Integrand size = 38, antiderivative size = 99 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}} \]
(1/2-1/2*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))/d/a^(1/2)+(A+I*B)*tan(d*x+c)^(1/2)/d/(a+I*a*tan(d*x+c))^(1/2)
Time = 1.37 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.12 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {\tan (c+d x)} \left (\frac {\sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {i a \tan (c+d x)}}+\frac {2 (A+i B)}{\sqrt {a+i a \tan (c+d x)}}\right )}{2 d} \]
(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[I*a*Tan[c + d*x]] + (2*(A + I*B)) /Sqrt[a + I*a*Tan[c + d*x]]))/(2*d)
Time = 0.42 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 4079, 27, 3042, 4027, 218}
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.
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\int \frac {a (A-i B) \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \frac {(A+i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {i a (A-i B) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\left (\frac {1}{2}-\frac {i}{2}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}\) |
((1/2 - I/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) + ((A + I*B)*Sqrt[Tan[c + d*x]])/(d*Sqr t[a + I*a*Tan[c + d*x]])
3.2.81.3.1 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[((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[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 632 vs. \(2 (79 ) = 158\).
Time = 0.17 (sec) , antiderivative size = 633, normalized size of antiderivative = 6.39
| method | result | size |
| derivativedivides | \(-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \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 )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )-2 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 )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \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 )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )-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 ) a +4 i A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+2 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 ) a \tan \left (d x +c \right )+4 i B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\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 )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -4 B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+4 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{4 d a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{2} \sqrt {-i a}}\) | \(633\) |
| default | \(-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \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 )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )-2 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 )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \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 )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )-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 ) a +4 i A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+2 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 ) a \tan \left (d x +c \right )+4 i B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\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 )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -4 B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+4 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{4 d a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{2} \sqrt {-i a}}\) | \(633\) |
| parts | \(-\frac {A \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (\sqrt {\tan }\left (d x +c \right )\right ) \left (i \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 ) \left (\tan ^{2}\left (d x +c \right )\right ) a -i \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 ) a +4 i \tan \left (d x +c \right ) \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+2 \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 ) \tan \left (d x +c \right ) a +4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{4 d a \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 )^{2}}+\frac {B \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (\sqrt {\tan }\left (d x +c \right )\right ) \left (2 i \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 ) \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 )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a -4 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \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 ) a +4 \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{4 d a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{2} \sqrt {-i a}}\) | \(698\) |
-1/4/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*(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)-I*a+3*a*tan(d*x+c)) /(tan(d*x+c)+I))*a*tan(d*x+c)^2-2*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)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*a *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)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^2-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)-I*a+3*a *tan(d*x+c))/(tan(d*x+c)+I))*a+4*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d *x+c)))^(1/2)*tan(d*x+c)+2*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)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+ c)+4*I*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-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)-I*a+3*a*tan(d *x+c))/(tan(d*x+c)+I))*a-4*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^ (1/2)*tan(d*x+c)+4*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2))/a /(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I)^2/(-I*a)^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 416 vs. \(2 (73) = 146\).
Time = 0.27 (sec) , antiderivative size = 416, normalized size of antiderivative = 4.20 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (\sqrt {2} a d \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {i \, \sqrt {2} a d \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - \sqrt {2} a d \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {-i \, \sqrt {2} a d \sqrt {-\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) + 2 \, \sqrt {2} {\left ({\left (A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \]
1/4*(sqrt(2)*a*d*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*d^2))*e^(I*d*x + I*c)*lo g((I*sqrt(2)*a*d*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*d^2))*e^(I*d*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*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)))/(4 *I*A + 4*B)) - sqrt(2)*a*d*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*d^2))*e^(I*d*x + I*c)*log((-I*sqrt(2)*a*d*sqrt(-(I*A^2 + 2*A*B - I*B^2)/(a*d^2))*e^(I*d* x + I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*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)))/(4*I*A + 4*B)) + 2*sqrt(2)*((A + I*B)*e^(2*I*d*x + 2*I*c) + A + I *B)*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)/(a*d)
\[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {A + B \tan {\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \sqrt {\tan {\left (c + d x \right )}}}\, dx \]
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: NotImplementedError} \]
Exception raised: NotImplementedError >> unable to parse Giac output: 2*(( 3*sqrt(abs(sageVARa))*sageVARB*sageVARa+3*i*abs(sageVARa)*sqrt(abs(sageVAR a))*sageVARB+3*i*sqrt(abs(sageVARa))*sageVARa*sageVARA-3*abs(sageVARa)*sqr t(abs(sageVARa))*sa
Time = 14.09 (sec) , antiderivative size = 426, normalized size of antiderivative = 4.30 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {B\,\ln \left (\frac {\sqrt {a}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (2-2{}\mathrm {i}\right )}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}}-\frac {a\,\mathrm {tan}\left (c+d\,x\right )}{{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}+1{}\mathrm {i}\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )}{\sqrt {a}\,d}+\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )\,\left (d\,1{}\mathrm {i}-\frac {a\,d\,\mathrm {tan}\left (c+d\,x\right )}{{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}\right )}-\frac {2\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )\,\left (d\,1{}\mathrm {i}-\frac {a\,d\,\mathrm {tan}\left (c+d\,x\right )}{{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}\right )}-\frac {\sqrt {\frac {1}{8}{}\mathrm {i}}\,B\,\ln \left (-\frac {a\,\mathrm {tan}\left (c+d\,x\right )}{{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}+\frac {2\,{\left (-1\right )}^{3/4}\,\sqrt {2}\,\sqrt {a}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}}+1{}\mathrm {i}\right )}{\sqrt {a}\,d}+\frac {2\,\sqrt {\frac {1}{8}{}\mathrm {i}}\,A\,\mathrm {atanh}\left (\frac {32\,\sqrt {\frac {1}{8}{}\mathrm {i}}\,A^2\,{\left (-a\right )}^{9/2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{\left (A^2\,a^4\,4{}\mathrm {i}-\frac {4\,A^2\,a^5\,\mathrm {tan}\left (c+d\,x\right )}{{\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}\right )\,\left (\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}\right )}{\sqrt {-a}\,d} \]
(B*log((a^(1/2)*tan(c + d*x)^(1/2)*(2 - 2i))/((a + a*tan(c + d*x)*1i)^(1/2 ) - a^(1/2)) - (a*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^ 2 + 1i)*(1/4 + 1i/4))/(a^(1/2)*d) + (A*tan(c + d*x)^(1/2)*2i)/(((a + a*tan (c + d*x)*1i)^(1/2) - a^(1/2))*(d*1i - (a*d*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2)) - (2*B*tan(c + d*x)^(1/2))/(((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*(d*1i - (a*d*tan(c + d*x))/((a + a*tan(c + d*x)* 1i)^(1/2) - a^(1/2))^2)) - ((1i/8)^(1/2)*B*log((2*(-1)^(3/4)*2^(1/2)*a^(1/ 2)*tan(c + d*x)^(1/2))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2)) - (a*tan( c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + 1i))/(a^(1/2)*d) + (2*(1i/8)^(1/2)*A*atanh((32*(1i/8)^(1/2)*A^2*(-a)^(9/2)*tan(c + d*x)^(1/2 ))/((A^2*a^4*4i - (4*A^2*a^5*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2)*((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2)))))/((-a)^(1/2)*d)