Integrand size = 43, antiderivative size = 252 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {5 (i A-13 B) c^{7/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a^3 f}+\frac {5 (i A-13 B) c^3 \sqrt {c-i c \tan (e+f x)}}{16 a^3 f}+\frac {5 (i A-13 B) c^2 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f (1+i \tan (e+f x))}-\frac {(i A-13 B) c (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]
-5/16*(I*A-13*B)*c^(7/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1 /2))/a^3/f*2^(1/2)+5/16*(I*A-13*B)*c^3*(c-I*c*tan(f*x+e))^(1/2)/a^3/f+5/48 *(I*A-13*B)*c^2*(c-I*c*tan(f*x+e))^(3/2)/a^3/f/(1+I*tan(f*x+e))-1/24*(I*A- 13*B)*c*(c-I*c*tan(f*x+e))^(5/2)/a^3/f/(1+I*tan(f*x+e))^2+1/6*(I*A-B)*(c-I *c*tan(f*x+e))^(7/2)/a^3/f/(1+I*tan(f*x+e))^3
Time = 6.71 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.77 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {c^3 \sec ^3(e+f x) \left (15 \sqrt {2} (A+13 i B) \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x)))-(3 (A+13 i B) \cos (e+f x)+(23 A+203 i B) \cos (3 (e+f x))+2 i (7 A+139 i B+(7 A+187 i B) \cos (2 (e+f x))) \sin (e+f x)) \sqrt {c-i c \tan (e+f x)}\right )}{48 a^3 f (-i+\tan (e+f x))^3} \]
(c^3*Sec[e + f*x]^3*(15*Sqrt[2]*(A + (13*I)*B)*Sqrt[c]*ArcTanh[Sqrt[c - I* c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])]*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)]) - (3*(A + (13*I)*B)*Cos[e + f*x] + (23*A + (203*I)*B)*Cos[3*(e + f*x)] + (2*I)*(7*A + (139*I)*B + (7*A + (187*I)*B)*Cos[2*(e + f*x)])*Sin[e + f*x]) *Sqrt[c - I*c*Tan[e + f*x]]))/(48*a^3*f*(-I + Tan[e + f*x])^3)
Time = 0.39 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.86, 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, 51, 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.
| \(\displaystyle \int \frac {(c-i c \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(a+i a \tan (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-i c \tan (e+f x))^{7/2} (A+B \tan (e+f x))}{(a+i a \tan (e+f x))^3}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{a^4 (i \tan (e+f x)+1)^4}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {c \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{(i \tan (e+f x)+1)^4}d\tan (e+f x)}{a^3 f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{6 c (1+i \tan (e+f x))^3}-\frac {1}{12} (A+13 i B) \int \frac {(c-i c \tan (e+f x))^{5/2}}{(i \tan (e+f x)+1)^3}d\tan (e+f x)\right )}{a^3 f}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{6 c (1+i \tan (e+f x))^3}-\frac {1}{12} (A+13 i B) \left (\frac {i (c-i c \tan (e+f x))^{5/2}}{2 (1+i \tan (e+f x))^2}-\frac {5}{4} c \int \frac {(c-i c \tan (e+f x))^{3/2}}{(i \tan (e+f x)+1)^2}d\tan (e+f x)\right )\right )}{a^3 f}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{6 c (1+i \tan (e+f x))^3}-\frac {1}{12} (A+13 i B) \left (\frac {i (c-i c \tan (e+f x))^{5/2}}{2 (1+i \tan (e+f x))^2}-\frac {5}{4} c \left (\frac {i (c-i c \tan (e+f x))^{3/2}}{1+i \tan (e+f x)}-\frac {3}{2} c \int \frac {\sqrt {c-i c \tan (e+f x)}}{i \tan (e+f x)+1}d\tan (e+f x)\right )\right )\right )}{a^3 f}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{6 c (1+i \tan (e+f x))^3}-\frac {1}{12} (A+13 i B) \left (\frac {i (c-i c \tan (e+f x))^{5/2}}{2 (1+i \tan (e+f x))^2}-\frac {5}{4} c \left (\frac {i (c-i c \tan (e+f x))^{3/2}}{1+i \tan (e+f x)}-\frac {3}{2} c \left (2 c \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)-2 i \sqrt {c-i c \tan (e+f x)}\right )\right )\right )\right )}{a^3 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{6 c (1+i \tan (e+f x))^3}-\frac {1}{12} (A+13 i B) \left (\frac {i (c-i c \tan (e+f x))^{5/2}}{2 (1+i \tan (e+f x))^2}-\frac {5}{4} c \left (\frac {i (c-i c \tan (e+f x))^{3/2}}{1+i \tan (e+f x)}-\frac {3}{2} c \left (4 i \int \frac {1}{2-\frac {c-i c \tan (e+f x)}{c}}d\sqrt {c-i c \tan (e+f x)}-2 i \sqrt {c-i c \tan (e+f x)}\right )\right )\right )\right )}{a^3 f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {c \left (\frac {(-B+i A) (c-i c \tan (e+f x))^{7/2}}{6 c (1+i \tan (e+f x))^3}-\frac {1}{12} (A+13 i B) \left (\frac {i (c-i c \tan (e+f x))^{5/2}}{2 (1+i \tan (e+f x))^2}-\frac {5}{4} c \left (\frac {i (c-i c \tan (e+f x))^{3/2}}{1+i \tan (e+f x)}-\frac {3}{2} c \left (2 i \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )-2 i \sqrt {c-i c \tan (e+f x)}\right )\right )\right )\right )}{a^3 f}\) |
(c*(((I*A - B)*(c - I*c*Tan[e + f*x])^(7/2))/(6*c*(1 + I*Tan[e + f*x])^3) - ((A + (13*I)*B)*(((I/2)*(c - I*c*Tan[e + f*x])^(5/2))/(1 + I*Tan[e + f*x ])^2 - (5*c*((I*(c - I*c*Tan[e + f*x])^(3/2))/(1 + I*Tan[e + f*x]) - (3*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))/4))/12))/(a^3*f)
3.8.80.3.1 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 = 0.24 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.67
| method | result | size |
| derivativedivides | \(\frac {2 i c^{3} \left (i \sqrt {c -i c \tan \left (f x +e \right )}\, B +c \left (\frac {8 \left (\frac {47 i B}{128}+\frac {11 A}{128}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+8 \left (-\frac {29}{24} i B c -\frac {5}{24} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+8 \left (\frac {33}{32} i B \,c^{2}+\frac {5}{32} c^{2} A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}-\frac {5 \left (\frac {13 i B}{8}+\frac {A}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}\right )\right )}{f \,a^{3}}\) | \(168\) |
| default | \(\frac {2 i c^{3} \left (i \sqrt {c -i c \tan \left (f x +e \right )}\, B +c \left (\frac {8 \left (\frac {47 i B}{128}+\frac {11 A}{128}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+8 \left (-\frac {29}{24} i B c -\frac {5}{24} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+8 \left (\frac {33}{32} i B \,c^{2}+\frac {5}{32} c^{2} A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}-\frac {5 \left (\frac {13 i B}{8}+\frac {A}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}\right )\right )}{f \,a^{3}}\) | \(168\) |
2*I/f/a^3*c^3*(I*(c-I*c*tan(f*x+e))^(1/2)*B+c*(8*((47/128*I*B+11/128*A)*(c -I*c*tan(f*x+e))^(5/2)+(-29/24*I*B*c-5/24*c*A)*(c-I*c*tan(f*x+e))^(3/2)+(3 3/32*I*B*c^2+5/32*c^2*A)*(c-I*c*tan(f*x+e))^(1/2))/(c+I*c*tan(f*x+e))^3-5/ 4*(13/8*I*B+1/8*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. 402 vs. \(2 (197) = 394\).
Time = 0.27 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.60 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {{\left (A^{2} + 26 i \, A B - 169 \, B^{2}\right )} c^{7}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {5 \, {\left ({\left (i \, A - 13 \, B\right )} c^{4} + \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {-\frac {{\left (A^{2} + 26 i \, A B - 169 \, B^{2}\right )} c^{7}}{a^{6} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{3} f}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {{\left (A^{2} + 26 i \, A B - 169 \, B^{2}\right )} c^{7}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {5 \, {\left ({\left (i \, A - 13 \, B\right )} c^{4} - \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {-\frac {{\left (A^{2} + 26 i \, A B - 169 \, B^{2}\right )} c^{7}}{a^{6} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a^{3} f}\right ) + \sqrt {2} {\left (15 \, {\left (-i \, A + 13 \, B\right )} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 5 \, {\left (-i \, A + 13 \, B\right )} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (i \, A - 13 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (-i \, A + B\right )} c^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{48 \, a^{3} f} \]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e))^3,x , algorithm="fricas")
-1/48*(15*sqrt(1/2)*a^3*f*sqrt(-(A^2 + 26*I*A*B - 169*B^2)*c^7/(a^6*f^2))* e^(6*I*f*x + 6*I*e)*log(-5/4*((I*A - 13*B)*c^4 + sqrt(2)*sqrt(1/2)*(a^3*f* e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(-(A^2 + 26*I*A*B - 169*B^2)*c^7/(a^6*f^2 ))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a^3*f)) - 15*sqrt( 1/2)*a^3*f*sqrt(-(A^2 + 26*I*A*B - 169*B^2)*c^7/(a^6*f^2))*e^(6*I*f*x + 6* I*e)*log(-5/4*((I*A - 13*B)*c^4 - sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2* I*e) + a^3*f)*sqrt(-(A^2 + 26*I*A*B - 169*B^2)*c^7/(a^6*f^2))*sqrt(c/(e^(2 *I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a^3*f)) + sqrt(2)*(15*(-I*A + 13* B)*c^3*e^(6*I*f*x + 6*I*e) + 5*(-I*A + 13*B)*c^3*e^(4*I*f*x + 4*I*e) + 2*( I*A - 13*B)*c^3*e^(2*I*f*x + 2*I*e) + 8*(-I*A + B)*c^3)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*f)
\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \left (\int \frac {A c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx + \int \left (- \frac {3 A c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\right )\, dx + \int \frac {B c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx + \int \left (- \frac {3 B c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\right )\, dx + \int \left (- \frac {3 i A c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\right )\, dx + \int \frac {i A c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx + \int \left (- \frac {3 i B c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\right )\, dx + \int \frac {i B c^{3} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx\right )}{a^{3}} \]
I*(Integral(A*c**3*sqrt(-I*c*tan(e + f*x) + c)/(tan(e + f*x)**3 - 3*I*tan( e + f*x)**2 - 3*tan(e + f*x) + I), x) + Integral(-3*A*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*ta n(e + f*x) + I), x) + Integral(B*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + I), x) + In tegral(-3*B*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3/(tan(e + f*x) **3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + I), x) + Integral(-3*I*A*c**3 *sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)/(tan(e + f*x)**3 - 3*I*tan(e + f *x)**2 - 3*tan(e + f*x) + I), x) + Integral(I*A*c**3*sqrt(-I*c*tan(e + f*x ) + c)*tan(e + f*x)**3/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + I), x) + Integral(-3*I*B*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f *x)**2/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + I), x) + Integral(I*B*c**3*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4/(tan(e + f*x )**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + I), x))/a**3
Time = 0.42 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \, {\left (\frac {15 \, \sqrt {2} {\left (A + 13 i \, B\right )} c^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3}} + \frac {192 i \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} B c^{4}}{a^{3}} - \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (11 \, A + 47 i \, B\right )} c^{5} - 16 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (5 \, A + 29 i \, B\right )} c^{6} + 12 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (5 \, A + 33 i \, B\right )} c^{7}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{3} c^{2} - 8 \, a^{3} c^{3}}\right )}}{96 \, c f} \]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e))^3,x , algorithm="maxima")
1/96*I*(15*sqrt(2)*(A + 13*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^3 + 192*I*sqrt(-I*c*tan(f*x + e) + c)*B*c^4/a^3 - 4*(3*(-I*c*tan(f*x + e) + c) ^(5/2)*(11*A + 47*I*B)*c^5 - 16*(-I*c*tan(f*x + e) + c)^(3/2)*(5*A + 29*I* B)*c^6 + 12*sqrt(-I*c*tan(f*x + e) + c)*(5*A + 33*I*B)*c^7)/((-I*c*tan(f*x + e) + c)^3*a^3 - 6*(-I*c*tan(f*x + e) + c)^2*a^3*c + 12*(-I*c*tan(f*x + e) + c)*a^3*c^2 - 8*a^3*c^3))/(c*f)
\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}} \,d x } \]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(7/2)/(a+I*a*tan(f*x+e))^3,x , algorithm="giac")
Time = 9.28 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.53 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\frac {A\,c^6\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,5{}\mathrm {i}}{2\,a^3\,f}-\frac {A\,c^5\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,10{}\mathrm {i}}{3\,a^3\,f}+\frac {A\,c^4\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,11{}\mathrm {i}}{8\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+8\,c^3}-\frac {\frac {33\,B\,c^6\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2}-\frac {58\,B\,c^5\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3}+\frac {47\,B\,c^4\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{8}}{8\,a^3\,c^3\,f-a^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+6\,a^3\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,a^3\,c^2\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}-\frac {2\,B\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{a^3\,f}+\frac {\sqrt {2}\,A\,{\left (-c\right )}^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,5{}\mathrm {i}}{16\,a^3\,f}+\frac {65\,\sqrt {2}\,B\,c^{7/2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{16\,a^3\,f} \]
((A*c^6*(c - c*tan(e + f*x)*1i)^(1/2)*5i)/(2*a^3*f) - (A*c^5*(c - c*tan(e + f*x)*1i)^(3/2)*10i)/(3*a^3*f) + (A*c^4*(c - c*tan(e + f*x)*1i)^(5/2)*11i )/(8*a^3*f))/(6*c*(c - c*tan(e + f*x)*1i)^2 - 12*c^2*(c - c*tan(e + f*x)*1 i) - (c - c*tan(e + f*x)*1i)^3 + 8*c^3) - ((33*B*c^6*(c - c*tan(e + f*x)*1 i)^(1/2))/2 - (58*B*c^5*(c - c*tan(e + f*x)*1i)^(3/2))/3 + (47*B*c^4*(c - c*tan(e + f*x)*1i)^(5/2))/8)/(8*a^3*c^3*f - a^3*f*(c - c*tan(e + f*x)*1i)^ 3 + 6*a^3*c*f*(c - c*tan(e + f*x)*1i)^2 - 12*a^3*c^2*f*(c - c*tan(e + f*x) *1i)) - (2*B*c^3*(c - c*tan(e + f*x)*1i)^(1/2))/(a^3*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)))*5i)/(16 *a^3*f) + (65*2^(1/2)*B*c^(7/2)*atanh((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/ 2))/(2*c^(1/2))))/(16*a^3*f)