\(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [98]

Optimal result

Integrand size = 26, antiderivative size = 184 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {11 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d} \] Output:

11/4*a^(3/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/d-2*2^(1/2)*a^(3/2) 
*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d-1/2*I*a^2*cot(d*x 
+c)/d/(a+I*a*tan(d*x+c))^(1/2)-1/2*a^2*cot(d*x+c)^2/d/(a+I*a*tan(d*x+c))^( 
1/2)-5/4*I*a*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d
 

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.64 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {-11 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )+8 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+a \cot (c+d x) (5 i+2 \cot (c+d x)) \sqrt {a+i a \tan (c+d x)}}{4 d} \] Input:

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

Output:

-1/4*(-11*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] + 8*Sqrt[2]* 
a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] + a*Cot[c + 
d*x]*(5*I + 2*Cot[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/d
 

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.08, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3042, 4036, 27, 3042, 4079, 3042, 4081, 27, 3042, 4083, 3042, 3961, 219, 4082, 73, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4036

\(\displaystyle -\frac {1}{2} \int -\frac {\cot ^2(c+d x) \left (7 i a^2-9 a^2 \tan (c+d x)\right )}{2 \sqrt {i \tan (c+d x) a+a}}dx-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int \frac {\cot ^2(c+d x) \left (7 i a^2-9 a^2 \tan (c+d x)\right )}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \frac {7 i a^2-9 a^2 \tan (c+d x)}{\tan (c+d x)^2 \sqrt {i \tan (c+d x) a+a}}dx-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4079

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4081

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\int -\frac {1}{2} \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (5 i \tan (c+d x) a^4+11 a^4\right )dx}{a}-\frac {5 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{a^2}-\frac {2 i a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {\int \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (5 i \tan (c+d x) a^4+11 a^4\right )dx}{2 a}-\frac {5 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{a^2}-\frac {2 i a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (5 i \tan (c+d x) a^4+11 a^4\right )}{\tan (c+d x)}dx}{2 a}-\frac {5 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{a^2}-\frac {2 i a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4083

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {16 i a^4 \int \sqrt {i \tan (c+d x) a+a}dx+11 a^3 \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 a}-\frac {5 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{a^2}-\frac {2 i a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {16 i a^4 \int \sqrt {i \tan (c+d x) a+a}dx+11 a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{2 a}-\frac {5 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{a^2}-\frac {2 i a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {\frac {32 a^5 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+11 a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{2 a}-\frac {5 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{a^2}-\frac {2 i a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {11 a^3 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx+\frac {16 \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {5 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{a^2}-\frac {2 i a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {\frac {11 a^5 \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}+\frac {16 \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {5 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{a^2}-\frac {2 i a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {\frac {16 \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {22 i a^4 \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}}{2 a}-\frac {5 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{a^2}-\frac {2 i a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {\frac {16 \sqrt {2} a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {22 a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}}{2 a}-\frac {5 i a^3 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{a^2}-\frac {2 i a^2 \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\)

Input:

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

Output:

-1/2*(a^2*Cot[c + d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((-2*I)*a^2*Co 
t[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (-1/2*((-22*a^(9/2)*ArcTanh[S 
qrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + (16*Sqrt[2]*a^(9/2)*ArcTanh[Sqrt[a 
 + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d)/a - ((5*I)*a^3*Cot[c + d*x]*Sq 
rt[a + I*a*Tan[c + d*x]])/d)/a^2)/4
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 221

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

rule 3042

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

rule 3961

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a 
, b, c, d}, x] && EqQ[a^2 + b^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 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 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 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 4083

Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)]))/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[( 
A*b + a*B)/(b*c + a*d)   Int[(a + b*Tan[e + f*x])^m, x], x] - Simp[(B*c - A 
*d)/(b*c + a*d)   Int[(a + b*Tan[e + f*x])^m*((a - b*Tan[e + f*x])/(c + d*T 
an[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - 
 a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
 

Maple [A] (verified)

Time = 1.51 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.64

method	result	size
derivativedivides	\(\frac {2 a^{4} \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{a^{\frac {5}{2}}}+\frac {-\frac {\frac {5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{8}-\frac {3 a \sqrt {a +i a \tan \left (d x +c \right )}}{8}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {11 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{2}}\right )}{d}\)	\(117\)
default	\(\frac {2 a^{4} \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{a^{\frac {5}{2}}}+\frac {-\frac {\frac {5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{8}-\frac {3 a \sqrt {a +i a \tan \left (d x +c \right )}}{8}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {11 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{2}}\right )}{d}\)	\(117\)

Input:

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

Output:

2/d*a^4*(-1/a^(5/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a 
^(1/2))+1/a^2*(-(5/8*(a+I*a*tan(d*x+c))^(3/2)-3/8*a*(a+I*a*tan(d*x+c))^(1/ 
2))/a^2/tan(d*x+c)^2+11/8/a^(1/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/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. 546 vs. \(2 (141) = 282\).

Time = 0.09 (sec) , antiderivative size = 546, normalized size of antiderivative = 2.97 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/16*(16*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sq 
rt(a^3/d^2)*log(4*(a^2*e^(I*d*x + I*c) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt( 
a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a) - 16*sqrt( 
2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*log 
(4*(a^2*e^(I*d*x + I*c) - (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*sqrt(a 
/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a) - 11*(d*e^(4*I*d*x + 4*I* 
c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*log(16*(3*a^2*e^(2*I*d*x + 
 2*I*c) + 2*sqrt(2)*(d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c))*sqrt(a^3/d 
^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) + a^2)*e^(-2*I*d*x - 2*I*c)) + 11*(d 
*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*log(16*( 
3*a^2*e^(2*I*d*x + 2*I*c) - 2*sqrt(2)*(d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x 
+ I*c))*sqrt(a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) + a^2)*e^(-2*I*d*x 
 - 2*I*c)) - 4*sqrt(2)*(7*a*e^(5*I*d*x + 5*I*c) + 4*a*e^(3*I*d*x + 3*I*c) 
- 3*a*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(4*I*d*x + 
4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)
 

Sympy [F]

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

integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**(3/2),x)
 

Output:

Integral((I*a*(tan(c + d*x) - I))**(3/2)*cot(c + d*x)**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.97 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {a^{2} {\left (\frac {8 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} - \frac {11 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{\sqrt {a}} + \frac {2 \, {\left (5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + a^{2}}\right )}}{8 \, d} \] Input:

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

Output:

1/8*a^2*(8*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sq 
rt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a)))/sqrt(a) - 11*log((sqrt(I*a*ta 
n(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/sqrt(a) 
 + 2*(5*(I*a*tan(d*x + c) + a)^(3/2) - 3*sqrt(I*a*tan(d*x + c) + a)*a)/((I 
*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2))/d
 

Giac [F(-2)]

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

integrate(cot(d*x+c)^3*(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 [B] (verification not implemented)

Time = 1.02 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.74 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {\mathrm {atan}\left (\frac {\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{a^2}\right )\,\sqrt {a^3}\,11{}\mathrm {i}}{4\,d}-\frac {5\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{4\,d\,{\mathrm {tan}\left (c+d\,x\right )}^2}+\frac {3\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{4\,d\,{\mathrm {tan}\left (c+d\,x\right )}^2}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,a^2}\right )\,\sqrt {a^3}\,2{}\mathrm {i}}{d} \] Input:

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

Output:

(3*a*(a + a*tan(c + d*x)*1i)^(1/2))/(4*d*tan(c + d*x)^2) - (5*(a + a*tan(c 
 + d*x)*1i)^(3/2))/(4*d*tan(c + d*x)^2) - (atan(((a^3)^(1/2)*(a + a*tan(c 
+ d*x)*1i)^(1/2)*1i)/a^2)*(a^3)^(1/2)*11i)/(4*d) + (2^(1/2)*atan((2^(1/2)* 
(a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^2))*(a^3)^(1/2)*2i)/d
 

Reduce [F]

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

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

Output:

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