Integrand size = 45, antiderivative size = 203 \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {2 a^{5/2} B \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{5/2} f}-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{5 f (c-i c \tan (e+f x))^{5/2}}+\frac {2 a B (a+i a \tan (e+f x))^{3/2}}{3 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 a^2 B \sqrt {a+i a \tan (e+f x)}}{c^2 f \sqrt {c-i c \tan (e+f x)}} \]
2*a^(5/2)*B*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))/c^(5/2)/f-2*a^2*B*(a+I*a*tan(f*x+e))^(1/2)/c^2/f/(c-I*c*tan(f* x+e))^(1/2)-1/5*(I*A+B)*(a+I*a*tan(f*x+e))^(5/2)/f/(c-I*c*tan(f*x+e))^(5/2 )+2/3*a*B*(a+I*a*tan(f*x+e))^(3/2)/c/f/(c-I*c*tan(f*x+e))^(3/2)
Time = 17.95 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00 \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {a^2 \cos ^2(e+f x) \left (\cos \left (\frac {1}{2} (e-2 f x)\right )-i \sin \left (\frac {1}{2} (e-2 f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e-2 f x)\right )+i \sin \left (\frac {1}{2} (e-2 f x)\right )\right ) \left (-10 B+(3 i A+33 B) \cos (2 (e+f x))-3 A \sin (2 (e+f x))-27 i B \sin (2 (e+f x))-30 B \arctan \left (e^{i (e+f x)}\right ) (\cos (3 (e+f x))-i \sin (3 (e+f x)))\right ) (-i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)}}{15 c^2 f \sqrt {c-i c \tan (e+f x)}} \]
Integrate[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan [e + f*x])^(5/2),x]
(a^2*Cos[e + f*x]^2*(Cos[(e - 2*f*x)/2] - I*Sin[(e - 2*f*x)/2])*(Cos[(e - 2*f*x)/2] + I*Sin[(e - 2*f*x)/2])*(-10*B + ((3*I)*A + 33*B)*Cos[2*(e + f*x )] - 3*A*Sin[2*(e + f*x)] - (27*I)*B*Sin[2*(e + f*x)] - 30*B*ArcTan[E^(I*( e + f*x))]*(Cos[3*(e + f*x)] - I*Sin[3*(e + f*x)]))*(-I + Tan[e + f*x])^2* Sqrt[a + I*a*Tan[e + f*x]])/(15*c^2*f*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.42 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3042, 4071, 87, 57, 57, 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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {(i \tan (e+f x) a+a)^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {a c \left (\frac {i B \int \frac {(i \tan (e+f x) a+a)^{3/2}}{(c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{c}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {a c \left (\frac {i B \left (-\frac {a \int \frac {\sqrt {i \tan (e+f x) a+a}}{(c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{c}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 57 |
| \(\displaystyle \frac {a c \left (\frac {i B \left (-\frac {a \left (-\frac {a \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{c}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{c}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 45 |
| \(\displaystyle \frac {a c \left (\frac {i B \left (-\frac {a \left (-\frac {2 a \int \frac {1}{i a+\frac {i c (i \tan (e+f x) a+a)}{c-i c \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}}{c}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{c}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {a c \left (\frac {i B \left (-\frac {a \left (\frac {2 i \sqrt {a} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{3/2}}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{c}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
(a*c*(-1/5*((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(a*c*(c - I*c*Tan[e + f*x])^(5/2)) + (I*B*((((-2*I)/3)*(a + I*a*Tan[e + f*x])^(3/2))/(c*(c - I*c *Tan[e + f*x])^(3/2)) - (a*(((2*I)*Sqrt[a]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Ta n[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/c^(3/2) - ((2*I)*Sqrt[ a + I*a*Tan[e + f*x]])/(c*Sqrt[c - I*c*Tan[e + f*x]])))/c))/c))/f
3.9.11.3.1 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 + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 506 vs. \(2 (166 ) = 332\).
Time = 0.35 (sec) , antiderivative size = 507, normalized size of antiderivative = 2.50
| method | result | size |
| parts | \(-\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (i-\tan \left (f x +e \right )\right )}{5 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}+\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (60 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right )^{3} a c +15 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right )^{4} a c -60 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right ) a c -97 i \tan \left (f x +e \right )^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}-90 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right )^{2} a c -43 \tan \left (f x +e \right )^{3} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}+23 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+15 a c \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right )+77 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{15 f \,c^{3} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{4} \sqrt {a c}}\) | \(507\) |
| 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} \left (-15 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 \tan \left (f x +e \right )^{4}+90 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 \tan \left (f x +e \right )^{2}+43 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}+60 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 \tan \left (f x +e \right )^{3}+3 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 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}-15 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 -77 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-60 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 \tan \left (f x +e \right )-97 B \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 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}-3 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+23 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{15 f \,c^{3} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{4} \sqrt {a c}}\) | \(555\) |
| 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} \left (-15 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 \tan \left (f x +e \right )^{4}+90 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 \tan \left (f x +e \right )^{2}+43 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}+60 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 \tan \left (f x +e \right )^{3}+3 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 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3}-15 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 -77 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-60 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 \tan \left (f x +e \right )-97 B \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 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}-3 A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+23 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{15 f \,c^{3} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{4} \sqrt {a c}}\) | \(555\) |
int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x,m ethod=_RETURNVERBOSE)
-1/5*A/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^2/c^3*(1 +tan(f*x+e)^2)*(I-tan(f*x+e))/(I+tan(f*x+e))^4+1/15*I*B/f*(a*(1+I*tan(f*x+ e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^2/c^3*(60*I*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))*tan(f*x+e)^3*a*c+15*l n((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*t an(f*x+e)^4*a*c-60*I*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))*tan(f*x+e)*a*c-97*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c )^(1/2)*tan(f*x+e)^2-90*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))*tan(f*x+e)^2*a*c-43*tan(f*x+e)^3*(a*c*(1+tan(f*x+e )^2))^(1/2)*(a*c)^(1/2)+23*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+15*a *c*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 ))+77*tan(f*x+e)*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1+tan(f*x +e)^2))^(1/2)/(I+tan(f*x+e))^4/(a*c)^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (159) = 318\).
Time = 0.27 (sec) , antiderivative size = 418, normalized size of antiderivative = 2.06 \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {15 \, c^{3} f \sqrt {-\frac {B^{2} a^{5}}{c^{5} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (B a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + B a^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{3} f\right )} \sqrt {-\frac {B^{2} a^{5}}{c^{5} f^{2}}}\right )}}{B a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{2}}\right ) - 15 \, c^{3} f \sqrt {-\frac {B^{2} a^{5}}{c^{5} f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (B a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + B a^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (c^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} - c^{3} f\right )} \sqrt {-\frac {B^{2} a^{5}}{c^{5} f^{2}}}\right )}}{B a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + B a^{2}}\right ) + 2 \, {\left (3 \, {\left (i \, A + B\right )} a^{2} e^{\left (7 i \, f x + 7 i \, e\right )} + {\left (3 i \, A - 7 \, B\right )} a^{2} e^{\left (5 i \, f x + 5 i \, e\right )} + 20 \, B a^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + 30 \, B a^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{30 \, c^{3} f} \]
integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/ 2),x, algorithm="fricas")
-1/30*(15*c^3*f*sqrt(-B^2*a^5/(c^5*f^2))*log(4*(2*(B*a^2*e^(3*I*f*x + 3*I* e) + B*a^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)) + (c^3*f*e^(2*I*f*x + 2*I*e) - c^3*f)*sqrt(-B^2*a^5/ (c^5*f^2)))/(B*a^2*e^(2*I*f*x + 2*I*e) + B*a^2)) - 15*c^3*f*sqrt(-B^2*a^5/ (c^5*f^2))*log(4*(2*(B*a^2*e^(3*I*f*x + 3*I*e) + B*a^2*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)) - (c^3*f *e^(2*I*f*x + 2*I*e) - c^3*f)*sqrt(-B^2*a^5/(c^5*f^2)))/(B*a^2*e^(2*I*f*x + 2*I*e) + B*a^2)) + 2*(3*(I*A + B)*a^2*e^(7*I*f*x + 7*I*e) + (3*I*A - 7*B )*a^2*e^(5*I*f*x + 5*I*e) + 20*B*a^2*e^(3*I*f*x + 3*I*e) + 30*B*a^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)))/(c^3*f)
\[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \left (A + B \tan {\left (e + f x \right )}\right )}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]
Integral((I*a*(tan(e + f*x) - I))**(5/2)*(A + B*tan(e + f*x))/(-I*c*(tan(e + f*x) + I))**(5/2), x)
Time = 0.43 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.05 \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {{\left (30 \, B a^{2} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 30 \, B a^{2} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - 6 \, {\left (i \, A + B\right )} a^{2} \cos \left (5 \, f x + 5 \, e\right ) + 20 \, B a^{2} \cos \left (3 \, f x + 3 \, e\right ) - 60 \, B a^{2} \cos \left (f x + e\right ) + 15 i \, B a^{2} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - 15 i \, B a^{2} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) + 6 \, {\left (A - i \, B\right )} a^{2} \sin \left (5 \, f x + 5 \, e\right ) + 20 i \, B a^{2} \sin \left (3 \, f x + 3 \, e\right ) - 60 i \, B a^{2} \sin \left (f x + e\right )\right )} \sqrt {a}}{30 \, c^{\frac {5}{2}} f} \]
integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/ 2),x, algorithm="maxima")
1/30*(30*B*a^2*arctan2(cos(f*x + e), sin(f*x + e) + 1) + 30*B*a^2*arctan2( cos(f*x + e), -sin(f*x + e) + 1) - 6*(I*A + B)*a^2*cos(5*f*x + 5*e) + 20*B *a^2*cos(3*f*x + 3*e) - 60*B*a^2*cos(f*x + e) + 15*I*B*a^2*log(cos(f*x + e )^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - 15*I*B*a^2*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) + 6*(A - I*B)*a^2*sin(5*f*x + 5*e) + 20*I*B*a^2*sin(3*f*x + 3*e) - 60*I*B*a^2*sin(f*x + e))*sqrt(a)/(c^(5/2) *f)
\[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/ 2),x, algorithm="giac")
integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(5/2)/(-I*c*tan(f*x + e) + c)^(5/2), x)
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]