3.118 \(\int \frac {(a+b \tan ^{-1}(c x))^2}{(1+i c x)^4} \, dx\)

Optimal. Leaf size=207 \[ \frac {i b \left (a+b \tan ^{-1}(c x)\right )}{12 c (-c x+i)}-\frac {b \left (a+b \tan ^{-1}(c x)\right )}{12 c (-c x+i)^2}-\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{9 c (-c x+i)^3}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{24 c}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{3 c (1+i c x)^3}+\frac {11 b^2}{144 c (-c x+i)}+\frac {5 i b^2}{144 c (-c x+i)^2}-\frac {b^2}{54 c (-c x+i)^3}-\frac {11 b^2 \tan ^{-1}(c x)}{144 c} \]

[Out]

-1/54*b^2/c/(I-c*x)^3+5/144*I*b^2/c/(I-c*x)^2+11/144*b^2/c/(I-c*x)-11/144*b^2*arctan(c*x)/c-1/9*I*b*(a+b*arcta
n(c*x))/c/(I-c*x)^3-1/12*b*(a+b*arctan(c*x))/c/(I-c*x)^2+1/12*I*b*(a+b*arctan(c*x))/c/(I-c*x)-1/24*I*(a+b*arct
an(c*x))^2/c+1/3*I*(a+b*arctan(c*x))^2/c/(1+I*c*x)^3

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Rubi [A]  time = 0.22, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4864, 4862, 627, 44, 203, 4884} \[ \frac {i b \left (a+b \tan ^{-1}(c x)\right )}{12 c (-c x+i)}-\frac {b \left (a+b \tan ^{-1}(c x)\right )}{12 c (-c x+i)^2}-\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{9 c (-c x+i)^3}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{24 c}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{3 c (1+i c x)^3}+\frac {11 b^2}{144 c (-c x+i)}+\frac {5 i b^2}{144 c (-c x+i)^2}-\frac {b^2}{54 c (-c x+i)^3}-\frac {11 b^2 \tan ^{-1}(c x)}{144 c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])^2/(1 + I*c*x)^4,x]

[Out]

-b^2/(54*c*(I - c*x)^3) + (((5*I)/144)*b^2)/(c*(I - c*x)^2) + (11*b^2)/(144*c*(I - c*x)) - (11*b^2*ArcTan[c*x]
)/(144*c) - ((I/9)*b*(a + b*ArcTan[c*x]))/(c*(I - c*x)^3) - (b*(a + b*ArcTan[c*x]))/(12*c*(I - c*x)^2) + ((I/1
2)*b*(a + b*ArcTan[c*x]))/(c*(I - c*x)) - ((I/24)*(a + b*ArcTan[c*x])^2)/c + ((I/3)*(a + b*ArcTan[c*x])^2)/(c*
(1 + I*c*x)^3)

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 )^2}{(1+i c x)^4} \, dx &=\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{3 c (1+i c x)^3}-\frac {1}{3} (2 i b) \int \left (\frac {a+b \tan ^{-1}(c x)}{2 (-i+c x)^4}+\frac {i \left (a+b \tan ^{-1}(c x)\right )}{4 (-i+c x)^3}-\frac {a+b \tan ^{-1}(c x)}{8 (-i+c x)^2}+\frac {a+b \tan ^{-1}(c x)}{8 \left (1+c^2 x^2\right )}\right ) \, dx\\ &=\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{3 c (1+i c x)^3}+\frac {1}{12} (i b) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx-\frac {1}{12} (i b) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx-\frac {1}{3} (i b) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^4} \, dx+\frac {1}{6} b \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx\\ &=-\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{9 c (i-c x)^3}-\frac {b \left (a+b \tan ^{-1}(c x)\right )}{12 c (i-c x)^2}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{12 c (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{24 c}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{3 c (1+i c x)^3}+\frac {1}{12} \left (i b^2\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx-\frac {1}{9} \left (i b^2\right ) \int \frac {1}{(-i+c x)^3 \left (1+c^2 x^2\right )} \, dx+\frac {1}{12} b^2 \int \frac {1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{9 c (i-c x)^3}-\frac {b \left (a+b \tan ^{-1}(c x)\right )}{12 c (i-c x)^2}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{12 c (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{24 c}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{3 c (1+i c x)^3}+\frac {1}{12} \left (i b^2\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx-\frac {1}{9} \left (i b^2\right ) \int \frac {1}{(-i+c x)^4 (i+c x)} \, dx+\frac {1}{12} b^2 \int \frac {1}{(-i+c x)^3 (i+c x)} \, dx\\ &=-\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{9 c (i-c x)^3}-\frac {b \left (a+b \tan ^{-1}(c x)\right )}{12 c (i-c x)^2}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{12 c (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{24 c}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{3 c (1+i c x)^3}+\frac {1}{12} \left (i b^2\right ) \int \left (-\frac {i}{2 (-i+c x)^2}+\frac {i}{2 \left (1+c^2 x^2\right )}\right ) \, dx-\frac {1}{9} \left (i b^2\right ) \int \left (-\frac {i}{2 (-i+c x)^4}+\frac {1}{4 (-i+c x)^3}+\frac {i}{8 (-i+c x)^2}-\frac {i}{8 \left (1+c^2 x^2\right )}\right ) \, dx+\frac {1}{12} b^2 \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\\ &=-\frac {b^2}{54 c (i-c x)^3}+\frac {5 i b^2}{144 c (i-c x)^2}+\frac {11 b^2}{144 c (i-c x)}-\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{9 c (i-c x)^3}-\frac {b \left (a+b \tan ^{-1}(c x)\right )}{12 c (i-c x)^2}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{12 c (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{24 c}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{3 c (1+i c x)^3}-\frac {1}{72} b^2 \int \frac {1}{1+c^2 x^2} \, dx-\frac {1}{48} b^2 \int \frac {1}{1+c^2 x^2} \, dx-\frac {1}{24} b^2 \int \frac {1}{1+c^2 x^2} \, dx\\ &=-\frac {b^2}{54 c (i-c x)^3}+\frac {5 i b^2}{144 c (i-c x)^2}+\frac {11 b^2}{144 c (i-c x)}-\frac {11 b^2 \tan ^{-1}(c x)}{144 c}-\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{9 c (i-c x)^3}-\frac {b \left (a+b \tan ^{-1}(c x)\right )}{12 c (i-c x)^2}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{12 c (i-c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{24 c}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{3 c (1+i c x)^3}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 155, normalized size = 0.75 \[ -\frac {144 a^2+12 a b \left (3 i c^2 x^2+9 c x-10 i\right )+3 b (c x+i) \tan ^{-1}(c x) \left (12 a \left (i c^2 x^2+4 c x-7 i\right )+b \left (11 c^2 x^2-32 i c x-29\right )\right )+b^2 \left (33 c^2 x^2-81 i c x-56\right )+18 b^2 \left (i c^3 x^3+3 c^2 x^2-3 i c x+7\right ) \tan ^{-1}(c x)^2}{432 c (c x-i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x])^2/(1 + I*c*x)^4,x]

[Out]

-1/432*(144*a^2 + 12*a*b*(-10*I + 9*c*x + (3*I)*c^2*x^2) + b^2*(-56 - (81*I)*c*x + 33*c^2*x^2) + 3*b*(I + c*x)
*(12*a*(-7*I + 4*c*x + I*c^2*x^2) + b*(-29 - (32*I)*c*x + 11*c^2*x^2))*ArcTan[c*x] + 18*b^2*(7 - (3*I)*c*x + 3
*c^2*x^2 + I*c^3*x^3)*ArcTan[c*x]^2)/(c*(-I + c*x)^3)

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fricas [A]  time = 0.80, size = 206, normalized size = 1.00 \[ -\frac {6 \, {\left (12 i \, a b + 11 \, b^{2}\right )} c^{2} x^{2} + {\left (216 \, a b - 162 i \, b^{2}\right )} c x - {\left (9 i \, b^{2} c^{3} x^{3} + 27 \, b^{2} c^{2} x^{2} - 27 i \, b^{2} c x + 63 \, b^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} + 288 \, a^{2} - 240 i \, a b - 112 \, b^{2} - {\left ({\left (36 \, a b - 33 i \, b^{2}\right )} c^{3} x^{3} - 9 \, {\left (12 i \, a b + 7 \, b^{2}\right )} c^{2} x^{2} - {\left (108 \, a b + 9 i \, b^{2}\right )} c x - 252 i \, a b - 87 \, b^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{864 \, c^{4} x^{3} - 2592 i \, c^{3} x^{2} - 2592 \, c^{2} x + 864 i \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^2/(1+I*c*x)^4,x, algorithm="fricas")

[Out]

-(6*(12*I*a*b + 11*b^2)*c^2*x^2 + (216*a*b - 162*I*b^2)*c*x - (9*I*b^2*c^3*x^3 + 27*b^2*c^2*x^2 - 27*I*b^2*c*x
+ 63*b^2)*log(-(c*x + I)/(c*x - I))^2 + 288*a^2 - 240*I*a*b - 112*b^2 - ((36*a*b - 33*I*b^2)*c^3*x^3 - 9*(12*
I*a*b + 7*b^2)*c^2*x^2 - (108*a*b + 9*I*b^2)*c*x - 252*I*a*b - 87*b^2)*log(-(c*x + I)/(c*x - I)))/(864*c^4*x^3
- 2592*I*c^3*x^2 - 2592*c^2*x + 864*I*c)

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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))^2/(1+I*c*x)^4,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.09, size = 404, normalized size = 1.95 \[ -\frac {i b^{2} \ln \left (-\frac {i \left (-c x +i\right )}{2}\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{48 c}-\frac {i b^{2} \ln \left (c x +i\right )^{2}}{96 c}+\frac {b^{2} \arctan \left (c x \right ) \ln \left (c x +i\right )}{24 c}-\frac {b^{2} \arctan \left (c x \right ) \ln \left (c x -i\right )}{24 c}-\frac {b^{2} \arctan \left (c x \right )}{12 c \left (c x -i\right )^{2}}+\frac {i b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{48 c}+\frac {i b^{2} \arctan \left (c x \right )}{9 c \left (c x -i\right )^{3}}-\frac {11 b^{2} \arctan \left (c x \right )}{144 c}+\frac {b^{2}}{54 c \left (c x -i\right )^{3}}-\frac {11 b^{2}}{144 c \left (c x -i\right )}-\frac {i b^{2} \arctan \left (c x \right )}{12 c \left (c x -i\right )}+\frac {i a b}{9 c \left (c x -i\right )^{3}}+\frac {2 i a b \arctan \left (c x \right )}{3 c \left (i c x +1\right )^{3}}-\frac {i a b \arctan \left (c x \right )}{12 c}+\frac {5 i b^{2}}{144 c \left (c x -i\right )^{2}}+\frac {i b^{2} \arctan \left (c x \right )^{2}}{3 c \left (i c x +1\right )^{3}}-\frac {i b^{2} \ln \left (c x -i\right )^{2}}{96 c}+\frac {i a^{2}}{3 c \left (i c x +1\right )^{3}}-\frac {a b}{12 c \left (c x -i\right )^{2}}+\frac {i b^{2} \ln \left (-\frac {i \left (-c x +i\right )}{2}\right ) \ln \left (c x +i\right )}{48 c}-\frac {i a b}{12 c \left (c x -i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x))^2/(1+I*c*x)^4,x)

[Out]

-1/48*I/c*b^2*ln(-1/2*I*(-c*x+I))*ln(-1/2*I*(I+c*x))-1/96*I/c*b^2*ln(I+c*x)^2+1/24/c*b^2*arctan(c*x)*ln(I+c*x)
-1/24/c*b^2*arctan(c*x)*ln(c*x-I)-1/12/c*b^2*arctan(c*x)/(c*x-I)^2+1/48*I/c*b^2*ln(c*x-I)*ln(-1/2*I*(I+c*x))+1
/9*I/c*b^2*arctan(c*x)/(c*x-I)^3-11/144*b^2*arctan(c*x)/c+1/54/c*b^2/(c*x-I)^3-11/144/c*b^2/(c*x-I)-1/12*I/c*b
^2*arctan(c*x)/(c*x-I)+1/9*I/c*a*b/(c*x-I)^3+2/3*I/c*a*b/(1+I*c*x)^3*arctan(c*x)-1/12*I/c*a*b*arctan(c*x)+5/14
4*I/c*b^2/(c*x-I)^2+1/3*I/c*b^2/(1+I*c*x)^3*arctan(c*x)^2-1/96*I/c*b^2*ln(c*x-I)^2+1/3*I/c*a^2/(1+I*c*x)^3-1/1
2/c*a*b/(c*x-I)^2+1/48*I/c*b^2*ln(-1/2*I*(-c*x+I))*ln(I+c*x)-1/12*I/c*a*b/(c*x-I)

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maxima [A]  time = 0.43, size = 181, normalized size = 0.87 \[ -\frac {{\left (36 i \, a b + 33 \, b^{2}\right )} c^{2} x^{2} + 27 \, {\left (4 \, a b - 3 i \, b^{2}\right )} c x + {\left (18 i \, b^{2} c^{3} x^{3} + 54 \, b^{2} c^{2} x^{2} - 54 i \, b^{2} c x + 126 \, b^{2}\right )} \arctan \left (c x\right )^{2} + 144 \, a^{2} - 120 i \, a b - 56 \, b^{2} + {\left ({\left (36 i \, a b + 33 \, b^{2}\right )} c^{3} x^{3} + 9 \, {\left (12 \, a b - 7 i \, b^{2}\right )} c^{2} x^{2} + {\left (-108 i \, a b + 9 \, b^{2}\right )} c x + 252 \, a b - 87 i \, b^{2}\right )} \arctan \left (c x\right )}{432 \, c^{4} x^{3} - 1296 i \, c^{3} x^{2} - 1296 \, c^{2} x + 432 i \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^2/(1+I*c*x)^4,x, algorithm="maxima")

[Out]

-((36*I*a*b + 33*b^2)*c^2*x^2 + 27*(4*a*b - 3*I*b^2)*c*x + (18*I*b^2*c^3*x^3 + 54*b^2*c^2*x^2 - 54*I*b^2*c*x +
126*b^2)*arctan(c*x)^2 + 144*a^2 - 120*I*a*b - 56*b^2 + ((36*I*a*b + 33*b^2)*c^3*x^3 + 9*(12*a*b - 7*I*b^2)*c
^2*x^2 + (-108*I*a*b + 9*b^2)*c*x + 252*a*b - 87*I*b^2)*arctan(c*x))/(432*c^4*x^3 - 1296*I*c^3*x^2 - 1296*c^2*
x + 432*I*c)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (1+c\,x\,1{}\mathrm {i}\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))^2/(c*x*1i + 1)^4,x)

[Out]

int((a + b*atan(c*x))^2/(c*x*1i + 1)^4, x)

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sympy [B]  time = 90.24, size = 549, normalized size = 2.65 \[ \frac {b \left (12 a - 11 i b\right ) \log {\left (- \frac {b \left (12 a - 11 i b\right )}{c} + x \left (12 i a b + 11 b^{2}\right ) \right )}}{288 c} - \frac {b \left (12 a - 11 i b\right ) \log {\left (\frac {b \left (12 a - 11 i b\right )}{c} + x \left (12 i a b + 11 b^{2}\right ) \right )}}{288 c} + \frac {\left (48 a b + 3 b^{2} c^{3} x^{3} \log {\left (i c x + 1 \right )} - 9 i b^{2} c^{2} x^{2} \log {\left (i c x + 1 \right )} + 6 i b^{2} c^{2} x^{2} - 9 b^{2} c x \log {\left (i c x + 1 \right )} + 18 b^{2} c x - 21 i b^{2} \log {\left (i c x + 1 \right )} - 20 i b^{2}\right ) \log {\left (- i c x + 1 \right )}}{144 i c^{4} x^{3} + 432 c^{3} x^{2} - 432 i c^{2} x - 144 c} + \frac {\left (- b^{2} c^{3} x^{3} + 3 i b^{2} c^{2} x^{2} + 3 b^{2} c x + 7 i b^{2}\right ) \log {\left (- i c x + 1 \right )}^{2}}{96 i c^{4} x^{3} + 288 c^{3} x^{2} - 288 i c^{2} x - 96 c} + \frac {\left (24 i a b - 3 b^{2} c^{2} x^{2} + 9 i b^{2} c x + 10 b^{2}\right ) \log {\left (i c x + 1 \right )}}{72 c^{4} x^{3} - 216 i c^{3} x^{2} - 216 c^{2} x + 72 i c} + \frac {\left (- i b^{2} c^{3} x^{3} - 3 b^{2} c^{2} x^{2} + 3 i b^{2} c x - 7 b^{2}\right ) \log {\left (i c x + 1 \right )}^{2}}{- 96 c^{4} x^{3} + 288 i c^{3} x^{2} + 288 c^{2} x - 96 i c} - \frac {- 144 a^{2} + 120 i a b + 56 b^{2} + x^{2} \left (- 36 i a b c^{2} - 33 b^{2} c^{2}\right ) + x \left (- 108 a b c + 81 i b^{2} c\right )}{- 432 c^{4} x^{3} + 1296 i c^{3} x^{2} + 1296 c^{2} x - 432 i c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))**2/(1+I*c*x)**4,x)

[Out]

b*(12*a - 11*I*b)*log(-b*(12*a - 11*I*b)/c + x*(12*I*a*b + 11*b**2))/(288*c) - b*(12*a - 11*I*b)*log(b*(12*a -
11*I*b)/c + x*(12*I*a*b + 11*b**2))/(288*c) + (48*a*b + 3*b**2*c**3*x**3*log(I*c*x + 1) - 9*I*b**2*c**2*x**2*
log(I*c*x + 1) + 6*I*b**2*c**2*x**2 - 9*b**2*c*x*log(I*c*x + 1) + 18*b**2*c*x - 21*I*b**2*log(I*c*x + 1) - 20*
I*b**2)*log(-I*c*x + 1)/(144*I*c**4*x**3 + 432*c**3*x**2 - 432*I*c**2*x - 144*c) + (-b**2*c**3*x**3 + 3*I*b**2
*c**2*x**2 + 3*b**2*c*x + 7*I*b**2)*log(-I*c*x + 1)**2/(96*I*c**4*x**3 + 288*c**3*x**2 - 288*I*c**2*x - 96*c)
+ (24*I*a*b - 3*b**2*c**2*x**2 + 9*I*b**2*c*x + 10*b**2)*log(I*c*x + 1)/(72*c**4*x**3 - 216*I*c**3*x**2 - 216*
c**2*x + 72*I*c) + (-I*b**2*c**3*x**3 - 3*b**2*c**2*x**2 + 3*I*b**2*c*x - 7*b**2)*log(I*c*x + 1)**2/(-96*c**4*
x**3 + 288*I*c**3*x**2 + 288*c**2*x - 96*I*c) - (-144*a**2 + 120*I*a*b + 56*b**2 + x**2*(-36*I*a*b*c**2 - 33*b
**2*c**2) + x*(-108*a*b*c + 81*I*b**2*c))/(-432*c**4*x**3 + 1296*I*c**3*x**2 + 1296*c**2*x - 432*I*c)

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