Integrand size = 43, antiderivative size = 274 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {35 (3 i A+B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{256 \sqrt {2} a^3 c^{3/2} f}-\frac {35 (3 i A+B)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {35 (3 i A+B)}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}} \]
35/512*(3*I*A+B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^3 /c^(3/2)/f*2^(1/2)-35/256*(3*I*A+B)/a^3/c/f/(c-I*c*tan(f*x+e))^(1/2)-35/38 4*(3*I*A+B)/a^3/f/(c-I*c*tan(f*x+e))^(3/2)+1/6*(I*A-B)/a^3/f/(1+I*tan(f*x+ e))^3/(c-I*c*tan(f*x+e))^(3/2)+1/16*(3*I*A+B)/a^3/f/(1+I*tan(f*x+e))^2/(c- I*c*tan(f*x+e))^(3/2)+7/64*(3*I*A+B)/a^3/f/(1+I*tan(f*x+e))/(c-I*c*tan(f*x +e))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 6.14 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.70 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sec ^4(e+f x) \left (105 (3 i A+B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {1}{2} i (i+\tan (e+f x))\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+2 \cos (e+f x) (3 (-55 i A+3 B) \cos (e+f x)+8 (i A+3 B) \cos (3 (e+f x))+(3 A-i B) (27 \sin (e+f x)-8 \sin (3 (e+f x))))\right )}{768 a^3 c f (-i+\tan (e+f x))^3 (i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \]
(Sec[e + f*x]^4*(105*((3*I)*A + B)*Hypergeometric2F1[-1/2, 1, 1/2, (-1/2*I )*(I + Tan[e + f*x])]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]) + 2*Cos[e + f*x]*(3*((-55*I)*A + 3*B)*Cos[e + f*x] + 8*(I*A + 3*B)*Cos[3*(e + f*x)] + (3*A - I*B)*(27*Sin[e + f*x] - 8*Sin[3*(e + f*x)]))))/(768*a^3*c*f*(-I + T an[e + f*x])^3*(I + Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.44 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.95, 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, 52, 52, 61, 61, 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.
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{a^4 (i \tan (e+f x)+1)^4 (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {c \int \frac {A+B \tan (e+f x)}{(i \tan (e+f x)+1)^4 (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{a^3 f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (3 A-i B) \int \frac {1}{(i \tan (e+f x)+1)^3 (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)+\frac {-B+i A}{6 c (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (3 A-i B) \left (\frac {7}{8} \int \frac {1}{(i \tan (e+f x)+1)^2 (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)+\frac {i}{4 c (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}\right )+\frac {-B+i A}{6 c (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (3 A-i B) \left (\frac {7}{8} \left (\frac {5}{4} \int \frac {1}{(i \tan (e+f x)+1) (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)+\frac {i}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}\right )+\frac {i}{4 c (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}\right )+\frac {-B+i A}{6 c (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (3 A-i B) \left (\frac {7}{8} \left (\frac {5}{4} \left (\frac {\int \frac {1}{(i \tan (e+f x)+1) (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{2 c}-\frac {i}{3 c (c-i c \tan (e+f x))^{3/2}}\right )+\frac {i}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}\right )+\frac {i}{4 c (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}\right )+\frac {-B+i A}{6 c (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (3 A-i B) \left (\frac {7}{8} \left (\frac {5}{4} \left (\frac {\frac {\int \frac {1}{(i \tan (e+f x)+1) \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{2 c}-\frac {i}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {i}{3 c (c-i c \tan (e+f x))^{3/2}}\right )+\frac {i}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}\right )+\frac {i}{4 c (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}\right )+\frac {-B+i A}{6 c (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (3 A-i B) \left (\frac {7}{8} \left (\frac {5}{4} \left (\frac {\frac {i \int \frac {1}{2-\frac {c-i c \tan (e+f x)}{c}}d\sqrt {c-i c \tan (e+f x)}}{c^2}-\frac {i}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {i}{3 c (c-i c \tan (e+f x))^{3/2}}\right )+\frac {i}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}\right )+\frac {i}{4 c (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}\right )+\frac {-B+i A}{6 c (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}\right )}{a^3 f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (3 A-i B) \left (\frac {7}{8} \left (\frac {5}{4} \left (\frac {\frac {i \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} c^{3/2}}-\frac {i}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {i}{3 c (c-i c \tan (e+f x))^{3/2}}\right )+\frac {i}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}\right )+\frac {i}{4 c (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}\right )+\frac {-B+i A}{6 c (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}\right )}{a^3 f}\) |
(c*((I*A - B)/(6*c*(1 + I*Tan[e + f*x])^3*(c - I*c*Tan[e + f*x])^(3/2)) + ((3*A - I*B)*((I/4)/(c*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2) ) + (7*((I/2)/(c*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) + (5*( (-1/3*I)/(c*(c - I*c*Tan[e + f*x])^(3/2)) + ((I*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*c^(3/2)) - I/(c*Sqrt[c - I*c*Tan[e + f*x]]))/(2*c)))/4))/8))/4))/(a^3*f)
3.8.85.3.1 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 + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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 = 0.23 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.75
| method | result | size |
| derivativedivides | \(\frac {2 i c^{3} \left (\frac {\frac {8 \left (-\frac {3 i B}{256}+\frac {41 A}{256}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+8 \left (\frac {1}{48} i B c -\frac {35}{48} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+8 \left (\frac {3}{64} i B \,c^{2}+\frac {55}{64} c^{2} A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {35 \left (\frac {3 A}{8}-\frac {i B}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}}{16 c^{4}}-\frac {-i B +2 A}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {-i B +A}{48 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,a^{3}}\) | \(205\) |
| default | \(\frac {2 i c^{3} \left (\frac {\frac {8 \left (-\frac {3 i B}{256}+\frac {41 A}{256}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+8 \left (\frac {1}{48} i B c -\frac {35}{48} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+8 \left (\frac {3}{64} i B \,c^{2}+\frac {55}{64} c^{2} A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {35 \left (\frac {3 A}{8}-\frac {i B}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}}{16 c^{4}}-\frac {-i B +2 A}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {-i B +A}{48 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,a^{3}}\) | \(205\) |
2*I/f/a^3*c^3*(1/16/c^4*(8*((-3/256*I*B+41/256*A)*(c-I*c*tan(f*x+e))^(5/2) +(1/48*I*B*c-35/48*c*A)*(c-I*c*tan(f*x+e))^(3/2)+(3/64*I*B*c^2+55/64*c^2*A )*(c-I*c*tan(f*x+e))^(1/2))/(c+I*c*tan(f*x+e))^3+35/8*(3/8*A-1/8*I*B)*2^(1 /2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))-1/16/c^ 4*(2*A-I*B)/(c-I*c*tan(f*x+e))^(1/2)-1/48/c^3*(A-I*B)/(c-I*c*tan(f*x+e))^( 3/2))
Time = 0.27 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.60 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {{\left (105 \, \sqrt {\frac {1}{2}} a^{3} c^{2} f \sqrt {-\frac {9 \, A^{2} - 6 i \, A B - B^{2}}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {35 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {9 \, A^{2} - 6 i \, A B - B^{2}}{a^{6} c^{3} f^{2}}} + 3 i \, A + B\right )} e^{\left (-i \, f x - i \, e\right )}}{128 \, a^{3} c f}\right ) - 105 \, \sqrt {\frac {1}{2}} a^{3} c^{2} f \sqrt {-\frac {9 \, A^{2} - 6 i \, A B - B^{2}}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {35 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {9 \, A^{2} - 6 i \, A B - B^{2}}{a^{6} c^{3} f^{2}}} - 3 i \, A - B\right )} e^{\left (-i \, f x - i \, e\right )}}{128 \, a^{3} c f}\right ) - \sqrt {2} {\left (16 \, {\left (i \, A + B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} + 32 \, {\left (7 i \, A + 4 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} - {\left (-43 i \, A - 121 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 5 \, {\left (-43 i \, A + 7 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-29 i \, A + 17 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, A + 8 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{1536 \, a^{3} c^{2} f} \]
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(3/2),x , algorithm="fricas")
1/1536*(105*sqrt(1/2)*a^3*c^2*f*sqrt(-(9*A^2 - 6*I*A*B - B^2)/(a^6*c^3*f^2 ))*e^(6*I*f*x + 6*I*e)*log(35/128*(sqrt(2)*sqrt(1/2)*(a^3*c*f*e^(2*I*f*x + 2*I*e) + a^3*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(9*A^2 - 6*I*A* B - B^2)/(a^6*c^3*f^2)) + 3*I*A + B)*e^(-I*f*x - I*e)/(a^3*c*f)) - 105*sqr t(1/2)*a^3*c^2*f*sqrt(-(9*A^2 - 6*I*A*B - B^2)/(a^6*c^3*f^2))*e^(6*I*f*x + 6*I*e)*log(-35/128*(sqrt(2)*sqrt(1/2)*(a^3*c*f*e^(2*I*f*x + 2*I*e) + a^3* c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(9*A^2 - 6*I*A*B - B^2)/(a^6* c^3*f^2)) - 3*I*A - B)*e^(-I*f*x - I*e)/(a^3*c*f)) - sqrt(2)*(16*(I*A + B) *e^(10*I*f*x + 10*I*e) + 32*(7*I*A + 4*B)*e^(8*I*f*x + 8*I*e) - (-43*I*A - 121*B)*e^(6*I*f*x + 6*I*e) + 5*(-43*I*A + 7*B)*e^(4*I*f*x + 4*I*e) + 2*(- 29*I*A + 17*B)*e^(2*I*f*x + 2*I*e) - 8*I*A + 8*B)*sqrt(c/(e^(2*I*f*x + 2*I *e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*c^2*f)
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {i \left (\int \frac {A}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 2 c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan {\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 2 c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx\right )}{a^{3}} \]
I*(Integral(A/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4 - 2*c*sqrt (-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 2*c*sqrt(-I*c*tan(e + f*x) + c)* tan(e + f*x) + I*c*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(B*tan(e + f *x)/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4 - 2*c*sqrt(-I*c*tan( e + f*x) + c)*tan(e + f*x)**3 - 2*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f* x) + I*c*sqrt(-I*c*tan(e + f*x) + c)), x))/a**3
Time = 0.39 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.96 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {i \, {\left (\frac {4 \, {\left (105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} {\left (3 \, A - i \, B\right )} - 560 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} {\left (3 \, A - i \, B\right )} c + 924 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (3 \, A - i \, B\right )} c^{2} - 384 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (3 \, A - i \, B\right )} c^{3} - 256 \, {\left (A - i \, B\right )} c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c^{2} - 8 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} c^{3}} + \frac {105 \, \sqrt {2} {\left (3 \, A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3} \sqrt {c}}\right )}}{3072 \, c f} \]
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(3/2),x , algorithm="maxima")
-1/3072*I*(4*(105*(-I*c*tan(f*x + e) + c)^4*(3*A - I*B) - 560*(-I*c*tan(f* x + e) + c)^3*(3*A - I*B)*c + 924*(-I*c*tan(f*x + e) + c)^2*(3*A - I*B)*c^ 2 - 384*(-I*c*tan(f*x + e) + c)*(3*A - I*B)*c^3 - 256*(A - I*B)*c^4)/((-I* c*tan(f*x + e) + c)^(9/2)*a^3 - 6*(-I*c*tan(f*x + e) + c)^(7/2)*a^3*c + 12 *(-I*c*tan(f*x + e) + c)^(5/2)*a^3*c^2 - 8*(-I*c*tan(f*x + e) + c)^(3/2)*a ^3*c^3) + 105*sqrt(2)*(3*A - I*B)*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*sqrt(c) ))/(c*f)
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=\int { \frac {B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(3/2),x , algorithm="giac")
Time = 9.49 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.62 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,35{}\mathrm {i}}{16\,a^3\,f}+\frac {A\,c^3\,1{}\mathrm {i}}{3\,a^3\,f}-\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,105{}\mathrm {i}}{256\,a^3\,c\,f}-\frac {A\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,231{}\mathrm {i}}{64\,a^3\,f}+\frac {A\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,a^3\,f}}{6\,c\,{\left (c-c\,\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 )}^{9/2}+8\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}-12\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\frac {B\,c^3}{3}+\frac {35\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3}{48}-\frac {77\,B\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{64}+\frac {B\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{2}-\frac {35\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4}{256\,c}}{a^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}-6\,a^3\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}-8\,a^3\,c^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}+12\,a^3\,c^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,105{}\mathrm {i}}{512\,a^3\,{\left (-c\right )}^{3/2}\,f}+\frac {35\,\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{512\,a^3\,c^{3/2}\,f} \]
((B*c^3)/3 + (35*B*(c - c*tan(e + f*x)*1i)^3)/48 - (77*B*c*(c - c*tan(e + f*x)*1i)^2)/64 + (B*c^2*(c - c*tan(e + f*x)*1i))/2 - (35*B*(c - c*tan(e + f*x)*1i)^4)/(256*c))/(a^3*f*(c - c*tan(e + f*x)*1i)^(9/2) - 6*a^3*c*f*(c - c*tan(e + f*x)*1i)^(7/2) - 8*a^3*c^3*f*(c - c*tan(e + f*x)*1i)^(3/2) + 12 *a^3*c^2*f*(c - c*tan(e + f*x)*1i)^(5/2)) - ((A*(c - c*tan(e + f*x)*1i)^3* 35i)/(16*a^3*f) + (A*c^3*1i)/(3*a^3*f) - (A*(c - c*tan(e + f*x)*1i)^4*105i )/(256*a^3*c*f) - (A*c*(c - c*tan(e + f*x)*1i)^2*231i)/(64*a^3*f) + (A*c^2 *(c - c*tan(e + f*x)*1i)*3i)/(2*a^3*f))/(6*c*(c - c*tan(e + f*x)*1i)^(7/2) - (c - c*tan(e + f*x)*1i)^(9/2) + 8*c^3*(c - c*tan(e + f*x)*1i)^(3/2) - 1 2*c^2*(c - c*tan(e + f*x)*1i)^(5/2)) + (2^(1/2)*A*atan((2^(1/2)*(c - c*tan (e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*105i)/(512*a^3*(-c)^(3/2)*f) + (35*2^ (1/2)*B*atanh((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(512*a ^3*c^(3/2)*f)