Integrand size = 43, antiderivative size = 275 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {7 (5 i A-13 B) c^{9/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a^2 f}-\frac {7 (5 i A-13 B) c^4 \sqrt {c-i c \tan (e+f x)}}{2 a^2 f}-\frac {7 (5 i A-13 B) c^3 (c-i c \tan (e+f x))^{3/2}}{12 a^2 f}-\frac {7 (5 i A-13 B) c^2 (c-i c \tan (e+f x))^{5/2}}{40 a^2 f}-\frac {(5 i A-13 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{9/2}}{4 a^2 f (1+i \tan (e+f x))^2} \]
7/2*(5*I*A-13*B)*c^(9/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1 /2))/a^2/f*2^(1/2)-7/2*(5*I*A-13*B)*c^4*(c-I*c*tan(f*x+e))^(1/2)/a^2/f-7/1 2*(5*I*A-13*B)*c^3*(c-I*c*tan(f*x+e))^(3/2)/a^2/f-7/40*(5*I*A-13*B)*c^2*(c -I*c*tan(f*x+e))^(5/2)/a^2/f-1/8*(5*I*A-13*B)*c*(c-I*c*tan(f*x+e))^(7/2)/a ^2/f/(1+I*tan(f*x+e))+1/4*(I*A-B)*(c-I*c*tan(f*x+e))^(9/2)/a^2/f/(1+I*tan( f*x+e))^2
Time = 7.92 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.71 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {105 \sqrt {2} (-5 i A+13 B) c^{9/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) \sec ^2(e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x)))-2 c^4 \sqrt {c-i c \tan (e+f x)} \left (-325 i A+851 B+5 (113 A+289 i B) \tan (e+f x)+(170 i A-478 B) \tan ^2(e+f x)+10 (A+5 i B) \tan ^3(e+f x)+6 B \tan ^4(e+f x)\right )}{30 a^2 f (-i+\tan (e+f x))^2} \]
(105*Sqrt[2]*((-5*I)*A + 13*B)*c^(9/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/ (Sqrt[2]*Sqrt[c])]*Sec[e + f*x]^2*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]) - 2*c^4*Sqrt[c - I*c*Tan[e + f*x]]*((-325*I)*A + 851*B + 5*(113*A + (289*I )*B)*Tan[e + f*x] + ((170*I)*A - 478*B)*Tan[e + f*x]^2 + 10*(A + (5*I)*B)* Tan[e + f*x]^3 + 6*B*Tan[e + f*x]^4))/(30*a^2*f*(-I + Tan[e + f*x])^2)
Time = 0.42 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.83, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {3042, 4071, 27, 87, 51, 60, 60, 60, 73, 219}
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.
| \(\displaystyle \int \frac {(c-i c \tan (e+f x))^{9/2} (A+B \tan (e+f x))}{(a+i a \tan (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-i c \tan (e+f x))^{9/2} (A+B \tan (e+f x))}{(a+i a \tan (e+f x))^2}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{a^3 (i \tan (e+f x)+1)^3}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {c \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(i \tan (e+f x)+1)^3}d\tan (e+f x)}{a^2 f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{9/2}}{4 c (1+i \tan (e+f x))^2}-\frac {1}{8} (5 A+13 i B) \int \frac {(c-i c \tan (e+f x))^{7/2}}{(i \tan (e+f x)+1)^2}d\tan (e+f x)\right )}{a^2 f}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{9/2}}{4 c (1+i \tan (e+f x))^2}-\frac {1}{8} (5 A+13 i B) \left (\frac {i (c-i c \tan (e+f x))^{7/2}}{1+i \tan (e+f x)}-\frac {7}{2} c \int \frac {(c-i c \tan (e+f x))^{5/2}}{i \tan (e+f x)+1}d\tan (e+f x)\right )\right )}{a^2 f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{9/2}}{4 c (1+i \tan (e+f x))^2}-\frac {1}{8} (5 A+13 i B) \left (\frac {i (c-i c \tan (e+f x))^{7/2}}{1+i \tan (e+f x)}-\frac {7}{2} c \left (2 c \int \frac {(c-i c \tan (e+f x))^{3/2}}{i \tan (e+f x)+1}d\tan (e+f x)-\frac {2}{5} i (c-i c \tan (e+f x))^{5/2}\right )\right )\right )}{a^2 f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{9/2}}{4 c (1+i \tan (e+f x))^2}-\frac {1}{8} (5 A+13 i B) \left (\frac {i (c-i c \tan (e+f x))^{7/2}}{1+i \tan (e+f x)}-\frac {7}{2} c \left (2 c \left (2 c \int \frac {\sqrt {c-i c \tan (e+f x)}}{i \tan (e+f x)+1}d\tan (e+f x)-\frac {2}{3} i (c-i c \tan (e+f x))^{3/2}\right )-\frac {2}{5} i (c-i c \tan (e+f x))^{5/2}\right )\right )\right )}{a^2 f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{9/2}}{4 c (1+i \tan (e+f x))^2}-\frac {1}{8} (5 A+13 i B) \left (\frac {i (c-i c \tan (e+f x))^{7/2}}{1+i \tan (e+f x)}-\frac {7}{2} c \left (2 c \left (2 c \left (2 c \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)-2 i \sqrt {c-i c \tan (e+f x)}\right )-\frac {2}{3} i (c-i c \tan (e+f x))^{3/2}\right )-\frac {2}{5} i (c-i c \tan (e+f x))^{5/2}\right )\right )\right )}{a^2 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{9/2}}{4 c (1+i \tan (e+f x))^2}-\frac {1}{8} (5 A+13 i B) \left (\frac {i (c-i c \tan (e+f x))^{7/2}}{1+i \tan (e+f x)}-\frac {7}{2} c \left (2 c \left (2 c \left (4 i \int \frac {1}{2-\frac {c-i c \tan (e+f x)}{c}}d\sqrt {c-i c \tan (e+f x)}-2 i \sqrt {c-i c \tan (e+f x)}\right )-\frac {2}{3} i (c-i c \tan (e+f x))^{3/2}\right )-\frac {2}{5} i (c-i c \tan (e+f x))^{5/2}\right )\right )\right )}{a^2 f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{9/2}}{4 c (1+i \tan (e+f x))^2}-\frac {1}{8} (5 A+13 i B) \left (\frac {i (c-i c \tan (e+f x))^{7/2}}{1+i \tan (e+f x)}-\frac {7}{2} c \left (2 c \left (2 c \left (2 i \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )-2 i \sqrt {c-i c \tan (e+f x)}\right )-\frac {2}{3} i (c-i c \tan (e+f x))^{3/2}\right )-\frac {2}{5} i (c-i c \tan (e+f x))^{5/2}\right )\right )\right )}{a^2 f}\) |
(c*(((I*A - B)*(c - I*c*Tan[e + f*x])^(9/2))/(4*c*(1 + I*Tan[e + f*x])^2) - ((5*A + (13*I)*B)*((I*(c - I*c*Tan[e + f*x])^(7/2))/(1 + I*Tan[e + f*x]) - (7*c*(((-2*I)/5)*(c - I*c*Tan[e + f*x])^(5/2) + 2*c*(((-2*I)/3)*(c - I* c*Tan[e + f*x])^(3/2) + 2*c*((2*I)*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Ta n[e + f*x]]/(Sqrt[2]*Sqrt[c])] - (2*I)*Sqrt[c - I*c*Tan[e + f*x]]))))/2))/ 8))/(a^2*f)
3.8.71.3.1 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_))^(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, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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 = 0.28 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.80
| method | result | size |
| derivativedivides | \(\frac {2 i c^{2} \left (-\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {5 i B c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {A c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-18 i \sqrt {c -i c \tan \left (f x +e \right )}\, B \,c^{2}-6 \sqrt {c -i c \tan \left (f x +e \right )}\, A \,c^{2}+8 c^{3} \left (\frac {4 \left (\frac {21 i B}{64}+\frac {13 A}{64}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+4 \left (-\frac {19}{32} i B c -\frac {11}{32} c A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {7 \left (\frac {13 i B}{4}+\frac {5 A}{4}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}\right )\right )}{f \,a^{2}}\) | \(220\) |
| default | \(\frac {2 i c^{2} \left (-\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {5 i B c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {A c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-18 i \sqrt {c -i c \tan \left (f x +e \right )}\, B \,c^{2}-6 \sqrt {c -i c \tan \left (f x +e \right )}\, A \,c^{2}+8 c^{3} \left (\frac {4 \left (\frac {21 i B}{64}+\frac {13 A}{64}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+4 \left (-\frac {19}{32} i B c -\frac {11}{32} c A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {7 \left (\frac {13 i B}{4}+\frac {5 A}{4}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}\right )\right )}{f \,a^{2}}\) | \(220\) |
2*I/f/a^2*c^2*(-1/5*I*B*(c-I*c*tan(f*x+e))^(5/2)-5/3*I*B*c*(c-I*c*tan(f*x+ e))^(3/2)-1/3*A*c*(c-I*c*tan(f*x+e))^(3/2)-18*I*(c-I*c*tan(f*x+e))^(1/2)*B *c^2-6*(c-I*c*tan(f*x+e))^(1/2)*A*c^2+8*c^3*(4*((21/64*I*B+13/64*A)*(c-I*c *tan(f*x+e))^(3/2)+(-19/32*I*B*c-11/32*c*A)*(c-I*c*tan(f*x+e))^(1/2))/(c+I *c*tan(f*x+e))^2+7/8*(13/4*I*B+5/4*A)*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*t an(f*x+e))^(1/2)*2^(1/2)/c^(1/2))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 513 vs. \(2 (216) = 432\).
Time = 0.28 (sec) , antiderivative size = 513, normalized size of antiderivative = 1.87 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {105 \, \sqrt {2} \sqrt {-\frac {{\left (25 \, A^{2} + 130 i \, A B - 169 \, B^{2}\right )} c^{9}}{a^{4} f^{2}}} {\left (a^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} + 2 \, a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \log \left (-\frac {14 \, {\left ({\left (-5 i \, A + 13 \, B\right )} c^{5} + \sqrt {-\frac {{\left (25 \, A^{2} + 130 i \, A B - 169 \, B^{2}\right )} c^{9}}{a^{4} f^{2}}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a^{2} f}\right ) - 105 \, \sqrt {2} \sqrt {-\frac {{\left (25 \, A^{2} + 130 i \, A B - 169 \, B^{2}\right )} c^{9}}{a^{4} f^{2}}} {\left (a^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} + 2 \, a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \log \left (-\frac {14 \, {\left ({\left (-5 i \, A + 13 \, B\right )} c^{5} - \sqrt {-\frac {{\left (25 \, A^{2} + 130 i \, A B - 169 \, B^{2}\right )} c^{9}}{a^{4} f^{2}}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a^{2} f}\right ) + 2 \, \sqrt {2} {\left (105 \, {\left (5 i \, A - 13 \, B\right )} c^{4} e^{\left (8 i \, f x + 8 i \, e\right )} + 245 \, {\left (5 i \, A - 13 \, B\right )} c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} + 161 \, {\left (5 i \, A - 13 \, B\right )} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} + 15 \, {\left (5 i \, A - 13 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 30 \, {\left (-i \, A + B\right )} c^{4}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, {\left (a^{2} f e^{\left (8 i \, f x + 8 i \, e\right )} + 2 \, a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )}} \]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(9/2)/(a+I*a*tan(f*x+e))^2,x , algorithm="fricas")
-1/60*(105*sqrt(2)*sqrt(-(25*A^2 + 130*I*A*B - 169*B^2)*c^9/(a^4*f^2))*(a^ 2*f*e^(8*I*f*x + 8*I*e) + 2*a^2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))*log(-14*((-5*I*A + 13*B)*c^5 + sqrt(-(25*A^2 + 130*I*A*B - 169*B^ 2)*c^9/(a^4*f^2))*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a^2*f)) - 105*sqrt(2)*sqrt(-(25*A^2 + 130 *I*A*B - 169*B^2)*c^9/(a^4*f^2))*(a^2*f*e^(8*I*f*x + 8*I*e) + 2*a^2*f*e^(6 *I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))*log(-14*((-5*I*A + 13*B)*c^5 - sqrt(-(25*A^2 + 130*I*A*B - 169*B^2)*c^9/(a^4*f^2))*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a^2*f )) + 2*sqrt(2)*(105*(5*I*A - 13*B)*c^4*e^(8*I*f*x + 8*I*e) + 245*(5*I*A - 13*B)*c^4*e^(6*I*f*x + 6*I*e) + 161*(5*I*A - 13*B)*c^4*e^(4*I*f*x + 4*I*e) + 15*(5*I*A - 13*B)*c^4*e^(2*I*f*x + 2*I*e) + 30*(-I*A + B)*c^4)*sqrt(c/( e^(2*I*f*x + 2*I*e) + 1)))/(a^2*f*e^(8*I*f*x + 8*I*e) + 2*a^2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))
Timed out. \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.89 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i \, {\left (\frac {105 \, \sqrt {2} {\left (5 \, A + 13 i \, B\right )} c^{\frac {11}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2}} - \frac {60 \, {\left ({\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (13 \, A + 21 i \, B\right )} c^{6} - 2 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (11 \, A + 19 i \, B\right )} c^{7}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} c + 4 \, a^{2} c^{2}} + \frac {8 \, {\left (3 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} B c^{3} + 5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A + 5 i \, B\right )} c^{4} + 90 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A + 3 i \, B\right )} c^{5}\right )}}{a^{2}}\right )}}{60 \, c f} \]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(9/2)/(a+I*a*tan(f*x+e))^2,x , algorithm="maxima")
-1/60*I*(105*sqrt(2)*(5*A + 13*I*B)*c^(11/2)*log(-(sqrt(2)*sqrt(c) - sqrt( -I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c)))/a ^2 - 60*((-I*c*tan(f*x + e) + c)^(3/2)*(13*A + 21*I*B)*c^6 - 2*sqrt(-I*c*t an(f*x + e) + c)*(11*A + 19*I*B)*c^7)/((-I*c*tan(f*x + e) + c)^2*a^2 - 4*( -I*c*tan(f*x + e) + c)*a^2*c + 4*a^2*c^2) + 8*(3*I*(-I*c*tan(f*x + e) + c) ^(5/2)*B*c^3 + 5*(-I*c*tan(f*x + e) + c)^(3/2)*(A + 5*I*B)*c^4 + 90*sqrt(- I*c*tan(f*x + e) + c)*(A + 3*I*B)*c^5)/a^2)/(c*f)
\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(9/2)/(a+I*a*tan(f*x+e))^2,x , algorithm="giac")
Time = 8.95 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.46 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {38\,B\,c^6\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-21\,B\,c^5\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{4\,a^2\,c^2\,f+a^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,a^2\,c\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}-\frac {\frac {A\,c^6\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,22{}\mathrm {i}}{a^2\,f}-\frac {A\,c^5\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,13{}\mathrm {i}}{a^2\,f}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+4\,c^2}-\frac {A\,c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,12{}\mathrm {i}}{a^2\,f}-\frac {A\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,a^2\,f}+\frac {36\,B\,c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{a^2\,f}+\frac {10\,B\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^2\,f}+\frac {2\,B\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,a^2\,f}+\frac {\sqrt {2}\,A\,{\left (-c\right )}^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,35{}\mathrm {i}}{2\,a^2\,f}+\frac {\sqrt {2}\,B\,c^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,91{}\mathrm {i}}{2\,a^2\,f} \]
(38*B*c^6*(c - c*tan(e + f*x)*1i)^(1/2) - 21*B*c^5*(c - c*tan(e + f*x)*1i) ^(3/2))/(4*a^2*c^2*f + a^2*f*(c - c*tan(e + f*x)*1i)^2 - 4*a^2*c*f*(c - c* tan(e + f*x)*1i)) - ((A*c^6*(c - c*tan(e + f*x)*1i)^(1/2)*22i)/(a^2*f) - ( A*c^5*(c - c*tan(e + f*x)*1i)^(3/2)*13i)/(a^2*f))/((c - c*tan(e + f*x)*1i) ^2 - 4*c*(c - c*tan(e + f*x)*1i) + 4*c^2) - (A*c^4*(c - c*tan(e + f*x)*1i) ^(1/2)*12i)/(a^2*f) - (A*c^3*(c - c*tan(e + f*x)*1i)^(3/2)*2i)/(3*a^2*f) + (36*B*c^4*(c - c*tan(e + f*x)*1i)^(1/2))/(a^2*f) + (10*B*c^3*(c - c*tan(e + f*x)*1i)^(3/2))/(3*a^2*f) + (2*B*c^2*(c - c*tan(e + f*x)*1i)^(5/2))/(5* a^2*f) + (2^(1/2)*A*(-c)^(9/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2) )/(2*(-c)^(1/2)))*35i)/(2*a^2*f) + (2^(1/2)*B*c^(9/2)*atan((2^(1/2)*(c - c *tan(e + f*x)*1i)^(1/2)*1i)/(2*c^(1/2)))*91i)/(2*a^2*f)