\(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [87]

Optimal result

Integrand size = 36, antiderivative size = 173 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {a^{5/2} (23 A-20 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {4 \sqrt {2} a^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 (7 i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d} \] Output:

1/4*a^(5/2)*(23*A-20*I*B)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/d-4*2^ 
(1/2)*a^(5/2)*(A-I*B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2) 
)/d-1/4*a^2*(7*I*A+4*B)*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-1/2*a*A*cot( 
d*x+c)^2*(a+I*a*tan(d*x+c))^(3/2)/d
 

Mathematica [A] (verified)

Time = 4.44 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.82 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {a^{5/2} (23 A-20 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )-16 \sqrt {2} a^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )-a^2 \cot (c+d x) (9 i A+4 B+2 A \cot (c+d x)) \sqrt {a+i a \tan (c+d x)}}{4 d} \] Input:

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

Output:

(a^(5/2)*(23*A - (20*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] - 1 
6*Sqrt[2]*a^(5/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sq 
rt[a])] - a^2*Cot[c + d*x]*((9*I)*A + 4*B + 2*A*Cot[c + d*x])*Sqrt[a + I*a 
*Tan[c + d*x]])/(4*d)
 

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 181, 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, 4076, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4076

\(\displaystyle \frac {1}{2} \int \frac {1}{2} \cot ^2(c+d x) (i \tan (c+d x) a+a)^{3/2} (a (7 i A+4 B)-a (A-4 i B) \tan (c+d x))dx-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int \cot ^2(c+d x) (i \tan (c+d x) a+a)^{3/2} (a (7 i A+4 B)-a (A-4 i B) \tan (c+d x))dx-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \frac {(i \tan (c+d x) a+a)^{3/2} (a (7 i A+4 B)-a (A-4 i B) \tan (c+d x))}{\tan (c+d x)^2}dx-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 4076

\(\displaystyle \frac {1}{4} \left (\int -\frac {1}{2} \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left ((23 A-20 i B) a^2+3 (3 i A+4 B) \tan (c+d x) a^2\right )dx-\frac {a^2 (4 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (-\frac {1}{2} \int \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left ((23 A-20 i B) a^2+3 (3 i A+4 B) \tan (c+d x) a^2\right )dx-\frac {a^2 (4 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (-\frac {1}{2} \int \frac {\sqrt {i \tan (c+d x) a+a} \left ((23 A-20 i B) a^2+3 (3 i A+4 B) \tan (c+d x) a^2\right )}{\tan (c+d x)}dx-\frac {a^2 (4 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 4083

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (-32 a^2 (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx-a (23 A-20 i B) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx\right )-\frac {a^2 (4 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (-32 a^2 (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx-a (23 A-20 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx\right )-\frac {a^2 (4 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {64 i a^3 (B+i A) \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}-a (23 A-20 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx\right )-\frac {a^2 (4 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {32 i \sqrt {2} a^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-a (23 A-20 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx\right )-\frac {a^2 (4 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {32 i \sqrt {2} a^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^3 (23 A-20 i B) \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}\right )-\frac {a^2 (4 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {2 i a^2 (23 A-20 i B) \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}+\frac {32 i \sqrt {2} a^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}\right )-\frac {a^2 (4 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\frac {2 a^{5/2} (23 A-20 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {32 i \sqrt {2} a^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}\right )-\frac {a^2 (4 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}\)

Input:

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

Output:

-1/2*(a*A*Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^(3/2))/d + (((2*a^(5/2)*(2 
3*A - (20*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + ((32*I)*S 
qrt[2]*a^(5/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[ 
a])])/d)/2 - (a^2*((7*I)*A + 4*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]]) 
/d)/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 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 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 = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\left (-\frac {i B}{2}+\frac {9 A}{8}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+\left (\frac {1}{2} i a B -\frac {7}{8} a A \right ) \sqrt {a +i a \tan \left (d x +c \right )}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {\left (-20 i B +23 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )}{d}\)	\(143\)
default	\(\frac {2 a^{3} \left (-\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\left (-\frac {i B}{2}+\frac {9 A}{8}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+\left (\frac {1}{2} i a B -\frac {7}{8} a A \right ) \sqrt {a +i a \tan \left (d x +c \right )}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {\left (-20 i B +23 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )}{d}\)	\(143\)

Input:

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

Output:

2/d*a^3*(-1/2*(-4*I*B+4*A)*2^(1/2)/a^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^ 
(1/2)*2^(1/2)/a^(1/2))-((-1/2*I*B+9/8*A)*(a+I*a*tan(d*x+c))^(3/2)+(1/2*I*a 
*B-7/8*a*A)*(a+I*a*tan(d*x+c))^(1/2))/a^2/tan(d*x+c)^2+1/8*(23*A-20*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. 774 vs. \(2 (134) = 268\).

Time = 0.11 (sec) , antiderivative size = 774, normalized size of antiderivative = 4.47 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/16*(32*sqrt(2)*sqrt((A^2 - 2*I*A*B - B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I* 
c) - 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)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a 
/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 32*sqrt( 
2)*sqrt((A^2 - 2*I*A*B - B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 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)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 
 2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) + sqrt((529*A^2 - 920*I* 
A*B - 400*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + 
 d)*log(-16*(3*(-23*I*A - 20*B)*a^3*e^(2*I*d*x + 2*I*c) + (-23*I*A - 20*B) 
*a^3 + 2*sqrt(2)*sqrt((529*A^2 - 920*I*A*B - 400*B^2)*a^5/d^2)*(I*d*e^(3*I 
*d*x + 3*I*c) + I*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)/((23*I*A + 20*B)*a)) - sqrt((529*A^2 - 920*I*A*B - 400* 
B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(-1 
6*(3*(-23*I*A - 20*B)*a^3*e^(2*I*d*x + 2*I*c) + (-23*I*A - 20*B)*a^3 + 2*s 
qrt(2)*sqrt((529*A^2 - 920*I*A*B - 400*B^2)*a^5/d^2)*(-I*d*e^(3*I*d*x + 3* 
I*c) - I*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)/((23*I*A + 20*B)*a)) - 4*sqrt(2)*((11*A - 4*I*B)*a^2*e^(5*I*d*x 
+ 5*I*c) + 4*A*a^2*e^(3*I*d*x + 3*I*c) - (7*A - 4*I*B)*a^2*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*...
 

Sympy [F(-1)]

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

integrate(cot(d*x+c)**3*(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 = 0.12 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.19 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {{\left (16 \, \sqrt {2} {\left (A - i \, B\right )} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - {\left (23 \, A - 20 i \, B\right )} \sqrt {a} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right ) + \frac {2 \, {\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (9 \, A - 4 i \, B\right )} a - \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (7 \, A - 4 i \, B\right )} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + a^{2}}\right )} a^{2}}{8 \, d} \] Input:

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

Output:

1/8*(16*sqrt(2)*(A - I*B)*sqrt(a)*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))) - (23*A - 20*I 
*B)*sqrt(a)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + 
 c) + a) + sqrt(a))) + 2*((I*a*tan(d*x + c) + a)^(3/2)*(9*A - 4*I*B)*a - s 
qrt(I*a*tan(d*x + c) + a)*(7*A - 4*I*B)*a^2)/((I*a*tan(d*x + c) + a)^2 - 2 
*(I*a*tan(d*x + c) + a)*a + a^2))*a^2/d
 

Giac [F(-2)]

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

integrate(cot(d*x+c)^3*(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 = 5.60 (sec) , antiderivative size = 2991, normalized size of antiderivative = 17.29 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

2*atanh((17*A^2*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((1041*A^2*a^5)/(128 
*d^2) - ((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22 
)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2)/(64*a^6) - ( 
57*B^2*a^5)/(8*d^2) - (A*B*a^5*243i)/(16*d^2))^(1/2))/(4*((663*A^3*a^11*d) 
/16 - B^3*a^11*d*252i - (7*A*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22) 
/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i 
)/d^4)^(1/2))/8 + (B*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - ( 
1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^( 
1/2)*1i)/2 + 507*A*B^2*a^11*d + (A^2*B*a^11*d*861i)/4)) - (3*d^4*(a + a*ta 
n(c + d*x)*1i)^(1/2)*((1041*A^2*a^5)/(128*d^2) - ((289*A^4*a^22)/(4*d^4) + 
 (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + 
(A^3*B*a^22*884i)/d^4)^(1/2)/(64*a^6) - (57*B^2*a^5)/(8*d^2) - (A*B*a^5*24 
3i)/(16*d^2))^(1/2)*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752* 
A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2)) 
/(2*((663*A^3*a^14*d)/16 - B^3*a^14*d*252i + 507*A*B^2*a^14*d + (A^2*B*a^1 
4*d*861i)/4 - (7*A*a^3*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - 
 (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4) 
^(1/2))/8 + (B*a^3*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (17 
52*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/ 
2)*1i)/2)) + (28*B^2*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((1041*A^2*a...
 

Reduce [F]

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

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