3.6.14 \(\int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [514]

3.6.14.1 Optimal result
3.6.14.2 Mathematica [C] (verified)
3.6.14.3 Rubi [A] (verified)
3.6.14.4 Maple [B] (verified)
3.6.14.5 Fricas [B] (verification not implemented)
3.6.14.6 Sympy [F(-1)]
3.6.14.7 Maxima [A] (verification not implemented)
3.6.14.8 Giac [F]
3.6.14.9 Mupad [F(-1)]

3.6.14.1 Optimal result

Integrand size = 36, antiderivative size = 146 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {8 \sqrt [4]{-1} a^3 (A-i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {16 a^3 (6 A-5 i B) \sqrt {\cot (c+d x)}}{15 d}-\frac {2 a A \sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^2}{5 d}-\frac {2 (9 i A+5 B) \sqrt {\cot (c+d x)} \left (i a^3+a^3 \cot (c+d x)\right )}{15 d} \]

output
8*(-1)^(1/4)*a^3*(A-I*B)*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/d+16/15*a^3* 
(6*A-5*I*B)*cot(d*x+c)^(1/2)/d-2/5*a*A*(I*a+a*cot(d*x+c))^2*cot(d*x+c)^(1/ 
2)/d-2/15*(9*I*A+5*B)*(I*a^3+a^3*cot(d*x+c))*cot(d*x+c)^(1/2)/d
 
3.6.14.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 3.58 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.77 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 a^3 \sqrt {\cot (c+d x)} \left (-15 i B-5 i A \cot (c+d x)-15 B \cot (c+d x)-9 A \cot ^2(c+d x)+12 A \cot ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},i \tan (c+d x)\right )+20 B \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},i \tan (c+d x)\right )\right )}{15 d} \]

input
Integrate[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]) 
,x]
 
output
(-2*a^3*Sqrt[Cot[c + d*x]]*((-15*I)*B - (5*I)*A*Cot[c + d*x] - 15*B*Cot[c 
+ d*x] - 9*A*Cot[c + d*x]^2 + 12*A*Cot[c + d*x]^2*Hypergeometric2F1[-5/2, 
1, -3/2, I*Tan[c + d*x]] + 20*B*Cot[c + d*x]*Hypergeometric2F1[-3/2, 1, -1 
/2, I*Tan[c + d*x]]))/(15*d)
 
3.6.14.3 Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.361, Rules used = {3042, 4064, 3042, 4077, 27, 3042, 4077, 27, 3042, 4075, 3042, 4016, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cot (c+d x)^{7/2} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 4064

\(\displaystyle \int \frac {(a \cot (c+d x)+i a)^3 (A \cot (c+d x)+B)}{\sqrt {\cot (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (-a \tan \left (c+d x+\frac {\pi }{2}\right )+i a\right )^3 \left (B-A \tan \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 4077

\(\displaystyle -\frac {2}{5} \int \frac {(\cot (c+d x) a+i a)^2 (a (A-5 i B)-a (9 i A+5 B) \cot (c+d x))}{2 \sqrt {\cot (c+d x)}}dx-\frac {2 a A \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)^2}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{5} \int \frac {(\cot (c+d x) a+i a)^2 (a (A-5 i B)-a (9 i A+5 B) \cot (c+d x))}{\sqrt {\cot (c+d x)}}dx-\frac {2 a A \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)^2}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {1}{5} \int \frac {\left (i a-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (a (A-5 i B)+a (9 i A+5 B) \tan \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a A \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)^2}{5 d}\)

\(\Big \downarrow \) 4077

\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int -\frac {2 (\cot (c+d x) a+i a) \left ((3 i A+5 B) a^2+2 (6 A-5 i B) \cot (c+d x) a^2\right )}{\sqrt {\cot (c+d x)}}dx-\frac {2 (5 B+9 i A) \sqrt {\cot (c+d x)} \left (a^3 \cot (c+d x)+i a^3\right )}{3 d}\right )-\frac {2 a A \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)^2}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (-\frac {4}{3} \int \frac {(\cot (c+d x) a+i a) \left ((3 i A+5 B) a^2+2 (6 A-5 i B) \cot (c+d x) a^2\right )}{\sqrt {\cot (c+d x)}}dx-\frac {2 (5 B+9 i A) \sqrt {\cot (c+d x)} \left (a^3 \cot (c+d x)+i a^3\right )}{3 d}\right )-\frac {2 a A \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)^2}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (-\frac {4}{3} \int \frac {\left (i a-a \tan \left (c+d x+\frac {\pi }{2}\right )\right ) \left (a^2 (3 i A+5 B)-2 a^2 (6 A-5 i B) \tan \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (5 B+9 i A) \sqrt {\cot (c+d x)} \left (a^3 \cot (c+d x)+i a^3\right )}{3 d}\right )-\frac {2 a A \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)^2}{5 d}\)

\(\Big \downarrow \) 4075

\(\displaystyle \frac {1}{5} \left (-\frac {4}{3} \left (\int \frac {15 a^3 (i A+B) \cot (c+d x)-15 a^3 (A-i B)}{\sqrt {\cot (c+d x)}}dx-\frac {4 a^3 (6 A-5 i B) \sqrt {\cot (c+d x)}}{d}\right )-\frac {2 (5 B+9 i A) \sqrt {\cot (c+d x)} \left (a^3 \cot (c+d x)+i a^3\right )}{3 d}\right )-\frac {2 a A \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)^2}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (-\frac {4}{3} \left (\int \frac {-15 (A-i B) a^3-15 (i A+B) \tan \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {4 a^3 (6 A-5 i B) \sqrt {\cot (c+d x)}}{d}\right )-\frac {2 (5 B+9 i A) \sqrt {\cot (c+d x)} \left (a^3 \cot (c+d x)+i a^3\right )}{3 d}\right )-\frac {2 a A \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)^2}{5 d}\)

\(\Big \downarrow \) 4016

\(\displaystyle \frac {1}{5} \left (-\frac {4}{3} \left (\frac {450 a^6 (A-i B)^2 \int \frac {1}{15 (A-i B) a^3+15 (i A+B) \cot (c+d x) a^3}d\sqrt {\cot (c+d x)}}{d}-\frac {4 a^3 (6 A-5 i B) \sqrt {\cot (c+d x)}}{d}\right )-\frac {2 (5 B+9 i A) \sqrt {\cot (c+d x)} \left (a^3 \cot (c+d x)+i a^3\right )}{3 d}\right )-\frac {2 a A \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)^2}{5 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{5} \left (-\frac {4}{3} \left (-\frac {30 \sqrt [4]{-1} a^3 (A-i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {4 a^3 (6 A-5 i B) \sqrt {\cot (c+d x)}}{d}\right )-\frac {2 (5 B+9 i A) \sqrt {\cot (c+d x)} \left (a^3 \cot (c+d x)+i a^3\right )}{3 d}\right )-\frac {2 a A \sqrt {\cot (c+d x)} (a \cot (c+d x)+i a)^2}{5 d}\)

input
Int[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]
 
output
(-2*a*A*Sqrt[Cot[c + d*x]]*(I*a + a*Cot[c + d*x])^2)/(5*d) + ((-4*((-30*(- 
1)^(1/4)*a^3*(A - I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (4*a^3* 
(6*A - (5*I)*B)*Sqrt[Cot[c + d*x]])/d))/3 - (2*((9*I)*A + 5*B)*Sqrt[Cot[c 
+ d*x]]*(I*a^3 + a^3*Cot[c + d*x]))/(3*d))/5
 

3.6.14.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4016
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2*(c^2/f)   Subst[Int[1/(b*c - d*x^2), x], x, Sqrt[b 
*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]
 

rule 4064
Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp 
[g^(m + n)   Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d + c 
*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !Integer 
Q[p] && IntegerQ[m] && IntegerQ[n]
 

rule 4075
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B 
*d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f* 
x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1]
 

rule 4077
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + 
n))), x] + Simp[1/(d*(m + n))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan 
[e + f*x])^n*Simp[a*A*d*(m + n) + B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - 
 a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*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] && 
GtQ[m, 1] &&  !LtQ[n, -1]
 
3.6.14.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (123 ) = 246\).

Time = 1.24 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.73

method result size
derivativedivides \(\frac {a^{3} \left (-\frac {2 A \cot \left (d x +c \right )^{\frac {5}{2}}}{5}-2 i A \cot \left (d x +c \right )^{\frac {3}{2}}-\frac {2 B \cot \left (d x +c \right )^{\frac {3}{2}}}{3}-6 i B \sqrt {\cot \left (d x +c \right )}+8 A \sqrt {\cot \left (d x +c \right )}-\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}-\frac {\left (-4 i A -4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}\right )}{d}\) \(252\)
default \(\frac {a^{3} \left (-\frac {2 A \cot \left (d x +c \right )^{\frac {5}{2}}}{5}-2 i A \cot \left (d x +c \right )^{\frac {3}{2}}-\frac {2 B \cot \left (d x +c \right )^{\frac {3}{2}}}{3}-6 i B \sqrt {\cot \left (d x +c \right )}+8 A \sqrt {\cot \left (d x +c \right )}-\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}-\frac {\left (-4 i A -4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4}\right )}{d}\) \(252\)

input
int(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x,method=_RETUR 
NVERBOSE)
 
output
a^3/d*(-2/5*A*cot(d*x+c)^(5/2)-2*I*A*cot(d*x+c)^(3/2)-2/3*B*cot(d*x+c)^(3/ 
2)-6*I*B*cot(d*x+c)^(1/2)+8*A*cot(d*x+c)^(1/2)-1/4*(4*A-4*I*B)*2^(1/2)*(ln 
((1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^ 
(1/2)))+2*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*cot(d*x+c 
)^(1/2)))-1/4*(-4*I*A-4*B)*2^(1/2)*(ln((1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1 
/2))/(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2)))+2*arctan(1+2^(1/2)*cot(d*x+c 
)^(1/2))+2*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))))
 
3.6.14.5 Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 449 vs. \(2 (118) = 236\).

Time = 0.27 (sec) , antiderivative size = 449, normalized size of antiderivative = 3.08 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (15 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{3}}\right ) - 15 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{3}}\right ) - 2 \, {\left ({\left (39 \, A - 25 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 3 \, {\left (19 \, A - 15 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, {\left (6 \, A - 5 i \, B\right )} a^{3}\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

input
integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algori 
thm="fricas")
 
output
-2/15*(15*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) - 
 2*d*e^(2*I*d*x + 2*I*c) + d)*log(2*((A - I*B)*a^3*e^(2*I*d*x + 2*I*c) - s 
qrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/d^2)*(I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqr 
t((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d*x - 2* 
I*c)/((-I*A - B)*a^3)) - 15*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/d^2)*(d*e^( 
4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(2*((A - I*B)*a^3*e^(2* 
I*d*x + 2*I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^6/d^2)*(-I*d*e^(2*I*d*x 
+ 2*I*c) + I*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1) 
))*e^(-2*I*d*x - 2*I*c)/((-I*A - B)*a^3)) - 2*((39*A - 25*I*B)*a^3*e^(4*I* 
d*x + 4*I*c) - 3*(19*A - 15*I*B)*a^3*e^(2*I*d*x + 2*I*c) + 4*(6*A - 5*I*B) 
*a^3)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(d*e^(4 
*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)
 
3.6.14.6 Sympy [F(-1)]

Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

input
integrate(cot(d*x+c)**(7/2)*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)
 
output
Timed out
 
3.6.14.7 Maxima [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.37 \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {15 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{3} + \frac {30 \, {\left (4 \, A - 3 i \, B\right )} a^{3}}{\sqrt {\tan \left (d x + c\right )}} + \frac {10 \, {\left (-3 i \, A - B\right )} a^{3}}{\tan \left (d x + c\right )^{\frac {3}{2}}} - \frac {6 \, A a^{3}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{15 \, d} \]

input
integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algori 
thm="maxima")
 
output
1/15*(15*(2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 
2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arctan(-1/2*sqr 
t(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + sqrt(2)*(-(I + 1)*A + (I - 1)*B)* 
log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - sqrt(2)*(-(I + 1)*A 
 + (I - 1)*B)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^3 + 
 30*(4*A - 3*I*B)*a^3/sqrt(tan(d*x + c)) + 10*(-3*I*A - B)*a^3/tan(d*x + c 
)^(3/2) - 6*A*a^3/tan(d*x + c)^(5/2))/d
 
3.6.14.8 Giac [F]

\[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \cot \left (d x + c\right )^{\frac {7}{2}} \,d x } \]

input
integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algori 
thm="giac")
 
output
integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^3*cot(d*x + c)^(7/2) 
, x)
 
3.6.14.9 Mupad [F(-1)]

Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \]

input
int(cot(c + d*x)^(7/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3,x)
 
output
int(cot(c + d*x)^(7/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3, x)