Integrand size = 45, antiderivative size = 279 \[ \int (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=-\frac {a^{7/2} (5 i A+2 B) c^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{4 f}+\frac {a^3 (5 A-2 i B) c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a^2 (5 i A+2 B) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{12 f}+\frac {a (5 i A+2 B) (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{20 f}+\frac {B (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}}{5 f} \]
-1/4*a^(7/2)*(5*I*A+2*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^3*(5*A-2*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^2*(5*I*A+2*B)*(a+I*a* tan(f*x+e))^(3/2)*(c-I*c*tan(f*x+e))^(3/2)/f+1/20*a*(5*I*A+2*B)*(a+I*a*tan (f*x+e))^(5/2)*(c-I*c*tan(f*x+e))^(3/2)/f+1/5*B*(a+I*a*tan(f*x+e))^(7/2)*( c-I*c*tan(f*x+e))^(3/2)/f
Time = 11.53 (sec) , antiderivative size = 458, normalized size of antiderivative = 1.64 \[ \int (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\frac {B (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}}{5 f}+\frac {-\frac {a (5 i A+2 B) c (a+i a \tan (e+f x))^{7/2} \sqrt {c-i c \tan (e+f x)}}{4 f}+\frac {-\frac {a^2 (5 i A+2 B) c^2 (a+i a \tan (e+f x))^{7/2}}{3 f \sqrt {c-i c \tan (e+f x)}}+\frac {-\frac {a^3 (5 i A+2 B) c^3 (a+i a \tan (e+f x))^{5/2}}{2 f \sqrt {c-i c \tan (e+f x)}}+\frac {-\frac {5 a^4 (5 i A+2 B) c^4 (a+i a \tan (e+f x))^{3/2}}{f \sqrt {c-i c \tan (e+f x)}}+\frac {30 i a^6 (5 A-2 i B) c^4 (1-i \tan (e+f x)) \left (\frac {1+i \tan (e+f x)}{1-i \tan (e+f x)}-\frac {\arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {1-i \tan (e+f x)}}\right )}{f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}}{2 c}}{3 c}}{4 a}}{5 a} \]
(B*(a + I*a*Tan[e + f*x])^(7/2)*(c - I*c*Tan[e + f*x])^(3/2))/(5*f) + (-1/ 4*(a*((5*I)*A + 2*B)*c*(a + I*a*Tan[e + f*x])^(7/2)*Sqrt[c - I*c*Tan[e + f *x]])/f + (-1/3*(a^2*((5*I)*A + 2*B)*c^2*(a + I*a*Tan[e + f*x])^(7/2))/(f* Sqrt[c - I*c*Tan[e + f*x]]) + (-1/2*(a^3*((5*I)*A + 2*B)*c^3*(a + I*a*Tan[ e + f*x])^(5/2))/(f*Sqrt[c - I*c*Tan[e + f*x]]) + ((-5*a^4*((5*I)*A + 2*B) *c^4*(a + I*a*Tan[e + f*x])^(3/2))/(f*Sqrt[c - I*c*Tan[e + f*x]]) + ((30*I )*a^6*(5*A - (2*I)*B)*c^4*(1 - I*Tan[e + f*x])*((1 + I*Tan[e + f*x])/(1 - I*Tan[e + f*x]) - (ArcSin[Sqrt[a + I*a*Tan[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Sq rt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[1 - I*Tan[e + f*x]])))/(f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]]))/(2*c))/(3*c))/(4*a))/(5*a)
Time = 0.44 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.95, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used = {3042, 4071, 90, 59, 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.
| \(\displaystyle \int (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2} (A+B \tan (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2} (A+B \tan (e+f x))dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int (i \tan (e+f x) a+a)^{5/2} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {a c \left (\frac {1}{5} (5 A-2 i B) \int (i \tan (e+f x) a+a)^{5/2} \sqrt {c-i c \tan (e+f x)}d\tan (e+f x)+\frac {B (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}}{5 a c}\right )}{f}\) |
| \(\Big \downarrow \) 59 |
| \(\displaystyle \frac {a c \left (\frac {1}{5} (5 A-2 i B) \left (\frac {5}{4} a \int (i \tan (e+f x) a+a)^{3/2} \sqrt {c-i c \tan (e+f x)}d\tan (e+f x)+\frac {i (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 c}\right )+\frac {B (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}}{5 a c}\right )}{f}\) |
| \(\Big \downarrow \) 59 |
| \(\displaystyle \frac {a c \left (\frac {1}{5} (5 A-2 i B) \left (\frac {5}{4} a \left (a \int \sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}d\tan (e+f x)+\frac {i (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 c}\right )+\frac {i (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 c}\right )+\frac {B (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}}{5 a c}\right )}{f}\) |
| \(\Big \downarrow \) 40 |
| \(\displaystyle \frac {a c \left (\frac {1}{5} (5 A-2 i B) \left (\frac {5}{4} a \left (a \left (\frac {1}{2} a c \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)+\frac {1}{2} \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}\right )+\frac {i (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 c}\right )+\frac {i (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 c}\right )+\frac {B (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}}{5 a c}\right )}{f}\) |
| \(\Big \downarrow \) 45 |
| \(\displaystyle \frac {a c \left (\frac {1}{5} (5 A-2 i B) \left (\frac {5}{4} a \left (a \left (a c \int \frac {1}{i a+\frac {i c (i \tan (e+f x) a+a)}{c-i c \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}+\frac {1}{2} \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}\right )+\frac {i (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 c}\right )+\frac {i (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 c}\right )+\frac {B (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}}{5 a c}\right )}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {a c \left (\frac {1}{5} (5 A-2 i B) \left (\frac {5}{4} a \left (a \left (\frac {1}{2} \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}-i \sqrt {a} \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )\right )+\frac {i (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 c}\right )+\frac {i (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}{4 c}\right )+\frac {B (a+i a \tan (e+f x))^{7/2} (c-i c \tan (e+f x))^{3/2}}{5 a c}\right )}{f}\) |
(a*c*((B*(a + I*a*Tan[e + f*x])^(7/2)*(c - I*c*Tan[e + f*x])^(3/2))/(5*a*c ) + ((5*A - (2*I)*B)*(((I/4)*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2))/c + (5*a*(((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*T an[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))/5))/f
3.9.19.3.1 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 = 0.32 (sec) , antiderivative size = 412, normalized size of antiderivative = 1.48
| method | result | size |
| derivativedivides | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} c \left (60 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}-24 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{4}+80 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}-30 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}-30 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +30 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+32 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}+80 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+75 A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +45 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+56 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{120 f \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}\) | \(412\) |
| default | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} c \left (60 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}-24 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{4}+80 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}-30 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}-30 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +30 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+32 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}+80 i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+75 A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +45 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+56 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{120 f \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}\) | \(412\) |
| parts | \(\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} c \left (16 i \tan \left (f x +e \right )^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}-6 \tan \left (f x +e \right )^{3} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}+16 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+15 a c \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right )+9 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{24 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}-\frac {B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} c \left (-30 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}+12 \tan \left (f x +e \right )^{4} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}+15 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c -15 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-16 \tan \left (f x +e \right )^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}-28 \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{60 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\) | \(465\) |
int((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2),x,m ethod=_RETURNVERBOSE)
1/120/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^3*c*(60*I *B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^3-24*B*(a*c)^(1/2)* (a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^4+80*I*A*(a*c)^(1/2)*(a*c*(1+tan(f *x+e)^2))^(1/2)*tan(f*x+e)^2-30*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2) *tan(f*x+e)^3-30*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+30*I*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*t an(f*x+e)+32*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^2+80*I* A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)+75*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+45*A*(a*c)^(1/2)*(a*c* (1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)+56*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. 682 vs. \(2 (213) = 426\).
Time = 0.29 (sec) , antiderivative size = 682, normalized size of antiderivative = 2.44 \[ \int (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=-\frac {15 \, \sqrt {\frac {{\left (25 \, A^{2} - 20 i \, A B - 4 \, B^{2}\right )} a^{7} c^{3}}{f^{2}}} {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (-\frac {4 \, {\left (2 \, {\left ({\left (-5 i \, A - 2 \, B\right )} a^{3} c e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-5 i \, A - 2 \, B\right )} a^{3} 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 (25 \, A^{2} - 20 i \, A B - 4 \, B^{2}\right )} a^{7} c^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (5 i \, A + 2 \, B\right )} a^{3} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (5 i \, A + 2 \, B\right )} a^{3} c}\right ) - 15 \, \sqrt {\frac {{\left (25 \, A^{2} - 20 i \, A B - 4 \, B^{2}\right )} a^{7} c^{3}}{f^{2}}} {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (-\frac {4 \, {\left (2 \, {\left ({\left (-5 i \, A - 2 \, B\right )} a^{3} c e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-5 i \, A - 2 \, B\right )} a^{3} 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 (25 \, A^{2} - 20 i \, A B - 4 \, B^{2}\right )} a^{7} c^{3}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} - f\right )}\right )}}{{\left (5 i \, A + 2 \, B\right )} a^{3} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (5 i \, A + 2 \, B\right )} a^{3} c}\right ) + 4 \, {\left (15 \, {\left (5 i \, A + 2 \, B\right )} a^{3} c e^{\left (9 i \, f x + 9 i \, e\right )} + 10 \, {\left (-29 i \, A - 50 \, B\right )} a^{3} c e^{\left (7 i \, f x + 7 i \, e\right )} + 128 \, {\left (-5 i \, A - 2 \, B\right )} a^{3} c e^{\left (5 i \, f x + 5 i \, e\right )} + 70 \, {\left (-5 i \, A - 2 \, B\right )} a^{3} c e^{\left (3 i \, f x + 3 i \, e\right )} + 15 \, {\left (-5 i \, A - 2 \, B\right )} a^{3} 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}}}{240 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/ 2),x, algorithm="fricas")
-1/240*(15*sqrt((25*A^2 - 20*I*A*B - 4*B^2)*a^7*c^3/f^2)*(f*e^(8*I*f*x + 8 *I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)*log(-4*(2*((-5*I*A - 2*B)*a^3*c*e^(3*I*f*x + 3*I*e) + (-5*I *A - 2*B)*a^3*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((25*A^2 - 20*I*A*B - 4*B^2)*a^7*c^3/f^2) *(f*e^(2*I*f*x + 2*I*e) - f))/((5*I*A + 2*B)*a^3*c*e^(2*I*f*x + 2*I*e) + ( 5*I*A + 2*B)*a^3*c)) - 15*sqrt((25*A^2 - 20*I*A*B - 4*B^2)*a^7*c^3/f^2)*(f *e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)*log(-4*(2*((-5*I*A - 2*B)*a^3*c*e^(3*I*f*x + 3*I*e) + (-5*I*A - 2*B)*a^3*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((25*A^2 - 20*I*A*B - 4*B^ 2)*a^7*c^3/f^2)*(f*e^(2*I*f*x + 2*I*e) - f))/((5*I*A + 2*B)*a^3*c*e^(2*I*f *x + 2*I*e) + (5*I*A + 2*B)*a^3*c)) + 4*(15*(5*I*A + 2*B)*a^3*c*e^(9*I*f*x + 9*I*e) + 10*(-29*I*A - 50*B)*a^3*c*e^(7*I*f*x + 7*I*e) + 128*(-5*I*A - 2*B)*a^3*c*e^(5*I*f*x + 5*I*e) + 70*(-5*I*A - 2*B)*a^3*c*e^(3*I*f*x + 3*I* e) + 15*(-5*I*A - 2*B)*a^3*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^(8*I*f*x + 8*I*e) + 4*f*e^(6 *I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I*f*x + 2*I*e) + f)
Timed out. \[ \int (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\text {Timed out} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1657 vs. \(2 (213) = 426\).
Time = 2.78 (sec) , antiderivative size = 1657, normalized size of antiderivative = 5.94 \[ \int (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\text {Too large to display} \]
integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/ 2),x, algorithm="maxima")
-480*(60*(5*A - 2*I*B)*a^3*c*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 40*(29*A - 50*I*B)*a^3*c*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2 *f*x + 2*e))) - 512*(5*A - 2*I*B)*a^3*c*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 280*(5*A - 2*I*B)*a^3*c*cos(3/2*arctan2(sin(2*f*x + 2 *e), cos(2*f*x + 2*e))) - 60*(5*A - 2*I*B)*a^3*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 60*(-5*I*A - 2*B)*a^3*c*sin(9/2*arctan2(sin( 2*f*x + 2*e), cos(2*f*x + 2*e))) - 40*(29*I*A + 50*B)*a^3*c*sin(7/2*arctan 2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 512*(5*I*A + 2*B)*a^3*c*sin(5/2*a rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 280*(5*I*A + 2*B)*a^3*c*sin( 3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 60*(5*I*A + 2*B)*a^3*c* sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 30*((5*A - 2*I*B)*a ^3*c*cos(10*f*x + 10*e) + 5*(5*A - 2*I*B)*a^3*c*cos(8*f*x + 8*e) + 10*(5*A - 2*I*B)*a^3*c*cos(6*f*x + 6*e) + 10*(5*A - 2*I*B)*a^3*c*cos(4*f*x + 4*e) + 5*(5*A - 2*I*B)*a^3*c*cos(2*f*x + 2*e) - (-5*I*A - 2*B)*a^3*c*sin(10*f* x + 10*e) - 5*(-5*I*A - 2*B)*a^3*c*sin(8*f*x + 8*e) - 10*(-5*I*A - 2*B)*a^ 3*c*sin(6*f*x + 6*e) - 10*(-5*I*A - 2*B)*a^3*c*sin(4*f*x + 4*e) - 5*(-5*I* A - 2*B)*a^3*c*sin(2*f*x + 2*e) + (5*A - 2*I*B)*a^3*c)*arctan2(cos(1/2*arc tan2(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) + 30*((5*A - 2*I*B)*a^3*c*cos(10*f*x + 10*e) + 5*(5*A - 2*I*B)*a^3*c*cos(8*f*x + 8*e) + 10*(5*A - 2*I*B)*a^3*c*cos(6*f...
\[ \int (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\int { {\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \]
integrate((a+I*a*tan(f*x+e))^(7/2)*(A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/ 2),x, algorithm="giac")
Timed out. \[ \int (a+i a \tan (e+f x))^{7/2} (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\int \left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]