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*((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)

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Rubi [A]  time = 0.789628, antiderivative size = 255, normalized size of antiderivative = 1., 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 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]

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 3475

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

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*}

Mathematica [B]  time = 7.2365, 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*(-Cos[2*c]/48 +
(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])*(Co
s[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]]*Cos[
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])*(-(Cos[2*c]*Log[Sin[c + d*x]^2])/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*((3
3*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] - (12*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]))*(Cos[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*(-Cos[4*c]/128 + (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*(-(A*Cos[
4*c])/2 - (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])/1
6)*(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*S
in[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]/48)*(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*Tan[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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Maple [A]  time = 0.136, size = 329, normalized size = 1.3 \begin{align*} -{\frac{A}{8\,{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{4}}}-{\frac{{\frac{7\,i}{12}}A}{{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{49\,B}{16\,{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{4\,iA}{{a}^{4}d\tan \left ( dx+c \right ) }}+{\frac{31\,A}{16\,{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{{\frac{111\,i}{16}}A}{{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{5\,B}{12\,{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}+{\frac{{\frac{17\,i}{16}}B}{{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{351\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{32\,{a}^{4}d}}-{\frac{{\frac{i}{32}}B\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{{a}^{4}d}}+{\frac{A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{32\,{a}^{4}d}}-{\frac{4\,iB\ln \left ( \tan \left ( dx+c \right ) \right ) }{{a}^{4}d}}-{\frac{A}{2\,{a}^{4}d \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{\frac{129\,i}{32}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{{a}^{4}d}}-11\,{\frac{A\ln \left ( \tan \left ( dx+c \right ) \right ) }{{a}^{4}d}}-{\frac{{\frac{i}{8}}B}{{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{4}}}-{\frac{B}{{a}^{4}d\tan \left ( dx+c \right ) }} \end{align*}

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]

-1/8/d/a^4/(tan(d*x+c)-I)^4*A-7/12*I/d/a^4/(tan(d*x+c)-I)^3*A-49/16/d/a^4/(tan(d*x+c)-I)*B+4*I/d/a^4/tan(d*x+c
)*A+31/16/d/a^4/(tan(d*x+c)-I)^2*A+111/16*I/d/a^4/(tan(d*x+c)-I)*A+5/12/d/a^4/(tan(d*x+c)-I)^3*B+17/16*I/d/a^4
/(tan(d*x+c)-I)^2*B+351/32/d/a^4*ln(tan(d*x+c)-I)*A-1/32*I/d/a^4*B*ln(tan(d*x+c)+I)+1/32/d/a^4*A*ln(tan(d*x+c)
+I)-4*I/d/a^4*B*ln(tan(d*x+c))-1/2/d/a^4*A/tan(d*x+c)^2+129/32*I/d/a^4*ln(tan(d*x+c)-I)*B-11/d/a^4*A*ln(tan(d*
x+c))-1/8*I/d/a^4/(tan(d*x+c)-I)^4*B-1/d/a^4/tan(d*x+c)*B

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

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

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Fricas [A]  time = 1.82941, size = 772, normalized size = 3.03 \begin{align*} \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 )}} \end{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.40691, size = 309, normalized size = 1.21 \begin{align*} \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 ({\left | \tan \left (d x + c\right ) \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} \end{align*}

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(abs(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) -
17404*B*tan(d*x + c) + 12455*A + 5017*I*B)/(a^4*(tan(d*x + c) - I)^4))/d