\(\int \cot ^5(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [24]

Optimal result

Integrand size = 34, antiderivative size = 157 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=4 a^3 (i A+B) x+\frac {4 a^3 (i A+B) \cot (c+d x)}{d}+\frac {a^3 (15 A-14 i B) \cot ^2(c+d x)}{12 d}+\frac {4 a^3 (A-i B) \log (\sin (c+d x))}{d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d}-\frac {(3 i A+2 B) \cot ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d} \]

[Out]

4*a^3*(I*A+B)*x+4*a^3*(I*A+B)*cot(d*x+c)/d+1/12*a^3*(15*A-14*I*B)*cot(d*x+c)^2/d+4*a^3*(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))^2/d-1/6*(3*I*A+2*B)*cot(d*x+c)^3*(a^3+I*a^3*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 157, 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} \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {a^3 (15 A-14 i B) \cot ^2(c+d x)}{12 d}+\frac {4 a^3 (B+i A) \cot (c+d x)}{d}+\frac {4 a^3 (A-i B) \log (\sin (c+d x))}{d}-\frac {(2 B+3 i A) \cot ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}+4 a^3 x (B+i A)-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d} \]

[In]

Int[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

4*a^3*(I*A + B)*x + (4*a^3*(I*A + B)*Cot[c + d*x])/d + (a^3*(15*A - (14*I)*B)*Cot[c + d*x]^2)/(12*d) + (4*a^3*
(A - I*B)*Log[Sin[c + d*x]])/d - (a*A*Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^2)/(4*d) - (((3*I)*A + 2*B)*Cot[c
+ d*x]^3*(a^3 + I*a^3*Tan[c + d*x]))/(6*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 ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d}+\frac {1}{4} \int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 (2 a (3 i A+2 B)-2 a (A-2 i B) \tan (c+d x)) \, dx \\ & = -\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d}-\frac {(3 i A+2 B) \cot ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}+\frac {1}{12} \int \cot ^3(c+d x) (a+i a \tan (c+d x)) \left (-2 a^2 (15 A-14 i B)-2 a^2 (9 i A+10 B) \tan (c+d x)\right ) \, dx \\ & = \frac {a^3 (15 A-14 i B) \cot ^2(c+d x)}{12 d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d}-\frac {(3 i A+2 B) \cot ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}+\frac {1}{12} \int \cot ^2(c+d x) \left (-48 a^3 (i A+B)+48 a^3 (A-i B) \tan (c+d x)\right ) \, dx \\ & = \frac {4 a^3 (i A+B) \cot (c+d x)}{d}+\frac {a^3 (15 A-14 i B) \cot ^2(c+d x)}{12 d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d}-\frac {(3 i A+2 B) \cot ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}+\frac {1}{12} \int \cot (c+d x) \left (48 a^3 (A-i B)+48 a^3 (i A+B) \tan (c+d x)\right ) \, dx \\ & = 4 a^3 (i A+B) x+\frac {4 a^3 (i A+B) \cot (c+d x)}{d}+\frac {a^3 (15 A-14 i B) \cot ^2(c+d x)}{12 d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d}-\frac {(3 i A+2 B) \cot ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}+\left (4 a^3 (A-i B)\right ) \int \cot (c+d x) \, dx \\ & = 4 a^3 (i A+B) x+\frac {4 a^3 (i A+B) \cot (c+d x)}{d}+\frac {a^3 (15 A-14 i B) \cot ^2(c+d x)}{12 d}+\frac {4 a^3 (A-i B) \log (\sin (c+d x))}{d}-\frac {a A \cot ^4(c+d x) (a+i a \tan (c+d x))^2}{4 d}-\frac {(3 i A+2 B) \cot ^3(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.45 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.71 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {i a^3 \left (-\left ((3 A-4 i B) (i+\cot (c+d x))^3\right )+3 i A \cot (c+d x) (i+\cot (c+d x))^3-6 i (A-i B) \left (6 i \cot (c+d x)+\cot ^2(c+d x)+8 \log (\tan (c+d x))-8 \log (i+\tan (c+d x))\right )\right )}{12 d} \]

[In]

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

((I/12)*a^3*(-((3*A - (4*I)*B)*(I + Cot[c + d*x])^3) + (3*I)*A*Cot[c + d*x]*(I + Cot[c + d*x])^3 - (6*I)*(A -
I*B)*((6*I)*Cot[c + d*x] + Cot[c + d*x]^2 + 8*Log[Tan[c + d*x]] - 8*Log[I + Tan[c + d*x]])))/d

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.69

method	result	size
parallelrisch	\(\frac {4 a^{3} \left (\left (-\frac {A}{2}+\frac {i B}{2}\right ) \ln \left (\sec ^{2}\left (d x +c \right )\right )+\left (-i B +A \right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {A \left (\cot ^{4}\left (d x +c \right )\right )}{16}+\left (\cot ^{3}\left (d x +c \right )\right ) \left (-\frac {i A}{4}-\frac {B}{12}\right )+\left (\cot ^{2}\left (d x +c \right )\right ) \left (-\frac {3 i B}{8}+\frac {A}{2}\right )+\cot \left (d x +c \right ) \left (i A +B \right )+\left (i A +B \right ) x d \right )}{d}\)	\(109\)
derivativedivides	\(\frac {a^{3} \left (-i A \left (\cot ^{3}\left (d x +c \right )\right )-\frac {A \left (\cot ^{4}\left (d x +c \right )\right )}{4}-\frac {3 i B \left (\cot ^{2}\left (d x +c \right )\right )}{2}-\frac {B \left (\cot ^{3}\left (d x +c \right )\right )}{3}+4 i A \cot \left (d x +c \right )+2 A \left (\cot ^{2}\left (d x +c \right )\right )+4 \cot \left (d x +c \right ) B +\frac {\left (4 i B -4 A \right ) \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2}+\left (-4 i A -4 B \right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )\right )}{d}\)	\(128\)
default	\(\frac {a^{3} \left (-i A \left (\cot ^{3}\left (d x +c \right )\right )-\frac {A \left (\cot ^{4}\left (d x +c \right )\right )}{4}-\frac {3 i B \left (\cot ^{2}\left (d x +c \right )\right )}{2}-\frac {B \left (\cot ^{3}\left (d x +c \right )\right )}{3}+4 i A \cot \left (d x +c \right )+2 A \left (\cot ^{2}\left (d x +c \right )\right )+4 \cot \left (d x +c \right ) B +\frac {\left (4 i B -4 A \right ) \ln \left (\cot ^{2}\left (d x +c \right )+1\right )}{2}+\left (-4 i A -4 B \right ) \left (\frac {\pi }{2}-\operatorname {arccot}\left (\cot \left (d x +c \right )\right )\right )\right )}{d}\)	\(128\)
risch	\(-\frac {8 a^{3} B c}{d}-\frac {8 i a^{3} A c}{d}+\frac {2 i a^{3} \left (36 i A \,{\mathrm e}^{6 i \left (d x +c \right )}+24 B \,{\mathrm e}^{6 i \left (d x +c \right )}-69 i A \,{\mathrm e}^{4 i \left (d x +c \right )}-57 B \,{\mathrm e}^{4 i \left (d x +c \right )}+54 i A \,{\mathrm e}^{2 i \left (d x +c \right )}+46 B \,{\mathrm e}^{2 i \left (d x +c \right )}-15 i A -13 B \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}-\frac {4 i a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}+\frac {4 A \,a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(171\)
norman	\(\frac {\left (4 i A \,a^{3}+4 B \,a^{3}\right ) x \left (\tan ^{4}\left (d x +c \right )\right )-\frac {A \,a^{3}}{4 d}+\frac {\left (-3 i B \,a^{3}+4 A \,a^{3}\right ) \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {\left (3 i A \,a^{3}+B \,a^{3}\right ) \tan \left (d x +c \right )}{3 d}+\frac {4 \left (i A \,a^{3}+B \,a^{3}\right ) \left (\tan ^{3}\left (d x +c \right )\right )}{d}}{\tan \left (d x +c \right )^{4}}+\frac {4 \left (-i B \,a^{3}+A \,a^{3}\right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {2 \left (-i B \,a^{3}+A \,a^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\)	\(176\)

[In]

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.45 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (12 \, {\left (3 \, A - 2 i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, {\left (23 \, A - 19 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (27 \, A - 23 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (15 \, A - 13 i \, B\right )} a^{3} - 6 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, {\left (A - i \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, {\left (A - i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, {\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A - i \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-2/3*(12*(3*A - 2*I*B)*a^3*e^(6*I*d*x + 6*I*c) - 3*(23*A - 19*I*B)*a^3*e^(4*I*d*x + 4*I*c) + 2*(27*A - 23*I*B)
*a^3*e^(2*I*d*x + 2*I*c) - (15*A - 13*I*B)*a^3 - 6*((A - I*B)*a^3*e^(8*I*d*x + 8*I*c) - 4*(A - I*B)*a^3*e^(6*I
*d*x + 6*I*c) + 6*(A - I*B)*a^3*e^(4*I*d*x + 4*I*c) - 4*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + (A - I*B)*a^3)*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 = 1.19 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.50 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {4 a^{3} \left (A - i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {30 A a^{3} - 26 i B a^{3} + \left (- 108 A a^{3} e^{2 i c} + 92 i B a^{3} e^{2 i c}\right ) e^{2 i d x} + \left (138 A a^{3} e^{4 i c} - 114 i B a^{3} e^{4 i c}\right ) e^{4 i d x} + \left (- 72 A a^{3} e^{6 i c} + 48 i B a^{3} e^{6 i c}\right ) e^{6 i d x}}{3 d e^{8 i c} e^{8 i d x} - 12 d e^{6 i c} e^{6 i d x} + 18 d e^{4 i c} e^{4 i d x} - 12 d e^{2 i c} e^{2 i d x} + 3 d} \]

[In]

integrate(cot(d*x+c)**5*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

4*a**3*(A - I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (30*A*a**3 - 26*I*B*a**3 + (-108*A*a**3*exp(2*I*c) + 92*I
*B*a**3*exp(2*I*c))*exp(2*I*d*x) + (138*A*a**3*exp(4*I*c) - 114*I*B*a**3*exp(4*I*c))*exp(4*I*d*x) + (-72*A*a**
3*exp(6*I*c) + 48*I*B*a**3*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 = 134, normalized size of antiderivative = 0.85 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {48 \, {\left (d x + c\right )} {\left (-i \, A - B\right )} a^{3} + 24 \, {\left (A - i \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 48 \, {\left (A - i \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )\right ) - \frac {48 \, {\left (i \, A + B\right )} a^{3} \tan \left (d x + c\right )^{3} + 6 \, {\left (4 \, A - 3 i \, B\right )} a^{3} \tan \left (d x + c\right )^{2} + 4 \, {\left (-3 i \, A - B\right )} a^{3} \tan \left (d x + c\right ) - 3 \, A a^{3}}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(48*(d*x + c)*(-I*A - B)*a^3 + 24*(A - I*B)*a^3*log(tan(d*x + c)^2 + 1) - 48*(A - I*B)*a^3*log(tan(d*x +
c)) - (48*(I*A + B)*a^3*tan(d*x + c)^3 + 6*(4*A - 3*I*B)*a^3*tan(d*x + c)^2 + 4*(-3*I*A - B)*a^3*tan(d*x + c)
- 3*A*a^3)/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. \(2 (137) = 274\).

Time = 0.83 (sec) , antiderivative size = 322, normalized size of antiderivative = 2.05 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 108 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 456 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 408 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1536 \, {\left (A a^{3} - i \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 768 \, {\left (A a^{3} - i \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {1600 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1600 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 456 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 408 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 108 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 72 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/192*(3*A*a^3*tan(1/2*d*x + 1/2*c)^4 - 24*I*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 8*B*a^3*tan(1/2*d*x + 1/2*c)^3 -
108*A*a^3*tan(1/2*d*x + 1/2*c)^2 + 72*I*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 456*I*A*a^3*tan(1/2*d*x + 1/2*c) + 408*
B*a^3*tan(1/2*d*x + 1/2*c) + 1536*(A*a^3 - I*B*a^3)*log(tan(1/2*d*x + 1/2*c) + I) - 768*(A*a^3 - I*B*a^3)*log(
tan(1/2*d*x + 1/2*c)) + (1600*A*a^3*tan(1/2*d*x + 1/2*c)^4 - 1600*I*B*a^3*tan(1/2*d*x + 1/2*c)^4 - 456*I*A*a^3
*tan(1/2*d*x + 1/2*c)^3 - 408*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 108*A*a^3*tan(1/2*d*x + 1/2*c)^2 + 72*I*B*a^3*tan
(1/2*d*x + 1/2*c)^2 + 24*I*A*a^3*tan(1/2*d*x + 1/2*c) + 8*B*a^3*tan(1/2*d*x + 1/2*c) + 3*A*a^3)/tan(1/2*d*x +
1/2*c)^4)/d

Mupad [B] (verification not implemented)

Time = 7.97 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.73 \[ \int \cot ^5(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (2\,A\,a^3-\frac {B\,a^3\,3{}\mathrm {i}}{2}\right )+{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (4\,B\,a^3+A\,a^3\,4{}\mathrm {i}\right )-\frac {A\,a^3}{4}-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {B\,a^3}{3}+A\,a^3\,1{}\mathrm {i}\right )}{d\,{\mathrm {tan}\left (c+d\,x\right )}^4}+\frac {8\,a^3\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{d} \]

[In]

int(cot(c + d*x)^5*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

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