\(\int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\) [787]

Optimal result

Integrand size = 28, antiderivative size = 258 \[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 \sqrt [4]{-1} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}-\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {i}{2 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Output:

2*(-1)^(1/4)*arctan((-1)^(3/4)*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+(1/8-1/8*I)*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)*ta 
n(d*x+c)^(1/2)/a^(5/2)/d-1/5/d/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(5/2)+1 
/2*I/a/d/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2)+7/4/a^2/d/cot(d*x+c)^(1 
/2)/(a+I*a*tan(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 3.92 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i e^{-6 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (1-8 e^{2 i (c+d x)}+48 e^{4 i (c+d x)}-41 e^{6 i (c+d x)}-5 e^{5 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \text {arctanh}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+40 \sqrt {2} e^{5 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \text {arctanh}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt {\cot (c+d x)}}{20 \sqrt {2} a^3 d} \] Input:

Integrate[1/(Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]
 

Output:

((I/20)*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(1 - 8*E^( 
(2*I)*(c + d*x)) + 48*E^((4*I)*(c + d*x)) - 41*E^((6*I)*(c + d*x)) - 5*E^( 
(5*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sq 
rt[-1 + E^((2*I)*(c + d*x))]] + 40*Sqrt[2]*E^((5*I)*(c + d*x))*Sqrt[-1 + E 
^((2*I)*(c + d*x))]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((2*I)*( 
c + d*x))]])*Sqrt[Cot[c + d*x]])/(Sqrt[2]*a^3*d*E^((6*I)*(c + d*x)))
 

Rubi [A] (verified)

Time = 1.51 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.98, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.679, Rules used = {3042, 4729, 3042, 4041, 27, 3042, 4078, 27, 3042, 4078, 27, 3042, 4084, 3042, 4027, 218, 4082, 65, 216}

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.

\(\displaystyle \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cot (c+d x)^{7/2} (a+i a \tan (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 4729

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(i \tan (c+d x) a+a)^{5/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan (c+d x)^{7/2}}{(i \tan (c+d x) a+a)^{5/2}}dx\)

\(\Big \downarrow \) 4041

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {-\frac {\int -\frac {\sqrt {i \tan (c+d x) a+a} \left (7 a^3-8 i a^3 \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)}}dx}{a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (7 a^3-8 i a^3 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (7 a^3-8 i a^3 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 4084

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {8 a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-a^3 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {8 a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-a^3 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 4027

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {8 a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {2 i a^5 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {8 a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {(1-i) a^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 4082

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {8 a^4 \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {(1-i) a^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 65

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {16 a^4 \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {(1-i) a^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {i a \tan ^{\frac {3}{2}}(c+d x)}{d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {-\frac {16 \sqrt [4]{-1} a^{7/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {(1-i) a^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {7 a^2 \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}}{2 a^2}-\frac {\tan ^{\frac {5}{2}}(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\right )\)

Input:

Int[1/(Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]
 

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-1/5*Tan[c + d*x]^(5/2)/(d*(a + I*a 
*Tan[c + d*x])^(5/2)) + ((I*a*Tan[c + d*x]^(3/2))/(d*(a + I*a*Tan[c + d*x] 
)^(3/2)) - (((-16*(-1)^(1/4)*a^(7/2)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c 
 + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - ((1 - I)*a^(7/2)*ArcTanh[((1 + 
I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d)/(2*a^2) - ( 
7*a^2*Sqrt[Tan[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]]))/(2*a^2))/(2*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 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 4041

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* 
((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m)   Int[(a + 
b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n 
 - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta 
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] 
 && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && GtQ[n, 1] && (In 
tegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4078

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

rule 4082

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

rule 4084

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*b + a*B)/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] 
 - Simp[B/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ 
e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - 
a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 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. 967 vs. \(2 (202 ) = 404\).

Time = 0.45 (sec) , antiderivative size = 968, normalized size of antiderivative = 3.75

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(968\)
default	\(\text {Expression too large to display}\)	\(968\)

Input:

int(1/cot(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/80/d/(1/tan(d*x+c))^(7/2)/tan(d*x+c)^3*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*( 
-320*I*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+ 
c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*tan(d*x+c)^3+196*I*(-I*a)^(1/2)*( 
I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+20*(I*a)^(1/ 
2)*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 
)+3*a*tan(d*x+c)-I*a)/(tan(d*x+c)+I))*a*tan(d*x+c)^3+5*I*2^(1/2)*(I*a)^(1/ 
2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+3*a*ta 
n(d*x+c)-I*a)/(tan(d*x+c)+I))*a+80*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2 
*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*tan(d 
*x+c)^4+516*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2) 
*(-I*a)^(1/2)-460*I*(-I*a)^(1/2)*(I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c) 
))^(1/2)*tan(d*x+c)+320*I*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d 
*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*tan(d*x+c)-20* 
(I*a)^(1/2)*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)+3*a*tan(d*x+c)-I*a)/(tan(d*x+c)+I))*a*tan(d*x+c)+5*I*2^(1/2)*(I 
*a)^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2) 
+3*a*tan(d*x+c)-I*a)/(tan(d*x+c)+I))*a*tan(d*x+c)^4-480*(-I*a)^(1/2)*ln(1/ 
2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a) 
/(I*a)^(1/2))*a*tan(d*x+c)^2-30*I*2^(1/2)*(I*a)^(1/2)*ln((2*2^(1/2)*(-I*a) 
^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+3*a*tan(d*x+c)-I*a)/(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. 662 vs. \(2 (192) = 384\).

Time = 0.10 (sec) , antiderivative size = 662, normalized size of antiderivative = 2.57 \[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(1/cot(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas 
")
 

Output:

-1/40*(10*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(2*sqrt( 
2)*(I*a^3*d*e^(2*I*d*x + 2*I*c) - I*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(-1/8*I 
/(a^5*d^2)) - I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 10*a^3*d*sqrt(-1/8* 
I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(2*sqrt(2)*(-I*a^3*d*e^(2*I*d*x + 
2*I*c) + I*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(-1/8*I/(a^5*d^2)) - I*a*e^(I*d* 
x + I*c))*e^(-I*d*x - I*c)) + 10*a^3*d*sqrt(-4*I/(a^5*d^2))*e^(5*I*d*x + 5 
*I*c)*log(-16*(sqrt(2)*(a^4*d*e^(3*I*d*x + 3*I*c) - a^4*d*e^(I*d*x + I*c)) 
*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(-4*I/(a^5*d^2)) + 3*a^2*e^(2*I*d*x + 2*I*c) - a^ 
2)*e^(-2*I*d*x - 2*I*c)) - 10*a^3*d*sqrt(-4*I/(a^5*d^2))*e^(5*I*d*x + 5*I* 
c)*log(16*(sqrt(2)*(a^4*d*e^(3*I*d*x + 3*I*c) - a^4*d*e^(I*d*x + I*c))*sqr 
t(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(-4*I/(a^5*d^2)) - 3*a^2*e^(2*I*d*x + 2*I*c) + a^2)*e 
^(-2*I*d*x - 2*I*c)) - sqrt(2)*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))*(-41*I*e^(6*I*d*x + 6*I 
*c) + 48*I*e^(4*I*d*x + 4*I*c) - 8*I*e^(2*I*d*x + 2*I*c) + I))*e^(-5*I*d*x 
 - 5*I*c)/(a^3*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(1/cot(d*x+c)**(7/2)/(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/cot(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima 
")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/cot(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="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. \[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:

int(1/(cot(c + d*x)^(7/2)*(a + a*tan(c + d*x)*1i)^(5/2)),x)
 

Output:

int(1/(cot(c + d*x)^(7/2)*(a + a*tan(c + d*x)*1i)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\int \frac {1}{\sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cot \left (d x +c \right )}\, \cot \left (d x +c \right )^{3} \tan \left (d x +c \right )^{2}-2 \sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cot \left (d x +c \right )}\, \cot \left (d x +c \right )^{3} \tan \left (d x +c \right ) i -\sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cot \left (d x +c \right )}\, \cot \left (d x +c \right )^{3}}d x}{\sqrt {a}\, a^{2}} \] Input:

int(1/cot(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

( - int(1/(sqrt(tan(c + d*x)*i + 1)*sqrt(cot(c + d*x))*cot(c + d*x)**3*tan 
(c + d*x)**2 - 2*sqrt(tan(c + d*x)*i + 1)*sqrt(cot(c + d*x))*cot(c + d*x)* 
*3*tan(c + d*x)*i - sqrt(tan(c + d*x)*i + 1)*sqrt(cot(c + d*x))*cot(c + d* 
x)**3),x))/(sqrt(a)*a**2)