\(\int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [25]

Optimal result

Integrand size = 34, antiderivative size = 180 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-4 a^3 (A-i B) x-\frac {4 a^3 (A-i B) \cot (c+d x)}{d}+\frac {2 a^3 (i A+B) \cot ^2(c+d x)}{d}+\frac {a^3 (47 A-45 i B) \cot ^3(c+d x)}{60 d}+\frac {4 a^3 (i A+B) \log (\sin (c+d x))}{d}-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(7 i A+5 B) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d} \]

[Out]

-4*a^3*(A-I*B)*x-4*a^3*(A-I*B)*cot(d*x+c)/d+2*a^3*(I*A+B)*cot(d*x+c)^2/d+1/60*a^3*(47*A-45*I*B)*cot(d*x+c)^3/d
+4*a^3*(I*A+B)*ln(sin(d*x+c))/d-1/5*a*A*cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2/d-1/20*(7*I*A+5*B)*cot(d*x+c)^4*(a^3
+I*a^3*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3674, 3672, 3610, 3612, 3556} \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {a^3 (47 A-45 i B) \cot ^3(c+d x)}{60 d}+\frac {2 a^3 (B+i A) \cot ^2(c+d x)}{d}-\frac {4 a^3 (A-i B) \cot (c+d x)}{d}+\frac {4 a^3 (B+i A) \log (\sin (c+d x))}{d}-\frac {(5 B+7 i A) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}-4 a^3 x (A-i B)-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d} \]

[In]

Int[Cot[c + d*x]^6*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

-4*a^3*(A - I*B)*x - (4*a^3*(A - I*B)*Cot[c + d*x])/d + (2*a^3*(I*A + B)*Cot[c + d*x]^2)/d + (a^3*(47*A - (45*
I)*B)*Cot[c + d*x]^3)/(60*d) + (4*a^3*(I*A + B)*Log[Sin[c + d*x]])/d - (a*A*Cot[c + d*x]^5*(a + I*a*Tan[c + d*
x])^2)/(5*d) - (((7*I)*A + 5*B)*Cot[c + d*x]^4*(a^3 + I*a^3*Tan[c + d*x]))/(20*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 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 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 ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac {1}{5} \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 (a (7 i A+5 B)-a (3 A-5 i B) \tan (c+d x)) \, dx \\ & = -\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(7 i A+5 B) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {1}{20} \int \cot ^4(c+d x) (a+i a \tan (c+d x)) \left (-a^2 (47 A-45 i B)-a^2 (33 i A+35 B) \tan (c+d x)\right ) \, dx \\ & = \frac {a^3 (47 A-45 i B) \cot ^3(c+d x)}{60 d}-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(7 i A+5 B) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {1}{20} \int \cot ^3(c+d x) \left (-80 a^3 (i A+B)+80 a^3 (A-i B) \tan (c+d x)\right ) \, dx \\ & = \frac {2 a^3 (i A+B) \cot ^2(c+d x)}{d}+\frac {a^3 (47 A-45 i B) \cot ^3(c+d x)}{60 d}-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(7 i A+5 B) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {1}{20} \int \cot ^2(c+d x) \left (80 a^3 (A-i B)+80 a^3 (i A+B) \tan (c+d x)\right ) \, dx \\ & = -\frac {4 a^3 (A-i B) \cot (c+d x)}{d}+\frac {2 a^3 (i A+B) \cot ^2(c+d x)}{d}+\frac {a^3 (47 A-45 i B) \cot ^3(c+d x)}{60 d}-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(7 i A+5 B) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\frac {1}{20} \int \cot (c+d x) \left (80 a^3 (i A+B)-80 a^3 (A-i B) \tan (c+d x)\right ) \, dx \\ & = -4 a^3 (A-i B) x-\frac {4 a^3 (A-i B) \cot (c+d x)}{d}+\frac {2 a^3 (i A+B) \cot ^2(c+d x)}{d}+\frac {a^3 (47 A-45 i B) \cot ^3(c+d x)}{60 d}-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(7 i A+5 B) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d}+\left (4 a^3 (i A+B)\right ) \int \cot (c+d x) \, dx \\ & = -4 a^3 (A-i B) x-\frac {4 a^3 (A-i B) \cot (c+d x)}{d}+\frac {2 a^3 (i A+B) \cot ^2(c+d x)}{d}+\frac {a^3 (47 A-45 i B) \cot ^3(c+d x)}{60 d}+\frac {4 a^3 (i A+B) \log (\sin (c+d x))}{d}-\frac {a A \cot ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {(7 i A+5 B) \cot ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{20 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.15 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=a^3 \left (-\frac {4 A \cot (c+d x)}{d}+\frac {4 i B \cot (c+d x)}{d}+\frac {2 i A \cot ^2(c+d x)}{d}+\frac {2 B \cot ^2(c+d x)}{d}+\frac {4 A \cot ^3(c+d x)}{3 d}-\frac {i B \cot ^3(c+d x)}{d}-\frac {3 i A \cot ^4(c+d x)}{4 d}-\frac {B \cot ^4(c+d x)}{4 d}-\frac {A \cot ^5(c+d x)}{5 d}+\frac {4 i A \log (\tan (c+d x))}{d}+\frac {4 B \log (\tan (c+d x))}{d}-\frac {4 i A \log (i+\tan (c+d x))}{d}-\frac {4 B \log (i+\tan (c+d x))}{d}\right ) \]

[In]

Integrate[Cot[c + d*x]^6*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

a^3*((-4*A*Cot[c + d*x])/d + ((4*I)*B*Cot[c + d*x])/d + ((2*I)*A*Cot[c + d*x]^2)/d + (2*B*Cot[c + d*x]^2)/d +
(4*A*Cot[c + d*x]^3)/(3*d) - (I*B*Cot[c + d*x]^3)/d - (((3*I)/4)*A*Cot[c + d*x]^4)/d - (B*Cot[c + d*x]^4)/(4*d
) - (A*Cot[c + d*x]^5)/(5*d) + ((4*I)*A*Log[Tan[c + d*x]])/d + (4*B*Log[Tan[c + d*x]])/d - ((4*I)*A*Log[I + Ta
n[c + d*x]])/d - (4*B*Log[I + Tan[c + d*x]])/d)

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.72

method	result	size
parallelrisch	\(\frac {4 \left (\left (-\frac {i A}{2}-\frac {B}{2}\right ) \ln \left (\sec ^{2}\left (d x +c \right )\right )+\left (i A +B \right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {A \left (\cot ^{5}\left (d x +c \right )\right )}{20}+\left (\cot ^{4}\left (d x +c \right )\right ) \left (-\frac {3 i A}{16}-\frac {B}{16}\right )+\left (\cot ^{3}\left (d x +c \right )\right ) \left (-\frac {i B}{4}+\frac {A}{3}\right )+\left (\cot ^{2}\left (d x +c \right )\right ) \left (\frac {i A}{2}+\frac {B}{2}\right )+\left (i B -A \right ) \cot \left (d x +c \right )+x d \left (i B -A \right )\right ) a^{3}}{d}\)	\(130\)
derivativedivides	\(\frac {a^{3} \left (-\frac {3 i A \left (\cot ^{4}\left (d x +c \right )\right )}{4}-\frac {A \left (\cot ^{5}\left (d x +c \right )\right )}{5}-i B \left (\cot ^{3}\left (d x +c \right )\right )-\frac {B \left (\cot ^{4}\left (d x +c \right )\right )}{4}+2 i A \left (\cot ^{2}\left (d x +c \right )\right )+\frac {4 A \left (\cot ^{3}\left (d x +c \right )\right )}{3}+4 i B \cot \left (d x +c \right )+2 B \left (\cot ^{2}\left (d x +c \right )\right )-4 A \cot \left (d x +c \right )+\frac {\left (-4 i A -4 B \right ) \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2}+\left (-4 i B +4 A \right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )\right )}{d}\)	\(151\)
default	\(\frac {a^{3} \left (-\frac {3 i A \left (\cot ^{4}\left (d x +c \right )\right )}{4}-\frac {A \left (\cot ^{5}\left (d x +c \right )\right )}{5}-i B \left (\cot ^{3}\left (d x +c \right )\right )-\frac {B \left (\cot ^{4}\left (d x +c \right )\right )}{4}+2 i A \left (\cot ^{2}\left (d x +c \right )\right )+\frac {4 A \left (\cot ^{3}\left (d x +c \right )\right )}{3}+4 i B \cot \left (d x +c \right )+2 B \left (\cot ^{2}\left (d x +c \right )\right )-4 A \cot \left (d x +c \right )+\frac {\left (-4 i A -4 B \right ) \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2}+\left (-4 i B +4 A \right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )\right )}{d}\)	\(151\)
risch	\(-\frac {8 i a^{3} B c}{d}+\frac {8 a^{3} A c}{d}-\frac {2 a^{3} \left (240 i A \,{\mathrm e}^{8 i \left (d x +c \right )}+180 B \,{\mathrm e}^{8 i \left (d x +c \right )}-585 i A \,{\mathrm e}^{6 i \left (d x +c \right )}-525 B \,{\mathrm e}^{6 i \left (d x +c \right )}+695 i A \,{\mathrm e}^{4 i \left (d x +c \right )}+615 B \,{\mathrm e}^{4 i \left (d x +c \right )}-385 i A \,{\mathrm e}^{2 i \left (d x +c \right )}-345 B \,{\mathrm e}^{2 i \left (d x +c \right )}+83 i A +75 B \right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{5}}+\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}+\frac {4 i a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A}{d}\)	\(195\)
norman	\(\frac {\left (4 i B \,a^{3}-4 A \,a^{3}\right ) x \left (\tan ^{5}\left (d x +c \right )\right )-\frac {A \,a^{3}}{5 d}+\frac {\left (-3 i B \,a^{3}+4 A \,a^{3}\right ) \left (\tan ^{2}\left (d x +c \right )\right )}{3 d}-\frac {\left (3 i A \,a^{3}+B \,a^{3}\right ) \tan \left (d x +c \right )}{4 d}-\frac {4 \left (-i B \,a^{3}+A \,a^{3}\right ) \left (\tan ^{4}\left (d x +c \right )\right )}{d}+\frac {2 \left (i A \,a^{3}+B \,a^{3}\right ) \left (\tan ^{3}\left (d x +c \right )\right )}{d}}{\tan \left (d x +c \right )^{5}}+\frac {4 \left (i A \,a^{3}+B \,a^{3}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {2 \left (i A \,a^{3}+B \,a^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(202\)

[In]

int(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

4*((-1/2*I*A-1/2*B)*ln(sec(d*x+c)^2)+(I*A+B)*ln(tan(d*x+c))-1/20*A*cot(d*x+c)^5+cot(d*x+c)^4*(-3/16*I*A-1/16*B
)+cot(d*x+c)^3*(-1/4*I*B+1/3*A)+cot(d*x+c)^2*(1/2*I*A+1/2*B)+(-A+I*B)*cot(d*x+c)+x*d*(-A+I*B))*a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.59 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (60 \, {\left (4 i \, A + 3 \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 15 \, {\left (-39 i \, A - 35 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 5 \, {\left (139 i \, A + 123 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, {\left (-77 i \, A - 69 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (83 i \, A + 75 \, B\right )} a^{3} + 30 \, {\left ({\left (-i \, A - B\right )} a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, {\left (i \, A + B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, {\left (-i \, A - B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, {\left (i \, A + B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, {\left (-i \, A - B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (i \, A + B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

[In]

integrate(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-2/15*(60*(4*I*A + 3*B)*a^3*e^(8*I*d*x + 8*I*c) + 15*(-39*I*A - 35*B)*a^3*e^(6*I*d*x + 6*I*c) + 5*(139*I*A + 1
23*B)*a^3*e^(4*I*d*x + 4*I*c) + 5*(-77*I*A - 69*B)*a^3*e^(2*I*d*x + 2*I*c) + (83*I*A + 75*B)*a^3 + 30*((-I*A -
B)*a^3*e^(10*I*d*x + 10*I*c) + 5*(I*A + B)*a^3*e^(8*I*d*x + 8*I*c) + 10*(-I*A - B)*a^3*e^(6*I*d*x + 6*I*c) +
10*(I*A + B)*a^3*e^(4*I*d*x + 4*I*c) + 5*(-I*A - B)*a^3*e^(2*I*d*x + 2*I*c) + (I*A + B)*a^3)*log(e^(2*I*d*x +
2*I*c) - 1))/(d*e^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x +
4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) - d)

Sympy [A] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.64 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {4 i a^{3} \left (A - i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 166 i A a^{3} - 150 B a^{3} + \left (770 i A a^{3} e^{2 i c} + 690 B a^{3} e^{2 i c}\right ) e^{2 i d x} + \left (- 1390 i A a^{3} e^{4 i c} - 1230 B a^{3} e^{4 i c}\right ) e^{4 i d x} + \left (1170 i A a^{3} e^{6 i c} + 1050 B a^{3} e^{6 i c}\right ) e^{6 i d x} + \left (- 480 i A a^{3} e^{8 i c} - 360 B a^{3} e^{8 i c}\right ) e^{8 i d x}}{15 d e^{10 i c} e^{10 i d x} - 75 d e^{8 i c} e^{8 i d x} + 150 d e^{6 i c} e^{6 i d x} - 150 d e^{4 i c} e^{4 i d x} + 75 d e^{2 i c} e^{2 i d x} - 15 d} \]

[In]

integrate(cot(d*x+c)**6*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

4*I*a**3*(A - I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-166*I*A*a**3 - 150*B*a**3 + (770*I*A*a**3*exp(2*I*c)
+ 690*B*a**3*exp(2*I*c))*exp(2*I*d*x) + (-1390*I*A*a**3*exp(4*I*c) - 1230*B*a**3*exp(4*I*c))*exp(4*I*d*x) + (1
170*I*A*a**3*exp(6*I*c) + 1050*B*a**3*exp(6*I*c))*exp(6*I*d*x) + (-480*I*A*a**3*exp(8*I*c) - 360*B*a**3*exp(8*
I*c))*exp(8*I*d*x))/(15*d*exp(10*I*c)*exp(10*I*d*x) - 75*d*exp(8*I*c)*exp(8*I*d*x) + 150*d*exp(6*I*c)*exp(6*I*
d*x) - 150*d*exp(4*I*c)*exp(4*I*d*x) + 75*d*exp(2*I*c)*exp(2*I*d*x) - 15*d)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.84 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {240 \, {\left (d x + c\right )} {\left (A - i \, B\right )} a^{3} + 120 \, {\left (i \, A + B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 240 \, {\left (-i \, A - B\right )} a^{3} \log \left (\tan \left (d x + c\right )\right ) + \frac {240 \, {\left (A - i \, B\right )} a^{3} \tan \left (d x + c\right )^{4} - 120 \, {\left (i \, A + B\right )} a^{3} \tan \left (d x + c\right )^{3} - 20 \, {\left (4 \, A - 3 i \, B\right )} a^{3} \tan \left (d x + c\right )^{2} - 15 \, {\left (-3 i \, A - B\right )} a^{3} \tan \left (d x + c\right ) + 12 \, A a^{3}}{\tan \left (d x + c\right )^{5}}}{60 \, d} \]

[In]

integrate(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(240*(d*x + c)*(A - I*B)*a^3 + 120*(I*A + B)*a^3*log(tan(d*x + c)^2 + 1) + 240*(-I*A - B)*a^3*log(tan(d*
x + c)) + (240*(A - I*B)*a^3*tan(d*x + c)^4 - 120*(I*A + B)*a^3*tan(d*x + c)^3 - 20*(4*A - 3*I*B)*a^3*tan(d*x
+ c)^2 - 15*(-3*I*A - B)*a^3*tan(d*x + c) + 12*A*a^3)/tan(d*x + c)^5)/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. 391 vs. \(2 (158) = 316\).

Time = 0.91 (sec) , antiderivative size = 391, normalized size of antiderivative = 2.17 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 45 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 190 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 660 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 540 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2460 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2280 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7680 \, {\left (i \, A a^{3} + B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 3840 \, {\left (-i \, A a^{3} - B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {-8768 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8768 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2460 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2280 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 660 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 540 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 190 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 120 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 45 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]

[In]

integrate(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/960*(6*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 45*I*A*a^3*tan(1/2*d*x + 1/2*c)^4 - 15*B*a^3*tan(1/2*d*x + 1/2*c)^4 -
190*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 120*I*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 660*I*A*a^3*tan(1/2*d*x + 1/2*c)^2 + 5
40*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 2460*A*a^3*tan(1/2*d*x + 1/2*c) - 2280*I*B*a^3*tan(1/2*d*x + 1/2*c) - 7680*(
I*A*a^3 + B*a^3)*log(tan(1/2*d*x + 1/2*c) + I) - 3840*(-I*A*a^3 - B*a^3)*log(tan(1/2*d*x + 1/2*c)) + (-8768*I*
A*a^3*tan(1/2*d*x + 1/2*c)^5 - 8768*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 2460*A*a^3*tan(1/2*d*x + 1/2*c)^4 + 2280*I*
B*a^3*tan(1/2*d*x + 1/2*c)^4 + 660*I*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 540*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 190*A*a
^3*tan(1/2*d*x + 1/2*c)^2 - 120*I*B*a^3*tan(1/2*d*x + 1/2*c)^2 - 45*I*A*a^3*tan(1/2*d*x + 1/2*c) - 15*B*a^3*ta
n(1/2*d*x + 1/2*c) - 6*A*a^3)/tan(1/2*d*x + 1/2*c)^5)/d

Mupad [B] (verification not implemented)

Time = 8.24 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.78 \[ \int \cot ^6(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {\frac {A\,a^3}{5}-{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {4\,A\,a^3}{3}-B\,a^3\,1{}\mathrm {i}\right )+{\mathrm {tan}\left (c+d\,x\right )}^4\,\left (4\,A\,a^3-B\,a^3\,4{}\mathrm {i}\right )-{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (2\,B\,a^3+A\,a^3\,2{}\mathrm {i}\right )+\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^3}{4}+\frac {A\,a^3\,3{}\mathrm {i}}{4}\right )}{d\,{\mathrm {tan}\left (c+d\,x\right )}^5}+\frac {a^3\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,8{}\mathrm {i}}{d} \]

[In]

int(cot(c + d*x)^6*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

(a^3*atan(2*tan(c + d*x) + 1i)*(A*1i + B)*8i)/d - (tan(c + d*x)^4*(4*A*a^3 - B*a^3*4i) - tan(c + d*x)^2*((4*A*
a^3)/3 - B*a^3*1i) - tan(c + d*x)^3*(A*a^3*2i + 2*B*a^3) + (A*a^3)/5 + tan(c + d*x)*((A*a^3*3i)/4 + (B*a^3)/4)
)/(d*tan(c + d*x)^5)