\(\int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx\) [66]

Optimal result

Integrand size = 34, antiderivative size = 255 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {5 (35 i A-13 B) x}{16 a^4}+\frac {5 (35 i A-13 B) \cot (c+d x)}{16 a^4 d}-\frac {(11 A+4 i B) \cot ^2(c+d x)}{2 a^4 d}-\frac {(11 A+4 i B) \log (\sin (c+d x))}{a^4 d}+\frac {(43 A+17 i B) \cot ^2(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {5 (35 A+13 i B) \cot ^2(c+d x)}{48 a^4 d (1+i \tan (c+d x))}+\frac {(A+i B) \cot ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {(2 A+i B) \cot ^2(c+d x)}{6 a d (a+i a \tan (c+d x))^3} \]

[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

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3677, 3610, 3612, 3556} \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=-\frac {(11 A+4 i B) \cot ^2(c+d x)}{2 a^4 d}+\frac {5 (-13 B+35 i A) \cot (c+d x)}{16 a^4 d}-\frac {(11 A+4 i B) \log (\sin (c+d x))}{a^4 d}+\frac {5 (35 A+13 i B) \cot ^2(c+d x)}{48 a^4 d (1+i \tan (c+d x))}+\frac {(43 A+17 i B) \cot ^2(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {5 x (-13 B+35 i A)}{16 a^4}+\frac {(2 A+i B) \cot ^2(c+d x)}{6 a d (a+i a \tan (c+d x))^3}+\frac {(A+i B) \cot ^2(c+d x)}{8 d (a+i a \tan (c+d x))^4} \]

[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 3556

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

Rule 3610

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 3612

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 3677

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*} \text {integral}& = \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 [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 4.60 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.80 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {\frac {6 (A+i B) \cot ^6(c+d x)}{(i+\cot (c+d x))^4}+\frac {8 (2 A+i B) \cot ^5(c+d x)}{(i+\cot (c+d x))^3}+\frac {(43 A+17 i B) \cot ^4(c+d x)}{(i+\cot (c+d x))^2}+\frac {5 (35 A+13 i B) \cot ^3(c+d x)}{i+\cot (c+d x)}+15 (35 i A-13 B) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )-24 (11 A+4 i B) \left (\cot ^2(c+d x)+2 (\log (\cos (c+d x))+\log (\tan (c+d x)))\right )}{48 a^4 d} \]

[In]

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

[Out]

((6*(A + I*B)*Cot[c + d*x]^6)/(I + Cot[c + d*x])^4 + (8*(2*A + I*B)*Cot[c + d*x]^5)/(I + Cot[c + d*x])^3 + ((4
3*A + (17*I)*B)*Cot[c + d*x]^4)/(I + Cot[c + d*x])^2 + (5*(35*A + (13*I)*B)*Cot[c + d*x]^3)/(I + Cot[c + d*x])
+ 15*((35*I)*A - 13*B)*Cot[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2] - 24*(11*A + (4*I)*B)*(C
ot[c + d*x]^2 + 2*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]])))/(48*a^4*d)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.11

method	result	size
risch	\(-\frac {129 x B}{16 a^{4}}-\frac {i {\mathrm e}^{-8 i \left (d x +c \right )} B}{128 d \,a^{4}}-\frac {2 i \left (-3 i A \,{\mathrm e}^{2 i \left (d x +c \right )}+B \,{\mathrm e}^{2 i \left (d x +c \right )}+4 i A -B \right )}{a^{4} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {75 \,{\mathrm e}^{-2 i \left (d x +c \right )} A}{16 d \,a^{4}}+\frac {22 i A c}{a^{4} d}-\frac {3 \,{\mathrm e}^{-4 i \left (d x +c \right )} A}{4 d \,a^{4}}-\frac {9 i {\mathrm e}^{-2 i \left (d x +c \right )} B}{4 d \,a^{4}}-\frac {5 \,{\mathrm e}^{-6 i \left (d x +c \right )} A}{48 d \,a^{4}}+\frac {351 i x A}{16 a^{4}}-\frac {{\mathrm e}^{-8 i \left (d x +c \right )} A}{128 d \,a^{4}}-\frac {15 i B \,{\mathrm e}^{-4 i \left (d x +c \right )}}{32 d \,a^{4}}-\frac {8 B c}{a^{4} d}-\frac {4 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{a^{4} d}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )} B}{12 d \,a^{4}}-\frac {11 A \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{4} d}\)	\(283\)
derivativedivides	\(-\frac {A}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {4 i A}{d \,a^{4} \tan \left (d x +c \right )}-\frac {49 B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {i B}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {31 A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {7 i A}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {5 B}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {2 i B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}}+\frac {11 A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{4}}+\frac {175 i A \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {4 i B \ln \left (\tan \left (d x +c \right )\right )}{d \,a^{4}}-\frac {65 B \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}+\frac {17 i B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {111 i A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {A}{2 d \,a^{4} \tan \left (d x +c \right )^{2}}-\frac {11 A \ln \left (\tan \left (d x +c \right )\right )}{d \,a^{4}}-\frac {B}{d \,a^{4} \tan \left (d x +c \right )}\)	\(325\)
default	\(-\frac {A}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {4 i A}{d \,a^{4} \tan \left (d x +c \right )}-\frac {49 B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {i B}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {31 A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {7 i A}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {5 B}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {2 i B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}}+\frac {11 A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{4}}+\frac {175 i A \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {4 i B \ln \left (\tan \left (d x +c \right )\right )}{d \,a^{4}}-\frac {65 B \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}+\frac {17 i B}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {111 i A}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {A}{2 d \,a^{4} \tan \left (d x +c \right )^{2}}-\frac {11 A \ln \left (\tan \left (d x +c \right )\right )}{d \,a^{4}}-\frac {B}{d \,a^{4} \tan \left (d x +c \right )}\)	\(325\)

[In]

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

[Out]

-129/16*x/a^4*B-1/128*I/d/a^4*exp(-8*I*(d*x+c))*B-2*I*(-3*I*A*exp(2*I*(d*x+c))+B*exp(2*I*(d*x+c))+4*I*A-B)/a^4
/d/(exp(2*I*(d*x+c))-1)^2-75/16/d/a^4*exp(-2*I*(d*x+c))*A+22*I/a^4/d*A*c-3/4/d/a^4*exp(-4*I*(d*x+c))*A-9/4*I/d
/a^4*exp(-2*I*(d*x+c))*B-5/48/d/a^4*exp(-6*I*(d*x+c))*A+351/16*I*x/a^4*A-1/128/d/a^4*exp(-8*I*(d*x+c))*A-15/32
*I/d/a^4*exp(-4*I*(d*x+c))*B-8/a^4/d*B*c-4*I/a^4/d*ln(exp(2*I*(d*x+c))-1)*B-1/12*I/d/a^4*exp(-6*I*(d*x+c))*B-1
1/a^4*A/d*ln(exp(2*I*(d*x+c))-1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.99 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=-\frac {72 \, {\left (-117 i \, A + 43 \, B\right )} d x e^{\left (12 i \, d x + 12 i \, c\right )} + 24 \, {\left (6 \, {\left (117 i \, A - 43 \, B\right )} d x + 171 \, A + 68 i \, B\right )} e^{\left (10 i \, d x + 10 i \, c\right )} + 12 \, {\left (6 \, {\left (-117 i \, A + 43 \, B\right )} d x - 532 \, A - 193 i \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 8 \, {\left (158 \, A + 67 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (211 \, A + 119 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (17 \, A + 13 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 384 \, {\left ({\left (11 \, A + 4 i \, B\right )} e^{\left (12 i \, d x + 12 i \, c\right )} - 2 \, {\left (11 \, A + 4 i \, B\right )} e^{\left (10 i \, d x + 10 i \, c\right )} + {\left (11 \, A + 4 i \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + 3 \, A + 3 i \, B}{384 \, {\left (a^{4} d e^{\left (12 i \, d x + 12 i \, c\right )} - 2 \, a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )}\right )}} \]

[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*(72*(-117*I*A + 43*B)*d*x*e^(12*I*d*x + 12*I*c) + 24*(6*(117*I*A - 43*B)*d*x + 171*A + 68*I*B)*e^(10*I*
d*x + 10*I*c) + 12*(6*(-117*I*A + 43*B)*d*x - 532*A - 193*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))

Sympy [A] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 466, normalized size of antiderivative = 1.83 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {8 A + 2 i B + \left (- 6 A e^{2 i c} - 2 i B e^{2 i c}\right ) e^{2 i d x}}{a^{4} d e^{4 i c} e^{4 i d x} - 2 a^{4} d e^{2 i c} e^{2 i d x} + a^{4} d} + \begin {cases} \frac {\left (\left (- 24576 A a^{12} d^{3} e^{12 i c} - 24576 i B a^{12} d^{3} e^{12 i c}\right ) e^{- 8 i d x} + \left (- 327680 A a^{12} d^{3} e^{14 i c} - 262144 i B a^{12} d^{3} e^{14 i c}\right ) e^{- 6 i d x} + \left (- 2359296 A a^{12} d^{3} e^{16 i c} - 1474560 i B a^{12} d^{3} e^{16 i c}\right ) e^{- 4 i d x} + \left (- 14745600 A a^{12} d^{3} e^{18 i c} - 7077888 i B a^{12} d^{3} e^{18 i c}\right ) e^{- 2 i d x}\right ) e^{- 20 i c}}{3145728 a^{16} d^{4}} & \text {for}\: a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (- \frac {351 i A - 129 B}{16 a^{4}} + \frac {\left (351 i A e^{8 i c} + 150 i A e^{6 i c} + 48 i A e^{4 i c} + 10 i A e^{2 i c} + i A - 129 B e^{8 i c} - 72 B e^{6 i c} - 30 B e^{4 i c} - 8 B e^{2 i c} - B\right ) e^{- 8 i c}}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (351 i A - 129 B\right )}{16 a^{4}} - \frac {\left (11 A + 4 i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{4} d} \]

[In]

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

[Out]

(8*A + 2*I*B + (-6*A*exp(2*I*c) - 2*I*B*exp(2*I*c))*exp(2*I*d*x))/(a**4*d*exp(4*I*c)*exp(4*I*d*x) - 2*a**4*d*e
xp(2*I*c)*exp(2*I*d*x) + a**4*d) + Piecewise((((-24576*A*a**12*d**3*exp(12*I*c) - 24576*I*B*a**12*d**3*exp(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(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*e
xp(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)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]

[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: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 1.08 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.89 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {\frac {12 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} + \frac {36 \, {\left (117 \, A + 43 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} - \frac {384 \, {\left (11 \, A + 4 i \, B\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{4}} + \frac {192 \, {\left (33 \, A \tan \left (d x + c\right )^{2} + 12 i \, B \tan \left (d x + c\right )^{2} + 8 i \, A \tan \left (d x + c\right ) - 2 \, B \tan \left (d x + c\right ) - A\right )}}{a^{4} \tan \left (d x + c\right )^{2}} - \frac {8775 \, A \tan \left (d x + c\right )^{4} + 3225 i \, B \tan \left (d x + c\right )^{4} - 37764 i \, A \tan \left (d x + c\right )^{3} + 14076 \, B \tan \left (d x + c\right )^{3} - 61386 \, A \tan \left (d x + c\right )^{2} - 23286 i \, B \tan \left (d x + c\right )^{2} + 44804 i \, A \tan \left (d x + c\right ) - 17404 \, B \tan \left (d x + c\right ) + 12455 \, A + 5017 i \, B}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 9.06 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.98 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (\frac {153\,A}{4\,a^4}+\frac {B\,57{}\mathrm {i}}{4\,a^4}\right )+{\mathrm {tan}\left (c+d\,x\right )}^5\,\left (-\frac {65\,B}{16\,a^4}+\frac {A\,175{}\mathrm {i}}{16\,a^4}\right )-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {271\,A}{12\,a^4}+\frac {B\,26{}\mathrm {i}}{3\,a^4}\right )-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (-\frac {851\,B}{48\,a^4}+\frac {A\,2269{}\mathrm {i}}{48\,a^4}\right )-\frac {A}{2\,a^4}+\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{a^4}+\frac {A\,2{}\mathrm {i}}{a^4}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^6-{\mathrm {tan}\left (c+d\,x\right )}^5\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^4+{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (11\,A+B\,4{}\mathrm {i}\right )}{a^4\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{32\,a^4\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (351\,A+B\,129{}\mathrm {i}\right )}{32\,a^4\,d} \]

[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)