\(\int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [567]

Optimal result

Integrand size = 38, antiderivative size = 214 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {A+i B}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {13 A+3 i B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {67 A-3 i B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Output:

(1/8-1/8*I)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+ 
c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/a^(5/2)/d+1/5*(A+I*B)/d/cot(d 
*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(5/2)+1/30*(13*A+3*I*B)/a/d/cot(d*x+c)^(1/2 
)/(a+I*a*tan(d*x+c))^(3/2)+1/60*(67*A-3*I*B)/a^2/d/cot(d*x+c)^(1/2)/(a+I*a 
*tan(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 3.60 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\cot ^{\frac {3}{2}}(c+d x) \sec ^2(c+d x) (i a \tan (c+d x))^{3/2} \left (2 (19 A+9 i B+(86 A+6 i B) \cos (2 (c+d x))+80 i A \sin (2 (c+d x))) \sqrt {i a \tan (c+d x)}+15 \sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}\right )}{120 a^4 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

(Cot[c + d*x]^(3/2)*Sec[c + d*x]^2*(I*a*Tan[c + d*x])^(3/2)*(2*(19*A + (9* 
I)*B + (86*A + (6*I)*B)*Cos[2*(c + d*x)] + (80*I)*A*Sin[2*(c + d*x)])*Sqrt 
[I*a*Tan[c + d*x]] + 15*Sqrt[2]*(A - I*B)*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c 
+ d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x) 
])*Sqrt[a + I*a*Tan[c + d*x]]))/(120*a^4*d*(-I + Tan[c + d*x])^2*Sqrt[a + 
I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.06, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 4729, 3042, 4079, 27, 3042, 4079, 27, 3042, 4079, 27, 3042, 4027, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

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

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {a (9 A-i B)-4 a (i A-B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {a (9 A-i B)-4 a (i A-B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 4079

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\int \frac {a^2 (41 A-9 i B)-2 a^2 (13 i A-3 B) \tan (c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\int \frac {a^2 (41 A-9 i B)-2 a^2 (13 i A-3 B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\int \frac {a^2 (41 A-9 i B)-2 a^2 (13 i A-3 B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 4079

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\frac {\int \frac {15 a^3 (A-i B) \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a^2}+\frac {a^2 (67 A-3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\frac {15}{2} a (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {a^2 (67 A-3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\frac {15}{2} a (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {a^2 (67 A-3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 4027

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\frac {a^2 (67 A-3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {15 i a^3 (A-i B) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{6 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\frac {\left (\frac {15}{2}-\frac {15 i}{2}\right ) a^{3/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (67 A-3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

Input:

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

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((A + I*B)*Sqrt[Tan[c + d*x]])/(5*d 
*(a + I*a*Tan[c + d*x])^(5/2)) + ((a*(13*A + (3*I)*B)*Sqrt[Tan[c + d*x]])/ 
(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + (((15/2 - (15*I)/2)*a^(3/2)*(A - I*B) 
*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]) 
/d + (a^2*(67*A - (3*I)*B)*Sqrt[Tan[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x 
]]))/(6*a^2))/(10*a^2))
 

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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 4027

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

rule 4079

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

rule 4729

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a 
+ b*x])^m*(c*Tan[a + b*x])^m   Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ[u, 
x]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1093 vs. \(2 (172 ) = 344\).

Time = 1.01 (sec) , antiderivative size = 1094, normalized size of antiderivative = 5.11

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1094\)
default	\(\text {Expression too large to display}\)	\(1094\)

Input:

int(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x,method=_R 
ETURNVERBOSE)
 

Output:

-1/240/d*(1/tan(d*x+c))^(1/2)*tan(d*x+c)*(a*(1+I*tan(d*x+c)))^(1/2)*(-12*I 
*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2-90*I*A* 
ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*t 
an(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)^2+15*B*2^(1/2)*ln((2*2^(1/ 
2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/ 
(tan(d*x+c)+I))*a*tan(d*x+c)^4-60*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2 
)*(-I*a)^(1/2)-60*I*B*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d* 
x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)^3+60 
*A*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2 
)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^3-1060*I*A*(a*tan(d*x+c 
)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)+60*I*B*ln((2*2^(1/2)*(-I 
*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d 
*x+c)+I))*2^(1/2)*a*tan(d*x+c)-90*B*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a* 
tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*t 
an(d*x+c)^2+12*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d* 
x+c)^3+268*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+ 
c)^3+15*I*A*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/ 
2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)^4-60*A*2^(1/2) 
*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a* 
tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)+908*A*(a*tan(d*x+c)*(1+I*tan(d...
 

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. 481 vs. \(2 (162) = 324\).

Time = 0.12 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.25 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} + {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} - {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + \sqrt {2} {\left ({\left (-83 i \, A + 3 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, {\left (-32 i \, A - 3 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (-8 i \, A + 3 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \] Input:

integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, al 
gorithm="fricas")
 

Output:

1/120*(15*sqrt(1/2)*a^3*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^5*d^2))*e^(5*I* 
d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) - a^3*d) 
*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2* 
I*d*x + 2*I*c) - 1))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^5*d^2)) + (A - I*B)* 
a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 15*sqrt(1/2)*a^3*d*sqrt(( 
-I*A^2 - 2*A*B + I*B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(4*(sqrt(2)*sqrt 
(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1) 
)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt((-I*A^2 
 - 2*A*B + I*B^2)/(a^5*d^2)) - (A - I*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I* 
c)/(I*A + B)) + sqrt(2)*((-83*I*A + 3*B)*e^(6*I*d*x + 6*I*c) - 2*(-32*I*A 
- 3*B)*e^(4*I*d*x + 4*I*c) - 2*(-8*I*A + 3*B)*e^(2*I*d*x + 2*I*c) + 3*I*A 
- 3*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/ 
(e^(2*I*d*x + 2*I*c) - 1)))*e^(-5*I*d*x - 5*I*c)/(a^3*d)
 

Sympy [F]

\[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {\cot {\left (c + d x \right )}}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(cot(d*x+c)**(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Integral((A + B*tan(c + d*x))*sqrt(cot(c + d*x))/(I*a*(tan(c + d*x) - I))* 
*(5/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, al 
gorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, al 
gorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {-\left (\int \frac {\sqrt {\cot \left (d x +c \right )}}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x \right ) a -\left (\int \frac {\sqrt {\cot \left (d x +c \right )}\, \tan \left (d x +c \right )}{\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}}d x \right ) b}{\sqrt {a}\, a^{2}} \] Input:

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

Output:

( - (int(sqrt(cot(c + d*x))/(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2 - 2* 
sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)),x)*a + 
 int((sqrt(cot(c + d*x))*tan(c + d*x))/(sqrt(tan(c + d*x)*i + 1)*tan(c + d 
*x)**2 - 2*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 
 1)),x)*b))/(sqrt(a)*a**2)