\(\int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [758]

Optimal result

Integrand size = 28, antiderivative size = 139 \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {(2-2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {2 i a \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \] Output:

(-2+2*I)*a^(3/2)*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-2*I*a*cot(d*x+c)^(1/2)*(a+I*a* 
tan(d*x+c))^(1/2)/d-2/3*cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(3/2)/d
 

Mathematica [A] (verified)

Time = 2.58 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.99 \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2 a \cot ^{\frac {3}{2}}(c+d x) \left (-3 \sqrt [4]{-1} a \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \tan ^{\frac {3}{2}}(c+d x) (-i+\tan (c+d x))-\frac {3 i \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) (i a \tan (c+d x))^{3/2} (-i+\tan (c+d x))}{\sqrt {a}}+\sqrt {1+i \tan (c+d x)} \left (\frac {3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (i a \tan (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{a}+a \left (-1-5 i \tan (c+d x)+4 \tan ^2(c+d x)\right )\right )\right )}{3 d \sqrt {1+i \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

(2*a*Cot[c + d*x]^(3/2)*(-3*(-1)^(1/4)*a*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d 
*x]]]*Tan[c + d*x]^(3/2)*(-I + Tan[c + d*x]) - ((3*I)*ArcSinh[Sqrt[I*a*Tan 
[c + d*x]]/Sqrt[a]]*(I*a*Tan[c + d*x])^(3/2)*(-I + Tan[c + d*x]))/Sqrt[a] 
+ Sqrt[1 + I*Tan[c + d*x]]*((3*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d 
*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*(I*a*Tan[c + d*x])^(3/2)*Sqrt[a + I*a*Ta 
n[c + d*x]])/a + a*(-1 - (5*I)*Tan[c + d*x] + 4*Tan[c + d*x]^2))))/(3*d*Sq 
rt[1 + I*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4729, 3042, 4031, 3042, 4028, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4031

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4028

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

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

Input:

Int[Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-2*(a + I*a*Tan[c + d*x])^(3/2))/( 
3*d*Tan[c + d*x]^(3/2)) + I*(((2 + 2*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*S 
qrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*a*Sqrt[a + I*a*Tan[ 
c + d*x]])/(d*Sqrt[Tan[c + d*x]])))
 

Defintions of rubi rules used

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 4028

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*b*(a + b*Tan[e + f*x])^(m - 1)*((c + 
 d*Tan[e + f*x])^(n + 1)/(f*(m - 1)*(a*c - b*d))), x] + Simp[2*(a^2/(a*c - 
b*d))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), 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] && EqQ[m + n, 0] && GtQ[m, 1/2]
 

rule 4031

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-d)*(a + b*Tan[e + f*x])^m*((c + d*Ta 
n[e + f*x])^(n + 1)/(f*m*(c^2 + d^2))), x] + Simp[a/(a*c - b*d)   Int[(a + 
b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d 
, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^ 
2, 0] && EqQ[m + n + 1, 0] &&  !LtQ[m, -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. 377 vs. \(2 (113 ) = 226\).

Time = 0.41 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.72

method	result	size
derivativedivides	\(-\frac {\left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {5}{2}} \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (3 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \tan \left (d x +c \right )^{2}+12 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{2}-3 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \tan \left (d x +c \right )^{2}+16 i \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{6 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(378\)
default	\(-\frac {\left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {5}{2}} \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (3 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \tan \left (d x +c \right )^{2}+12 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )^{2}-3 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \tan \left (d x +c \right )^{2}+16 i \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{6 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(378\)

Input:

int(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/6/d*(1/tan(d*x+c))^(5/2)*tan(d*x+c)*(a*(1+I*tan(d*x+c)))^(1/2)*a*(3*I*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))*(I*a)^(1/2)*a*tan(d*x+c)^2+12*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)^2-3*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))*(I*a)^(1/2)*a*tan(d*x+c)^2+16*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)+4*(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. 362 vs. \(2 (105) = 210\).

Time = 0.09 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.60 \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {8 \, \sqrt {2} {\left (5 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 3 i \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} + 3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {32 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {-\frac {32 i \, a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} + 8 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) - 3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {32 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {-\frac {32 i \, a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} + 8 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right )}{12 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \] Input:

integrate(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-1/12*(8*sqrt(2)*(5*I*a*e^(3*I*d*x + 3*I*c) - 3*I*a*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)) + 3*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-32*I*a^3/d^2)*log(1/2 
*(sqrt(2)*(I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(-32*I*a^3/d^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)) + 8*I*a^2*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) - 3*(d*e^(2*I*d*x + 
 2*I*c) - d)*sqrt(-32*I*a^3/d^2)*log(1/2*(sqrt(2)*(-I*d*e^(2*I*d*x + 2*I*c 
) + I*d)*sqrt(-32*I*a^3/d^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)) + 8*I*a^2*e^(I*d*x + I*c) 
)*e^(-I*d*x - I*c)/a))/(d*e^(2*I*d*x + 2*I*c) - d)
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:

integrate(cot(d*x+c)**(5/2)*(a+I*a*tan(d*x+c))**(3/2),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. 995 vs. \(2 (105) = 210\).

Time = 0.32 (sec) , antiderivative size = 995, normalized size of antiderivative = 7.16 \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Too large to display} \] Input:

integrate(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

1/3*(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*I - 3)*a*cos(3*d*x + 3*c) + (I - 1)*a*cos(d*x + c) + (3*I + 3)*a 
*sin(3*d*x + 3*c) - (I + 1)*a*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x + 2* 
c), cos(2*d*x + 2*c) - 1)) + (-(3*I + 3)*a*cos(3*d*x + 3*c) + (I + 1)*a*co 
s(d*x + c) - (3*I - 3)*a*sin(3*d*x + 3*c) + (I - 1)*a*sin(d*x + c))*sin(3/ 
2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqrt(a) + 3*(2*(-(I + 
1)*a*cos(2*d*x + 2*c)^2 - (I + 1)*a*sin(2*d*x + 2*c)^2 + (2*I + 2)*a*cos(2 
*d*x + 2*c) - (I + 1)*a)*arctan2(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^ 
2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2* 
d*x + 2*c) - 1)) + 2*sin(d*x + c), 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c 
)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos( 
2*d*x + 2*c) - 1)) + 2*cos(d*x + c)) + ((I - 1)*a*cos(2*d*x + 2*c)^2 + (I 
- 1)*a*sin(2*d*x + 2*c)^2 - (2*I - 2)*a*cos(2*d*x + 2*c) + (I - 1)*a)*log( 
4*cos(d*x + c)^2 + 4*sin(d*x + c)^2 + 4*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d* 
x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*d*x + 2*c), co 
s(2*d*x + 2*c) - 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c 
) - 1))^2) + 8*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2* 
c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 
2*c) - 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2* 
c) - 1)))))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*...
 

Giac [F(-2)]

Exception generated. \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(3/2),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. \[ \int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \] Input:

int(cot(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^(3/2),x)
                                                                                    
                                                                                    
 

Output:

int(cot(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^(3/2), x)
 

Reduce [F]

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

int(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(3/2),x)
 

Output:

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