\(\int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\) [33]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 34, antiderivative size = 177 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=8 a^4 (i A+B) x+\frac {a^4 (67 i A+64 B) \cot (c+d x)}{12 d}+\frac {8 a^4 (A-i B) \log (\sin (c+d x))}{d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d}-\frac {(7 i A+4 B) \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{12 d}+\frac {(19 A-16 i B) \cot ^2(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{12 d} \]

[Out]

8*a^4*(I*A+B)*x+1/12*a^4*(67*I*A+64*B)*cot(d*x+c)/d+8*a^4*(A-I*B)*ln(sin(d*x+c))/d-1/4*a*A*cot(d*x+c)^4*(a+I*a
*tan(d*x+c))^3/d-1/12*(7*I*A+4*B)*cot(d*x+c)^3*(a^2+I*a^2*tan(d*x+c))^2/d+1/12*(19*A-16*I*B)*cot(d*x+c)^2*(a^4
+I*a^4*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3674, 3672, 3612, 3556} \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {a^4 (64 B+67 i A) \cot (c+d x)}{12 d}+\frac {8 a^4 (A-i B) \log (\sin (c+d x))}{d}+\frac {(19 A-16 i B) \cot ^2(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{12 d}+8 a^4 x (B+i A)-\frac {(4 B+7 i A) \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{12 d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d} \]

[In]

Int[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

8*a^4*(I*A + B)*x + (a^4*((67*I)*A + 64*B)*Cot[c + d*x])/(12*d) + (8*a^4*(A - I*B)*Log[Sin[c + d*x]])/d - (a*A
*Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^3)/(4*d) - (((7*I)*A + 4*B)*Cot[c + d*x]^3*(a^2 + I*a^2*Tan[c + d*x])^2
)/(12*d) + ((19*A - (16*I)*B)*Cot[c + d*x]^2*(a^4 + I*a^4*Tan[c + d*x]))/(12*d)

Rule 3556

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

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 3672

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

Rule 3674

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^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x]
)^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c
 + d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m
 - 1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && E
qQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d}+\frac {1}{4} \int \cot ^4(c+d x) (a+i a \tan (c+d x))^3 (a (7 i A+4 B)-a (A-4 i B) \tan (c+d x)) \, dx \\ & = -\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d}-\frac {(7 i A+4 B) \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{12 d}+\frac {1}{12} \int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 \left (-2 a^2 (19 A-16 i B)-2 a^2 (5 i A+8 B) \tan (c+d x)\right ) \, dx \\ & = -\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d}-\frac {(7 i A+4 B) \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{12 d}+\frac {(19 A-16 i B) \cot ^2(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{12 d}+\frac {1}{24} \int \cot ^2(c+d x) (a+i a \tan (c+d x)) \left (-2 a^3 (67 i A+64 B)+2 a^3 (29 A-32 i B) \tan (c+d x)\right ) \, dx \\ & = \frac {a^4 (67 i A+64 B) \cot (c+d x)}{12 d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d}-\frac {(7 i A+4 B) \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{12 d}+\frac {(19 A-16 i B) \cot ^2(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{12 d}+\frac {1}{24} \int \cot (c+d x) \left (192 a^4 (A-i B)+192 a^4 (i A+B) \tan (c+d x)\right ) \, dx \\ & = 8 a^4 (i A+B) x+\frac {a^4 (67 i A+64 B) \cot (c+d x)}{12 d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d}-\frac {(7 i A+4 B) \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{12 d}+\frac {(19 A-16 i B) \cot ^2(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{12 d}+\left (8 a^4 (A-i B)\right ) \int \cot (c+d x) \, dx \\ & = 8 a^4 (i A+B) x+\frac {a^4 (67 i A+64 B) \cot (c+d x)}{12 d}+\frac {8 a^4 (A-i B) \log (\sin (c+d x))}{d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^3}{4 d}-\frac {(7 i A+4 B) \cot ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{12 d}+\frac {(19 A-16 i B) \cot ^2(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{12 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.51 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {a^4 \left (-3 A (i+\cot (c+d x))^4+4 (A-i B) \left (21 i \cot (c+d x)+6 \cot ^2(c+d x)-i \cot ^3(c+d x)+24 (\log (\tan (c+d x))-\log (i+\tan (c+d x)))\right )\right )}{12 d} \]

[In]

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

(a^4*(-3*A*(I + Cot[c + d*x])^4 + 4*(A - I*B)*((21*I)*Cot[c + d*x] + 6*Cot[c + d*x]^2 - I*Cot[c + d*x]^3 + 24*
(Log[Tan[c + d*x]] - Log[I + Tan[c + d*x]]))))/(12*d)

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.63

method result size
parallelrisch \(\frac {8 a^{4} \left (\left (-\frac {A}{2}+\frac {i B}{2}\right ) \ln \left (\sec ^{2}\left (d x +c \right )\right )+\left (-i B +A \right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {A \left (\cot ^{4}\left (d x +c \right )\right )}{32}+\left (\cot ^{3}\left (d x +c \right )\right ) \left (-\frac {i A}{6}-\frac {B}{24}\right )+\left (\cot ^{2}\left (d x +c \right )\right ) \left (-\frac {i B}{4}+\frac {7 A}{16}\right )+\cot \left (d x +c \right ) \left (i A +\frac {7 B}{8}\right )+\left (i A +B \right ) x d \right )}{d}\) \(111\)
derivativedivides \(\frac {a^{4} \left (-\frac {4 i A \left (\cot ^{3}\left (d x +c \right )\right )}{3}-\frac {A \left (\cot ^{4}\left (d x +c \right )\right )}{4}-2 i B \left (\cot ^{2}\left (d x +c \right )\right )-\frac {B \left (\cot ^{3}\left (d x +c \right )\right )}{3}+8 i A \cot \left (d x +c \right )+\frac {7 A \left (\cot ^{2}\left (d x +c \right )\right )}{2}+7 \cot \left (d x +c \right ) B +\frac {\left (8 i B -8 A \right ) \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2}+\left (-8 i A -8 B \right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )\right )}{d}\) \(128\)
default \(\frac {a^{4} \left (-\frac {4 i A \left (\cot ^{3}\left (d x +c \right )\right )}{3}-\frac {A \left (\cot ^{4}\left (d x +c \right )\right )}{4}-2 i B \left (\cot ^{2}\left (d x +c \right )\right )-\frac {B \left (\cot ^{3}\left (d x +c \right )\right )}{3}+8 i A \cot \left (d x +c \right )+\frac {7 A \left (\cot ^{2}\left (d x +c \right )\right )}{2}+7 \cot \left (d x +c \right ) B +\frac {\left (8 i B -8 A \right ) \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2}+\left (-8 i A -8 B \right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )\right )}{d}\) \(128\)
risch \(-\frac {16 a^{4} B c}{d}-\frac {16 i a^{4} A c}{d}+\frac {4 i a^{4} \left (30 i A \,{\mathrm e}^{6 i \left (d x +c \right )}+18 B \,{\mathrm e}^{6 i \left (d x +c \right )}-63 i A \,{\mathrm e}^{4 i \left (d x +c \right )}-45 B \,{\mathrm e}^{4 i \left (d x +c \right )}+50 i A \,{\mathrm e}^{2 i \left (d x +c \right )}+38 B \,{\mathrm e}^{2 i \left (d x +c \right )}-14 i A -11 B \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}-\frac {8 i a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}+\frac {8 A \,a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) \(171\)
norman \(\frac {\frac {\left (8 i A \,a^{4}+7 B \,a^{4}\right ) \left (\tan ^{3}\left (d x +c \right )\right )}{d}+\left (8 i A \,a^{4}+8 B \,a^{4}\right ) x \left (\tan ^{4}\left (d x +c \right )\right )-\frac {A \,a^{4}}{4 d}+\frac {\left (-4 i B \,a^{4}+7 A \,a^{4}\right ) \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {\left (4 i A \,a^{4}+B \,a^{4}\right ) \tan \left (d x +c \right )}{3 d}}{\tan \left (d x +c \right )^{4}}+\frac {8 \left (-i B \,a^{4}+A \,a^{4}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {4 \left (-i B \,a^{4}+A \,a^{4}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\) \(176\)

[In]

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

8*a^4*((-1/2*A+1/2*I*B)*ln(sec(d*x+c)^2)+(A-I*B)*ln(tan(d*x+c))-1/32*A*cot(d*x+c)^4+cot(d*x+c)^3*(-1/6*I*A-1/2
4*B)+cot(d*x+c)^2*(-1/4*I*B+7/16*A)+cot(d*x+c)*(I*A+7/8*B)+(I*A+B)*x*d)/d

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.29 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-\frac {4 \, {\left (6 \, {\left (5 \, A - 3 i \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} - 9 \, {\left (7 \, A - 5 i \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (25 \, A - 19 i \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (14 \, A - 11 i \, B\right )} a^{4} - 6 \, {\left ({\left (A - i \, B\right )} a^{4} e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, {\left (A - i \, B\right )} a^{4} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, {\left (A - i \, B\right )} a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, {\left (A - i \, B\right )} a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A - i \, B\right )} a^{4}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-4/3*(6*(5*A - 3*I*B)*a^4*e^(6*I*d*x + 6*I*c) - 9*(7*A - 5*I*B)*a^4*e^(4*I*d*x + 4*I*c) + 2*(25*A - 19*I*B)*a^
4*e^(2*I*d*x + 2*I*c) - (14*A - 11*I*B)*a^4 - 6*((A - I*B)*a^4*e^(8*I*d*x + 8*I*c) - 4*(A - I*B)*a^4*e^(6*I*d*
x + 6*I*c) + 6*(A - I*B)*a^4*e^(4*I*d*x + 4*I*c) - 4*(A - I*B)*a^4*e^(2*I*d*x + 2*I*c) + (A - I*B)*a^4)*log(e^
(2*I*d*x + 2*I*c) - 1))/(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*
I*d*x + 2*I*c) + d)

Sympy [A] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.33 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {8 a^{4} \left (A - i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {56 A a^{4} - 44 i B a^{4} + \left (- 200 A a^{4} e^{2 i c} + 152 i B a^{4} e^{2 i c}\right ) e^{2 i d x} + \left (252 A a^{4} e^{4 i c} - 180 i B a^{4} e^{4 i c}\right ) e^{4 i d x} + \left (- 120 A a^{4} e^{6 i c} + 72 i B a^{4} e^{6 i c}\right ) e^{6 i d x}}{3 d e^{8 i c} e^{8 i d x} - 12 d e^{6 i c} e^{6 i d x} + 18 d e^{4 i c} e^{4 i d x} - 12 d e^{2 i c} e^{2 i d x} + 3 d} \]

[In]

integrate(cot(d*x+c)**5*(a+I*a*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)

[Out]

8*a**4*(A - I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (56*A*a**4 - 44*I*B*a**4 + (-200*A*a**4*exp(2*I*c) + 152*
I*B*a**4*exp(2*I*c))*exp(2*I*d*x) + (252*A*a**4*exp(4*I*c) - 180*I*B*a**4*exp(4*I*c))*exp(4*I*d*x) + (-120*A*a
**4*exp(6*I*c) + 72*I*B*a**4*exp(6*I*c))*exp(6*I*d*x))/(3*d*exp(8*I*c)*exp(8*I*d*x) - 12*d*exp(6*I*c)*exp(6*I*
d*x) + 18*d*exp(4*I*c)*exp(4*I*d*x) - 12*d*exp(2*I*c)*exp(2*I*d*x) + 3*d)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.77 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-\frac {96 \, {\left (d x + c\right )} {\left (-i \, A - B\right )} a^{4} + 48 \, {\left (A - i \, B\right )} a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 96 \, {\left (A - i \, B\right )} a^{4} \log \left (\tan \left (d x + c\right )\right ) - \frac {12 \, {\left (8 i \, A + 7 \, B\right )} a^{4} \tan \left (d x + c\right )^{3} + 6 \, {\left (7 \, A - 4 i \, B\right )} a^{4} \tan \left (d x + c\right )^{2} + 4 \, {\left (-4 i \, A - B\right )} a^{4} \tan \left (d x + c\right ) - 3 \, A a^{4}}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(96*(d*x + c)*(-I*A - B)*a^4 + 48*(A - I*B)*a^4*log(tan(d*x + c)^2 + 1) - 96*(A - I*B)*a^4*log(tan(d*x +
 c)) - (12*(8*I*A + 7*B)*a^4*tan(d*x + c)^3 + 6*(7*A - 4*I*B)*a^4*tan(d*x + c)^2 + 4*(-4*I*A - B)*a^4*tan(d*x
+ c) - 3*A*a^4)/tan(d*x + c)^4)/d

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (153) = 306\).

Time = 1.12 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.82 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=-\frac {3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 32 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 180 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 96 i \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 864 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 696 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3072 \, {\left (A a^{4} - i \, B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 1536 \, {\left (A a^{4} - i \, B a^{4}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {3200 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3200 i \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 864 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 696 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 180 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 96 i \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 32 i \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a^{4}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/192*(3*A*a^4*tan(1/2*d*x + 1/2*c)^4 - 32*I*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 8*B*a^4*tan(1/2*d*x + 1/2*c)^3 -
180*A*a^4*tan(1/2*d*x + 1/2*c)^2 + 96*I*B*a^4*tan(1/2*d*x + 1/2*c)^2 + 864*I*A*a^4*tan(1/2*d*x + 1/2*c) + 696*
B*a^4*tan(1/2*d*x + 1/2*c) + 3072*(A*a^4 - I*B*a^4)*log(tan(1/2*d*x + 1/2*c) + I) - 1536*(A*a^4 - I*B*a^4)*log
(tan(1/2*d*x + 1/2*c)) + (3200*A*a^4*tan(1/2*d*x + 1/2*c)^4 - 3200*I*B*a^4*tan(1/2*d*x + 1/2*c)^4 - 864*I*A*a^
4*tan(1/2*d*x + 1/2*c)^3 - 696*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 180*A*a^4*tan(1/2*d*x + 1/2*c)^2 + 96*I*B*a^4*ta
n(1/2*d*x + 1/2*c)^2 + 32*I*A*a^4*tan(1/2*d*x + 1/2*c) + 8*B*a^4*tan(1/2*d*x + 1/2*c) + 3*A*a^4)/tan(1/2*d*x +
 1/2*c)^4)/d

Mupad [B] (verification not implemented)

Time = 7.90 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.64 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {7\,A\,a^4}{2}-B\,a^4\,2{}\mathrm {i}\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (7\,B\,a^4+A\,a^4\,8{}\mathrm {i}\right )-\frac {A\,a^4}{4}-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^4}{3}+\frac {A\,a^4\,4{}\mathrm {i}}{3}\right )}{d\,{\mathrm {tan}\left (c+d\,x\right )}^4}+\frac {16\,a^4\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{d} \]

[In]

int(cot(c + d*x)^5*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^4,x)

[Out]

(tan(c + d*x)^2*((7*A*a^4)/2 - B*a^4*2i) + tan(c + d*x)^3*(A*a^4*8i + 7*B*a^4) - (A*a^4)/4 - tan(c + d*x)*((A*
a^4*4i)/3 + (B*a^4)/3))/(d*tan(c + d*x)^4) + (16*a^4*atan(2*tan(c + d*x) + 1i)*(A*1i + B))/d