Integrand size = 41, antiderivative size = 183 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\frac {(5 A+i B) x}{16 a^2 c^3}-\frac {i A-B}{32 a^2 c^3 f (i-\tan (e+f x))^2}-\frac {2 A+i B}{16 a^2 c^3 f (i-\tan (e+f x))}-\frac {A-i B}{24 a^2 c^3 f (i+\tan (e+f x))^3}+\frac {3 i A+B}{32 a^2 c^3 f (i+\tan (e+f x))^2}+\frac {3 A}{16 a^2 c^3 f (i+\tan (e+f x))} \]
1/16*(5*A+I*B)*x/a^2/c^3+1/32*(-I*A+B)/a^2/c^3/f/(I-tan(f*x+e))^2+1/16*(-2 *A-I*B)/a^2/c^3/f/(I-tan(f*x+e))+1/24*(-A+I*B)/a^2/c^3/f/(I+tan(f*x+e))^3+ 1/32*(3*I*A+B)/a^2/c^3/f/(I+tan(f*x+e))^2+3/16*A/a^2/c^3/f/(I+tan(f*x+e))
Time = 6.00 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.95 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\frac {\sec ^4(e+f x) (47 A-5 i B-2 (7 A+11 i B) \cos (2 (e+f x))-A \cos (4 (e+f x))-5 i B \cos (4 (e+f x))+40 i A \sin (2 (e+f x))-8 B \sin (2 (e+f x))+5 i A \sin (4 (e+f x))-B \sin (4 (e+f x))+12 (5 A+i B) \arctan (\tan (e+f x)) (i+\tan (e+f x)))}{192 a^2 c^3 f (-i+\tan (e+f x))^2 (i+\tan (e+f x))^3} \]
(Sec[e + f*x]^4*(47*A - (5*I)*B - 2*(7*A + (11*I)*B)*Cos[2*(e + f*x)] - A* Cos[4*(e + f*x)] - (5*I)*B*Cos[4*(e + f*x)] + (40*I)*A*Sin[2*(e + f*x)] - 8*B*Sin[2*(e + f*x)] + (5*I)*A*Sin[4*(e + f*x)] - B*Sin[4*(e + f*x)] + 12* (5*A + I*B)*ArcTan[Tan[e + f*x]]*(I + Tan[e + f*x])))/(192*a^2*c^3*f*(-I + Tan[e + f*x])^2*(I + Tan[e + f*x])^3)
Time = 0.41 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.81, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3042, 4071, 27, 86, 2009}
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.
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{a^3 c^4 (1-i \tan (e+f x))^4 (i \tan (e+f x)+1)^3}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {A+B \tan (e+f x)}{(1-i \tan (e+f x))^4 (i \tan (e+f x)+1)^3}d\tan (e+f x)}{a^2 c^3 f}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {\int \left (-\frac {3 A}{16 (\tan (e+f x)+i)^2}+\frac {5 A+i B}{16 \left (\tan ^2(e+f x)+1\right )}+\frac {-2 A-i B}{16 (\tan (e+f x)-i)^2}+\frac {i (A+i B)}{16 (\tan (e+f x)-i)^3}-\frac {i (3 A-i B)}{16 (\tan (e+f x)+i)^3}+\frac {A-i B}{8 (\tan (e+f x)+i)^4}\right )d\tan (e+f x)}{a^2 c^3 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{16} (5 A+i B) \arctan (\tan (e+f x))-\frac {2 A+i B}{16 (-\tan (e+f x)+i)}-\frac {-B+i A}{32 (-\tan (e+f x)+i)^2}+\frac {B+3 i A}{32 (\tan (e+f x)+i)^2}-\frac {A-i B}{24 (\tan (e+f x)+i)^3}+\frac {3 A}{16 (\tan (e+f x)+i)}}{a^2 c^3 f}\) |
(((5*A + I*B)*ArcTan[Tan[e + f*x]])/16 - (I*A - B)/(32*(I - Tan[e + f*x])^ 2) - (2*A + I*B)/(16*(I - Tan[e + f*x])) - (A - I*B)/(24*(I + Tan[e + f*x] )^3) + ((3*I)*A + B)/(32*(I + Tan[e + f*x])^2) + (3*A)/(16*(I + Tan[e + f* x])))/(a^2*c^3*f)
3.8.24.3.1 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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 209, normalized size of antiderivative = 1.14
| method | result | size |
| norman | \(\frac {\frac {\left (i B +5 A \right ) x}{16 a c}-\frac {i A +B}{6 a c f}+\frac {\left (-i B +11 A \right ) \tan \left (f x +e \right )}{16 a c f}+\frac {\left (i B +5 A \right ) \tan \left (f x +e \right )^{3}}{6 a c f}+\frac {\left (i B +5 A \right ) \tan \left (f x +e \right )^{5}}{16 a c f}+\frac {3 \left (i B +5 A \right ) x \tan \left (f x +e \right )^{2}}{16 a c}+\frac {3 \left (i B +5 A \right ) x \tan \left (f x +e \right )^{4}}{16 a c}+\frac {\left (i B +5 A \right ) x \tan \left (f x +e \right )^{6}}{16 a c}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3} a \,c^{2}}\) | \(209\) |
| risch | \(\frac {i x B}{16 a^{2} c^{3}}+\frac {5 x A}{16 a^{2} c^{3}}-\frac {{\mathrm e}^{6 i \left (f x +e \right )} B}{192 a^{2} c^{3} f}-\frac {i {\mathrm e}^{6 i \left (f x +e \right )} A}{192 a^{2} c^{3} f}-\frac {\cos \left (4 f x +4 e \right ) B}{32 a^{2} c^{3} f}-\frac {i \cos \left (4 f x +4 e \right ) A}{32 a^{2} c^{3} f}-\frac {i \sin \left (4 f x +4 e \right ) B}{64 a^{2} c^{3} f}+\frac {3 \sin \left (4 f x +4 e \right ) A}{64 a^{2} c^{3} f}-\frac {5 \cos \left (2 f x +2 e \right ) B}{64 a^{2} c^{3} f}-\frac {5 i \cos \left (2 f x +2 e \right ) A}{64 a^{2} c^{3} f}+\frac {i \sin \left (2 f x +2 e \right ) B}{64 a^{2} c^{3} f}+\frac {15 \sin \left (2 f x +2 e \right ) A}{64 a^{2} c^{3} f}\) | \(238\) |
| derivativedivides | \(\frac {i B}{16 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {A}{8 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {i B \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{2} c^{3}}+\frac {5 A \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{2} c^{3}}+\frac {i B}{24 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {B}{32 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {i A}{32 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {A}{24 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {3 i A}{32 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {3 A}{16 a^{2} c^{3} f \left (i+\tan \left (f x +e \right )\right )}+\frac {B}{32 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}\) | \(252\) |
| default | \(\frac {i B}{16 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {A}{8 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {i B \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{2} c^{3}}+\frac {5 A \arctan \left (\tan \left (f x +e \right )\right )}{16 f \,a^{2} c^{3}}+\frac {i B}{24 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {B}{32 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {i A}{32 f \,a^{2} c^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {A}{24 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {3 i A}{32 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {3 A}{16 a^{2} c^{3} f \left (i+\tan \left (f x +e \right )\right )}+\frac {B}{32 f \,a^{2} c^{3} \left (i+\tan \left (f x +e \right )\right )^{2}}\) | \(252\) |
(1/16*(5*A+I*B)/a/c*x-1/6*(I*A+B)/a/c/f+1/16*(-I*B+11*A)/a/c/f*tan(f*x+e)+ 1/6*(5*A+I*B)/a/c/f*tan(f*x+e)^3+1/16*(5*A+I*B)/a/c/f*tan(f*x+e)^5+3/16*(5 *A+I*B)/a/c*x*tan(f*x+e)^2+3/16*(5*A+I*B)/a/c*x*tan(f*x+e)^4+1/16*(5*A+I*B )/a/c*x*tan(f*x+e)^6)/(1+tan(f*x+e)^2)^3/a/c^2
Time = 0.25 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.63 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\frac {{\left (24 \, {\left (5 \, A + i \, B\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, {\left (i \, A + B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} - 3 \, {\left (5 i \, A + 3 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} - 12 \, {\left (5 i \, A + B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - 6 \, {\left (-5 i \, A + 3 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, A - 3 \, B\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{384 \, a^{2} c^{3} f} \]
1/384*(24*(5*A + I*B)*f*x*e^(4*I*f*x + 4*I*e) - 2*(I*A + B)*e^(10*I*f*x + 10*I*e) - 3*(5*I*A + 3*B)*e^(8*I*f*x + 8*I*e) - 12*(5*I*A + B)*e^(6*I*f*x + 6*I*e) - 6*(-5*I*A + 3*B)*e^(2*I*f*x + 2*I*e) + 3*I*A - 3*B)*e^(-4*I*f*x - 4*I*e)/(a^2*c^3*f)
Time = 0.42 (sec) , antiderivative size = 454, normalized size of antiderivative = 2.48 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\begin {cases} \frac {\left (\left (50331648 i A a^{8} c^{12} f^{4} e^{2 i e} - 50331648 B a^{8} c^{12} f^{4} e^{2 i e}\right ) e^{- 4 i f x} + \left (503316480 i A a^{8} c^{12} f^{4} e^{4 i e} - 301989888 B a^{8} c^{12} f^{4} e^{4 i e}\right ) e^{- 2 i f x} + \left (- 1006632960 i A a^{8} c^{12} f^{4} e^{8 i e} - 201326592 B a^{8} c^{12} f^{4} e^{8 i e}\right ) e^{2 i f x} + \left (- 251658240 i A a^{8} c^{12} f^{4} e^{10 i e} - 150994944 B a^{8} c^{12} f^{4} e^{10 i e}\right ) e^{4 i f x} + \left (- 33554432 i A a^{8} c^{12} f^{4} e^{12 i e} - 33554432 B a^{8} c^{12} f^{4} e^{12 i e}\right ) e^{6 i f x}\right ) e^{- 6 i e}}{6442450944 a^{10} c^{15} f^{5}} & \text {for}\: a^{10} c^{15} f^{5} e^{6 i e} \neq 0 \\x \left (- \frac {5 A + i B}{16 a^{2} c^{3}} + \frac {\left (A e^{10 i e} + 5 A e^{8 i e} + 10 A e^{6 i e} + 10 A e^{4 i e} + 5 A e^{2 i e} + A - i B e^{10 i e} - 3 i B e^{8 i e} - 2 i B e^{6 i e} + 2 i B e^{4 i e} + 3 i B e^{2 i e} + i B\right ) e^{- 4 i e}}{32 a^{2} c^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (5 A + i B\right )}{16 a^{2} c^{3}} \]
Piecewise((((50331648*I*A*a**8*c**12*f**4*exp(2*I*e) - 50331648*B*a**8*c** 12*f**4*exp(2*I*e))*exp(-4*I*f*x) + (503316480*I*A*a**8*c**12*f**4*exp(4*I *e) - 301989888*B*a**8*c**12*f**4*exp(4*I*e))*exp(-2*I*f*x) + (-1006632960 *I*A*a**8*c**12*f**4*exp(8*I*e) - 201326592*B*a**8*c**12*f**4*exp(8*I*e))* exp(2*I*f*x) + (-251658240*I*A*a**8*c**12*f**4*exp(10*I*e) - 150994944*B*a **8*c**12*f**4*exp(10*I*e))*exp(4*I*f*x) + (-33554432*I*A*a**8*c**12*f**4* exp(12*I*e) - 33554432*B*a**8*c**12*f**4*exp(12*I*e))*exp(6*I*f*x))*exp(-6 *I*e)/(6442450944*a**10*c**15*f**5), Ne(a**10*c**15*f**5*exp(6*I*e), 0)), (x*(-(5*A + I*B)/(16*a**2*c**3) + (A*exp(10*I*e) + 5*A*exp(8*I*e) + 10*A*e xp(6*I*e) + 10*A*exp(4*I*e) + 5*A*exp(2*I*e) + A - I*B*exp(10*I*e) - 3*I*B *exp(8*I*e) - 2*I*B*exp(6*I*e) + 2*I*B*exp(4*I*e) + 3*I*B*exp(2*I*e) + I*B )*exp(-4*I*e)/(32*a**2*c**3)), True)) + x*(5*A + I*B)/(16*a**2*c**3)
Exception generated. \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.79 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.12 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=-\frac {\frac {6 \, {\left (-5 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{2} c^{3}} + \frac {6 \, {\left (5 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{2} c^{3}} + \frac {3 \, {\left (15 i \, A \tan \left (f x + e\right )^{2} - 3 \, B \tan \left (f x + e\right )^{2} + 38 \, A \tan \left (f x + e\right ) + 10 i \, B \tan \left (f x + e\right ) - 25 i \, A + 9 \, B\right )}}{a^{2} c^{3} {\left (i \, \tan \left (f x + e\right ) + 1\right )}^{2}} + \frac {55 i \, A \tan \left (f x + e\right )^{3} - 11 \, B \tan \left (f x + e\right )^{3} - 201 \, A \tan \left (f x + e\right )^{2} - 33 i \, B \tan \left (f x + e\right )^{2} - 255 i \, A \tan \left (f x + e\right ) + 27 \, B \tan \left (f x + e\right ) + 117 \, A - 3 i \, B}{a^{2} c^{3} {\left (\tan \left (f x + e\right ) + i\right )}^{3}}}{192 \, f} \]
-1/192*(6*(-5*I*A + B)*log(tan(f*x + e) + I)/(a^2*c^3) + 6*(5*I*A - B)*log (tan(f*x + e) - I)/(a^2*c^3) + 3*(15*I*A*tan(f*x + e)^2 - 3*B*tan(f*x + e) ^2 + 38*A*tan(f*x + e) + 10*I*B*tan(f*x + e) - 25*I*A + 9*B)/(a^2*c^3*(I*t an(f*x + e) + 1)^2) + (55*I*A*tan(f*x + e)^3 - 11*B*tan(f*x + e)^3 - 201*A *tan(f*x + e)^2 - 33*I*B*tan(f*x + e)^2 - 255*I*A*tan(f*x + e) + 27*B*tan( f*x + e) + 117*A - 3*I*B)/(a^2*c^3*(tan(f*x + e) + I)^3))/f
Time = 9.09 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.14 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^3} \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (-\frac {5\,B}{48\,a^2\,c^3}+\frac {A\,25{}\mathrm {i}}{48\,a^2\,c^3}\right )+{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (-\frac {B}{16\,a^2\,c^3}+\frac {A\,5{}\mathrm {i}}{16\,a^2\,c^3}\right )+{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (\frac {5\,A}{16\,a^2\,c^3}+\frac {B\,1{}\mathrm {i}}{16\,a^2\,c^3}\right )+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {25\,A}{48\,a^2\,c^3}+\frac {B\,5{}\mathrm {i}}{48\,a^2\,c^3}\right )+\frac {A}{6\,a^2\,c^3}-\frac {B\,1{}\mathrm {i}}{6\,a^2\,c^3}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^5+{\mathrm {tan}\left (e+f\,x\right )}^4\,1{}\mathrm {i}+2\,{\mathrm {tan}\left (e+f\,x\right )}^3+{\mathrm {tan}\left (e+f\,x\right )}^2\,2{}\mathrm {i}+\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {x\,\left (-B+A\,5{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^2\,c^3} \]
(tan(e + f*x)*((A*25i)/(48*a^2*c^3) - (5*B)/(48*a^2*c^3)) + tan(e + f*x)^3 *((A*5i)/(16*a^2*c^3) - B/(16*a^2*c^3)) + tan(e + f*x)^4*((5*A)/(16*a^2*c^ 3) + (B*1i)/(16*a^2*c^3)) + tan(e + f*x)^2*((25*A)/(48*a^2*c^3) + (B*5i)/( 48*a^2*c^3)) + A/(6*a^2*c^3) - (B*1i)/(6*a^2*c^3))/(f*(tan(e + f*x) + tan( e + f*x)^2*2i + 2*tan(e + f*x)^3 + tan(e + f*x)^4*1i + tan(e + f*x)^5 + 1i )) - (x*(A*5i - B)*1i)/(16*a^2*c^3)