\(\int \cot ^{\frac {9}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [762]

Optimal result

Integrand size = 28, antiderivative size = 222 \[ \int \cot ^{\frac {9}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {(4-4 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {104 i a^2 \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{21 d}+\frac {32 a^2 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{21 d}-\frac {6 i a^2 \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {2 a^2 \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d} \] Output:

(4-4*I)*a^(5/2)*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)/d+104/21*I*a^2*cot(d*x+c)^(1/2)*( 
a+I*a*tan(d*x+c))^(1/2)/d+32/21*a^2*cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(1 
/2)/d-6/7*I*a^2*cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(1/2)/d-2/7*a^2*cot(d* 
x+c)^(7/2)*(a+I*a*tan(d*x+c))^(1/2)/d
 

Mathematica [A] (verified)

Time = 3.35 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.75 \[ \int \cot ^{\frac {9}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {4 i a^2 e^{-i (c+d x)} \left (-21 e^{i (c+d x)}+70 e^{3 i (c+d x)}-77 e^{5 i (c+d x)}+40 e^{7 i (c+d x)}-21 \left (-1+e^{2 i (c+d x)}\right )^{7/2} \text {arctanh}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{21 d \left (-1+e^{2 i (c+d x)}\right )^3} \] Input:

Integrate[Cot[c + d*x]^(9/2)*(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

(((4*I)/21)*a^2*(-21*E^(I*(c + d*x)) + 70*E^((3*I)*(c + d*x)) - 77*E^((5*I 
)*(c + d*x)) + 40*E^((7*I)*(c + d*x)) - 21*(-1 + E^((2*I)*(c + d*x)))^(7/2 
)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Cot[c + d* 
x]]*Sqrt[a + I*a*Tan[c + d*x]])/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x 
)))^3)
 

Rubi [A] (verified)

Time = 1.30 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.07, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.607, Rules used = {3042, 4729, 3042, 4036, 27, 3042, 4081, 27, 3042, 4081, 27, 3042, 4081, 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.

\(\displaystyle \int \cot ^{\frac {9}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cot (c+d x)^{9/2} (a+i a \tan (c+d x))^{5/2}dx\)

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4036

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2}{7} \int -\frac {\sqrt {i \tan (c+d x) a+a} \left (15 i a^2-13 a^2 \tan (c+d x)\right )}{2 \tan ^{\frac {7}{2}}(c+d x)}dx-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \int \frac {\sqrt {i \tan (c+d x) a+a} \left (15 i a^2-13 a^2 \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)}dx-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \int \frac {\sqrt {i \tan (c+d x) a+a} \left (15 i a^2-13 a^2 \tan (c+d x)\right )}{\tan (c+d x)^{7/2}}dx-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4081

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (\frac {2 \int -\frac {10 \sqrt {i \tan (c+d x) a+a} \left (3 i \tan (c+d x) a^3+4 a^3\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{5 a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (-\frac {4 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (3 i \tan (c+d x) a^3+4 a^3\right )}{\tan ^{\frac {5}{2}}(c+d x)}dx}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (-\frac {4 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (3 i \tan (c+d x) a^3+4 a^3\right )}{\tan (c+d x)^{5/2}}dx}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4081

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (-\frac {4 \left (\frac {2 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (13 i a^4-8 a^4 \tan (c+d x)\right )}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (-\frac {4 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (13 i a^4-8 a^4 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (-\frac {4 \left (\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (13 i a^4-8 a^4 \tan (c+d x)\right )}{\tan (c+d x)^{3/2}}dx}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4081

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (-\frac {4 \left (\frac {\frac {2 \int -\frac {21 a^5 \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (-\frac {4 \left (\frac {-21 a^4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (-\frac {4 \left (\frac {-21 a^4 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 4027

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (-\frac {4 \left (\frac {\frac {42 i a^6 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{7} \left (-\frac {6 i a^2 \sqrt {a+i a \tan (c+d x)}}{d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 \left (\frac {-\frac {(21-21 i) a^{9/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {26 i a^4 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 a^3 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\right )}{a}\right )-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{7 d \tan ^{\frac {7}{2}}(c+d x)}\right )\)

Input:

Int[Cot[c + d*x]^(9/2)*(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-2*a^2*Sqrt[a + I*a*Tan[c + d*x]]) 
/(7*d*Tan[c + d*x]^(7/2)) + (((-6*I)*a^2*Sqrt[a + I*a*Tan[c + d*x]])/(d*Ta 
n[c + d*x]^(5/2)) - (4*((-8*a^3*Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Tan[c + d 
*x]^(3/2)) + (((-21 + 21*I)*a^(9/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + 
d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - ((26*I)*a^4*Sqrt[a + I*a*Tan[c + d 
*x]])/(d*Sqrt[Tan[c + d*x]]))/(3*a)))/a)/7)
 

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 4036

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x] 
)^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] + Si 
mp[a/(d*(b*c + a*d)*(n + 1))   Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[ 
e + f*x])^(n + 1)*Simp[b*(b*c*(m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) 
 + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[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] && GtQ[m, 1] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4081

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

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. 466 vs. \(2 (179 ) = 358\).

Time = 0.63 (sec) , antiderivative size = 467, normalized size of antiderivative = 2.10

method	result	size
derivativedivides	\(\frac {\left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {9}{2}} \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (21 i \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, \sqrt {i a}\, a \tan \left (d x +c \right )^{4}+84 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )^{4}-21 \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, \sqrt {i a}\, a \tan \left (d x +c \right )^{4}+32 \tan \left (d x +c \right )^{2} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+104 i \sqrt {-i a}\, \sqrt {i a}\, \tan \left (d x +c \right )^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-18 i \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-6 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{21 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(467\)
default	\(\frac {\left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {9}{2}} \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (21 i \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, \sqrt {i a}\, a \tan \left (d x +c \right )^{4}+84 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )^{4}-21 \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, \sqrt {i a}\, a \tan \left (d x +c \right )^{4}+32 \tan \left (d x +c \right )^{2} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+104 i \sqrt {-i a}\, \sqrt {i a}\, \tan \left (d x +c \right )^{3} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-18 i \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-6 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{21 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(467\)

Input:

int(cot(d*x+c)^(9/2)*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/21/d*(1/tan(d*x+c))^(9/2)*tan(d*x+c)*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(21* 
I*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)*(I*a)^(1/2)*a*tan(d*x+c)^4+84*I*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))*(-I*a)^(1/2)*a*tan(d*x+c)^4-21*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)*(I*a)^(1/2)*a*tan(d*x+c)^4+32*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*ta 
n(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+104*I*(-I*a)^(1/2)*(I*a)^(1/2)*t 
an(d*x+c)^3*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-18*I*tan(d*x+c)*(a*tan(d 
*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)-6*(a*tan(d*x+c)*(1+ 
I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+ 
c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)
 

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. 466 vs. \(2 (168) = 336\).

Time = 0.13 (sec) , antiderivative size = 466, normalized size of antiderivative = 2.10 \[ \int \cot ^{\frac {9}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {16 \, \sqrt {2} {\left (-40 i \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} + 77 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 70 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 21 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} - 21 \, \sqrt {-\frac {128 i \, a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (16 i \, a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {-\frac {128 i \, a^{5}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right ) + 21 \, \sqrt {-\frac {128 i \, a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (16 i \, a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {-\frac {128 i \, a^{5}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right )}{84 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \] Input:

integrate(cot(d*x+c)^(9/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

-1/84*(16*sqrt(2)*(-40*I*a^2*e^(7*I*d*x + 7*I*c) + 77*I*a^2*e^(5*I*d*x + 5 
*I*c) - 70*I*a^2*e^(3*I*d*x + 3*I*c) + 21*I*a^2*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)) - 21*sqrt(-128*I*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d 
*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)*log(1/4*(16*I*a^3*e^(I*d*x + I* 
c) + sqrt(2)*sqrt(-128*I*a^5/d^2)*(I*d*e^(2*I*d*x + 2*I*c) - I*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)))*e^(-I*d*x - I*c)/a^2) + 21*sqrt(-128*I*a^5/d^2)*(d*e^(6*I*d*x 
 + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)*log(1/4 
*(16*I*a^3*e^(I*d*x + I*c) + sqrt(2)*sqrt(-128*I*a^5/d^2)*(-I*d*e^(2*I*d*x 
 + 2*I*c) + I*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)))*e^(-I*d*x - I*c)/a^2))/(d*e^(6*I*d*x 
 + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^{\frac {9}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)**(9/2)*(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3165 vs. \(2 (168) = 336\).

Time = 0.46 (sec) , antiderivative size = 3165, normalized size of antiderivative = 14.26 \[ \int \cot ^{\frac {9}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \] Input:

integrate(cot(d*x+c)^(9/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

-2/105*(2*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c 
) + 1)*(3*(-(35*I - 35)*a^2*cos(7*d*x + 7*c) + (35*I - 35)*a^2*cos(5*d*x + 
 5*c) - (21*I - 21)*a^2*cos(3*d*x + 3*c) + (I - 1)*a^2*cos(d*x + c) + (35* 
I + 35)*a^2*sin(7*d*x + 7*c) - (35*I + 35)*a^2*sin(5*d*x + 5*c) + (21*I + 
21)*a^2*sin(3*d*x + 3*c) - (I + 1)*a^2*sin(d*x + c))*cos(7/2*arctan2(sin(2 
*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 5*(13*((I - 1)*a^2*cos(d*x + c) - (I 
 + 1)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (13*I - 13)*a^2*cos(d*x + c) 
+ 13*((I - 1)*a^2*cos(d*x + c) - (I + 1)*a^2*sin(d*x + c))*sin(2*d*x + 2*c 
)^2 - (13*I + 13)*a^2*sin(d*x + c) + 21*(-(I - 1)*a^2*cos(2*d*x + 2*c)^2 - 
 (I - 1)*a^2*sin(2*d*x + 2*c)^2 + (2*I - 2)*a^2*cos(2*d*x + 2*c) - (I - 1) 
*a^2)*cos(3*d*x + 3*c) + 26*(-(I - 1)*a^2*cos(d*x + c) + (I + 1)*a^2*sin(d 
*x + c))*cos(2*d*x + 2*c) + 21*((I + 1)*a^2*cos(2*d*x + 2*c)^2 + (I + 1)*a 
^2*sin(2*d*x + 2*c)^2 - (2*I + 2)*a^2*cos(2*d*x + 2*c) + (I + 1)*a^2)*sin( 
3*d*x + 3*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 3 
*(-(35*I + 35)*a^2*cos(7*d*x + 7*c) + (35*I + 35)*a^2*cos(5*d*x + 5*c) - ( 
21*I + 21)*a^2*cos(3*d*x + 3*c) + (I + 1)*a^2*cos(d*x + c) - (35*I - 35)*a 
^2*sin(7*d*x + 7*c) + (35*I - 35)*a^2*sin(5*d*x + 5*c) - (21*I - 21)*a^2*s 
in(3*d*x + 3*c) + (I - 1)*a^2*sin(d*x + c))*sin(7/2*arctan2(sin(2*d*x + 2* 
c), cos(2*d*x + 2*c) - 1)) + 5*(13*((I + 1)*a^2*cos(d*x + c) + (I - 1)*a^2 
*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (13*I + 13)*a^2*cos(d*x + c) + 13*(...
 

Giac [F(-2)]

Exception generated. \[ \int \cot ^{\frac {9}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(d*x+c)^(9/2)*(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. \[ \int \cot ^{\frac {9}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{9/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \] Input:

int(cot(c + d*x)^(9/2)*(a + a*tan(c + d*x)*1i)^(5/2),x)
 

Output:

int(cot(c + d*x)^(9/2)*(a + a*tan(c + d*x)*1i)^(5/2), x)
 

Reduce [F]

\[ \int \cot ^{\frac {9}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (-\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cot \left (d x +c \right )}\, \cot \left (d x +c \right )^{4} \tan \left (d x +c \right )^{2}d x \right )+2 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cot \left (d x +c \right )}\, \cot \left (d x +c \right )^{4} \tan \left (d x +c \right )d x \right ) i +\int \sqrt {\tan \left (d x +c \right ) i +1}\, \sqrt {\cot \left (d x +c \right )}\, \cot \left (d x +c \right )^{4}d x \right ) \] Input:

int(cot(d*x+c)^(9/2)*(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

sqrt(a)*a**2*( - int(sqrt(tan(c + d*x)*i + 1)*sqrt(cot(c + d*x))*cot(c + d 
*x)**4*tan(c + d*x)**2,x) + 2*int(sqrt(tan(c + d*x)*i + 1)*sqrt(cot(c + d* 
x))*cot(c + d*x)**4*tan(c + d*x),x)*i + int(sqrt(tan(c + d*x)*i + 1)*sqrt( 
cot(c + d*x))*cot(c + d*x)**4,x))