\(\int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [86]

Optimal result

Integrand size = 36, antiderivative size = 158 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {a^{5/2} (5 i A+2 B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {4 \sqrt {2} a^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {a^2 (i A-2 B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^{3/2}}{d} \] Output:

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

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.82 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {-i a^{5/2} (5 A-2 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )+4 \sqrt {2} a^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+a^2 (-2 B-A \cot (c+d x)) \sqrt {a+i a \tan (c+d x)}}{d} \] Input:

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

Output:

((-I)*a^(5/2)*(5*A - (2*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] 
+ 4*Sqrt[2]*a^(5/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]* 
Sqrt[a])] + a^2*(-2*B - A*Cot[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/d
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3042, 4076, 27, 3042, 4077, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4076

\(\displaystyle \int \frac {1}{2} \cot (c+d x) (i \tan (c+d x) a+a)^{3/2} (a (5 i A+2 B)+a (A+2 i B) \tan (c+d x))dx-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int \cot (c+d x) (i \tan (c+d x) a+a)^{3/2} (a (5 i A+2 B)+a (A+2 i B) \tan (c+d x))dx-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \frac {(i \tan (c+d x) a+a)^{3/2} (a (5 i A+2 B)+a (A+2 i B) \tan (c+d x))}{\tan (c+d x)}dx-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^{3/2}}{d}\)

\(\Big \downarrow \) 4077

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4083

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3961

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-((a*A*Cot[c + d*x]*(a + I*a*Tan[c + d*x])^(3/2))/d) + ((-2*a^(5/2)*((5*I) 
*A + 2*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + ((8*I)*Sqrt[2]* 
a^(5/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d 
 + (2*a^2*(I*A - 2*B)*Sqrt[a + I*a*Tan[c + d*x]])/d)/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 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 4076

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

rule 4077

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*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + 
n))), x] + Simp[1/(d*(m + n))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan 
[e + f*x])^n*Simp[a*A*d*(m + n) + B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - 
 a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a 
, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && 
GtQ[m, 1] &&  !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 = 0.22 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {2 i a^{2} \left (i B \sqrt {a +i a \tan \left (d x +c \right )}+2 \sqrt {a}\, \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-a \left (-\frac {i A \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {\left (-2 i B +5 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\right )}{d}\)	\(131\)
default	\(\frac {2 i a^{2} \left (i B \sqrt {a +i a \tan \left (d x +c \right )}+2 \sqrt {a}\, \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-a \left (-\frac {i A \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {\left (-2 i B +5 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\right )}{d}\)	\(131\)

Input:

int(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x,method=_RETUR 
NVERBOSE)
 

Output:

2*I/d*a^2*(I*B*(a+I*a*tan(d*x+c))^(1/2)+2*a^(1/2)*(A-I*B)*2^(1/2)*arctanh( 
1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-a*(-1/2*I*A*(a+I*a*tan(d*x+c 
))^(1/2)/a/tan(d*x+c)+1/2*(5*A-2*I*B)/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. 705 vs. \(2 (125) = 250\).

Time = 0.09 (sec) , antiderivative size = 705, normalized size of antiderivative = 4.46 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx =\text {Too large to display} \] Input:

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algori 
thm="fricas")
 

Output:

-1/4*(8*sqrt(2)*sqrt(-(A^2 - 2*I*A*B - B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c 
) - d)*log(4*((-I*A - B)*a^3*e^(I*d*x + I*c) + sqrt(-(A^2 - 2*I*A*B - B^2) 
*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e 
^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 8*sqrt(2)*sqrt(-(A^2 - 2*I*A*B - B^2)* 
a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(4*((-I*A - B)*a^3*e^(I*d*x + I*c) 
 - sqrt(-(A^2 - 2*I*A*B - B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a 
/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - sqrt(-(2 
5*A^2 - 20*I*A*B - 4*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(-16*(3* 
(-5*I*A - 2*B)*a^3*e^(2*I*d*x + 2*I*c) + (-5*I*A - 2*B)*a^3 + 2*sqrt(2)*sq 
rt(-(25*A^2 - 20*I*A*B - 4*B^2)*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)))*e^(-2*I*d*x - 2*I*c)/((5*I*A 
 + 2*B)*a)) + sqrt(-(25*A^2 - 20*I*A*B - 4*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2 
*I*c) - d)*log(-16*(3*(-5*I*A - 2*B)*a^3*e^(2*I*d*x + 2*I*c) + (-5*I*A - 2 
*B)*a^3 - 2*sqrt(2)*sqrt(-(25*A^2 - 20*I*A*B - 4*B^2)*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)))*e^(-2* 
I*d*x - 2*I*c)/((5*I*A + 2*B)*a)) + 4*sqrt(2)*((I*A + 2*B)*a^2*e^(3*I*d*x 
+ 3*I*c) + (I*A - 2*B)*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 
1)))/(d*e^(2*I*d*x + 2*I*c) - d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.03 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {i \, {\left (4 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - {\left (5 \, A - 2 i \, B\right )} a^{\frac {3}{2}} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right ) - 4 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} B a - \frac {2 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} A a}{\tan \left (d x + c\right )}\right )} a}{2 \, d} \] Input:

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algori 
thm="maxima")
 

Output:

-1/2*I*(4*sqrt(2)*(A - I*B)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d 
*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) - (5*A - 2*I 
*B)*a^(3/2)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + 
 c) + a) + sqrt(a))) - 4*I*sqrt(I*a*tan(d*x + c) + a)*B*a - 2*I*sqrt(I*a*t 
an(d*x + c) + a)*A*a/tan(d*x + c))*a/d
 

Giac [F(-2)]

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

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algori 
thm="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 TypeError: Bad Argument TypeError: Bad Argument TypeError: Bad Ar 
gument Ty
 

Mupad [B] (verification not implemented)

Time = 5.48 (sec) , antiderivative size = 2947, normalized size of antiderivative = 18.65 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

atan((d^4*(a + a*tan(c + d*x)*1i)^(1/2)*(((49*A^4*a^22)/d^4 + (784*B^4*a^2 
2)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*61 
6i)/d^4)^(1/2)/(8*a^6) - (57*A^2*a^5)/(8*d^2) + (9*B^2*a^5)/(2*d^2) + (A*B 
*a^5*21i)/(2*d^2))^(1/2)*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A 
^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)*6 
i)/(A^3*a^14*d*126i - 336*B^3*a^14*d - A*B^2*a^14*d*1032i + 876*A^2*B*a^14 
*d + A*a^3*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^2 
2)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)*2i - 4*B*a^ 
3*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + 
(A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)) + (A^2*a^8*d^2*(a + 
 a*tan(c + d*x)*1i)^(1/2)*(((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328 
*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2) 
/(8*a^6) - (57*A^2*a^5)/(8*d^2) + (9*B^2*a^5)/(2*d^2) + (A*B*a^5*21i)/(2*d 
^2))^(1/2)*14i)/(A^3*a^11*d*126i - 336*B^3*a^11*d + A*d^3*((49*A^4*a^22)/d 
^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 
 - (A^3*B*a^22*616i)/d^4)^(1/2)*2i - 4*B*d^3*((49*A^4*a^22)/d^4 + (784*B^4 
*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^2 
2*616i)/d^4)^(1/2) - A*B^2*a^11*d*1032i + 876*A^2*B*a^11*d) - (B^2*a^8*d^2 
*(a + a*tan(c + d*x)*1i)^(1/2)*(((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - 
(2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^...
 

Reduce [F]

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

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

Output:

sqrt(a)*a**2*( - int(sqrt(tan(c + d*x)*i + 1)*cot(c + d*x)**2*tan(c + d*x) 
**3,x)*b - int(sqrt(tan(c + d*x)*i + 1)*cot(c + d*x)**2*tan(c + d*x)**2,x) 
*a + 2*int(sqrt(tan(c + d*x)*i + 1)*cot(c + d*x)**2*tan(c + d*x)**2,x)*b*i 
 + 2*int(sqrt(tan(c + d*x)*i + 1)*cot(c + d*x)**2*tan(c + d*x),x)*a*i + in 
t(sqrt(tan(c + d*x)*i + 1)*cot(c + d*x)**2*tan(c + d*x),x)*b + int(sqrt(ta 
n(c + d*x)*i + 1)*cot(c + d*x)**2,x)*a)