Integrand size = 45, antiderivative size = 217 \[ \int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {\sqrt {a} (3 i A-2 B) c^{5/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}-\frac {(3 i A-2 B) c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {(3 i A-2 B) c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{6 f}+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{3 f} \]
-(3*I*A-2*B)*c^(5/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I* c*tan(f*x+e))^(1/2))*a^(1/2)/f-1/2*(3*I*A-2*B)*c^2*(a+I*a*tan(f*x+e))^(1/2 )*(c-I*c*tan(f*x+e))^(1/2)/f-1/6*(3*I*A-2*B)*c*(a+I*a*tan(f*x+e))^(1/2)*(c -I*c*tan(f*x+e))^(3/2)/f+1/3*B*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e)) ^(5/2)/f
Time = 5.31 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.69 \[ \int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx=\frac {6 \sqrt {a} (-3 i A+2 B) c^{5/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )-c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)} \left (12 i A-10 B+3 (A+2 i B) \tan (e+f x)+2 B \tan ^2(e+f x)\right )}{6 f} \]
(6*Sqrt[a]*((-3*I)*A + 2*B)*c^(5/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f *x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])] - c^2*Sqrt[a + I*a*Tan[e + f*x ]]*Sqrt[c - I*c*Tan[e + f*x]]*((12*I)*A - 10*B + 3*(A + (2*I)*B)*Tan[e + f *x] + 2*B*Tan[e + f*x]^2))/(6*f)
Time = 0.42 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3042, 4071, 90, 60, 60, 45, 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.
| \(\displaystyle \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} (A+B \tan (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} (A+B \tan (e+f x))dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{\sqrt {i \tan (e+f x) a+a}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {a c \left (\frac {1}{3} (3 A+2 i B) \int \frac {(c-i c \tan (e+f x))^{3/2}}{\sqrt {i \tan (e+f x) a+a}}d\tan (e+f x)+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{3 a c}\right )}{f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a c \left (\frac {1}{3} (3 A+2 i B) \left (\frac {3}{2} c \int \frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}d\tan (e+f x)-\frac {i \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 a}\right )+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{3 a c}\right )}{f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {a c \left (\frac {1}{3} (3 A+2 i B) \left (\frac {3}{2} c \left (c \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)-\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a}\right )-\frac {i \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 a}\right )+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{3 a c}\right )}{f}\) |
| \(\Big \downarrow \) 45 |
| \(\displaystyle \frac {a c \left (\frac {1}{3} (3 A+2 i B) \left (\frac {3}{2} c \left (2 c \int \frac {1}{i a+\frac {i c (i \tan (e+f x) a+a)}{c-i c \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}-\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a}\right )-\frac {i \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 a}\right )+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{3 a c}\right )}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {a c \left (\frac {1}{3} (3 A+2 i B) \left (\frac {3}{2} c \left (-\frac {2 i \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {a}}-\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{a}\right )-\frac {i \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 a}\right )+\frac {B \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}{3 a c}\right )}{f}\) |
(a*c*((B*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(5/2))/(3*a*c) + ((3*A + (2*I)*B)*(((-1/2*I)*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(3/2))/a + (3*c*(((-2*I)*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/Sqrt[a] - (I*Sqrt[a + I*a* Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/a))/2))/3))/f
3.8.88.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && !GtQ[c, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 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[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x], x , Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
Time = 0.31 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.31
| method | result | size |
| derivativedivides | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, c^{2} \left (-6 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +6 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+2 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}+12 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}-9 A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +3 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-10 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{6 f \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}\) | \(285\) |
| default | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, c^{2} \left (-6 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +6 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+2 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}+12 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}-9 A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +3 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-10 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{6 f \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}\) | \(285\) |
| parts | \(-\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, c^{2} \left (4 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+\tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}-3 a c \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right )\right )}{2 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}+\frac {B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, c^{2} \left (3 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c -3 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-\tan \left (f x +e \right )^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}+5 \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{3 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\) | \(340\) |
int((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2),x,m ethod=_RETURNVERBOSE)
-1/6/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*c^2*(-6*I*B* ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))* a*c+6*I*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)+2*B*(a*c)^(1 /2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^2+12*I*A*(a*c)^(1/2)*(a*c*(1+t an(f*x+e)^2))^(1/2)-9*A*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^ 2))^(1/2))/(a*c)^(1/2))*a*c+3*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*t an(f*x+e)-10*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2)/(a*c* (1+tan(f*x+e)^2))^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 547 vs. \(2 (164) = 328\).
Time = 0.27 (sec) , antiderivative size = 547, normalized size of antiderivative = 2.52 \[ \int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx=\frac {3 \, \sqrt {\frac {{\left (9 \, A^{2} + 12 i \, A B - 4 \, B^{2}\right )} a c^{5}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (-\frac {4 \, {\left (2 \, {\left ({\left (3 i \, A - 2 \, B\right )} c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (3 i \, A - 2 \, B\right )} c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + \sqrt {\frac {{\left (9 \, A^{2} + 12 i \, A B - 4 \, B^{2}\right )} a c^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (-3 i \, A + 2 \, B\right )} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-3 i \, A + 2 \, B\right )} c^{2}}\right ) - 3 \, \sqrt {\frac {{\left (9 \, A^{2} + 12 i \, A B - 4 \, B^{2}\right )} a c^{5}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (-\frac {4 \, {\left (2 \, {\left ({\left (3 i \, A - 2 \, B\right )} c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (3 i \, A - 2 \, B\right )} c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {{\left (9 \, A^{2} + 12 i \, A B - 4 \, B^{2}\right )} a c^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (-3 i \, A + 2 \, B\right )} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-3 i \, A + 2 \, B\right )} c^{2}}\right ) - 4 \, {\left (3 \, {\left (3 i \, A - 2 \, B\right )} c^{2} e^{\left (5 i \, f x + 5 i \, e\right )} + 8 \, {\left (3 i \, A - 2 \, B\right )} c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + 3 \, {\left (5 i \, A - 6 \, B\right )} c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
integrate((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/ 2),x, algorithm="fricas")
1/12*(3*sqrt((9*A^2 + 12*I*A*B - 4*B^2)*a*c^5/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(-4*(2*((3*I*A - 2*B)*c^2*e^(3*I*f*x + 3 *I*e) + (3*I*A - 2*B)*c^2*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1 ))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) + sqrt((9*A^2 + 12*I*A*B - 4*B^2)*a*c ^5/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-3*I*A + 2*B)*c^2*e^(2*I*f*x + 2*I* e) + (-3*I*A + 2*B)*c^2)) - 3*sqrt((9*A^2 + 12*I*A*B - 4*B^2)*a*c^5/f^2)*( f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)*log(-4*(2*((3*I*A - 2 *B)*c^2*e^(3*I*f*x + 3*I*e) + (3*I*A - 2*B)*c^2*e^(I*f*x + I*e))*sqrt(a/(e ^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - sqrt((9*A^2 + 12*I*A*B - 4*B^2)*a*c^5/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((-3*I*A + 2*B) *c^2*e^(2*I*f*x + 2*I*e) + (-3*I*A + 2*B)*c^2)) - 4*(3*(3*I*A - 2*B)*c^2*e ^(5*I*f*x + 5*I*e) + 8*(3*I*A - 2*B)*c^2*e^(3*I*f*x + 3*I*e) + 3*(5*I*A - 6*B)*c^2*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I *f*x + 2*I*e) + 1)))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)
\[ \int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx=\int \sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}} \left (A + B \tan {\left (e + f x \right )}\right )\, dx \]
Integral(sqrt(I*a*(tan(e + f*x) - I))*(-I*c*(tan(e + f*x) + I))**(5/2)*(A + B*tan(e + f*x)), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1085 vs. \(2 (164) = 328\).
Time = 0.85 (sec) , antiderivative size = 1085, normalized size of antiderivative = 5.00 \[ \int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx=\text {Too large to display} \]
integrate((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/ 2),x, algorithm="maxima")
-6*(12*(3*A + 2*I*B)*c^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e ))) + 32*(3*A + 2*I*B)*c^2*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2 *e))) + 12*(5*A + 6*I*B)*c^2*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*(3*I*A - 2*B)*c^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 32*(3*I*A - 2*B)*c^2*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f *x + 2*e))) + 12*(5*I*A - 6*B)*c^2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2 *f*x + 2*e))) + 6*((3*A + 2*I*B)*c^2*cos(6*f*x + 6*e) + 3*(3*A + 2*I*B)*c^ 2*cos(4*f*x + 4*e) + 3*(3*A + 2*I*B)*c^2*cos(2*f*x + 2*e) + (3*I*A - 2*B)* c^2*sin(6*f*x + 6*e) + 3*(3*I*A - 2*B)*c^2*sin(4*f*x + 4*e) + 3*(3*I*A - 2 *B)*c^2*sin(2*f*x + 2*e) + (3*A + 2*I*B)*c^2)*arctan2(cos(1/2*arctan2(sin( 2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2* f*x + 2*e))) + 1) + 6*((3*A + 2*I*B)*c^2*cos(6*f*x + 6*e) + 3*(3*A + 2*I*B )*c^2*cos(4*f*x + 4*e) + 3*(3*A + 2*I*B)*c^2*cos(2*f*x + 2*e) + (3*I*A - 2 *B)*c^2*sin(6*f*x + 6*e) + 3*(3*I*A - 2*B)*c^2*sin(4*f*x + 4*e) + 3*(3*I*A - 2*B)*c^2*sin(2*f*x + 2*e) + (3*A + 2*I*B)*c^2)*arctan2(cos(1/2*arctan2( sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(sin(2*f*x + 2*e), c os(2*f*x + 2*e))) + 1) + 3*((3*I*A - 2*B)*c^2*cos(6*f*x + 6*e) + 3*(3*I*A - 2*B)*c^2*cos(4*f*x + 4*e) + 3*(3*I*A - 2*B)*c^2*cos(2*f*x + 2*e) - (3*A + 2*I*B)*c^2*sin(6*f*x + 6*e) - 3*(3*A + 2*I*B)*c^2*sin(4*f*x + 4*e) - 3*( 3*A + 2*I*B)*c^2*sin(2*f*x + 2*e) + (3*I*A - 2*B)*c^2)*log(cos(1/2*arct...
\[ \int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx=\int { {\left (B \tan \left (f x + e\right ) + A\right )} \sqrt {i \, a \tan \left (f x + e\right ) + a} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \]
integrate((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/ 2),x, algorithm="giac")
Timed out. \[ \int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx=\int \left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]