\(\int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) [16]

Optimal result

Integrand size = 34, antiderivative size = 139 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=2 a^2 (i A+B) x+\frac {2 a^2 (i A+B) \cot (c+d x)}{d}+\frac {a^2 (A-i B) \cot ^2(c+d x)}{d}-\frac {a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}+\frac {2 a^2 (A-i B) \log (\sin (c+d x))}{d}-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d} \]

[Out]

2*a^2*(I*A+B)*x+2*a^2*(I*A+B)*cot(d*x+c)/d+a^2*(A-I*B)*cot(d*x+c)^2/d-1/12*a^2*(5*I*A+4*B)*cot(d*x+c)^3/d+2*a^
2*(A-I*B)*ln(sin(d*x+c))/d-1/4*A*cot(d*x+c)^4*(a^2+I*a^2*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3674, 3672, 3610, 3612, 3556} \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {a^2 (4 B+5 i A) \cot ^3(c+d x)}{12 d}+\frac {a^2 (A-i B) \cot ^2(c+d x)}{d}+\frac {2 a^2 (B+i A) \cot (c+d x)}{d}+\frac {2 a^2 (A-i B) \log (\sin (c+d x))}{d}+2 a^2 x (B+i A)-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d} \]

[In]

Int[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

2*a^2*(I*A + B)*x + (2*a^2*(I*A + B)*Cot[c + d*x])/d + (a^2*(A - I*B)*Cot[c + d*x]^2)/d - (a^2*((5*I)*A + 4*B)
*Cot[c + d*x]^3)/(12*d) + (2*a^2*(A - I*B)*Log[Sin[c + d*x]])/d - (A*Cot[c + d*x]^4*(a^2 + I*a^2*Tan[c + d*x])
)/(4*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 \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac {1}{4} \int \cot ^4(c+d x) (a+i a \tan (c+d x)) (a (5 i A+4 B)-a (3 A-4 i B) \tan (c+d x)) \, dx \\ & = -\frac {a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac {1}{4} \int \cot ^3(c+d x) \left (-8 a^2 (A-i B)-8 a^2 (i A+B) \tan (c+d x)\right ) \, dx \\ & = \frac {a^2 (A-i B) \cot ^2(c+d x)}{d}-\frac {a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac {1}{4} \int \cot ^2(c+d x) \left (-8 a^2 (i A+B)+8 a^2 (A-i B) \tan (c+d x)\right ) \, dx \\ & = \frac {2 a^2 (i A+B) \cot (c+d x)}{d}+\frac {a^2 (A-i B) \cot ^2(c+d x)}{d}-\frac {a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\frac {1}{4} \int \cot (c+d x) \left (8 a^2 (A-i B)+8 a^2 (i A+B) \tan (c+d x)\right ) \, dx \\ & = 2 a^2 (i A+B) x+\frac {2 a^2 (i A+B) \cot (c+d x)}{d}+\frac {a^2 (A-i B) \cot ^2(c+d x)}{d}-\frac {a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d}+\left (2 a^2 (A-i B)\right ) \int \cot (c+d x) \, dx \\ & = 2 a^2 (i A+B) x+\frac {2 a^2 (i A+B) \cot (c+d x)}{d}+\frac {a^2 (A-i B) \cot ^2(c+d x)}{d}-\frac {a^2 (5 i A+4 B) \cot ^3(c+d x)}{12 d}+\frac {2 a^2 (A-i B) \log (\sin (c+d x))}{d}-\frac {A \cot ^4(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.25 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=a^2 \left (\frac {2 i A \cot (c+d x)}{d}+\frac {2 B \cot (c+d x)}{d}+\frac {A \cot ^2(c+d x)}{d}-\frac {i B \cot ^2(c+d x)}{d}-\frac {2 i A \cot ^3(c+d x)}{3 d}-\frac {B \cot ^3(c+d x)}{3 d}-\frac {A \cot ^4(c+d x)}{4 d}+\frac {2 A \log (\tan (c+d x))}{d}-\frac {2 i B \log (\tan (c+d x))}{d}-\frac {2 A \log (i+\tan (c+d x))}{d}+\frac {2 i B \log (i+\tan (c+d x))}{d}\right ) \]

[In]

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.78

method	result	size
parallelrisch	\(\frac {2 a^{2} \left (\left (-\frac {A}{2}+\frac {i B}{2}\right ) \ln \left (\sec ^{2}\left (d x +c \right )\right )+\left (-i B +A \right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {A \left (\cot ^{4}\left (d x +c \right )\right )}{8}+\left (\cot ^{3}\left (d x +c \right )\right ) \left (-\frac {i A}{3}-\frac {B}{6}\right )+\left (\cot ^{2}\left (d x +c \right )\right ) \left (\frac {A}{2}-\frac {i B}{2}\right )+\cot \left (d x +c \right ) \left (i A +B \right )+\left (i A +B \right ) x d \right )}{d}\)	\(109\)
derivativedivides	\(\frac {-A \,a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )-B \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+2 i A \,a^{2} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+2 i B \,a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+A \,a^{2} \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+B \,a^{2} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )}{d}\)	\(168\)
default	\(\frac {-A \,a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )-B \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+2 i A \,a^{2} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )+2 i B \,a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+A \,a^{2} \left (-\frac {\left (\cot ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+B \,a^{2} \left (-\frac {\left (\cot ^{3}\left (d x +c \right )\right )}{3}+\cot \left (d x +c \right )+d x +c \right )}{d}\)	\(168\)
risch	\(-\frac {4 a^{2} B c}{d}-\frac {4 i a^{2} A c}{d}+\frac {2 i a^{2} \left (21 i A \,{\mathrm e}^{6 i \left (d x +c \right )}+15 B \,{\mathrm e}^{6 i \left (d x +c \right )}-36 i A \,{\mathrm e}^{4 i \left (d x +c \right )}-33 B \,{\mathrm e}^{4 i \left (d x +c \right )}+29 i A \,{\mathrm e}^{2 i \left (d x +c \right )}+25 B \,{\mathrm e}^{2 i \left (d x +c \right )}-8 i A -7 B \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}-\frac {2 i a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}+\frac {2 A \,a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(171\)
norman	\(\frac {\frac {\left (-i B \,a^{2}+A \,a^{2}\right ) \left (\tan ^{2}\left (d x +c \right )\right )}{d}+\left (2 i A \,a^{2}+2 B \,a^{2}\right ) x \left (\tan ^{4}\left (d x +c \right )\right )-\frac {A \,a^{2}}{4 d}+\frac {2 \left (i A \,a^{2}+B \,a^{2}\right ) \left (\tan ^{3}\left (d x +c \right )\right )}{d}-\frac {\left (2 i A \,a^{2}+B \,a^{2}\right ) \tan \left (d x +c \right )}{3 d}}{\tan \left (d x +c \right )^{4}}+\frac {2 \left (-i B \,a^{2}+A \,a^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (-i B \,a^{2}+A \,a^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(174\)

[In]

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.63 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (3 \, {\left (7 \, A - 5 i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, {\left (12 \, A - 11 i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (29 \, A - 25 i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (8 \, A - 7 i \, B\right )} a^{2} - 3 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, {\left (A - i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, {\left (A - i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A - i \, B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-2/3*(3*(7*A - 5*I*B)*a^2*e^(6*I*d*x + 6*I*c) - 3*(12*A - 11*I*B)*a^2*e^(4*I*d*x + 4*I*c) + (29*A - 25*I*B)*a^
2*e^(2*I*d*x + 2*I*c) - (8*A - 7*I*B)*a^2 - 3*((A - I*B)*a^2*e^(8*I*d*x + 8*I*c) - 4*(A - I*B)*a^2*e^(6*I*d*x
+ 6*I*c) + 6*(A - I*B)*a^2*e^(4*I*d*x + 4*I*c) - 4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) + (A - I*B)*a^2)*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 = 0.54 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.69 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {2 a^{2} \left (A - i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {16 A a^{2} - 14 i B a^{2} + \left (- 58 A a^{2} e^{2 i c} + 50 i B a^{2} e^{2 i c}\right ) e^{2 i d x} + \left (72 A a^{2} e^{4 i c} - 66 i B a^{2} e^{4 i c}\right ) e^{4 i d x} + \left (- 42 A a^{2} e^{6 i c} + 30 i B a^{2} e^{6 i c}\right ) e^{6 i d x}}{3 d e^{8 i c} e^{8 i d x} - 12 d e^{6 i c} e^{6 i d x} + 18 d e^{4 i c} e^{4 i d x} - 12 d e^{2 i c} e^{2 i d x} + 3 d} \]

[In]

integrate(cot(d*x+c)**5*(a+I*a*tan(d*x+c))**2*(A+B*tan(d*x+c)),x)

[Out]

2*a**2*(A - I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (16*A*a**2 - 14*I*B*a**2 + (-58*A*a**2*exp(2*I*c) + 50*I*
B*a**2*exp(2*I*c))*exp(2*I*d*x) + (72*A*a**2*exp(4*I*c) - 66*I*B*a**2*exp(4*I*c))*exp(4*I*d*x) + (-42*A*a**2*e
xp(6*I*c) + 30*I*B*a**2*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 = 0.30 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.95 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {24 \, {\left (d x + c\right )} {\left (-i \, A - B\right )} a^{2} + 12 \, {\left (A - i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 24 \, {\left (A - i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )\right ) - \frac {24 \, {\left (i \, A + B\right )} a^{2} \tan \left (d x + c\right )^{3} + 12 \, {\left (A - i \, B\right )} a^{2} \tan \left (d x + c\right )^{2} + 4 \, {\left (-2 i \, A - B\right )} a^{2} \tan \left (d x + c\right ) - 3 \, A a^{2}}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(24*(d*x + c)*(-I*A - B)*a^2 + 12*(A - I*B)*a^2*log(tan(d*x + c)^2 + 1) - 24*(A - I*B)*a^2*log(tan(d*x +
c)) - (24*(I*A + B)*a^2*tan(d*x + c)^3 + 12*(A - I*B)*a^2*tan(d*x + c)^2 + 4*(-2*I*A - B)*a^2*tan(d*x + c) -
3*A*a^2)/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. \(2 (123) = 246\).

Time = 1.10 (sec) , antiderivative size = 322, normalized size of antiderivative = 2.32 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 16 i \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 48 i \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 240 i \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 216 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 768 \, {\left (A a^{2} - i \, B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 384 \, {\left (A a^{2} - i \, B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {800 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 800 i \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 240 i \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 216 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 48 i \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 16 i \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/192*(3*A*a^2*tan(1/2*d*x + 1/2*c)^4 - 16*I*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 8*B*a^2*tan(1/2*d*x + 1/2*c)^3 -
60*A*a^2*tan(1/2*d*x + 1/2*c)^2 + 48*I*B*a^2*tan(1/2*d*x + 1/2*c)^2 + 240*I*A*a^2*tan(1/2*d*x + 1/2*c) + 216*B
*a^2*tan(1/2*d*x + 1/2*c) + 768*(A*a^2 - I*B*a^2)*log(tan(1/2*d*x + 1/2*c) + I) - 384*(A*a^2 - I*B*a^2)*log(ta
n(1/2*d*x + 1/2*c)) + (800*A*a^2*tan(1/2*d*x + 1/2*c)^4 - 800*I*B*a^2*tan(1/2*d*x + 1/2*c)^4 - 240*I*A*a^2*tan
(1/2*d*x + 1/2*c)^3 - 216*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 60*A*a^2*tan(1/2*d*x + 1/2*c)^2 + 48*I*B*a^2*tan(1/2*
d*x + 1/2*c)^2 + 16*I*A*a^2*tan(1/2*d*x + 1/2*c) + 8*B*a^2*tan(1/2*d*x + 1/2*c) + 3*A*a^2)/tan(1/2*d*x + 1/2*c
)^4)/d

Mupad [B] (verification not implemented)

Time = 7.92 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.81 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (A\,a^2-B\,a^2\,1{}\mathrm {i}\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (2\,B\,a^2+A\,a^2\,2{}\mathrm {i}\right )-\frac {A\,a^2}{4}-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^2}{3}+\frac {A\,a^2\,2{}\mathrm {i}}{3}\right )}{d\,{\mathrm {tan}\left (c+d\,x\right )}^4}+\frac {4\,a^2\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{d} \]

[In]

int(cot(c + d*x)^5*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

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