\(\int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx\) [766]

Optimal result

Integrand size = 28, antiderivative size = 179 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx=\frac {5 \sqrt [4]{-1} a^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(4-4 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \] Output:

5*(-1)^(1/4)*a^(5/2)*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)/d+(4-4*I)*a^(5/2)*arctan 
h((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-a^2*(a+I*a*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(1/2)
 

Mathematica [A] (verified)

Time = 2.89 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.25 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {i a^2 \sqrt {\cot (c+d x)} \left (-5 i \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {i a \tan (c+d x)} (-i+\tan (c+d x))+\sqrt {1+i \tan (c+d x)} \left (a \tan (c+d x) (-i+\tan (c+d x))+4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}\right )\right )}{d \sqrt {1+i \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.93, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.464, Rules used = {3042, 4729, 3042, 4039, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4039

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4084

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 65

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{2} a \left (\frac {(8-8 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {10 a^2 \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}\right )-\frac {a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {1}{2} a \left (\frac {10 \sqrt [4]{-1} a^{3/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(8-8 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\right )-\frac {a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )\)

Input:

Int[Sqrt[Cot[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*((10*(-1)^(1/4)*a^(3/2)*ArcTan[( 
(-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + (( 
8 - 8*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a 
*Tan[c + d*x]]])/d))/2 - (a^2*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x] 
])/d)
 

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 4039

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + 
 d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[a/(d*(m + n - 1)) 
Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + 
a*d*(m + 2*n) + (a*c*(m - 2) + b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x 
] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] 
 && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

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. 370 vs. \(2 (144 ) = 288\).

Time = 0.40 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.07

method	result	size
derivativedivides	\(\frac {\sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (2 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a -2 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +8 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -5 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \sqrt {i a}}\)	\(371\)
default	\(\frac {\sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (2 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a -2 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +8 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -5 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \sqrt {i a}}\)	\(371\)

Input:

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

Output:

1/2/d*(1/tan(d*x+c))^(1/2)*tan(d*x+c)*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*(2*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)+3 
*a*tan(d*x+c)-I*a)/(tan(d*x+c)+I))*(I*a)^(1/2)*a-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))*(I*a)^(1/2)*a+8*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-5*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-2*(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. 648 vs. \(2 (137) = 274\).

Time = 0.11 (sec) , antiderivative size = 648, normalized size of antiderivative = 3.62 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/4*(4*sqrt(2)*(-I*a^2*e^(3*I*d*x + 3*I*c) + 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)) + sqrt(-25*I*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(-16/5 
*(15*a^3*e^(2*I*d*x + 2*I*c) - 5*a^3 + 2*sqrt(2)*sqrt(-25*I*a^5/d^2)*(d*e^ 
(3*I*d*x + 3*I*c) - d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*s 
qrt((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)/a) - sqrt(-25*I*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(-16/5*(15* 
a^3*e^(2*I*d*x + 2*I*c) - 5*a^3 - 2*sqrt(2)*sqrt(-25*I*a^5/d^2)*(d*e^(3*I* 
d*x + 3*I*c) - 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)))*e^(-2*I*d*x - 2*I*c 
)/a) - sqrt(-128*I*a^5/d^2)*(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) + sqrt(-128*I*a^5/d^2)*(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^(2*I*d*x + 2*I*c) + d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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