Integrand size = 43, antiderivative size = 238 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {5 (3 i A-11 B) c^{7/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{2 \sqrt {2} a^2 f}-\frac {5 (3 i A-11 B) c^3 \sqrt {c-i c \tan (e+f x)}}{4 a^2 f}-\frac {5 (3 i A-11 B) c^2 (c-i c \tan (e+f x))^{3/2}}{24 a^2 f}-\frac {(3 i A-11 B) c (c-i c \tan (e+f x))^{5/2}}{8 a^2 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{4 a^2 f (1+i \tan (e+f x))^2} \]
5/4*(3*I*A-11*B)*c^(7/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1 /2))/a^2/f*2^(1/2)-5/4*(3*I*A-11*B)*c^3*(c-I*c*tan(f*x+e))^(1/2)/a^2/f-5/2 4*(3*I*A-11*B)*c^2*(c-I*c*tan(f*x+e))^(3/2)/a^2/f-1/8*(3*I*A-11*B)*c*(c-I* c*tan(f*x+e))^(5/2)/a^2/f/(1+I*tan(f*x+e))+1/4*(I*A-B)*(c-I*c*tan(f*x+e))^ (7/2)/a^2/f/(1+I*tan(f*x+e))^2
Time = 6.89 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.76 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {15 \sqrt {2} (-3 i A+11 B) c^{7/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) \sec ^2(e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+2 c^3 \sqrt {c-i c \tan (e+f x)} \left (27 i A-103 B-(51 A+175 i B) \tan (e+f x)+4 (-3 i A+14 B) \tan ^2(e+f x)-4 i B \tan ^3(e+f x)\right )}{12 a^2 f (-i+\tan (e+f x))^2} \]
(15*Sqrt[2]*((-3*I)*A + 11*B)*c^(7/2)*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^3*Sqrt[c - I*c*Tan[e + f*x]]*((27*I)*A - 103*B - (51*A + (175*I)*B)*T an[e + f*x] + 4*((-3*I)*A + 14*B)*Tan[e + f*x]^2 - (4*I)*B*Tan[e + f*x]^3) )/(12*a^2*f*(-I + Tan[e + f*x])^2)
Time = 0.41 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.85, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {3042, 4071, 27, 87, 51, 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.
| \(\displaystyle \int \frac {(c-i c \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(a+i a \tan (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-i c \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(a+i a \tan (e+f x))^2}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{a^3 (i \tan (e+f x)+1)^3}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {c \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{(i \tan (e+f x)+1)^3}d\tan (e+f x)}{a^2 f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{4 c (1+i \tan (e+f x))^2}-\frac {1}{8} (3 A+11 i B) \int \frac {(c-i c \tan (e+f x))^{5/2}}{(i \tan (e+f x)+1)^2}d\tan (e+f x)\right )}{a^2 f}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{4 c (1+i \tan (e+f x))^2}-\frac {1}{8} (3 A+11 i B) \left (\frac {i (c-i c \tan (e+f x))^{5/2}}{1+i \tan (e+f x)}-\frac {5}{2} c \int \frac {(c-i c \tan (e+f x))^{3/2}}{i \tan (e+f x)+1}d\tan (e+f x)\right )\right )}{a^2 f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{4 c (1+i \tan (e+f x))^2}-\frac {1}{8} (3 A+11 i B) \left (\frac {i (c-i c \tan (e+f x))^{5/2}}{1+i \tan (e+f x)}-\frac {5}{2} c \left (2 c \int \frac {\sqrt {c-i c \tan (e+f x)}}{i \tan (e+f x)+1}d\tan (e+f x)-\frac {2}{3} i (c-i c \tan (e+f x))^{3/2}\right )\right )\right )}{a^2 f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{4 c (1+i \tan (e+f x))^2}-\frac {1}{8} (3 A+11 i B) \left (\frac {i (c-i c \tan (e+f x))^{5/2}}{1+i \tan (e+f x)}-\frac {5}{2} c \left (2 c \left (2 c \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)-2 i \sqrt {c-i c \tan (e+f x)}\right )-\frac {2}{3} i (c-i c \tan (e+f x))^{3/2}\right )\right )\right )}{a^2 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{4 c (1+i \tan (e+f x))^2}-\frac {1}{8} (3 A+11 i B) \left (\frac {i (c-i c \tan (e+f x))^{5/2}}{1+i \tan (e+f x)}-\frac {5}{2} c \left (2 c \left (4 i \int \frac {1}{2-\frac {c-i c \tan (e+f x)}{c}}d\sqrt {c-i c \tan (e+f x)}-2 i \sqrt {c-i c \tan (e+f x)}\right )-\frac {2}{3} i (c-i c \tan (e+f x))^{3/2}\right )\right )\right )}{a^2 f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{4 c (1+i \tan (e+f x))^2}-\frac {1}{8} (3 A+11 i B) \left (\frac {i (c-i c \tan (e+f x))^{5/2}}{1+i \tan (e+f x)}-\frac {5}{2} c \left (2 c \left (2 i \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )-2 i \sqrt {c-i c \tan (e+f x)}\right )-\frac {2}{3} i (c-i c \tan (e+f x))^{3/2}\right )\right )\right )}{a^2 f}\) |
(c*(((I*A - B)*(c - I*c*Tan[e + f*x])^(7/2))/(4*c*(1 + I*Tan[e + f*x])^2) - ((3*A + (11*I)*B)*((I*(c - I*c*Tan[e + f*x])^(5/2))/(1 + I*Tan[e + f*x]) - (5*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*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])] - (2*I)*Sqrt[c - I*c*Tan[e + f*x]])))/2))/8))/(a^2*f)
3.8.72.3.1 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 = 0.29 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.75
| method | result | size |
| derivativedivides | \(\frac {2 i c^{2} \left (-\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-5 i \sqrt {c -i c \tan \left (f x +e \right )}\, B c -\sqrt {c -i c \tan \left (f x +e \right )}\, c A +2 c^{2} \left (\frac {4 \left (\frac {17 i B}{32}+\frac {9 A}{32}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+4 \left (-\frac {15}{16} i B c -\frac {7}{16} c A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {5 \left (\frac {11 i B}{4}+\frac {3 A}{4}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}\right )\right )}{f \,a^{2}}\) | \(179\) |
| default | \(\frac {2 i c^{2} \left (-\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-5 i \sqrt {c -i c \tan \left (f x +e \right )}\, B c -\sqrt {c -i c \tan \left (f x +e \right )}\, c A +2 c^{2} \left (\frac {4 \left (\frac {17 i B}{32}+\frac {9 A}{32}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+4 \left (-\frac {15}{16} i B c -\frac {7}{16} c A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {5 \left (\frac {11 i B}{4}+\frac {3 A}{4}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}\right )\right )}{f \,a^{2}}\) | \(179\) |
2*I/f/a^2*c^2*(-1/3*I*B*(c-I*c*tan(f*x+e))^(3/2)-5*I*(c-I*c*tan(f*x+e))^(1 /2)*B*c-(c-I*c*tan(f*x+e))^(1/2)*c*A+2*c^2*(4*((17/32*I*B+9/32*A)*(c-I*c*t an(f*x+e))^(3/2)+(-15/16*I*B*c-7/16*c*A)*(c-I*c*tan(f*x+e))^(1/2))/(c+I*c* tan(f*x+e))^2+5/4*(11/4*I*B+3/4*A)*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan( 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. 458 vs. \(2 (185) = 370\).
Time = 0.27 (sec) , antiderivative size = 458, normalized size of antiderivative = 1.92 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {15 \, \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {-\frac {{\left (9 \, A^{2} + 66 i \, A B - 121 \, B^{2}\right )} c^{7}}{a^{4} f^{2}}} \log \left (-\frac {5 \, {\left ({\left (-3 i \, A + 11 \, B\right )} c^{4} + \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {-\frac {{\left (9 \, A^{2} + 66 i \, A B - 121 \, B^{2}\right )} c^{7}}{a^{4} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a^{2} f}\right ) - 15 \, \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )} \sqrt {-\frac {{\left (9 \, A^{2} + 66 i \, A B - 121 \, B^{2}\right )} c^{7}}{a^{4} f^{2}}} \log \left (-\frac {5 \, {\left ({\left (-3 i \, A + 11 \, B\right )} c^{4} - \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {-\frac {{\left (9 \, A^{2} + 66 i \, A B - 121 \, B^{2}\right )} c^{7}}{a^{4} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a^{2} f}\right ) + \sqrt {2} {\left (15 \, {\left (3 i \, A - 11 \, B\right )} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 20 \, {\left (3 i \, A - 11 \, B\right )} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, {\left (3 i \, A - 11 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 \, {\left (-i \, A + B\right )} c^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (a^{2} f e^{\left (6 i \, f x + 6 i \, e\right )} + a^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}\right )}} \]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e))^2,x , algorithm="fricas")
-1/12*(15*sqrt(1/2)*(a^2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e) )*sqrt(-(9*A^2 + 66*I*A*B - 121*B^2)*c^7/(a^4*f^2))*log(-5*((-3*I*A + 11*B )*c^4 + sqrt(2)*sqrt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(-(9*A^2 + 66*I*A*B - 121*B^2)*c^7/(a^4*f^2))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e ^(-I*f*x - I*e)/(a^2*f)) - 15*sqrt(1/2)*(a^2*f*e^(6*I*f*x + 6*I*e) + a^2*f *e^(4*I*f*x + 4*I*e))*sqrt(-(9*A^2 + 66*I*A*B - 121*B^2)*c^7/(a^4*f^2))*lo g(-5*((-3*I*A + 11*B)*c^4 - sqrt(2)*sqrt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(-(9*A^2 + 66*I*A*B - 121*B^2)*c^7/(a^4*f^2))*sqrt(c/(e^(2*I*f *x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a^2*f)) + sqrt(2)*(15*(3*I*A - 11*B)* c^3*e^(6*I*f*x + 6*I*e) + 20*(3*I*A - 11*B)*c^3*e^(4*I*f*x + 4*I*e) + 3*(3 *I*A - 11*B)*c^3*e^(2*I*f*x + 2*I*e) + 6*(-I*A + B)*c^3)*sqrt(c/(e^(2*I*f* x + 2*I*e) + 1)))/(a^2*f*e^(6*I*f*x + 6*I*e) + a^2*f*e^(4*I*f*x + 4*I*e))
\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {A c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \left (- \frac {3 A c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\right )\, dx + \int \frac {B c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \left (- \frac {3 B c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\right )\, dx + \int \left (- \frac {3 i A c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\right )\, dx + \int \frac {i A c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \left (- \frac {3 i B c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\right )\, dx + \int \frac {i B c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \]
-(Integral(A*c**3*sqrt(-I*c*tan(e + f*x) + c)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) + Integral(-3*A*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) + Integral(B*c**3*sqr t(-I*c*tan(e + f*x) + c)*tan(e + f*x)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) + Integral(-3*B*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3/ (tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) + Integral(-3*I*A*c**3*sqrt(- I*c*tan(e + f*x) + c)*tan(e + f*x)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1 ), x) + Integral(I*A*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3/(tan (e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) + Integral(-3*I*B*c**3*sqrt(-I*c* tan(e + f*x) + c)*tan(e + f*x)**2/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1) , x) + Integral(I*B*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4/(tan( e + f*x)**2 - 2*I*tan(e + f*x) - 1), x))/a**2
Time = 0.30 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i \, {\left (\frac {15 \, \sqrt {2} {\left (3 \, A + 11 i \, B\right )} c^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2}} - \frac {12 \, {\left ({\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (9 \, A + 17 i \, B\right )} c^{5} - 2 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (7 \, A + 15 i \, B\right )} c^{6}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} c + 4 \, a^{2} c^{2}} + \frac {16 \, {\left (i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} B c^{3} + 3 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A + 5 i \, B\right )} c^{4}\right )}}{a^{2}}\right )}}{24 \, c f} \]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e))^2,x , algorithm="maxima")
-1/24*I*(15*sqrt(2)*(3*A + 11*I*B)*c^(9/2)*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 - 12*((-I*c*tan(f*x + e) + c)^(3/2)*(9*A + 17*I*B)*c^5 - 2*sqrt(-I*c*tan( f*x + e) + c)*(7*A + 15*I*B)*c^6)/((-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) + 16*(I*(-I*c*tan(f*x + e) + c)^(3/2 )*B*c^3 + 3*sqrt(-I*c*tan(f*x + e) + c)*(A + 5*I*B)*c^4)/a^2)/(c*f)
\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e))^2,x , algorithm="giac")
Time = 8.91 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.47 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {15\,B\,c^5\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\frac {17\,B\,c^4\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{2}}{4\,a^2\,c^2\,f+a^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,a^2\,c\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}-\frac {\frac {A\,c^5\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,7{}\mathrm {i}}{a^2\,f}-\frac {A\,c^4\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,9{}\mathrm {i}}{2\,a^2\,f}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+4\,c^2}-\frac {A\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a^2\,f}+\frac {10\,B\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{a^2\,f}+\frac {2\,B\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^2\,f}-\frac {\sqrt {2}\,A\,{\left (-c\right )}^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,15{}\mathrm {i}}{4\,a^2\,f}+\frac {\sqrt {2}\,B\,c^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,55{}\mathrm {i}}{4\,a^2\,f} \]
(15*B*c^5*(c - c*tan(e + f*x)*1i)^(1/2) - (17*B*c^4*(c - c*tan(e + f*x)*1i )^(3/2))/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^5*(c - c*tan(e + f*x)*1i)^(1/2)*7i)/(a^2*f) - (A*c^4*(c - c*tan(e + f*x)*1i)^(3/2)*9i)/(2*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^3*(c - c*tan(e + f*x) *1i)^(1/2)*2i)/(a^2*f) + (10*B*c^3*(c - c*tan(e + f*x)*1i)^(1/2))/(a^2*f) + (2*B*c^2*(c - c*tan(e + f*x)*1i)^(3/2))/(3*a^2*f) - (2^(1/2)*A*(-c)^(7/2 )*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*15i)/(4*a^2 *f) + (2^(1/2)*B*c^(7/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2)*1i)/( 2*c^(1/2)))*55i)/(4*a^2*f)