\(\int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\) [58]

Optimal result

Integrand size = 34, antiderivative size = 216 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {5 (11 i A-5 B) x}{8 a^3}+\frac {5 (11 i A-5 B) \cot (c+d x)}{8 a^3 d}-\frac {(7 A+3 i B) \cot ^2(c+d x)}{2 a^3 d}-\frac {(7 A+3 i B) \log (\sin (c+d x))}{a^3 d}+\frac {(A+i B) \cot ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(13 A+7 i B) \cot ^2(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {5 (11 A+5 i B) \cot ^2(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

5/8*(11*I*A-5*B)*x/a^3+5/8*(11*I*A-5*B)*cot(d*x+c)/a^3/d-1/2*(7*A+3*I*B)*cot(d*x+c)^2/a^3/d-(7*A+3*I*B)*ln(sin
(d*x+c))/a^3/d+1/6*(A+I*B)*cot(d*x+c)^2/d/(a+I*a*tan(d*x+c))^3+1/24*(13*A+7*I*B)*cot(d*x+c)^2/a/d/(a+I*a*tan(d
*x+c))^2+5/24*(11*A+5*I*B)*cot(d*x+c)^2/d/(a^3+I*a^3*tan(d*x+c))

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3677, 3610, 3612, 3556} \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {(7 A+3 i B) \cot ^2(c+d x)}{2 a^3 d}+\frac {5 (-5 B+11 i A) \cot (c+d x)}{8 a^3 d}-\frac {(7 A+3 i B) \log (\sin (c+d x))}{a^3 d}+\frac {5 (11 A+5 i B) \cot ^2(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {5 x (-5 B+11 i A)}{8 a^3}+\frac {(13 A+7 i B) \cot ^2(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {(A+i B) \cot ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3} \]

[In]

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

[Out]

(5*((11*I)*A - 5*B)*x)/(8*a^3) + (5*((11*I)*A - 5*B)*Cot[c + d*x])/(8*a^3*d) - ((7*A + (3*I)*B)*Cot[c + d*x]^2
)/(2*a^3*d) - ((7*A + (3*I)*B)*Log[Sin[c + d*x]])/(a^3*d) + ((A + I*B)*Cot[c + d*x]^2)/(6*d*(a + I*a*Tan[c + d
*x])^3) + ((13*A + (7*I)*B)*Cot[c + d*x]^2)/(24*a*d*(a + I*a*Tan[c + d*x])^2) + (5*(11*A + (5*I)*B)*Cot[c + d*
x]^2)/(24*d*(a^3 + I*a^3*Tan[c + d*x]))

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)}{6 d (a+i a \tan (c+d x))^3}+\frac {\int \frac {\cot ^3(c+d x) (2 a (4 A+i B)-5 a (i A-B) \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {(A+i B) \cot ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(13 A+7 i B) \cot ^2(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\cot ^3(c+d x) \left (2 a^2 (29 A+11 i B)-4 a^2 (13 i A-7 B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4} \\ & = \frac {(A+i B) \cot ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(13 A+7 i B) \cot ^2(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {5 (11 A+5 i B) \cot ^2(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \cot ^3(c+d x) \left (48 a^3 (7 A+3 i B)-30 a^3 (11 i A-5 B) \tan (c+d x)\right ) \, dx}{48 a^6} \\ & = -\frac {(7 A+3 i B) \cot ^2(c+d x)}{2 a^3 d}+\frac {(A+i B) \cot ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(13 A+7 i B) \cot ^2(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {5 (11 A+5 i B) \cot ^2(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \cot ^2(c+d x) \left (-30 a^3 (11 i A-5 B)-48 a^3 (7 A+3 i B) \tan (c+d x)\right ) \, dx}{48 a^6} \\ & = \frac {5 (11 i A-5 B) \cot (c+d x)}{8 a^3 d}-\frac {(7 A+3 i B) \cot ^2(c+d x)}{2 a^3 d}+\frac {(A+i B) \cot ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(13 A+7 i B) \cot ^2(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {5 (11 A+5 i B) \cot ^2(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int \cot (c+d x) \left (-48 a^3 (7 A+3 i B)+30 a^3 (11 i A-5 B) \tan (c+d x)\right ) \, dx}{48 a^6} \\ & = \frac {5 (11 i A-5 B) x}{8 a^3}+\frac {5 (11 i A-5 B) \cot (c+d x)}{8 a^3 d}-\frac {(7 A+3 i B) \cot ^2(c+d x)}{2 a^3 d}+\frac {(A+i B) \cot ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(13 A+7 i B) \cot ^2(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {5 (11 A+5 i B) \cot ^2(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {(7 A+3 i B) \int \cot (c+d x) \, dx}{a^3} \\ & = \frac {5 (11 i A-5 B) x}{8 a^3}+\frac {5 (11 i A-5 B) \cot (c+d x)}{8 a^3 d}-\frac {(7 A+3 i B) \cot ^2(c+d x)}{2 a^3 d}-\frac {(7 A+3 i B) \log (\sin (c+d x))}{a^3 d}+\frac {(A+i B) \cot ^2(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(13 A+7 i B) \cot ^2(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac {5 (11 A+5 i B) \cot ^2(c+d x)}{24 d \left (a^3+i a^3 \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 = 3.57 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.80 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {4 (A+i B) \cot ^5(c+d x)}{(i+\cot (c+d x))^3}+\frac {(13 A+7 i B) \cot ^4(c+d x)}{(i+\cot (c+d x))^2}+\frac {5 (11 A+5 i B) \cot ^3(c+d x)}{i+\cot (c+d x)}+15 (11 i A-5 B) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )-12 (7 A+3 i B) \left (\cot ^2(c+d x)+2 (\log (\cos (c+d x))+\log (\tan (c+d x)))\right )}{24 a^3 d} \]

[In]

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

[Out]

((4*(A + I*B)*Cot[c + d*x]^5)/(I + Cot[c + d*x])^3 + ((13*A + (7*I)*B)*Cot[c + d*x]^4)/(I + Cot[c + d*x])^2 +
(5*(11*A + (5*I)*B)*Cot[c + d*x]^3)/(I + Cot[c + d*x]) + 15*((11*I)*A - 5*B)*Cot[c + d*x]*Hypergeometric2F1[-1
/2, 1, 1/2, -Tan[c + d*x]^2] - 12*(7*A + (3*I)*B)*(Cot[c + d*x]^2 + 2*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]]))
)/(24*a^3*d)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.14

method	result	size
risch	\(-\frac {49 x B}{8 a^{3}}+\frac {111 i x A}{8 a^{3}}-\frac {23 i {\mathrm e}^{-2 i \left (d x +c \right )} B}{16 a^{3} d}-\frac {39 \,{\mathrm e}^{-2 i \left (d x +c \right )} A}{16 a^{3} d}-\frac {7 i {\mathrm e}^{-4 i \left (d x +c \right )} B}{32 a^{3} d}-\frac {9 \,{\mathrm e}^{-4 i \left (d x +c \right )} A}{32 a^{3} d}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )} B}{48 a^{3} d}-\frac {{\mathrm e}^{-6 i \left (d x +c \right )} A}{48 a^{3} d}-\frac {6 B c}{a^{3} d}+\frac {14 i A c}{a^{3} d}-\frac {2 i \left (-2 i A \,{\mathrm e}^{2 i \left (d x +c \right )}+B \,{\mathrm e}^{2 i \left (d x +c \right )}+3 i A -B \right )}{a^{3} d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {3 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{a^{3} d}-\frac {7 A \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{3} d}\)	\(246\)
derivativedivides	\(-\frac {A \left (\cot ^{2}\left (d x +c \right )\right )}{2 a^{3} d}+\frac {9 i B}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{2}}-\frac {B \cot \left (d x +c \right )}{a^{3} d}+\frac {7 A \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2 a^{3} d}-\frac {i A}{6 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{3}}+\frac {3 i B \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2 a^{3} d}+\frac {25 B \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{8 a^{3} d}+\frac {3 i A \cot \left (d x +c \right )}{a^{3} d}-\frac {31 B}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )}-\frac {55 i A \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{8 a^{3} d}+\frac {49 i A}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )}+\frac {B}{6 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{3}}+\frac {11 A}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{2}}\)	\(259\)
default	\(-\frac {A \left (\cot ^{2}\left (d x +c \right )\right )}{2 a^{3} d}+\frac {9 i B}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{2}}-\frac {B \cot \left (d x +c \right )}{a^{3} d}+\frac {7 A \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2 a^{3} d}-\frac {i A}{6 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{3}}+\frac {3 i B \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2 a^{3} d}+\frac {25 B \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{8 a^{3} d}+\frac {3 i A \cot \left (d x +c \right )}{a^{3} d}-\frac {31 B}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )}-\frac {55 i A \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )}{8 a^{3} d}+\frac {49 i A}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )}+\frac {B}{6 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{3}}+\frac {11 A}{8 a^{3} d \left (i+\cot \left (d x +c \right )\right )^{2}}\)	\(259\)

[In]

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

[Out]

-49/8*x/a^3*B+111/8*I*x/a^3*A-23/16*I/a^3/d*exp(-2*I*(d*x+c))*B-39/16/a^3/d*exp(-2*I*(d*x+c))*A-7/32*I/a^3/d*e
xp(-4*I*(d*x+c))*B-9/32/a^3/d*exp(-4*I*(d*x+c))*A-1/48*I/a^3/d*exp(-6*I*(d*x+c))*B-1/48/a^3/d*exp(-6*I*(d*x+c)
)*A-6*B/a^3/d*c+14*I/a^3/d*A*c-2*I*(-2*I*A*exp(2*I*(d*x+c))+B*exp(2*I*(d*x+c))+3*I*A-B)/a^3/d/(exp(2*I*(d*x+c)
)-1)^2-3*I/a^3/d*ln(exp(2*I*(d*x+c))-1)*B-7*A/a^3/d*ln(exp(2*I*(d*x+c))-1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.09 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {12 \, {\left (-111 i \, A + 49 \, B\right )} d x e^{\left (10 i \, d x + 10 i \, c\right )} + 6 \, {\left (4 \, {\left (111 i \, A - 49 \, B\right )} d x + 103 \, A + 55 i \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 3 \, {\left (4 \, {\left (-111 i \, A + 49 \, B\right )} d x - 339 \, A - 149 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 14 \, {\left (13 \, A + 7 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (23 \, A + 17 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 96 \, {\left ({\left (7 \, A + 3 i \, B\right )} e^{\left (10 i \, d x + 10 i \, c\right )} - 2 \, {\left (7 \, A + 3 i \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (7 \, A + 3 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) + 2 \, A + 2 i \, B}{96 \, {\left (a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} - 2 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )}\right )}} \]

[In]

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

[Out]

-1/96*(12*(-111*I*A + 49*B)*d*x*e^(10*I*d*x + 10*I*c) + 6*(4*(111*I*A - 49*B)*d*x + 103*A + 55*I*B)*e^(8*I*d*x
+ 8*I*c) + 3*(4*(-111*I*A + 49*B)*d*x - 339*A - 149*I*B)*e^(6*I*d*x + 6*I*c) + 14*(13*A + 7*I*B)*e^(4*I*d*x +
4*I*c) + (23*A + 17*I*B)*e^(2*I*d*x + 2*I*c) + 96*((7*A + 3*I*B)*e^(10*I*d*x + 10*I*c) - 2*(7*A + 3*I*B)*e^(8
*I*d*x + 8*I*c) + (7*A + 3*I*B)*e^(6*I*d*x + 6*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) + 2*A + 2*I*B)/(a^3*d*e^(10*
I*d*x + 10*I*c) - 2*a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))

Sympy [A] (verification not implemented)

Time = 0.92 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.84 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {6 A + 2 i B + \left (- 4 A e^{2 i c} - 2 i B e^{2 i c}\right ) e^{2 i d x}}{a^{3} d e^{4 i c} e^{4 i d x} - 2 a^{3} d e^{2 i c} e^{2 i d x} + a^{3} d} + \begin {cases} \frac {\left (\left (- 512 A a^{6} d^{2} e^{6 i c} - 512 i B a^{6} d^{2} e^{6 i c}\right ) e^{- 6 i d x} + \left (- 6912 A a^{6} d^{2} e^{8 i c} - 5376 i B a^{6} d^{2} e^{8 i c}\right ) e^{- 4 i d x} + \left (- 59904 A a^{6} d^{2} e^{10 i c} - 35328 i B a^{6} d^{2} e^{10 i c}\right ) e^{- 2 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (- \frac {111 i A - 49 B}{8 a^{3}} + \frac {\left (111 i A e^{6 i c} + 39 i A e^{4 i c} + 9 i A e^{2 i c} + i A - 49 B e^{6 i c} - 23 B e^{4 i c} - 7 B e^{2 i c} - B\right ) e^{- 6 i c}}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (111 i A - 49 B\right )}{8 a^{3}} - \frac {\left (7 A + 3 i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{3} d} \]

[In]

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

[Out]

(6*A + 2*I*B + (-4*A*exp(2*I*c) - 2*I*B*exp(2*I*c))*exp(2*I*d*x))/(a**3*d*exp(4*I*c)*exp(4*I*d*x) - 2*a**3*d*e
xp(2*I*c)*exp(2*I*d*x) + a**3*d) + Piecewise((((-512*A*a**6*d**2*exp(6*I*c) - 512*I*B*a**6*d**2*exp(6*I*c))*ex
p(-6*I*d*x) + (-6912*A*a**6*d**2*exp(8*I*c) - 5376*I*B*a**6*d**2*exp(8*I*c))*exp(-4*I*d*x) + (-59904*A*a**6*d*
*2*exp(10*I*c) - 35328*I*B*a**6*d**2*exp(10*I*c))*exp(-2*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(a**9*d**3*
exp(12*I*c), 0)), (x*(-(111*I*A - 49*B)/(8*a**3) + (111*I*A*exp(6*I*c) + 39*I*A*exp(4*I*c) + 9*I*A*exp(2*I*c)
+ I*A - 49*B*exp(6*I*c) - 23*B*exp(4*I*c) - 7*B*exp(2*I*c) - B)*exp(-6*I*c)/(8*a**3)), True)) + x*(111*I*A - 4
9*B)/(8*a**3) - (7*A + 3*I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/(a**3*d)

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 1.30 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.96 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {6 \, {\left (A - i \, B\right )} \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} + \frac {6 \, {\left (111 \, A + 49 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac {96 \, {\left (7 \, A + 3 i \, B\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{3}} + \frac {48 \, {\left (21 \, A \tan \left (d x + c\right )^{2} + 9 i \, B \tan \left (d x + c\right )^{2} + 6 i \, A \tan \left (d x + c\right ) - 2 \, B \tan \left (d x + c\right ) - A\right )}}{a^{3} \tan \left (d x + c\right )^{2}} + \frac {1221 i \, A \tan \left (d x + c\right )^{3} - 539 \, B \tan \left (d x + c\right )^{3} + 4035 \, A \tan \left (d x + c\right )^{2} + 1821 i \, B \tan \left (d x + c\right )^{2} - 4491 i \, A \tan \left (d x + c\right ) + 2085 \, B \tan \left (d x + c\right ) - 1693 \, A - 819 i \, B}{a^{3} {\left (i \, \tan \left (d x + c\right ) + 1\right )}^{3}}}{96 \, d} \]

[In]

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

[Out]

1/96*(6*(A - I*B)*log(tan(d*x + c) + I)/a^3 + 6*(111*A + 49*I*B)*log(tan(d*x + c) - I)/a^3 - 96*(7*A + 3*I*B)*
log(tan(d*x + c))/a^3 + 48*(21*A*tan(d*x + c)^2 + 9*I*B*tan(d*x + c)^2 + 6*I*A*tan(d*x + c) - 2*B*tan(d*x + c)
- A)/(a^3*tan(d*x + c)^2) + (1221*I*A*tan(d*x + c)^3 - 539*B*tan(d*x + c)^3 + 4035*A*tan(d*x + c)^2 + 1821*I*
B*tan(d*x + c)^2 - 4491*I*A*tan(d*x + c) + 2085*B*tan(d*x + c) - 1693*A - 819*I*B)/(a^3*(I*tan(d*x + c) + 1)^3
))/d

Mupad [B] (verification not implemented)

Time = 8.25 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.02 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\frac {A}{2\,a^3}+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (-\frac {63\,B}{8\,a^3}+\frac {A\,137{}\mathrm {i}}{8\,a^3}\right )+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {149\,A}{12\,a^3}+\frac {B\,71{}\mathrm {i}}{12\,a^3}\right )-{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (\frac {55\,A}{8\,a^3}+\frac {B\,25{}\mathrm {i}}{8\,a^3}\right )-\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{a^3}+\frac {A\,3{}\mathrm {i}}{2\,a^3}\right )}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^5\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^4+{\mathrm {tan}\left (c+d\,x\right )}^3\,3{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (7\,A+B\,3{}\mathrm {i}\right )}{a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (111\,A+B\,49{}\mathrm {i}\right )}{16\,a^3\,d} \]

[In]

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

[Out]

(log(tan(c + d*x) + 1i)*(A - B*1i))/(16*a^3*d) - (log(tan(c + d*x))*(7*A + B*3i))/(a^3*d) - (tan(c + d*x)^3*((
A*137i)/(8*a^3) - (63*B)/(8*a^3)) - tan(c + d*x)^4*((55*A)/(8*a^3) + (B*25i)/(8*a^3)) + tan(c + d*x)^2*((149*A
)/(12*a^3) + (B*71i)/(12*a^3)) + A/(2*a^3) - tan(c + d*x)*((A*3i)/(2*a^3) - B/a^3))/(d*(tan(c + d*x)^2 + tan(c
+ d*x)^3*3i - 3*tan(c + d*x)^4 - tan(c + d*x)^5*1i)) + (log(tan(c + d*x) - 1i)*(111*A + B*49i))/(16*a^3*d)