\(\int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) [73]

Optimal result

Integrand size = 36, antiderivative size = 169 \[ \int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a} (7 A-4 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {\sqrt {2} \sqrt {a} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {(i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d} \] Output:

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

Mathematica [A] (verified)

Time = 1.87 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.82 \[ \int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {a} (7 A-4 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )-4 \sqrt {2} \sqrt {a} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )-\cot (c+d x) (i A+4 B+2 A \cot (c+d x)) \sqrt {a+i a \tan (c+d x)}}{4 d} \] Input:

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

Output:

(Sqrt[a]*(7*A - (4*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] - 4*S 
qrt[2]*Sqrt[a]*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[ 
a])] - Cot[c + d*x]*(I*A + 4*B + 2*A*Cot[c + d*x])*Sqrt[a + I*a*Tan[c + d* 
x]])/(4*d)
 

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.09, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3042, 4081, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4081

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4081

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4083

\(\displaystyle \frac {-\frac {8 a^2 (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx+a (7 A-4 i B) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{2 a}-\frac {a (4 B+i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{4 a}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {8 a^2 (B+i A) \int \sqrt {i \tan (c+d x) a+a}dx+a (7 A-4 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{2 a}-\frac {a (4 B+i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{4 a}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3961

\(\displaystyle \frac {-\frac {a (7 A-4 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {16 i a^3 (B+i A) \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}}{2 a}-\frac {a (4 B+i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{4 a}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {a (7 A-4 i B) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {8 i \sqrt {2} a^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {a (4 B+i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{4 a}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {-\frac {\frac {a^3 (7 A-4 i B) \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {8 i \sqrt {2} a^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {a (4 B+i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{4 a}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {-\frac {-\frac {2 i a^2 (7 A-4 i B) \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}-\frac {8 i \sqrt {2} a^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {a (4 B+i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{4 a}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {-\frac {2 a^{5/2} (7 A-4 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {8 i \sqrt {2} a^{5/2} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{2 a}-\frac {a (4 B+i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}}{4 a}-\frac {A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\)

Input:

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

Output:

-1/2*(A*Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/d + (-1/2*((-2*a^(5/2)* 
(7*A - (4*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d - ((8*I)*Sq 
rt[2]*a^(5/2)*(I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a 
])])/d)/a - (a*(I*A + 4*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d)/(4* 
a)
 

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 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 = 0.27 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-\frac {\left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 a^{\frac {5}{2}}}+\frac {-\frac {\left (-\frac {i B}{2}+\frac {A}{8}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+\left (\frac {1}{2} i a B +\frac {1}{8} a A \right ) \sqrt {a +i a \tan \left (d x +c \right )}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {\left (-4 i B +7 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{2}}\right )}{d}\)	\(146\)
default	\(\frac {2 a^{3} \left (-\frac {\left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 a^{\frac {5}{2}}}+\frac {-\frac {\left (-\frac {i B}{2}+\frac {A}{8}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+\left (\frac {1}{2} i a B +\frac {1}{8} a A \right ) \sqrt {a +i a \tan \left (d x +c \right )}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {\left (-4 i B +7 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{2}}\right )}{d}\)	\(146\)

Input:

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x,method=_RETUR 
NVERBOSE)
 

Output:

2/d*a^3*(-1/2*(A-I*B)/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*(-((-1/2*I*B+1/8*A)*(a+I*a*tan(d*x+c))^(3/2)+(1/2* 
I*a*B+1/8*a*A)*(a+I*a*tan(d*x+c))^(1/2))/a^2/tan(d*x+c)^2+1/8*(7*A-4*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. 730 vs. \(2 (130) = 260\).

Time = 0.11 (sec) , antiderivative size = 730, normalized size of antiderivative = 4.32 \[ \int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx =\text {Too large to display} \] Input:

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algori 
thm="fricas")
 

Output:

-1/16*(8*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqr 
t((A^2 - 2*I*A*B - B^2)*a/d^2)*log(-4*((-I*A - B)*a*e^(I*d*x + I*c) - (I*d 
*e^(2*I*d*x + 2*I*c) + I*d)*sqrt((A^2 - 2*I*A*B - B^2)*a/d^2)*sqrt(a/(e^(2 
*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(I*A + B)) - 8*sqrt(2)*(d*e^(4*I*d 
*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((A^2 - 2*I*A*B - B^2)*a/d^ 
2)*log(-4*((-I*A - B)*a*e^(I*d*x + I*c) - (-I*d*e^(2*I*d*x + 2*I*c) - I*d) 
*sqrt((A^2 - 2*I*A*B - B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(- 
I*d*x - I*c)/(I*A + B)) + (d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) 
 + d)*sqrt((49*A^2 - 56*I*A*B - 16*B^2)*a/d^2)*log(-16*(3*(-7*I*A - 4*B)*a 
^2*e^(2*I*d*x + 2*I*c) + (-7*I*A - 4*B)*a^2 + 2*sqrt(2)*(I*a*d*e^(3*I*d*x 
+ 3*I*c) + I*a*d*e^(I*d*x + I*c))*sqrt((49*A^2 - 56*I*A*B - 16*B^2)*a/d^2) 
*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/(7*I*A + 4*B)) - 
(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((49*A^2 - 56*I* 
A*B - 16*B^2)*a/d^2)*log(-16*(3*(-7*I*A - 4*B)*a^2*e^(2*I*d*x + 2*I*c) + ( 
-7*I*A - 4*B)*a^2 + 2*sqrt(2)*(-I*a*d*e^(3*I*d*x + 3*I*c) - I*a*d*e^(I*d*x 
 + I*c))*sqrt((49*A^2 - 56*I*A*B - 16*B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I 
*c) + 1)))*e^(-2*I*d*x - 2*I*c)/(7*I*A + 4*B)) - 4*sqrt(2)*((3*A - 4*I*B)* 
e^(5*I*d*x + 5*I*c) + 4*A*e^(3*I*d*x + 3*I*c) + (A + 4*I*B)*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]

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

integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**(1/2)*(A+B*tan(d*x+c)),x)
 

Output:

Integral(sqrt(I*a*(tan(c + d*x) - I))*(A + B*tan(c + d*x))*cot(c + d*x)**3 
, x)
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.20 \[ \int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {a^{2} {\left (\frac {4 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}} - \frac {{\left (7 \, A - 4 i \, B\right )} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2 \, {\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (A - 4 i \, B\right )} + \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A + 4 i \, B\right )} a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} + a^{3}}\right )}}{8 \, d} \] Input:

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algori 
thm="maxima")
 

Output:

1/8*a^2*(4*sqrt(2)*(A - I*B)*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)))/a^(3/2) - (7*A - 4* 
I*B)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a 
) + sqrt(a)))/a^(3/2) + 2*((I*a*tan(d*x + c) + a)^(3/2)*(A - 4*I*B) + sqrt 
(I*a*tan(d*x + c) + a)*(A + 4*I*B)*a)/((I*a*tan(d*x + c) + a)^2*a - 2*(I*a 
*tan(d*x + c) + a)*a^2 + a^3))/d
 

Giac [F(-2)]

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

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)*(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 = 4.11 (sec) , antiderivative size = 702, normalized size of antiderivative = 4.15 \[ \int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx =\text {Too large to display} \] Input:

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

Output:

(((A*a^2 + B*a^2*4i)*(a + a*tan(c + d*x)*1i)^(1/2))/(4*d) + ((A*a - B*a*4i 
)*(a + a*tan(c + d*x)*1i)^(3/2))/(4*d))/((a + a*tan(c + d*x)*1i)^2 - 2*a*( 
a + a*tan(c + d*x)*1i) + a^2) - (atan((17*A^3*a^4*d*(-a/2)^(1/2)*(a + a*ta 
n(c + d*x)*1i)^(1/2))/(17*A^3*a^5*d - B^3*a^5*d*16i + 24*A*B^2*a^5*d - A^2 
*B*a^5*d*9i) - (B^3*a^4*d*(-a/2)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*16i)/ 
(17*A^3*a^5*d - B^3*a^5*d*16i + 24*A*B^2*a^5*d - A^2*B*a^5*d*9i) + (24*A*B 
^2*a^4*d*(-a/2)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(17*A^3*a^5*d - B^3*a 
^5*d*16i + 24*A*B^2*a^5*d - A^2*B*a^5*d*9i) - (A^2*B*a^4*d*(-a/2)^(1/2)*(a 
 + a*tan(c + d*x)*1i)^(1/2)*9i)/(17*A^3*a^5*d - B^3*a^5*d*16i + 24*A*B^2*a 
^5*d - A^2*B*a^5*d*9i))*(A*1i + B)*(-a/2)^(1/2)*2i)/d + ((-a)^(1/2)*atan(( 
119*A^3*(-a)^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2))/(4*((119*A^3*a^5*d)/4 
- B^3*a^5*d*16i + 36*A*B^2*a^5*d - A^2*B*a^5*d*3i)) - (B^3*(-a)^(9/2)*d*(a 
 + a*tan(c + d*x)*1i)^(1/2)*16i)/((119*A^3*a^5*d)/4 - B^3*a^5*d*16i + 36*A 
*B^2*a^5*d - A^2*B*a^5*d*3i) + (36*A*B^2*(-a)^(9/2)*d*(a + a*tan(c + d*x)* 
1i)^(1/2))/((119*A^3*a^5*d)/4 - B^3*a^5*d*16i + 36*A*B^2*a^5*d - A^2*B*a^5 
*d*3i) - (A^2*B*(-a)^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2)*3i)/((119*A^3*a 
^5*d)/4 - B^3*a^5*d*16i + 36*A*B^2*a^5*d - A^2*B*a^5*d*3i))*(A*7i + 4*B)*1 
i)/(4*d)
 

Reduce [F]

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

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

Output:

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