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

Optimal. Leaf size=182 \[ \frac {3 b^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c d^2 (-c x+i)}+\frac {3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^2 (-c x+i)}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2}{4 c d^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{c d^2 (1+i c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^2}-\frac {3 i b^3}{4 c d^2 (-c x+i)}+\frac {3 i b^3 \tan ^{-1}(c x)}{4 c d^2} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4864, 4862, 627, 44, 203, 4884} \[ \frac {3 b^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c d^2 (-c x+i)}+\frac {3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^2 (-c x+i)}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2}{4 c d^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{c d^2 (1+i c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^2}-\frac {3 i b^3}{4 c d^2 (-c x+i)}+\frac {3 i b^3 \tan ^{-1}(c x)}{4 c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 4864

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

Rule 4884

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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tan ^{-1}(c x)\right )^3}{(d+i c d x)^2} \, dx &=\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{c d^2 (1+i c x)}-\frac {(3 i b) \int \left (-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d (-i+c x)^2}+\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d \left (1+c^2 x^2\right )}\right ) \, dx}{d}\\ &=\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{c d^2 (1+i c x)}+\frac {(3 i b) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^2} \, dx}{2 d^2}-\frac {(3 i b) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{2 d^2}\\ &=\frac {3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^2 (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{c d^2 (1+i c x)}+\frac {\left (3 i b^2\right ) \int \left (-\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 (-i+c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{d^2}\\ &=\frac {3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^2 (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{c d^2 (1+i c x)}+\frac {\left (3 b^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{2 d^2}-\frac {\left (3 b^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{2 d^2}\\ &=\frac {3 b^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c d^2 (i-c x)}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2}{4 c d^2}+\frac {3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^2 (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{c d^2 (1+i c x)}+\frac {\left (3 b^3\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{2 d^2}\\ &=\frac {3 b^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c d^2 (i-c x)}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2}{4 c d^2}+\frac {3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^2 (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{c d^2 (1+i c x)}+\frac {\left (3 b^3\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{2 d^2}\\ &=\frac {3 b^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c d^2 (i-c x)}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2}{4 c d^2}+\frac {3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^2 (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{c d^2 (1+i c x)}+\frac {\left (3 b^3\right ) \int \left (-\frac {i}{2 (-i+c x)^2}+\frac {i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{2 d^2}\\ &=-\frac {3 i b^3}{4 c d^2 (i-c x)}+\frac {3 b^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c d^2 (i-c x)}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2}{4 c d^2}+\frac {3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^2 (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{c d^2 (1+i c x)}+\frac {\left (3 i b^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{4 d^2}\\ &=-\frac {3 i b^3}{4 c d^2 (i-c x)}+\frac {3 i b^3 \tan ^{-1}(c x)}{4 c d^2}+\frac {3 b^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c d^2 (i-c x)}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2}{4 c d^2}+\frac {3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^2 (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{c d^2 (1+i c x)}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 121, normalized size = 0.66 \[ \frac {4 a^3+3 i b \left (-2 a^2+2 i a b+b^2\right ) (c x+i) \tan ^{-1}(c x)-6 i a^2 b-3 b^2 (b+2 i a) (c x+i) \tan ^{-1}(c x)^2-6 a b^2+2 b^3 (1-i c x) \tan ^{-1}(c x)^3+3 i b^3}{4 c d^2 (c x-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^3 - (6*I)*a^2*b - 6*a*b^2 + (3*I)*b^3 + (3*I)*b*(-2*a^2 + (2*I)*a*b + b^2)*(I + c*x)*ArcTan[c*x] - 3*b^2*
((2*I)*a + b)*(I + c*x)*ArcTan[c*x]^2 + 2*b^3*(1 - I*c*x)*ArcTan[c*x]^3)/(4*c*d^2*(-I + c*x))

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fricas [A]  time = 0.67, size = 176, normalized size = 0.97 \[ -\frac {{\left (b^{3} c x + i \, b^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{3} - 16 \, a^{3} + 24 i \, a^{2} b + 24 \, a b^{2} - 12 i \, b^{3} + {\left (6 \, a b^{2} - 3 i \, b^{3} + 3 \, {\left (-2 i \, a b^{2} - b^{3}\right )} c x\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - {\left (12 i \, a^{2} b + 12 \, a b^{2} - 6 i \, b^{3} + 6 \, {\left (2 \, a^{2} b - 2 i \, a b^{2} - b^{3}\right )} c x\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{16 \, c^{2} d^{2} x - 16 i \, c d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((b^3*c*x + I*b^3)*log(-(c*x + I)/(c*x - I))^3 - 16*a^3 + 24*I*a^2*b + 24*a*b^2 - 12*I*b^3 + (6*a*b^2 - 3*I*b
^3 + 3*(-2*I*a*b^2 - b^3)*c*x)*log(-(c*x + I)/(c*x - I))^2 - (12*I*a^2*b + 12*a*b^2 - 6*I*b^3 + 6*(2*a^2*b - 2
*I*a*b^2 - b^3)*c*x)*log(-(c*x + I)/(c*x - I)))/(16*c^2*d^2*x - 16*I*c*d^2)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.89, size = 551, normalized size = 3.03 \[ \frac {3 i b^{3}}{4 c \,d^{2} \left (c x -i\right )}-\frac {3 i a^{2} b \arctan \left (c x \right )}{2 c \,d^{2}}-\frac {b^{3} \arctan \left (c x \right )^{3}}{2 c \,d^{2} \left (c x -i\right )}+\frac {i b^{3} \arctan \left (c x \right )^{3}}{c \,d^{2} \left (i c x +1\right )}-\frac {3 b^{3} \arctan \left (c x \right )^{2} x}{4 d^{2} \left (c x -i\right )}+\frac {3 i a \,b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{4 c \,d^{2}}-\frac {3 b^{3} \arctan \left (c x \right )}{4 c \,d^{2} \left (c x -i\right )}+\frac {i a^{3}}{c \,d^{2} \left (i c x +1\right )}+\frac {3 i a^{2} b \arctan \left (c x \right )}{c \,d^{2} \left (i c x +1\right )}+\frac {3 i b^{3} \arctan \left (c x \right ) x}{4 d^{2} \left (c x -i\right )}+\frac {3 a \,b^{2} \arctan \left (c x \right ) \ln \left (c x +i\right )}{2 c \,d^{2}}-\frac {3 a \,b^{2} \arctan \left (c x \right ) \ln \left (c x -i\right )}{2 c \,d^{2}}-\frac {3 i b^{3} \arctan \left (c x \right )^{2}}{4 c \,d^{2} \left (c x -i\right )}-\frac {3 i a \,b^{2} \arctan \left (c x \right )}{c \,d^{2} \left (c x -i\right )}-\frac {3 i a \,b^{2} \ln \left (c x +i\right )^{2}}{8 c \,d^{2}}-\frac {3 a \,b^{2} \arctan \left (c x \right )}{2 c \,d^{2}}-\frac {3 a \,b^{2}}{2 c \,d^{2} \left (c x -i\right )}-\frac {3 i a^{2} b}{2 c \,d^{2} \left (c x -i\right )}-\frac {3 i a \,b^{2} \ln \left (-\frac {i \left (-c x +i\right )}{2}\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{4 c \,d^{2}}-\frac {i b^{3} \arctan \left (c x \right )^{3} x}{2 d^{2} \left (c x -i\right )}+\frac {3 i a \,b^{2} \ln \left (-\frac {i \left (-c x +i\right )}{2}\right ) \ln \left (c x +i\right )}{4 c \,d^{2}}+\frac {3 i a \,b^{2} \arctan \left (c x \right )^{2}}{c \,d^{2} \left (i c x +1\right )}-\frac {3 i a \,b^{2} \ln \left (c x -i\right )^{2}}{8 c \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*I/c*b^3/d^2/(c*x-I)-3/2*I/c*a^2*b/d^2*arctan(c*x)-1/2/c*b^3/d^2/(c*x-I)*arctan(c*x)^3+I/c*b^3/d^2/(1+I*c*x
)*arctan(c*x)^3-3/4*b^3/d^2/(c*x-I)*arctan(c*x)^2*x+3/4*I/c*a*b^2/d^2*ln(c*x-I)*ln(-1/2*I*(I+c*x))-3/4/c*b^3/d
^2/(c*x-I)*arctan(c*x)+I/c*a^3/d^2/(1+I*c*x)+3*I/c*a^2*b/d^2/(1+I*c*x)*arctan(c*x)+3/4*I*b^3/d^2/(c*x-I)*arcta
n(c*x)*x+3/2/c*a*b^2/d^2*arctan(c*x)*ln(I+c*x)-3/2/c*a*b^2/d^2*arctan(c*x)*ln(c*x-I)-3/4*I/c*b^3/d^2/(c*x-I)*a
rctan(c*x)^2-3*I/c*a*b^2/d^2*arctan(c*x)/(c*x-I)-3/8*I/c*a*b^2/d^2*ln(I+c*x)^2-3/2/c*a*b^2/d^2*arctan(c*x)-3/2
/c*a*b^2/d^2/(c*x-I)-3/2*I/c*a^2*b/d^2/(c*x-I)-3/4*I/c*a*b^2/d^2*ln(-1/2*I*(-c*x+I))*ln(-1/2*I*(I+c*x))-1/2*I*
b^3/d^2/(c*x-I)*arctan(c*x)^3*x+3/4*I/c*a*b^2/d^2*ln(-1/2*I*(-c*x+I))*ln(I+c*x)+3*I/c*a*b^2/d^2/(1+I*c*x)*arct
an(c*x)^2-3/8*I/c*a*b^2/d^2*ln(c*x-I)^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 37.13, size = 627, normalized size = 3.45 \[ - \frac {3 i b \left (a \left (1 - i\right ) - b\right ) \left (a \left (1 - i\right ) - i b\right ) \log {\left (- \frac {3 b \left (a \left (1 - i\right ) - b\right ) \left (a \left (1 - i\right ) - i b\right )}{c} + x \left (- 6 a^{2} b + 6 i a b^{2} + 3 b^{3}\right ) \right )}}{8 c d^{2}} + \frac {3 i b \left (a \left (1 - i\right ) - b\right ) \left (a \left (1 - i\right ) - i b\right ) \log {\left (\frac {3 b \left (a \left (1 - i\right ) - b\right ) \left (a \left (1 - i\right ) - i b\right )}{c} + x \left (- 6 a^{2} b + 6 i a b^{2} + 3 b^{3}\right ) \right )}}{8 c d^{2}} + \frac {\left (- i b^{3} c x + b^{3}\right ) \log {\left (- i c x + 1 \right )}^{3}}{16 i c^{2} d^{2} x + 16 c d^{2}} + \frac {\left (i b^{3} c x - b^{3}\right ) \log {\left (i c x + 1 \right )}^{3}}{16 i c^{2} d^{2} x + 16 c d^{2}} + \frac {\left (- 6 a b^{2} c x - 6 i a b^{2} + 3 i b^{3} c x \log {\left (i c x + 1 \right )} + 3 i b^{3} c x - 3 b^{3} \log {\left (i c x + 1 \right )} - 3 b^{3}\right ) \log {\left (- i c x + 1 \right )}^{2}}{16 i c^{2} d^{2} x + 16 c d^{2}} + \frac {\left (- 24 a^{2} b + 12 a b^{2} c x \log {\left (i c x + 1 \right )} + 12 i a b^{2} \log {\left (i c x + 1 \right )} + 24 i a b^{2} - 3 i b^{3} c x \log {\left (i c x + 1 \right )}^{2} - 6 i b^{3} c x \log {\left (i c x + 1 \right )} + 3 b^{3} \log {\left (i c x + 1 \right )}^{2} + 6 b^{3} \log {\left (i c x + 1 \right )} + 12 b^{3}\right ) \log {\left (- i c x + 1 \right )}}{16 i c^{2} d^{2} x + 16 c d^{2}} + \frac {\left (6 a^{2} b - 6 i a b^{2} - 3 b^{3}\right ) \log {\left (i c x + 1 \right )}}{4 i c^{2} d^{2} x + 4 c d^{2}} + \frac {\left (6 a b^{2} c x + 6 i a b^{2} - 3 i b^{3} c x + 3 b^{3}\right ) \log {\left (i c x + 1 \right )}^{2}}{- 16 i c^{2} d^{2} x - 16 c d^{2}} - \frac {- 4 a^{3} + 6 i a^{2} b + 6 a b^{2} - 3 i b^{3}}{4 c^{2} d^{2} x - 4 i c d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*I*b*(a*(1 - I) - b)*(a*(1 - I) - I*b)*log(-3*b*(a*(1 - I) - b)*(a*(1 - I) - I*b)/c + x*(-6*a**2*b + 6*I*a*b
**2 + 3*b**3))/(8*c*d**2) + 3*I*b*(a*(1 - I) - b)*(a*(1 - I) - I*b)*log(3*b*(a*(1 - I) - b)*(a*(1 - I) - I*b)/
c + x*(-6*a**2*b + 6*I*a*b**2 + 3*b**3))/(8*c*d**2) + (-I*b**3*c*x + b**3)*log(-I*c*x + 1)**3/(16*I*c**2*d**2*
x + 16*c*d**2) + (I*b**3*c*x - b**3)*log(I*c*x + 1)**3/(16*I*c**2*d**2*x + 16*c*d**2) + (-6*a*b**2*c*x - 6*I*a
*b**2 + 3*I*b**3*c*x*log(I*c*x + 1) + 3*I*b**3*c*x - 3*b**3*log(I*c*x + 1) - 3*b**3)*log(-I*c*x + 1)**2/(16*I*
c**2*d**2*x + 16*c*d**2) + (-24*a**2*b + 12*a*b**2*c*x*log(I*c*x + 1) + 12*I*a*b**2*log(I*c*x + 1) + 24*I*a*b*
*2 - 3*I*b**3*c*x*log(I*c*x + 1)**2 - 6*I*b**3*c*x*log(I*c*x + 1) + 3*b**3*log(I*c*x + 1)**2 + 6*b**3*log(I*c*
x + 1) + 12*b**3)*log(-I*c*x + 1)/(16*I*c**2*d**2*x + 16*c*d**2) + (6*a**2*b - 6*I*a*b**2 - 3*b**3)*log(I*c*x
+ 1)/(4*I*c**2*d**2*x + 4*c*d**2) + (6*a*b**2*c*x + 6*I*a*b**2 - 3*I*b**3*c*x + 3*b**3)*log(I*c*x + 1)**2/(-16
*I*c**2*d**2*x - 16*c*d**2) - (-4*a**3 + 6*I*a**2*b + 6*a*b**2 - 3*I*b**3)/(4*c**2*d**2*x - 4*I*c*d**2)

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