Integrand size = 41, antiderivative size = 153 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\frac {(2 A-i B) x}{8 a^3 c}+\frac {A+i B}{12 a^3 c f (i-\tan (e+f x))^3}-\frac {i A}{8 a^3 c f (i-\tan (e+f x))^2}-\frac {3 A-i B}{16 a^3 c f (i-\tan (e+f x))}+\frac {A-i B}{16 a^3 c f (i+\tan (e+f x))} \]
1/8*(2*A-I*B)*x/a^3/c+1/12*(A+I*B)/a^3/c/f/(I-tan(f*x+e))^3-1/8*I*A/a^3/c/ f/(I-tan(f*x+e))^2+1/16*(-3*A+I*B)/a^3/c/f/(I-tan(f*x+e))+1/16*(A-I*B)/a^3 /c/f/(I+tan(f*x+e))
Time = 5.83 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.97 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\frac {\sec ^3(e+f x) (-9 i A \cos (e+f x)+(-1+2 \cos (2 (e+f x))) ((i A+2 B) \cos (e+f x)+(-2 A+i B) \sin (e+f x))-3 (2 A-i B) \arctan (\tan (e+f x)) \sec (e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x))))}{24 a^3 c f (-i+\tan (e+f x))^3 (i+\tan (e+f x))} \]
(Sec[e + f*x]^3*((-9*I)*A*Cos[e + f*x] + (-1 + 2*Cos[2*(e + f*x)])*((I*A + 2*B)*Cos[e + f*x] + (-2*A + I*B)*Sin[e + f*x]) - 3*(2*A - I*B)*ArcTan[Tan [e + f*x]]*Sec[e + f*x]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])))/(24*a^3* c*f*(-I + Tan[e + f*x])^3*(I + Tan[e + f*x]))
Time = 0.39 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.83, 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.
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{a^4 c^2 (1-i \tan (e+f x))^2 (i \tan (e+f x)+1)^4}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {A+B \tan (e+f x)}{(1-i \tan (e+f x))^2 (i \tan (e+f x)+1)^4}d\tan (e+f x)}{a^3 c f}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {\int \left (\frac {i A}{4 (\tan (e+f x)-i)^3}+\frac {2 A-i B}{8 \left (\tan ^2(e+f x)+1\right )}+\frac {i B-3 A}{16 (\tan (e+f x)-i)^2}+\frac {i B-A}{16 (\tan (e+f x)+i)^2}+\frac {A+i B}{4 (\tan (e+f x)-i)^4}\right )d\tan (e+f x)}{a^3 c f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{8} (2 A-i B) \arctan (\tan (e+f x))-\frac {3 A-i B}{16 (-\tan (e+f x)+i)}+\frac {A-i B}{16 (\tan (e+f x)+i)}+\frac {A+i B}{12 (-\tan (e+f x)+i)^3}-\frac {i A}{8 (-\tan (e+f x)+i)^2}}{a^3 c f}\) |
(((2*A - I*B)*ArcTan[Tan[e + f*x]])/8 + (A + I*B)/(12*(I - Tan[e + f*x])^3 ) - ((I/8)*A)/(I - Tan[e + f*x])^2 - (3*A - I*B)/(16*(I - Tan[e + f*x])) + (A - I*B)/(16*(I + Tan[e + f*x])))/(a^3*c*f)
3.8.34.3.1 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 = 195, normalized size of antiderivative = 1.27
| method | result | size |
| risch | \(-\frac {i x B}{8 a^{3} c}+\frac {x A}{4 a^{3} c}-\frac {{\mathrm e}^{-4 i \left (f x +e \right )} B}{32 a^{3} c f}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )} A}{16 a^{3} c f}-\frac {{\mathrm e}^{-6 i \left (f x +e \right )} B}{96 a^{3} c f}+\frac {i {\mathrm e}^{-6 i \left (f x +e \right )} A}{96 a^{3} c f}-\frac {\cos \left (2 f x +2 e \right ) B}{32 a^{3} c f}+\frac {5 i \cos \left (2 f x +2 e \right ) A}{32 a^{3} c f}-\frac {i \sin \left (2 f x +2 e \right ) B}{32 a^{3} c f}+\frac {7 \sin \left (2 f x +2 e \right ) A}{32 a^{3} c f}\) | \(195\) |
| derivativedivides | \(-\frac {i A}{8 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {3 A}{16 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )}-\frac {i B}{16 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )}-\frac {A}{12 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {i B}{12 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {i B \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3} c}+\frac {A \arctan \left (\tan \left (f x +e \right )\right )}{4 f \,a^{3} c}+\frac {A}{16 f \,a^{3} c \left (i+\tan \left (f x +e \right )\right )}-\frac {i B}{16 f \,a^{3} c \left (i+\tan \left (f x +e \right )\right )}\) | \(206\) |
| default | \(-\frac {i A}{8 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {3 A}{16 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )}-\frac {i B}{16 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )}-\frac {A}{12 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {i B}{12 f \,a^{3} c \left (-i+\tan \left (f x +e \right )\right )^{3}}-\frac {i B \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{3} c}+\frac {A \arctan \left (\tan \left (f x +e \right )\right )}{4 f \,a^{3} c}+\frac {A}{16 f \,a^{3} c \left (i+\tan \left (f x +e \right )\right )}-\frac {i B}{16 f \,a^{3} c \left (i+\tan \left (f x +e \right )\right )}\) | \(206\) |
| norman | \(\frac {\frac {\left (-i B +2 A \right ) x}{8 a c}-\frac {-4 i A +B}{12 a c f}+\frac {B \tan \left (f x +e \right )^{2}}{4 a c f}+\frac {\left (-i B +2 A \right ) \tan \left (f x +e \right )^{3}}{3 a c f}+\frac {\left (-i B +2 A \right ) \tan \left (f x +e \right )^{5}}{8 a c f}+\frac {3 \left (-i B +2 A \right ) x \tan \left (f x +e \right )^{2}}{8 a c}+\frac {3 \left (-i B +2 A \right ) x \tan \left (f x +e \right )^{4}}{8 a c}+\frac {\left (-i B +2 A \right ) x \tan \left (f x +e \right )^{6}}{8 a c}+\frac {\left (i B +6 A \right ) \tan \left (f x +e \right )}{8 a c f}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3} a^{2}}\) | \(226\) |
-1/8*I*x/a^3/c*B+1/4*x/a^3/c*A-1/32/a^3/c/f*exp(-4*I*(f*x+e))*B+1/16*I/a^3 /c/f*exp(-4*I*(f*x+e))*A-1/96/a^3/c/f*exp(-6*I*(f*x+e))*B+1/96*I/a^3/c/f*e xp(-6*I*(f*x+e))*A-1/32/a^3/c/f*cos(2*f*x+2*e)*B+5/32*I/a^3/c/f*cos(2*f*x+ 2*e)*A-1/32*I/a^3/c/f*sin(2*f*x+2*e)*B+7/32/a^3/c/f*sin(2*f*x+2*e)*A
Time = 0.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.59 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\frac {{\left (12 \, {\left (2 \, A - i \, B\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, {\left (i \, A + B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 18 i \, A e^{\left (4 i \, f x + 4 i \, e\right )} - 3 \, {\left (-2 i \, A + B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} c f} \]
1/96*(12*(2*A - I*B)*f*x*e^(6*I*f*x + 6*I*e) - 3*(I*A + B)*e^(8*I*f*x + 8* I*e) + 18*I*A*e^(4*I*f*x + 4*I*e) - 3*(-2*I*A + B)*e^(2*I*f*x + 2*I*e) + I *A - B)*e^(-6*I*f*x - 6*I*e)/(a^3*c*f)
Time = 0.35 (sec) , antiderivative size = 340, normalized size of antiderivative = 2.22 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\begin {cases} \frac {\left (294912 i A a^{9} c^{3} f^{3} e^{10 i e} e^{- 2 i f x} + \left (16384 i A a^{9} c^{3} f^{3} e^{6 i e} - 16384 B a^{9} c^{3} f^{3} e^{6 i e}\right ) e^{- 6 i f x} + \left (98304 i A a^{9} c^{3} f^{3} e^{8 i e} - 49152 B a^{9} c^{3} f^{3} e^{8 i e}\right ) e^{- 4 i f x} + \left (- 49152 i A a^{9} c^{3} f^{3} e^{14 i e} - 49152 B a^{9} c^{3} f^{3} e^{14 i e}\right ) e^{2 i f x}\right ) e^{- 12 i e}}{1572864 a^{12} c^{4} f^{4}} & \text {for}\: a^{12} c^{4} f^{4} e^{12 i e} \neq 0 \\x \left (- \frac {2 A - i B}{8 a^{3} c} + \frac {\left (A e^{8 i e} + 4 A e^{6 i e} + 6 A e^{4 i e} + 4 A e^{2 i e} + A - i B e^{8 i e} - 2 i B e^{6 i e} + 2 i B e^{2 i e} + i B\right ) e^{- 6 i e}}{16 a^{3} c}\right ) & \text {otherwise} \end {cases} + \frac {x \left (2 A - i B\right )}{8 a^{3} c} \]
Piecewise(((294912*I*A*a**9*c**3*f**3*exp(10*I*e)*exp(-2*I*f*x) + (16384*I *A*a**9*c**3*f**3*exp(6*I*e) - 16384*B*a**9*c**3*f**3*exp(6*I*e))*exp(-6*I *f*x) + (98304*I*A*a**9*c**3*f**3*exp(8*I*e) - 49152*B*a**9*c**3*f**3*exp( 8*I*e))*exp(-4*I*f*x) + (-49152*I*A*a**9*c**3*f**3*exp(14*I*e) - 49152*B*a **9*c**3*f**3*exp(14*I*e))*exp(2*I*f*x))*exp(-12*I*e)/(1572864*a**12*c**4* f**4), Ne(a**12*c**4*f**4*exp(12*I*e), 0)), (x*(-(2*A - I*B)/(8*a**3*c) + (A*exp(8*I*e) + 4*A*exp(6*I*e) + 6*A*exp(4*I*e) + 4*A*exp(2*I*e) + A - I*B *exp(8*I*e) - 2*I*B*exp(6*I*e) + 2*I*B*exp(2*I*e) + I*B)*exp(-6*I*e)/(16*a **3*c)), True)) + x*(2*A - I*B)/(8*a**3*c)
Exception generated. \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.63 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.18 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=-\frac {\frac {6 \, {\left (-2 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a^{3} c} + \frac {6 \, {\left (2 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3} c} + \frac {6 \, {\left (2 i \, A \tan \left (f x + e\right ) + B \tan \left (f x + e\right ) - 3 \, A + 2 i \, B\right )}}{a^{3} c {\left (\tan \left (f x + e\right ) + i\right )}} + \frac {-22 i \, A \tan \left (f x + e\right )^{3} - 11 \, B \tan \left (f x + e\right )^{3} - 84 \, A \tan \left (f x + e\right )^{2} + 39 i \, B \tan \left (f x + e\right )^{2} + 114 i \, A \tan \left (f x + e\right ) + 45 \, B \tan \left (f x + e\right ) + 60 \, A - 9 i \, B}{a^{3} c {\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{96 \, f} \]
-1/96*(6*(-2*I*A - B)*log(tan(f*x + e) + I)/(a^3*c) + 6*(2*I*A + B)*log(ta n(f*x + e) - I)/(a^3*c) + 6*(2*I*A*tan(f*x + e) + B*tan(f*x + e) - 3*A + 2 *I*B)/(a^3*c*(tan(f*x + e) + I)) + (-22*I*A*tan(f*x + e)^3 - 11*B*tan(f*x + e)^3 - 84*A*tan(f*x + e)^2 + 39*I*B*tan(f*x + e)^2 + 114*I*A*tan(f*x + e ) + 45*B*tan(f*x + e) + 60*A - 9*I*B)/(a^3*c*(tan(f*x + e) - I)^3))/f
Time = 8.73 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))} \, dx=\frac {\frac {A}{3\,a^3\,c}+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {A}{2\,a^3\,c}-\frac {B\,1{}\mathrm {i}}{4\,a^3\,c}\right )+{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {B}{8\,a^3\,c}+\frac {A\,1{}\mathrm {i}}{4\,a^3\,c}\right )-\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {B}{24\,a^3\,c}+\frac {A\,1{}\mathrm {i}}{12\,a^3\,c}\right )+\frac {B\,1{}\mathrm {i}}{12\,a^3\,c}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^4\,1{}\mathrm {i}+2\,{\mathrm {tan}\left (e+f\,x\right )}^3+2\,\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )}-\frac {x\,\left (B+A\,2{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^3\,c} \]