Integrand size = 45, antiderivative size = 222 \[ \int (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=-\frac {a^{5/2} (4 i A+B) c^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{4 f}+\frac {a^2 (4 A-i B) c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a (4 i A+B) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}+\frac {B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 f} \]
-1/4*a^(5/2)*(4*I*A+B)*c^(3/2)*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))/f+1/8*a^2*(4*A-I*B)*c*(a+I*a*tan(f*x+e))^(1 /2)*(c-I*c*tan(f*x+e))^(1/2)*tan(f*x+e)/f+1/12*a*(4*I*A+B)*(a+I*a*tan(f*x+ e))^(3/2)*(c-I*c*tan(f*x+e))^(3/2)/f+1/4*B*(a+I*a*tan(f*x+e))^(5/2)*(c-I*c *tan(f*x+e))^(3/2)/f
Time = 7.93 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.00 \[ \int (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\frac {a^{5/2} c^2 (i+\tan (e+f x)) \left (-6 (4 A-i B) \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \sqrt {a+i a \tan (e+f x)}+\sqrt {a} \sqrt {1-i \tan (e+f x)} (-i+\tan (e+f x)) \left (8 (i A+B)+3 (4 A+i B) \tan (e+f x)+8 (i A+B) \tan ^2(e+f x)+6 i B \tan ^3(e+f x)\right )\right )}{24 f \sqrt {1-i \tan (e+f x)} \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
(a^(5/2)*c^2*(I + Tan[e + f*x])*(-6*(4*A - I*B)*ArcSin[Sqrt[a + I*a*Tan[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Sqrt[a + I*a*Tan[e + f*x]] + Sqrt[a]*Sqrt[1 - I *Tan[e + f*x]]*(-I + Tan[e + f*x])*(8*(I*A + B) + 3*(4*A + I*B)*Tan[e + f* x] + 8*(I*A + B)*Tan[e + f*x]^2 + (6*I)*B*Tan[e + f*x]^3)))/(24*f*Sqrt[1 - I*Tan[e + f*x]]*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.41 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.97, 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, 59, 40, 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.
| \(\displaystyle \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2} (A+B \tan (e+f x))dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int (i \tan (e+f x) a+a)^{3/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {a c \left (\frac {1}{4} (4 A-i B) \int (i \tan (e+f x) a+a)^{3/2} \sqrt {c-i c \tan (e+f x)}d\tan (e+f x)+\frac {B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 a c}\right )}{f}\) |
| \(\Big \downarrow \) 59 |
| \(\displaystyle \frac {a c \left (\frac {1}{4} (4 A-i B) \left (a \int \sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}d\tan (e+f x)+\frac {i (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 c}\right )+\frac {B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 a c}\right )}{f}\) |
| \(\Big \downarrow \) 40 |
| \(\displaystyle \frac {a c \left (\frac {1}{4} (4 A-i B) \left (a \left (\frac {1}{2} a c \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)+\frac {1}{2} \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}\right )+\frac {i (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 c}\right )+\frac {B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 a c}\right )}{f}\) |
| \(\Big \downarrow \) 45 |
| \(\displaystyle \frac {a c \left (\frac {1}{4} (4 A-i B) \left (a \left (a c \int \frac {1}{i a+\frac {i c (i \tan (e+f x) a+a)}{c-i c \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}+\frac {1}{2} \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}\right )+\frac {i (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 c}\right )+\frac {B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 a c}\right )}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {a c \left (\frac {1}{4} (4 A-i B) \left (a \left (\frac {1}{2} \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}-i \sqrt {a} \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )\right )+\frac {i (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 c}\right )+\frac {B (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 a c}\right )}{f}\) |
(a*c*((B*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2))/(4*a*c ) + ((4*A - I*B)*(((I/3)*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x ])^(3/2))/c + a*((-I)*Sqrt[a]*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])] + (Tan[e + f*x]*Sqrt[a + I*a *Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/2)))/4))/f
3.9.7.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[x* (a + b*x)^m*((c + d*x)^m/(2*m + 1)), x] + Simp[2*a*c*(m/(2*m + 1)) Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ b*c + a*d, 0] && IGtQ[m + 1/2, 0]
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[2*c*(n/(m + n + 1) ) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0] && IGtQ[n + 1/2, 0] && LtQ[m, n]
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 = 0.38 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.58
| method | result | size |
| derivativedivides | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} c \left (6 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}+8 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}-3 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +3 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+8 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}+8 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+12 A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +12 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+8 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{24 f \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}\) | \(350\) |
| default | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} c \left (6 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}+8 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}-3 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +3 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+8 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}+8 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+12 A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +12 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+8 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{24 f \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}\) | \(350\) |
| parts | \(\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} c \left (2 i \tan \left (f x +e \right )^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}+2 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+3 a c \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right )+3 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{6 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}-\frac {B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} c \left (-6 i \tan \left (f x +e \right )^{3} \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+3 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c -3 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-8 \tan \left (f x +e \right )^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}-8 \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{24 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\) | \(405\) |
int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2),x,m ethod=_RETURNVERBOSE)
1/24/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^2*c*(6*I*B *(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^3+8*I*A*(a*c)^(1/2)*( a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^2-3*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+3*I*B*(a*c)^(1/2)*(a*c* (1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)+8*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^ (1/2)*tan(f*x+e)^2+8*I*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)+12*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 +12*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)+8*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. 614 vs. \(2 (168) = 336\).
Time = 0.29 (sec) , antiderivative size = 614, normalized size of antiderivative = 2.77 \[ \int (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=-\frac {3 \, \sqrt {\frac {{\left (16 \, A^{2} - 8 i \, A B - B^{2}\right )} a^{5} c^{3}}{f^{2}}} {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (-\frac {4 \, {\left (2 \, {\left ({\left (-4 i \, A - B\right )} a^{2} c e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-4 i \, A - B\right )} a^{2} c 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 (16 \, A^{2} - 8 i \, A B - B^{2}\right )} a^{5} c^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (4 i \, A + B\right )} a^{2} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (4 i \, A + B\right )} a^{2} c}\right ) - 3 \, \sqrt {\frac {{\left (16 \, A^{2} - 8 i \, A B - B^{2}\right )} a^{5} c^{3}}{f^{2}}} {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (-\frac {4 \, {\left (2 \, {\left ({\left (-4 i \, A - B\right )} a^{2} c e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-4 i \, A - B\right )} a^{2} c 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 (16 \, A^{2} - 8 i \, A B - B^{2}\right )} a^{5} c^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (4 i \, A + B\right )} a^{2} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (4 i \, A + B\right )} a^{2} c}\right ) + 4 \, {\left (3 \, {\left (4 i \, A + B\right )} a^{2} c e^{\left (7 i \, f x + 7 i \, e\right )} - {\left (20 i \, A + 53 \, B\right )} a^{2} c e^{\left (5 i \, f x + 5 i \, e\right )} + 11 \, {\left (-4 i \, A - B\right )} a^{2} c e^{\left (3 i \, f x + 3 i \, e\right )} + 3 \, {\left (-4 i \, A - B\right )} a^{2} c 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}}}{48 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/ 2),x, algorithm="fricas")
-1/48*(3*sqrt((16*A^2 - 8*I*A*B - B^2)*a^5*c^3/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)*log(-4*(2*((-4*I *A - B)*a^2*c*e^(3*I*f*x + 3*I*e) + (-4*I*A - B)*a^2*c*e^(I*f*x + I*e))*sq rt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) + sqrt(( 16*A^2 - 8*I*A*B - B^2)*a^5*c^3/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((4*I*A + B)*a^2*c*e^(2*I*f*x + 2*I*e) + (4*I*A + B)*a^2*c)) - 3*sqrt((16*A^2 - 8* I*A*B - B^2)*a^5*c^3/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)*log(-4*(2*((-4*I*A - B)*a^2*c*e^(3*I*f*x + 3*I*e) + (-4*I*A - B)*a^2*c*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((16*A^2 - 8*I*A*B - B^2)*a^ 5*c^3/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((4*I*A + B)*a^2*c*e^(2*I*f*x + 2* I*e) + (4*I*A + B)*a^2*c)) + 4*(3*(4*I*A + B)*a^2*c*e^(7*I*f*x + 7*I*e) - (20*I*A + 53*B)*a^2*c*e^(5*I*f*x + 5*I*e) + 11*(-4*I*A - B)*a^2*c*e^(3*I*f *x + 3*I*e) + 3*(-4*I*A - B)*a^2*c*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^(6*I*f*x + 6*I*e) + 3* f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)
Timed out. \[ \int (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\text {Timed out} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1375 vs. \(2 (168) = 336\).
Time = 1.10 (sec) , antiderivative size = 1375, normalized size of antiderivative = 6.19 \[ \int (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\text {Too large to display} \]
integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/ 2),x, algorithm="maxima")
-96*(12*(4*A - I*B)*a^2*c*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2* e))) - 4*(20*A - 53*I*B)*a^2*c*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 44*(4*A - I*B)*a^2*c*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f *x + 2*e))) - 12*(4*A - I*B)*a^2*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2 *f*x + 2*e))) - 12*(-4*I*A - B)*a^2*c*sin(7/2*arctan2(sin(2*f*x + 2*e), co s(2*f*x + 2*e))) - 4*(20*I*A + 53*B)*a^2*c*sin(5/2*arctan2(sin(2*f*x + 2*e ), cos(2*f*x + 2*e))) - 44*(4*I*A + B)*a^2*c*sin(3/2*arctan2(sin(2*f*x + 2 *e), cos(2*f*x + 2*e))) - 12*(4*I*A + B)*a^2*c*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 6*((4*A - I*B)*a^2*c*cos(8*f*x + 8*e) + 4*(4*A - I*B)*a^2*c*cos(6*f*x + 6*e) + 6*(4*A - I*B)*a^2*c*cos(4*f*x + 4*e) + 4* (4*A - I*B)*a^2*c*cos(2*f*x + 2*e) - (-4*I*A - B)*a^2*c*sin(8*f*x + 8*e) - 4*(-4*I*A - B)*a^2*c*sin(6*f*x + 6*e) - 6*(-4*I*A - B)*a^2*c*sin(4*f*x + 4*e) - 4*(-4*I*A - B)*a^2*c*sin(2*f*x + 2*e) + (4*A - I*B)*a^2*c)*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*((4*A - I*B)*a^2*c*cos(8*f*x + 8 *e) + 4*(4*A - I*B)*a^2*c*cos(6*f*x + 6*e) + 6*(4*A - I*B)*a^2*c*cos(4*f*x + 4*e) + 4*(4*A - I*B)*a^2*c*cos(2*f*x + 2*e) - (-4*I*A - B)*a^2*c*sin(8* f*x + 8*e) - 4*(-4*I*A - B)*a^2*c*sin(6*f*x + 6*e) - 6*(-4*I*A - B)*a^2*c* sin(4*f*x + 4*e) - 4*(-4*I*A - B)*a^2*c*sin(2*f*x + 2*e) + (4*A - I*B)*a^2 *c)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(...
\[ \int (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\int { {\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \]
integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/ 2),x, algorithm="giac")
Timed out. \[ \int (a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\int \left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]