I*c^2*d^3*x^3 - 6*c*d^3*x^2 + 2*I*d^3*x), 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))/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{b \arctan \left (c x\right ) + a}{{\left (i \, c d x + d\right )}^{3} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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