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]

-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)

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Rubi [A]  time = 0.222115, antiderivative size = 207, normalized size of antiderivative = 1., 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 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 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 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*}

Mathematica [A]  time = 0.2092, 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]

-(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)/(432*c*(-I + c*x)^3)

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Maple [B]  time = 0.075, size = 404, normalized size = 2. \begin{align*}{\frac{{\frac{i}{9}}ab}{c \left ( cx-i \right ) ^{3}}}+{\frac{{\frac{2\,i}{3}}ab\arctan \left ( cx \right ) }{c \left ( 1+icx \right ) ^{3}}}-{\frac{{b}^{2}\arctan \left ( cx \right ) \ln \left ( cx-i \right ) }{24\,c}}-{\frac{{b}^{2}\arctan \left ( cx \right ) }{12\,c \left ( cx-i \right ) ^{2}}}+{\frac{{\frac{i}{9}}{b}^{2}\arctan \left ( cx \right ) }{c \left ( cx-i \right ) ^{3}}}-{\frac{{\frac{i}{12}}ab}{c \left ( cx-i \right ) }}+{\frac{{b}^{2}\arctan \left ( cx \right ) \ln \left ( cx+i \right ) }{24\,c}}+{\frac{{b}^{2}}{54\,c \left ( cx-i \right ) ^{3}}}-{\frac{11\,{b}^{2}}{144\,c \left ( cx-i \right ) }}-{\frac{{\frac{i}{12}}{b}^{2}\arctan \left ( cx \right ) }{c \left ( cx-i \right ) }}-{\frac{11\,{b}^{2}\arctan \left ( cx \right ) }{144\,c}}-{\frac{{\frac{i}{48}}{b}^{2}\ln \left ( -{\frac{i}{2}} \left ( -cx+i \right ) \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{c}}+{\frac{{\frac{i}{3}}{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}}{c \left ( 1+icx \right ) ^{3}}}-{\frac{{\frac{i}{96}}{b}^{2} \left ( \ln \left ( cx+i \right ) \right ) ^{2}}{c}}-{\frac{{\frac{i}{12}}ab\arctan \left ( cx \right ) }{c}}+{\frac{{\frac{i}{48}}{b}^{2}\ln \left ( cx-i \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{c}}+{\frac{{\frac{i}{48}}{b}^{2}\ln \left ( -{\frac{i}{2}} \left ( -cx+i \right ) \right ) \ln \left ( cx+i \right ) }{c}}-{\frac{{\frac{i}{96}}{b}^{2} \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{c}}-{\frac{ab}{12\,c \left ( cx-i \right ) ^{2}}}+{\frac{{\frac{i}{3}}{a}^{2}}{c \left ( 1+icx \right ) ^{3}}}+{\frac{{\frac{5\,i}{144}}{b}^{2}}{c \left ( cx-i \right ) ^{2}}} \end{align*}

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]

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

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Maxima [A]  time = 1.32133, size = 248, normalized size = 1.2 \begin{align*} \frac{3 \,{\left (-12 i \, a b - 11 \, b^{2}\right )} c^{2} x^{2} -{\left (108 \, a b - 81 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 (3 \,{\left (-12 i \, a b - 11 \, b^{2}\right )} c^{3} x^{3} -{\left (108 \, a b - 63 i \, b^{2}\right )} c^{2} x^{2} + 9 \,{\left (12 i \, a b - 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} \end{align*}

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]

(3*(-12*I*a*b - 11*b^2)*c^2*x^2 - (108*a*b - 81*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 + (3*(-12*I*a*b - 11*b^2)*c^3*x^3 - (108*a*b - 63*I*b
^2)*c^2*x^2 + 9*(12*I*a*b - 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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Fricas [A]  time = 2.29416, size = 502, normalized size = 2.43 \begin{align*} \frac{{\left (-72 i \, a b - 66 \, b^{2}\right )} c^{2} x^{2} - 54 \,{\left (4 \, a b - 3 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 (3 \,{\left (12 \, a b - 11 i \, b^{2}\right )} c^{3} x^{3} +{\left (-108 i \, a b - 63 \, b^{2}\right )} c^{2} x^{2} - 9 \,{\left (12 \, a b + 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} \end{align*}

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]

((-72*I*a*b - 66*b^2)*c^2*x^2 - 54*(4*a*b - 3*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 + (3*(12*a*b - 11*I*b^2)*c^3*x^3 + (-108*I
*a*b - 63*b^2)*c^2*x^2 - 9*(12*a*b + 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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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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 )}^{2}}{{\left (i \, c x + 1\right )}^{4}}\,{d x} \end{align*}

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]

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