3.66 \(\int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=255 \[ -\frac {(11 A+4 i B) \cot ^2(c+d x)}{2 a^4 d}+\frac {5 (-13 B+35 i A) \cot (c+d x)}{16 a^4 d}-\frac {(11 A+4 i B) \log (\sin (c+d x))}{a^4 d}+\frac {5 (35 A+13 i B) \cot ^2(c+d x)}{48 a^4 d (1+i \tan (c+d x))}+\frac {(43 A+17 i B) \cot ^2(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {5 x (-13 B+35 i A)}{16 a^4}+\frac {(2 A+i B) \cot ^2(c+d x)}{6 a d (a+i a \tan (c+d x))^3}+\frac {(A+i B) \cot ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4} \]

[Out]

5/16*(35*I*A-13*B)*x/a^4+5/16*(35*I*A-13*B)*cot(d*x+c)/a^4/d-1/2*(11*A+4*I*B)*cot(d*x+c)^2/a^4/d-(11*A+4*I*B)*
ln(sin(d*x+c))/a^4/d+1/48*(43*A+17*I*B)*cot(d*x+c)^2/a^4/d/(1+I*tan(d*x+c))^2+5/48*(35*A+13*I*B)*cot(d*x+c)^2/
a^4/d/(1+I*tan(d*x+c))+1/8*(A+I*B)*cot(d*x+c)^2/d/(a+I*a*tan(d*x+c))^4+1/6*(2*A+I*B)*cot(d*x+c)^2/a/d/(a+I*a*t
an(d*x+c))^3

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Rubi [A]  time = 0.79, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3596, 3529, 3531, 3475} \[ -\frac {(11 A+4 i B) \cot ^2(c+d x)}{2 a^4 d}+\frac {5 (-13 B+35 i A) \cot (c+d x)}{16 a^4 d}-\frac {(11 A+4 i B) \log (\sin (c+d x))}{a^4 d}+\frac {5 (35 A+13 i B) \cot ^2(c+d x)}{48 a^4 d (1+i \tan (c+d x))}+\frac {(43 A+17 i B) \cot ^2(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {5 x (-13 B+35 i A)}{16 a^4}+\frac {(2 A+i B) \cot ^2(c+d x)}{6 a d (a+i a \tan (c+d x))^3}+\frac {(A+i B) \cot ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(5*((35*I)*A - 13*B)*x)/(16*a^4) + (5*((35*I)*A - 13*B)*Cot[c + d*x])/(16*a^4*d) - ((11*A + (4*I)*B)*Cot[c + d
*x]^2)/(2*a^4*d) - ((11*A + (4*I)*B)*Log[Sin[c + d*x]])/(a^4*d) + ((43*A + (17*I)*B)*Cot[c + d*x]^2)/(48*a^4*d
*(1 + I*Tan[c + d*x])^2) + (5*(35*A + (13*I)*B)*Cot[c + d*x]^2)/(48*a^4*d*(1 + I*Tan[c + d*x])) + ((A + I*B)*C
ot[c + d*x]^2)/(8*d*(a + I*a*Tan[c + d*x])^4) + ((2*A + I*B)*Cot[c + d*x]^2)/(6*a*d*(a + I*a*Tan[c + d*x])^3)

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3596

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
0]

Rubi steps

\begin {align*} \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx &=\frac {(A+i B) \cot ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {\int \frac {\cot ^3(c+d x) (2 a (5 A+i B)-6 a (i A-B) \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx}{8 a^2}\\ &=\frac {(A+i B) \cot ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(2 A+i B) \cot ^2(c+d x)}{6 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\cot ^3(c+d x) \left (4 a^2 (23 A+7 i B)-40 a^2 (2 i A-B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{48 a^4}\\ &=\frac {(43 A+17 i B) \cot ^2(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(A+i B) \cot ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(2 A+i B) \cot ^2(c+d x)}{6 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\cot ^3(c+d x) \left (8 a^3 (89 A+31 i B)-16 a^3 (43 i A-17 B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{192 a^6}\\ &=\frac {(43 A+17 i B) \cot ^2(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(A+i B) \cot ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(2 A+i B) \cot ^2(c+d x)}{6 a d (a+i a \tan (c+d x))^3}+\frac {5 (35 A+13 i B) \cot ^2(c+d x)}{48 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\int \cot ^3(c+d x) \left (384 a^4 (11 A+4 i B)-120 a^4 (35 i A-13 B) \tan (c+d x)\right ) \, dx}{384 a^8}\\ &=-\frac {(11 A+4 i B) \cot ^2(c+d x)}{2 a^4 d}+\frac {(43 A+17 i B) \cot ^2(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(A+i B) \cot ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(2 A+i B) \cot ^2(c+d x)}{6 a d (a+i a \tan (c+d x))^3}+\frac {5 (35 A+13 i B) \cot ^2(c+d x)}{48 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\int \cot ^2(c+d x) \left (-120 a^4 (35 i A-13 B)-384 a^4 (11 A+4 i B) \tan (c+d x)\right ) \, dx}{384 a^8}\\ &=\frac {5 (35 i A-13 B) \cot (c+d x)}{16 a^4 d}-\frac {(11 A+4 i B) \cot ^2(c+d x)}{2 a^4 d}+\frac {(43 A+17 i B) \cot ^2(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(A+i B) \cot ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(2 A+i B) \cot ^2(c+d x)}{6 a d (a+i a \tan (c+d x))^3}+\frac {5 (35 A+13 i B) \cot ^2(c+d x)}{48 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\int \cot (c+d x) \left (-384 a^4 (11 A+4 i B)+120 a^4 (35 i A-13 B) \tan (c+d x)\right ) \, dx}{384 a^8}\\ &=\frac {5 (35 i A-13 B) x}{16 a^4}+\frac {5 (35 i A-13 B) \cot (c+d x)}{16 a^4 d}-\frac {(11 A+4 i B) \cot ^2(c+d x)}{2 a^4 d}+\frac {(43 A+17 i B) \cot ^2(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(A+i B) \cot ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(2 A+i B) \cot ^2(c+d x)}{6 a d (a+i a \tan (c+d x))^3}+\frac {5 (35 A+13 i B) \cot ^2(c+d x)}{48 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {(11 A+4 i B) \int \cot (c+d x) \, dx}{a^4}\\ &=\frac {5 (35 i A-13 B) x}{16 a^4}+\frac {5 (35 i A-13 B) \cot (c+d x)}{16 a^4 d}-\frac {(11 A+4 i B) \cot ^2(c+d x)}{2 a^4 d}-\frac {(11 A+4 i B) \log (\sin (c+d x))}{a^4 d}+\frac {(43 A+17 i B) \cot ^2(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {(A+i B) \cot ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(2 A+i B) \cot ^2(c+d x)}{6 a d (a+i a \tan (c+d x))^3}+\frac {5 (35 A+13 i B) \cot ^2(c+d x)}{48 d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 7.63, size = 1625, normalized size = 6.37 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-3*(8*A + (5*I)*B)*Cos[4*d*x]*Sec[c + d*x]^3*(Cos[d*x] + I*Sin[d*x])^4*(A + B*Tan[c + d*x]))/(32*d*(A*Cos[c +
d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^4) + ((5*A + (4*I)*B)*Cos[6*d*x]*Sec[c + d*x]^3*(-1/48*Cos[2*c]
+ (I/48)*Sin[2*c])*(Cos[d*x] + I*Sin[d*x])^4*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a +
I*a*Tan[c + d*x])^4) + ((25*A + (12*I)*B)*Cos[2*d*x]*Sec[c + d*x]^3*((-3*Cos[2*c])/16 - ((3*I)/16)*Sin[2*c])*(
Cos[d*x] + I*Sin[d*x])^4*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^4)
+ (Sec[c + d*x]^3*(11*A*Cos[2*c] + (4*I)*B*Cos[2*c] + (11*I)*A*Sin[2*c] - 4*B*Sin[2*c])*(I*ArcTan[Tan[d*x]]*Co
s[2*c] - ArcTan[Tan[d*x]]*Sin[2*c])*(Cos[d*x] + I*Sin[d*x])^4*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin
[c + d*x])*(a + I*a*Tan[c + d*x])^4) + (Sec[c + d*x]^3*(11*A*Cos[2*c] + (4*I)*B*Cos[2*c] + (11*I)*A*Sin[2*c] -
4*B*Sin[2*c])*(-1/2*(Cos[2*c]*Log[Sin[c + d*x]^2]) - (I/2)*Log[Sin[c + d*x]^2]*Sin[2*c])*(Cos[d*x] + I*Sin[d*
x])^4*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^4) + (x*Sec[c + d*x]^3
*((33*I)*A*Cos[c]^2 - 12*B*Cos[c]^2 + 11*A*Cos[c]^2*Cot[c] + (4*I)*B*Cos[c]^2*Cot[c] - 33*A*Cos[c]*Sin[c] - (1
2*I)*B*Cos[c]*Sin[c] - (11*I)*A*Sin[c]^2 + 4*B*Sin[c]^2 + (11*A + (4*I)*B)*Cot[c]*(-Cos[4*c] - I*Sin[4*c]))*(C
os[d*x] + I*Sin[d*x])^4*(A + B*Tan[c + d*x]))/((A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^4) + (
(A + I*B)*Cos[8*d*x]*Sec[c + d*x]^3*(-1/128*Cos[4*c] + (I/128)*Sin[4*c])*(Cos[d*x] + I*Sin[d*x])^4*(A + B*Tan[
c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^4) + (Csc[c + d*x]^2*Sec[c + d*x]^3*(-1
/2*(A*Cos[4*c]) - (I/2)*A*Sin[4*c])*(Cos[d*x] + I*Sin[d*x])^4*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin
[c + d*x])*(a + I*a*Tan[c + d*x])^4) + ((35*A + (13*I)*B)*Sec[c + d*x]^3*(((5*I)/16)*d*x*Cos[4*c] - (5*d*x*Sin
[4*c])/16)*(Cos[d*x] + I*Sin[d*x])^4*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c
+ d*x])^4) + ((25*A + (12*I)*B)*Sec[c + d*x]^3*(((3*I)/16)*Cos[2*c] - (3*Sin[2*c])/16)*(Cos[d*x] + I*Sin[d*x]
)^4*Sin[2*d*x]*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^4) + (((3*I)/
32)*(8*A + (5*I)*B)*Sec[c + d*x]^3*(Cos[d*x] + I*Sin[d*x])^4*Sin[4*d*x]*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*
x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^4) + ((5*A + (4*I)*B)*Sec[c + d*x]^3*((I/48)*Cos[2*c] + Sin[2*c]/4
8)*(Cos[d*x] + I*Sin[d*x])^4*Sin[6*d*x]*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Ta
n[c + d*x])^4) + ((A + I*B)*Sec[c + d*x]^3*((I/128)*Cos[4*c] + Sin[4*c]/128)*(Cos[d*x] + I*Sin[d*x])^4*Sin[8*d
*x]*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^4) + (Csc[c]*Csc[c + d*x
]*Sec[c + d*x]^3*(Cos[d*x] + I*Sin[d*x])^4*(2*A*Cos[4*c - d*x] + (I/2)*B*Cos[4*c - d*x] - 2*A*Cos[4*c + d*x] -
(I/2)*B*Cos[4*c + d*x] + (2*I)*A*Sin[4*c - d*x] - (B*Sin[4*c - d*x])/2 - (2*I)*A*Sin[4*c + d*x] + (B*Sin[4*c
+ d*x])/2)*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^4)

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fricas [A]  time = 0.75, size = 249, normalized size = 0.98 \[ \frac {{\left (8424 i \, A - 3096 \, B\right )} d x e^{\left (12 i \, d x + 12 i \, c\right )} + {\left ({\left (-16848 i \, A + 6192 \, B\right )} d x - 4104 \, A - 1632 i \, B\right )} e^{\left (10 i \, d x + 10 i \, c\right )} + {\left ({\left (8424 i \, A - 3096 \, B\right )} d x + 6384 \, A + 2316 i \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} - 8 \, {\left (158 \, A + 67 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - {\left (211 \, A + 119 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (17 \, A + 13 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 384 \, {\left ({\left (11 \, A + 4 i \, B\right )} e^{\left (12 i \, d x + 12 i \, c\right )} - 2 \, {\left (11 \, A + 4 i \, B\right )} e^{\left (10 i \, d x + 10 i \, c\right )} + {\left (11 \, A + 4 i \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 3 \, A - 3 i \, B}{384 \, {\left (a^{4} d e^{\left (12 i \, d x + 12 i \, c\right )} - 2 \, a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/384*((8424*I*A - 3096*B)*d*x*e^(12*I*d*x + 12*I*c) + ((-16848*I*A + 6192*B)*d*x - 4104*A - 1632*I*B)*e^(10*I
*d*x + 10*I*c) + ((8424*I*A - 3096*B)*d*x + 6384*A + 2316*I*B)*e^(8*I*d*x + 8*I*c) - 8*(158*A + 67*I*B)*e^(6*I
*d*x + 6*I*c) - (211*A + 119*I*B)*e^(4*I*d*x + 4*I*c) - 2*(17*A + 13*I*B)*e^(2*I*d*x + 2*I*c) - 384*((11*A + 4
*I*B)*e^(12*I*d*x + 12*I*c) - 2*(11*A + 4*I*B)*e^(10*I*d*x + 10*I*c) + (11*A + 4*I*B)*e^(8*I*d*x + 8*I*c))*log
(e^(2*I*d*x + 2*I*c) - 1) - 3*A - 3*I*B)/(a^4*d*e^(12*I*d*x + 12*I*c) - 2*a^4*d*e^(10*I*d*x + 10*I*c) + a^4*d*
e^(8*I*d*x + 8*I*c))

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giac [A]  time = 5.27, size = 228, normalized size = 0.89 \[ \frac {\frac {12 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} + \frac {36 \, {\left (117 \, A + 43 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} - \frac {384 \, {\left (11 \, A + 4 i \, B\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{4}} + \frac {192 \, {\left (33 \, A \tan \left (d x + c\right )^{2} + 12 i \, B \tan \left (d x + c\right )^{2} + 8 i \, A \tan \left (d x + c\right ) - 2 \, B \tan \left (d x + c\right ) - A\right )}}{a^{4} \tan \left (d x + c\right )^{2}} - \frac {8775 \, A \tan \left (d x + c\right )^{4} + 3225 i \, B \tan \left (d x + c\right )^{4} - 37764 i \, A \tan \left (d x + c\right )^{3} + 14076 \, B \tan \left (d x + c\right )^{3} - 61386 \, A \tan \left (d x + c\right )^{2} - 23286 i \, B \tan \left (d x + c\right )^{2} + 44804 i \, A \tan \left (d x + c\right ) - 17404 \, B \tan \left (d x + c\right ) + 12455 \, A + 5017 i \, B}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/384*(12*(A - I*B)*log(tan(d*x + c) + I)/a^4 + 36*(117*A + 43*I*B)*log(tan(d*x + c) - I)/a^4 - 384*(11*A + 4*
I*B)*log(tan(d*x + c))/a^4 + 192*(33*A*tan(d*x + c)^2 + 12*I*B*tan(d*x + c)^2 + 8*I*A*tan(d*x + c) - 2*B*tan(d
*x + c) - A)/(a^4*tan(d*x + c)^2) - (8775*A*tan(d*x + c)^4 + 3225*I*B*tan(d*x + c)^4 - 37764*I*A*tan(d*x + c)^
3 + 14076*B*tan(d*x + c)^3 - 61386*A*tan(d*x + c)^2 - 23286*I*B*tan(d*x + c)^2 + 44804*I*A*tan(d*x + c) - 1740
4*B*tan(d*x + c) + 12455*A + 5017*I*B)/(a^4*(tan(d*x + c) - I)^4))/d

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maple [A]  time = 0.79, size = 329, normalized size = 1.29 \[ \frac {A \ln \left (\tan \left (d x +c \right )+i\right )}{32 d \,a^{4}}-\frac {i B}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {A}{2 a^{4} d \tan \left (d x +c \right )^{2}}-\frac {7 i A}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {11 A \ln \left (\tan \left (d x +c \right )\right )}{a^{4} d}+\frac {129 i \ln \left (\tan \left (d x +c \right )-i\right ) B}{32 d \,a^{4}}-\frac {B}{a^{4} d \tan \left (d x +c \right )}-\frac {A}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {i B \ln \left (\tan \left (d x +c \right )+i\right )}{32 d \,a^{4}}-\frac {49 B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}+\frac {4 i A}{a^{4} d \tan \left (d x +c \right )}+\frac {31 A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {111 i A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}+\frac {5 B}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {17 i B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {351 \ln \left (\tan \left (d x +c \right )-i\right ) A}{32 d \,a^{4}}-\frac {4 i B \ln \left (\tan \left (d x +c \right )\right )}{a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,x)

[Out]

1/32/d/a^4*A*ln(tan(d*x+c)+I)-1/8*I/d/a^4/(tan(d*x+c)-I)^4*B-1/2/a^4/d*A/tan(d*x+c)^2-7/12*I/a^4/d/(tan(d*x+c)
-I)^3*A-11/a^4/d*A*ln(tan(d*x+c))+129/32*I/a^4/d*ln(tan(d*x+c)-I)*B-1/a^4/d/tan(d*x+c)*B-1/8/d/a^4/(tan(d*x+c)
-I)^4*A-1/32*I/d/a^4*B*ln(tan(d*x+c)+I)-49/16/d/a^4/(tan(d*x+c)-I)*B+4*I/a^4/d/tan(d*x+c)*A+31/16/d/a^4/(tan(d
*x+c)-I)^2*A+111/16*I/a^4/d/(tan(d*x+c)-I)*A+5/12/d/a^4/(tan(d*x+c)-I)^3*B+17/16*I/a^4/d/(tan(d*x+c)-I)^2*B+35
1/32/d/a^4*ln(tan(d*x+c)-I)*A-4*I/a^4/d*B*ln(tan(d*x+c))

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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(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^4,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 [B]  time = 7.64, size = 251, normalized size = 0.98 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (\frac {153\,A}{4\,a^4}+\frac {B\,57{}\mathrm {i}}{4\,a^4}\right )+{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (-\frac {65\,B}{16\,a^4}+\frac {A\,175{}\mathrm {i}}{16\,a^4}\right )-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {271\,A}{12\,a^4}+\frac {B\,26{}\mathrm {i}}{3\,a^4}\right )-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (-\frac {851\,B}{48\,a^4}+\frac {A\,2269{}\mathrm {i}}{48\,a^4}\right )-\frac {A}{2\,a^4}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{a^4}+\frac {A\,2{}\mathrm {i}}{a^4}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^6-{\mathrm {tan}\left (c+d\,x\right )}^5\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^4+{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (11\,A+B\,4{}\mathrm {i}\right )}{a^4\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{32\,a^4\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (351\,A+B\,129{}\mathrm {i}\right )}{32\,a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cot(c + d*x)^3*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

(tan(c + d*x)^4*((153*A)/(4*a^4) + (B*57i)/(4*a^4)) + tan(c + d*x)^5*((A*175i)/(16*a^4) - (65*B)/(16*a^4)) - t
an(c + d*x)^2*((271*A)/(12*a^4) + (B*26i)/(3*a^4)) - tan(c + d*x)^3*((A*2269i)/(48*a^4) - (851*B)/(48*a^4)) -
A/(2*a^4) + tan(c + d*x)*((A*2i)/a^4 - B/a^4))/(d*(tan(c + d*x)^2 + tan(c + d*x)^3*4i - 6*tan(c + d*x)^4 - tan
(c + d*x)^5*4i + tan(c + d*x)^6)) - (log(tan(c + d*x))*(11*A + B*4i))/(a^4*d) + (log(tan(c + d*x) + 1i)*(A - B
*1i))/(32*a^4*d) + (log(tan(c + d*x) - 1i)*(351*A + B*129i))/(32*a^4*d)

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sympy [A]  time = 2.12, size = 471, normalized size = 1.85 \[ \frac {8 i A - 2 B + \left (- 6 i A e^{2 i c} + 2 B e^{2 i c}\right ) e^{2 i d x}}{i a^{4} d e^{4 i c} e^{4 i d x} - 2 i a^{4} d e^{2 i c} e^{2 i d x} + i a^{4} d} + \begin {cases} \frac {\left (\left (- 24576 A a^{12} d^{3} e^{12 i c} - 24576 i B a^{12} d^{3} e^{12 i c}\right ) e^{- 8 i d x} + \left (- 327680 A a^{12} d^{3} e^{14 i c} - 262144 i B a^{12} d^{3} e^{14 i c}\right ) e^{- 6 i d x} + \left (- 2359296 A a^{12} d^{3} e^{16 i c} - 1474560 i B a^{12} d^{3} e^{16 i c}\right ) e^{- 4 i d x} + \left (- 14745600 A a^{12} d^{3} e^{18 i c} - 7077888 i B a^{12} d^{3} e^{18 i c}\right ) e^{- 2 i d x}\right ) e^{- 20 i c}}{3145728 a^{16} d^{4}} & \text {for}\: 3145728 a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (- \frac {351 i A - 129 B}{16 a^{4}} + \frac {\left (351 i A e^{8 i c} + 150 i A e^{6 i c} + 48 i A e^{4 i c} + 10 i A e^{2 i c} + i A - 129 B e^{8 i c} - 72 B e^{6 i c} - 30 B e^{4 i c} - 8 B e^{2 i c} - B\right ) e^{- 8 i c}}{16 a^{4}}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- 351 i A + 129 B\right )}{16 a^{4}} - \frac {\left (11 A + 4 i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**4,x)

[Out]

(8*I*A - 2*B + (-6*I*A*exp(2*I*c) + 2*B*exp(2*I*c))*exp(2*I*d*x))/(I*a**4*d*exp(4*I*c)*exp(4*I*d*x) - 2*I*a**4
*d*exp(2*I*c)*exp(2*I*d*x) + I*a**4*d) + Piecewise((((-24576*A*a**12*d**3*exp(12*I*c) - 24576*I*B*a**12*d**3*e
xp(12*I*c))*exp(-8*I*d*x) + (-327680*A*a**12*d**3*exp(14*I*c) - 262144*I*B*a**12*d**3*exp(14*I*c))*exp(-6*I*d*
x) + (-2359296*A*a**12*d**3*exp(16*I*c) - 1474560*I*B*a**12*d**3*exp(16*I*c))*exp(-4*I*d*x) + (-14745600*A*a**
12*d**3*exp(18*I*c) - 7077888*I*B*a**12*d**3*exp(18*I*c))*exp(-2*I*d*x))*exp(-20*I*c)/(3145728*a**16*d**4), Ne
(3145728*a**16*d**4*exp(20*I*c), 0)), (x*(-(351*I*A - 129*B)/(16*a**4) + (351*I*A*exp(8*I*c) + 150*I*A*exp(6*I
*c) + 48*I*A*exp(4*I*c) + 10*I*A*exp(2*I*c) + I*A - 129*B*exp(8*I*c) - 72*B*exp(6*I*c) - 30*B*exp(4*I*c) - 8*B
*exp(2*I*c) - B)*exp(-8*I*c)/(16*a**4)), True)) - x*(-351*I*A + 129*B)/(16*a**4) - (11*A + 4*I*B)*log(exp(2*I*
d*x) - exp(-2*I*c))/(a**4*d)

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