Integrand size = 38, antiderivative size = 152 \[ \int \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {2 (-1)^{3/4} \sqrt {a} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(1-i) \sqrt {a} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d} \]
-2*(-1)^(3/4)*B*arctan((-1)^(3/4)*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+(1-I)*(A-I*B)*arcta nh((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
Time = 3.44 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.06 \[ \int \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {a \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\frac {\sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}}{\sqrt {i a \tan (c+d x)}}+\frac {2 \sqrt [4]{-1} B \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {1+i \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \]
(a*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((Sqrt[2]*(A - I*B)*ArcTanh[(Sqrt [2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Tan[c + d*x]] )/Sqrt[I*a*Tan[c + d*x]] + (2*(-1)^(1/4)*B*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sqrt[1 + I*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]))/d
Time = 0.81 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.88, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 4729, 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.
| \(\displaystyle \int \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x))dx\) |
| \(\Big \downarrow \) 4729 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\sqrt {i \tan (c+d x) a+a} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\sqrt {i \tan (c+d x) a+a} (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4084 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left ((A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {i B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left ((A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {i B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{a}\right )\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {i B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {2 i a^2 (A-i B) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {i B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{a}+\frac {(1-i) \sqrt {a} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\right )\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {i a B \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}+\frac {(1-i) \sqrt {a} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\right )\) |
| \(\Big \downarrow \) 65 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 i a B \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}+\frac {(1-i) \sqrt {a} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(1-i) \sqrt {a} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 (-1)^{3/4} \sqrt {a} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\right )\) |
((-2*(-1)^(3/4)*Sqrt[a]*B*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/S qrt[a + I*a*Tan[c + d*x]]])/d + ((1 - I)*Sqrt[a]*(A - I*B)*ArcTanh[((1 + I )*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d)*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]
3.6.39.3.1 Defintions of rubi rules used
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*(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. 505 vs. \(2 (121 ) = 242\).
Time = 0.56 (sec) , antiderivative size = 506, normalized size of antiderivative = 3.33
| method | result | size |
| derivativedivides | \(\frac {\sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (i A \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \tan \left (d x +c \right )-i B \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}-2 i B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \tan \left (d x +c \right )+B \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \tan \left (d x +c \right )+A \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}-2 B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right )\right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \left (-\tan \left (d x +c \right )+i\right ) \sqrt {-i a}}\) | \(506\) |
| default | \(\frac {\sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (i A \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \tan \left (d x +c \right )-i B \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}-2 i B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \tan \left (d x +c \right )+B \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \tan \left (d x +c \right )+A \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+3 a \tan \left (d x +c \right )-i a}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}-2 B \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right )\right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \left (-\tan \left (d x +c \right )+i\right ) \sqrt {-i a}}\) | \(506\) |
1/2/d*(1/tan(d*x+c))^(1/2)*tan(d*x+c)*(a*(1+I*tan(d*x+c)))^(1/2)*a*(I*A*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)+3*a *tan(d*x+c)-I*a)/(tan(d*x+c)+I))*(I*a)^(1/2)*tan(d*x+c)-I*B*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)+3*a*tan(d*x+c)- I*a)/(tan(d*x+c)+I))*(I*a)^(1/2)-2*I*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))*tan (d*x+c)+B*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)+3*a*tan(d*x+c)-I*a)/(tan(d*x+c)+I))*(I*a)^(1/2)*tan(d*x+c)+A*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)+3*a*t an(d*x+c)-I*a)/(tan(d*x+c)+I))*(I*a)^(1/2)-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*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-tan(d*x+c)+I)/(- I*a)^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 576 vs. \(2 (115) = 230\).
Time = 0.26 (sec) , antiderivative size = 576, normalized size of antiderivative = 3.79 \[ \int \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {1}{2} \, \sqrt {2} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a}{d^{2}}} \log \left (-\frac {4 \, {\left ({\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \frac {1}{2} \, \sqrt {2} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a}{d^{2}}} \log \left (-\frac {4 \, {\left ({\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \frac {1}{4} \, \sqrt {\frac {4 i \, B^{2} a}{d^{2}}} \log \left (-\frac {16 \, {\left (3 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{2} + \sqrt {2} {\left (i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {4 i \, B^{2} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{B}\right ) + \frac {1}{4} \, \sqrt {\frac {4 i \, B^{2} a}{d^{2}}} \log \left (-\frac {16 \, {\left (3 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{2} + \sqrt {2} {\left (-i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {4 i \, B^{2} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{B}\right ) \]
1/2*sqrt(2)*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) + (d*e^(2*I*d*x + 2*I*c) - 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)) - 1/2*sqrt(2)*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) - (d*e^( 2*I*d*x + 2*I*c) - 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)) - 1/4*sqrt(4*I*B^2*a/d^2)*log(-16*(3*B*a ^2*e^(2*I*d*x + 2*I*c) - B*a^2 + sqrt(2)*(I*a*d*e^(3*I*d*x + 3*I*c) - I*a* d*e^(I*d*x + I*c))*sqrt(4*I*B^2*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*s qrt((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)/B) + 1/4*sqrt(4*I*B^2*a/d^2)*log(-16*(3*B*a^2*e^(2*I*d*x + 2*I*c) - B*a^2 + sqrt(2)*(-I*a*d*e^(3*I*d*x + 3*I*c) + I*a*d*e^(I*d*x + I*c))*sqrt (4*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)/B)
\[ \int \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {\cot {\left (c + d x \right )}}\, dx \]
\[ \int \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {i \, a \tan \left (d x + c\right ) + a} \sqrt {\cot \left (d x + c\right )} \,d x } \]
\[ \int \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {i \, a \tan \left (d x + c\right ) + a} \sqrt {\cot \left (d x + c\right )} \,d x } \]
Timed out. \[ \int \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]