3.2.31 \(\int \frac {(a+b \text {ArcTan}(c x))^3}{x^2 (d+i c d x)} \, dx\) [131]

Optimal. Leaf size=263 \[ -\frac {i c (a+b \text {ArcTan}(c x))^3}{d}-\frac {(a+b \text {ArcTan}(c x))^3}{d x}+\frac {3 b c (a+b \text {ArcTan}(c x))^2 \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {i c (a+b \text {ArcTan}(c x))^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {3 i b^2 c (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )}{d}+\frac {3 b c (a+b \text {ArcTan}(c x))^2 \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{2 d}+\frac {3 b^3 c \text {PolyLog}\left (3,-1+\frac {2}{1-i c x}\right )}{2 d}-\frac {3 i b^2 c (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {3 b^3 c \text {PolyLog}\left (4,-1+\frac {2}{1+i c x}\right )}{4 d} \]

[Out]

-I*c*(a+b*arctan(c*x))^3/d-(a+b*arctan(c*x))^3/d/x+3*b*c*(a+b*arctan(c*x))^2*ln(2-2/(1-I*c*x))/d-I*c*(a+b*arct
an(c*x))^3*ln(2-2/(1+I*c*x))/d-3*I*b^2*c*(a+b*arctan(c*x))*polylog(2,-1+2/(1-I*c*x))/d+3/2*b*c*(a+b*arctan(c*x
))^2*polylog(2,-1+2/(1+I*c*x))/d+3/2*b^3*c*polylog(3,-1+2/(1-I*c*x))/d-3/2*I*b^2*c*(a+b*arctan(c*x))*polylog(3
,-1+2/(1+I*c*x))/d-3/4*b^3*c*polylog(4,-1+2/(1+I*c*x))/d

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Rubi [A]
time = 0.43, antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4990, 4946, 5044, 4988, 5004, 5112, 6745, 5114, 5118} \begin {gather*} -\frac {3 i b^2 c \text {Li}_2\left (\frac {2}{1-i c x}-1\right ) (a+b \text {ArcTan}(c x))}{d}-\frac {3 i b^2 c \text {Li}_3\left (\frac {2}{i c x+1}-1\right ) (a+b \text {ArcTan}(c x))}{2 d}+\frac {3 b c \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) (a+b \text {ArcTan}(c x))^2}{2 d}-\frac {i c (a+b \text {ArcTan}(c x))^3}{d}-\frac {(a+b \text {ArcTan}(c x))^3}{d x}+\frac {3 b c \log \left (2-\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))^2}{d}-\frac {i c \log \left (2-\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^3}{d}+\frac {3 b^3 c \text {Li}_3\left (\frac {2}{1-i c x}-1\right )}{2 d}-\frac {3 b^3 c \text {Li}_4\left (\frac {2}{i c x+1}-1\right )}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4946

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]

Rule 4988

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 4990

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 5004

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 5044

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]

Rule 5112

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[I*(a + b*ArcTa
n[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 5114

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
*(I/(I - c*x)))^2, 0]

Rule 5118

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[I*(a +
b*ArcTan[c*x])^p*(PolyLog[k + 1, u]/(2*c*d)), x] - Dist[b*p*(I/2), Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[k
+ 1, u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[u^2 - (1 -
2*(I/(I - c*x)))^2, 0]

Rule 6745

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 )^3}{x^2 (d+i c d x)} \, dx &=-\left ((i c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^3}{x (d+i c d x)} \, dx\right )+\frac {\int \frac {\left (a+b \tan ^{-1}(c x)\right )^3}{x^2} \, dx}{d}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d x}-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {(3 b c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x \left (1+c^2 x^2\right )} \, dx}{d}+\frac {\left (3 i b c^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d x}-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{2 d}+\frac {(3 i b c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x (i+c x)} \, dx}{d}-\frac {\left (3 b^2 c^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d x}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {\left (6 b^2 c^2\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 {\left (3 i b^3 c^2\right ) \int \frac {\text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{2 d}\\ &=-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d x}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {3 b^3 c \text {Li}_4\left (-1+\frac {2}{1+i c x}\right )}{4 d}+\frac {\left (3 i b^3 c^2\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d x}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{2 d}+\frac {3 b^3 c \text {Li}_3\left (-1+\frac {2}{1-i c x}\right )}{2 d}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {3 b^3 c \text {Li}_4\left (-1+\frac {2}{1+i c x}\right )}{4 d}\\ \end {align*}

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Mathematica [A]
time = 0.91, size = 436, normalized size = 1.66 \begin {gather*} -\frac {\frac {2 a^3}{x}+2 a^3 c \text {ArcTan}(c x)+2 i a^3 c \log (x)-i a^3 c \log \left (1+c^2 x^2\right )+3 a^2 b c \left (2 \left (\text {ArcTan}(c x)^2+\text {ArcTan}(c x) \left (\frac {1}{c x}+i \log \left (1-e^{2 i \text {ArcTan}(c x)}\right )\right )-\log \left (\frac {c x}{\sqrt {1+c^2 x^2}}\right )\right )+\text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c x)}\right )\right )+6 i a b^2 c \left (-\frac {i \pi ^3}{24}+\text {ArcTan}(c x)^2-\frac {i \text {ArcTan}(c x)^2}{c x}+\text {ArcTan}(c x)^2 \log \left (1-e^{-2 i \text {ArcTan}(c x)}\right )+2 i \text {ArcTan}(c x) \log \left (1-e^{2 i \text {ArcTan}(c x)}\right )+i \text {ArcTan}(c x) \text {PolyLog}\left (2,e^{-2 i \text {ArcTan}(c x)}\right )+\text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c x)}\right )+\frac {1}{2} \text {PolyLog}\left (3,e^{-2 i \text {ArcTan}(c x)}\right )\right )+2 i b^3 c \left (\frac {\pi ^3}{8}-\frac {i \pi ^4}{64}-\text {ArcTan}(c x)^3-\frac {i \text {ArcTan}(c x)^3}{c x}+3 i \text {ArcTan}(c x)^2 \log \left (1-e^{-2 i \text {ArcTan}(c x)}\right )+\text {ArcTan}(c x)^3 \log \left (1-e^{-2 i \text {ArcTan}(c x)}\right )+\frac {3}{2} i \text {ArcTan}(c x) (2 i+\text {ArcTan}(c x)) \text {PolyLog}\left (2,e^{-2 i \text {ArcTan}(c x)}\right )+\frac {3}{2} (i+\text {ArcTan}(c x)) \text {PolyLog}\left (3,e^{-2 i \text {ArcTan}(c x)}\right )-\frac {3}{4} i \text {PolyLog}\left (4,e^{-2 i \text {ArcTan}(c x)}\right )\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.61, size = 11097, normalized size = 42.19

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(11097\)
default	\(\text {Expression too large to display}\)	\(11097\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*(I*c*log(I*c*x + 1)/d - I*c*log(x)/d - 1/(d*x)) - 1/512*(64*b^3*c*x*arctan(c*x)^4 - 4*b^3*c*x*log(c^2*x^2
+ 1)^4 + 64*b^3*arctan(c*x)^3 - 48*b^3*arctan(c*x)*log(c^2*x^2 + 1)^2 - 8*(-2*I*b^3*c*x*arctan(c*x) + I*b^3)*l
og(c^2*x^2 + 1)^3 - (48*b^3*c*arctan(c*x)^4/d - 6144*b^3*c^3*integrate(1/64*x^3*arctan(c*x)^2*log(c^2*x^2 + 1)
/(c^2*d*x^4 + d*x^2), x) - 3*b^3*c*log(c^2*x^2 + 1)^4/d + 3072*b^3*c^2*integrate(1/64*x^2*arctan(c*x)*log(c^2*
x^2 + 1)^2/(c^2*d*x^4 + d*x^2), x) - 12288*b^3*c^2*integrate(1/64*x^2*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*d*x^4
+ d*x^2), x) - 6144*b^3*c*integrate(1/64*x*arctan(c*x)^2*log(c^2*x^2 + 1)/(c^2*d*x^4 + d*x^2), x) + 512*b^3*c*
integrate(1/64*x*log(c^2*x^2 + 1)^3/(c^2*d*x^4 + d*x^2), x) + 12288*b^3*c*integrate(1/64*x*arctan(c*x)^2/(c^2*
d*x^4 + d*x^2), x) - 3072*b^3*c*integrate(1/64*x*log(c^2*x^2 + 1)^2/(c^2*d*x^4 + d*x^2), x) + 28672*b^3*integr
ate(1/64*arctan(c*x)^3/(c^2*d*x^4 + d*x^2), x) + 3072*b^3*integrate(1/64*arctan(c*x)*log(c^2*x^2 + 1)^2/(c^2*d
*x^4 + d*x^2), x) + 98304*a*b^2*integrate(1/64*arctan(c*x)^2/(c^2*d*x^4 + d*x^2), x) + 98304*a^2*b*integrate(1
/64*arctan(c*x)/(c^2*d*x^4 + d*x^2), x))*d*x - 64*I*(192*b^3*c^3*integrate(1/64*x^3*arctan(c*x)^3/(c^2*d*x^4 +
d*x^2), x) + 48*b^3*c^3*integrate(1/64*x^3*arctan(c*x)*log(c^2*x^2 + 1)^2/(c^2*d*x^4 + d*x^2), x) + b^3*c*arc
tan(c*x)^3/d + 96*b^3*c^2*integrate(1/64*x^2*arctan(c*x)^2*log(c^2*x^2 + 1)/(c^2*d*x^4 + d*x^2), x) + 24*b^3*c
^2*integrate(1/64*x^2*log(c^2*x^2 + 1)^3/(c^2*d*x^4 + d*x^2), x) - 48*b^3*c^2*integrate(1/64*x^2*log(c^2*x^2 +
1)^2/(c^2*d*x^4 + d*x^2), x) - 448*b^3*c*integrate(1/64*x*arctan(c*x)^3/(c^2*d*x^4 + d*x^2), x) - 48*b^3*c*in
tegrate(1/64*x*arctan(c*x)*log(c^2*x^2 + 1)^2/(c^2*d*x^4 + d*x^2), x) - 1536*a*b^2*c*integrate(1/64*x*arctan(c
*x)^2/(c^2*d*x^4 + d*x^2), x) + 192*b^3*c*integrate(1/64*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*d*x^4 + d*x^2), x
) - 1536*a^2*b*c*integrate(1/64*x*arctan(c*x)/(c^2*d*x^4 + d*x^2), x) - 96*b^3*integrate(1/64*arctan(c*x)^2*lo
g(c^2*x^2 + 1)/(c^2*d*x^4 + d*x^2), x) + 8*b^3*integrate(1/64*log(c^2*x^2 + 1)^3/(c^2*d*x^4 + d*x^2), x))*d*x
- 32*(-2*I*b^3*c*x*arctan(c*x)^3 - 3*I*b^3*arctan(c*x)^2)*log(c^2*x^2 + 1))/(d*x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(b^3*c*x*log(2*c*x/(c*x - I))*log(-(c*x + I)/(c*x - I))^3 + 3*b^3*c*x*dilog(-2*c*x/(c*x - I) + 1)*log(-(c
*x + I)/(c*x - I))^2 - 6*b^3*c*x*log(-(c*x + I)/(c*x - I))*polylog(3, -(c*x + I)/(c*x - I)) - I*b^3*log(-(c*x
+ I)/(c*x - I))^3 + 6*b^3*c*x*polylog(4, -(c*x + I)/(c*x - I)) - 8*d*x*integral(1/4*(-4*I*a^3*c*x + 4*a^3 - 3*
(a*b^2 + (-I*a*b^2 + b^3)*c*x)*log(-(c*x + I)/(c*x - I))^2 + 6*(a^2*b*c*x + I*a^2*b)*log(-(c*x + I)/(c*x - I))
)/(c^2*d*x^4 + d*x^2), x))/(d*x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \left (\int \frac {a^{3}}{c x^{3} - i x^{2}}\, dx + \int \frac {b^{3} \operatorname {atan}^{3}{\left (c x \right )}}{c x^{3} - i x^{2}}\, dx + \int \frac {3 a b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{c x^{3} - i x^{2}}\, dx + \int \frac {3 a^{2} b \operatorname {atan}{\left (c x \right )}}{c x^{3} - i x^{2}}\, dx\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3}{x^2\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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