Integrand size = 38, antiderivative size = 192 \[ \int \frac {\sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=-\frac {(-1)^{3/4} \sqrt {a} (2 A-i B) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {(1+i) \sqrt {a} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {B \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \]
-(-1)^(3/4)*(2*A-I*B)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*ta n(d*x+c))^(1/2))*a^(1/2)*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d-(1+I)*(A-I*B) *arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*a^(1/2)* cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d+B*(a+I*a*tan(d*x+c))^(1/2)/d/cot(d*x+c )^(1/2)
Time = 3.75 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}+\frac {-\frac {i \sqrt {2} \left (\frac {1}{2} a^2 (2 A-i B)-\frac {1}{2} i a^2 B\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}}{d \sqrt {i a \tan (c+d x)}}+\frac {\sqrt [4]{-1} a^2 (2 A-i B) \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {1+i \tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{a}\right ) \]
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((B*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a* Tan[c + d*x]])/d + (((-I)*Sqrt[2]*((a^2*(2*A - I*B))/2 - (I/2)*a^2*B)*ArcT anh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Tan[ c + d*x]])/(d*Sqrt[I*a*Tan[c + d*x]]) + ((-1)^(1/4)*a^2*(2*A - I*B)*ArcSin h[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sqrt[1 + I*Tan[c + d*x]])/(d*Sqrt[a + I*a *Tan[c + d*x]]))/a)
Time = 1.11 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.93, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.342, Rules used = {3042, 4729, 3042, 4080, 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.
| \(\displaystyle \int \frac {\sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}}dx\) |
| \(\Big \downarrow \) 4729 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 4080 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int -\frac {\sqrt {i \tan (c+d x) a+a} (a B-a (2 A-i B) \tan (c+d x))}{2 \sqrt {\tan (c+d x)}}dx}{a}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} (a B-a (2 A-i B) \tan (c+d x))}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} (a B-a (2 A-i B) \tan (c+d x))}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )\) |
| \(\Big \downarrow \) 4084 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-(B+2 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 a (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-(B+2 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {-\frac {4 i a^3 (B+i A) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-(B+2 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {(2-2 i) a^{3/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-(B+2 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}\right )\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {(2-2 i) a^{3/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {a^2 (B+2 i A) \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}}{2 a}\right )\) |
| \(\Big \downarrow \) 65 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {(2-2 i) a^{3/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (B+2 i A) \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\frac {2 \sqrt [4]{-1} a^{3/2} (B+2 i A) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(2-2 i) a^{3/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a}\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-1/2*((2*(-1)^(1/4)*a^(3/2)*((2*I)* A + B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + ((2 - 2*I)*a^(3/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Ta n[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d)/a + (B*Sqrt[Tan[c + d*x]]*Sqr t[a + I*a*Tan[c + d*x]])/d)
3.6.40.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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])
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]
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]
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*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 1/(a*(m + n)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T an[e + f*x], 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] && GtQ[n, 0]
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]
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]
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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 315 vs. \(2 (155 ) = 310\).
Time = 0.68 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.65
| method | result | size |
| derivativedivides | \(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a -2 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+2 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}-\sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \right )}{2 d \sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \left (1+i \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) \sqrt {i a}\, a \sqrt {-i a}}\) | \(316\) |
| default | \(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a -2 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+2 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}-\sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \right )}{2 d \sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \left (1+i \tan \left (d x +c \right )\right ) \tan \left (d x +c \right ) \sqrt {i a}\, a \sqrt {-i a}}\) | \(316\) |
-1/2/d*(a*(1+I*tan(d*x+c)))^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(B *(-I*a)^(1/2)*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))*a-2*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/ 2)*(I*a)^(1/2)*(-I*a)^(1/2)+2*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))*a*(-I*a)^(1/2)-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)+I*a-3*a *tan(d*x+c))/(tan(d*x+c)+I))*(I*a)^(1/2)*a)/(1/tan(d*x+c))^(1/2)/(1+I*tan( d*x+c))/tan(d*x+c)/(I*a)^(1/2)/a/(-I*a)^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 793 vs. \(2 (148) = 296\).
Time = 0.26 (sec) , antiderivative size = 793, normalized size of antiderivative = 4.13 \[ \int \frac {\sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\text {Too large to display} \]
-1/4*(2*sqrt(2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-I*A^2 - 2*A*B + I*B^2) *a/d^2)*log(-4*((A - I*B)*a*e^(I*d*x + I*c) - (I*d*e^(2*I*d*x + 2*I*c) - I *d)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a/d^2)*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)/(I*A + B)) - 2*sqrt(2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-I*A^2 - 2 *A*B + I*B^2)*a/d^2)*log(-4*((A - I*B)*a*e^(I*d*x + I*c) - (-I*d*e^(2*I*d* x + 2*I*c) + I*d)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a/d^2)*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)/(I*A + B)) + 4*sqrt(2)*(I*B*e^(3*I*d*x + 3*I*c) - I*B* 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)) - (d*e^(2*I*d*x + 2*I*c) + d)*sqrt((4 *I*A^2 + 4*A*B - I*B^2)*a/d^2)*log(-16*(3*(2*I*A + B)*a^2*e^(2*I*d*x + 2*I *c) + (-2*I*A - B)*a^2 + 2*sqrt(2)*(a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c))*sqrt((4*I*A^2 + 4*A*B - I*B^2)*a/d^2)*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)/(2*I*A + B)) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt((4*I*A^2 + 4*A*B - I*B^2)*a/d^2)*log(-16*(3*(2*I*A + B)*a^2*e^(2*I*d*x + 2*I*c) + (-2 *I*A - B)*a^2 - 2*sqrt(2)*(a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c))* sqrt((4*I*A^2 + 4*A*B - I*B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sq rt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d*x ...
\[ \int \frac {\sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int \frac {\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \left (A + B \tan {\left (c + d x \right )}\right )}{\sqrt {\cot {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {\cot \left (d x + c\right )}} \,d x } \]
\[ \int \frac {\sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {\cot \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}} \,d x \]