Integrand size = 41, antiderivative size = 113 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^2} \, dx=\frac {(3 A+i B) x}{8 a c^2}-\frac {A+i B}{8 a c^2 f (i-\tan (e+f x))}+\frac {i A+B}{8 a c^2 f (i+\tan (e+f x))^2}+\frac {A}{4 a c^2 f (i+\tan (e+f x))} \]
1/8*(3*A+I*B)*x/a/c^2+1/8*(-A-I*B)/a/c^2/f/(I-tan(f*x+e))+1/8*(I*A+B)/a/c^ 2/f/(I+tan(f*x+e))^2+1/4*A/a/c^2/f/(I+tan(f*x+e))
Time = 5.63 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^2} \, dx=\frac {\sec ^2(e+f x) (5 A-i B-(A+3 i B) \cos (2 (e+f x))+3 i A \sin (2 (e+f x))-B \sin (2 (e+f x))+2 (3 A+i B) \arctan (\tan (e+f x)) (i+\tan (e+f x)))}{16 a c^2 f (-i+\tan (e+f x)) (i+\tan (e+f x))^2} \]
(Sec[e + f*x]^2*(5*A - I*B - (A + (3*I)*B)*Cos[2*(e + f*x)] + (3*I)*A*Sin[ 2*(e + f*x)] - B*Sin[2*(e + f*x)] + 2*(3*A + I*B)*ArcTan[Tan[e + f*x]]*(I + Tan[e + f*x])))/(16*a*c^2*f*(-I + Tan[e + f*x])*(I + Tan[e + f*x])^2)
Time = 0.37 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.85, 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.
\(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^2}dx\) |
\(\Big \downarrow \) 4071 |
\(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{a^2 c^3 (1-i \tan (e+f x))^3 (i \tan (e+f x)+1)^2}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {A+B \tan (e+f x)}{(1-i \tan (e+f x))^3 (i \tan (e+f x)+1)^2}d\tan (e+f x)}{a c^2 f}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\int \left (-\frac {A}{4 (\tan (e+f x)+i)^2}+\frac {3 A+i B}{8 \left (\tan ^2(e+f x)+1\right )}+\frac {-A-i B}{8 (\tan (e+f x)-i)^2}-\frac {i (A-i B)}{4 (\tan (e+f x)+i)^3}\right )d\tan (e+f x)}{a c^2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{8} (3 A+i B) \arctan (\tan (e+f x))-\frac {A+i B}{8 (-\tan (e+f x)+i)}+\frac {B+i A}{8 (\tan (e+f x)+i)^2}+\frac {A}{4 (\tan (e+f x)+i)}}{a c^2 f}\) |
(((3*A + I*B)*ArcTan[Tan[e + f*x]])/8 - (A + I*B)/(8*(I - Tan[e + f*x])) + (I*A + B)/(8*(I + Tan[e + f*x])^2) + A/(4*(I + Tan[e + f*x])))/(a*c^2*f)
3.8.12.3.1 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 = 0.19 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.15
method | result | size |
risch | \(\frac {i x B}{8 a \,c^{2}}+\frac {3 x A}{8 a \,c^{2}}-\frac {{\mathrm e}^{4 i \left (f x +e \right )} B}{32 a \,c^{2} f}-\frac {i {\mathrm e}^{4 i \left (f x +e \right )} A}{32 a \,c^{2} f}-\frac {\cos \left (2 f x +2 e \right ) B}{8 a \,c^{2} f}-\frac {i \cos \left (2 f x +2 e \right ) A}{8 a \,c^{2} f}+\frac {A \sin \left (2 f x +2 e \right )}{4 a \,c^{2} f}\) | \(130\) |
norman | \(\frac {\frac {\left (i B +3 A \right ) x}{8 a c}-\frac {i A +B}{4 a c f}+\frac {\left (i B +3 A \right ) \tan \left (f x +e \right )^{3}}{8 a c f}+\frac {\left (i B +3 A \right ) x \tan \left (f x +e \right )^{2}}{4 a c}+\frac {\left (i B +3 A \right ) x \tan \left (f x +e \right )^{4}}{8 a c}+\frac {\left (-i B +5 A \right ) \tan \left (f x +e \right )}{8 a c f}}{c \left (1+\tan \left (f x +e \right )^{2}\right )^{2}}\) | \(154\) |
derivativedivides | \(\frac {3 A \arctan \left (\tan \left (f x +e \right )\right )}{8 f a \,c^{2}}+\frac {i B \arctan \left (\tan \left (f x +e \right )\right )}{8 f a \,c^{2}}+\frac {A}{4 a \,c^{2} f \left (i+\tan \left (f x +e \right )\right )}+\frac {i A}{8 f a \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {B}{8 f a \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {A}{8 f a \,c^{2} \left (-i+\tan \left (f x +e \right )\right )}+\frac {i B}{8 f a \,c^{2} \left (-i+\tan \left (f x +e \right )\right )}\) | \(158\) |
default | \(\frac {3 A \arctan \left (\tan \left (f x +e \right )\right )}{8 f a \,c^{2}}+\frac {i B \arctan \left (\tan \left (f x +e \right )\right )}{8 f a \,c^{2}}+\frac {A}{4 a \,c^{2} f \left (i+\tan \left (f x +e \right )\right )}+\frac {i A}{8 f a \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {B}{8 f a \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {A}{8 f a \,c^{2} \left (-i+\tan \left (f x +e \right )\right )}+\frac {i B}{8 f a \,c^{2} \left (-i+\tan \left (f x +e \right )\right )}\) | \(158\) |
1/8*I*x/a/c^2*B+3/8*x/a/c^2*A-1/32/a/c^2/f*exp(4*I*(f*x+e))*B-1/32*I/a/c^2 /f*exp(4*I*(f*x+e))*A-1/8/a/c^2/f*cos(2*f*x+2*e)*B-1/8*I/a/c^2/f*cos(2*f*x +2*e)*A+1/4*A/a/c^2/f*sin(2*f*x+2*e)
Time = 0.25 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.71 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^2} \, dx=\frac {{\left (4 \, {\left (3 \, A + i \, B\right )} f x e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A - B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - 2 \, {\left (3 i \, A + B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 i \, A - 2 \, B\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{32 \, a c^{2} f} \]
1/32*(4*(3*A + I*B)*f*x*e^(2*I*f*x + 2*I*e) + (-I*A - B)*e^(6*I*f*x + 6*I* e) - 2*(3*I*A + B)*e^(4*I*f*x + 4*I*e) + 2*I*A - 2*B)*e^(-2*I*f*x - 2*I*e) /(a*c^2*f)
Time = 0.27 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.51 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^2} \, dx=\begin {cases} \frac {\left (\left (512 i A a^{2} c^{4} f^{2} - 512 B a^{2} c^{4} f^{2}\right ) e^{- 2 i f x} + \left (- 1536 i A a^{2} c^{4} f^{2} e^{4 i e} - 512 B a^{2} c^{4} f^{2} e^{4 i e}\right ) e^{2 i f x} + \left (- 256 i A a^{2} c^{4} f^{2} e^{6 i e} - 256 B a^{2} c^{4} f^{2} e^{6 i e}\right ) e^{4 i f x}\right ) e^{- 2 i e}}{8192 a^{3} c^{6} f^{3}} & \text {for}\: a^{3} c^{6} f^{3} e^{2 i e} \neq 0 \\x \left (- \frac {3 A + i B}{8 a c^{2}} + \frac {\left (A e^{6 i e} + 3 A e^{4 i e} + 3 A e^{2 i e} + A - i B e^{6 i e} - i B e^{4 i e} + i B e^{2 i e} + i B\right ) e^{- 2 i e}}{8 a c^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (3 A + i B\right )}{8 a c^{2}} \]
Piecewise((((512*I*A*a**2*c**4*f**2 - 512*B*a**2*c**4*f**2)*exp(-2*I*f*x) + (-1536*I*A*a**2*c**4*f**2*exp(4*I*e) - 512*B*a**2*c**4*f**2*exp(4*I*e))* exp(2*I*f*x) + (-256*I*A*a**2*c**4*f**2*exp(6*I*e) - 256*B*a**2*c**4*f**2* exp(6*I*e))*exp(4*I*f*x))*exp(-2*I*e)/(8192*a**3*c**6*f**3), Ne(a**3*c**6* f**3*exp(2*I*e), 0)), (x*(-(3*A + I*B)/(8*a*c**2) + (A*exp(6*I*e) + 3*A*ex p(4*I*e) + 3*A*exp(2*I*e) + A - I*B*exp(6*I*e) - I*B*exp(4*I*e) + I*B*exp( 2*I*e) + I*B)*exp(-2*I*e)/(8*a*c**2)), True)) + x*(3*A + I*B)/(8*a*c**2)
Exception generated. \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.55 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.41 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^2} \, dx=\frac {\frac {2 \, {\left (3 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{a c^{2}} + \frac {2 \, {\left (-3 i \, A + B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a c^{2}} - \frac {2 \, {\left (3 \, A \tan \left (f x + e\right ) + i \, B \tan \left (f x + e\right ) - 5 i \, A + 3 \, B\right )}}{a c^{2} {\left (i \, \tan \left (f x + e\right ) + 1\right )}} + \frac {-9 i \, A \tan \left (f x + e\right )^{2} + 3 \, B \tan \left (f x + e\right )^{2} + 26 \, A \tan \left (f x + e\right ) + 6 i \, B \tan \left (f x + e\right ) + 21 i \, A + B}{a c^{2} {\left (\tan \left (f x + e\right ) + i\right )}^{2}}}{32 \, f} \]
1/32*(2*(3*I*A - B)*log(tan(f*x + e) + I)/(a*c^2) + 2*(-3*I*A + B)*log(tan (f*x + e) - I)/(a*c^2) - 2*(3*A*tan(f*x + e) + I*B*tan(f*x + e) - 5*I*A + 3*B)/(a*c^2*(I*tan(f*x + e) + 1)) + (-9*I*A*tan(f*x + e)^2 + 3*B*tan(f*x + e)^2 + 26*A*tan(f*x + e) + 6*I*B*tan(f*x + e) + 21*I*A + B)/(a*c^2*(tan(f *x + e) + I)^2))/f
Time = 9.06 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.14 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^2} \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (-\frac {B}{8\,a\,c^2}+\frac {A\,3{}\mathrm {i}}{8\,a\,c^2}\right )+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {3\,A}{8\,a\,c^2}+\frac {B\,1{}\mathrm {i}}{8\,a\,c^2}\right )+\frac {A}{4\,a\,c^2}-\frac {B\,1{}\mathrm {i}}{4\,a\,c^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^3+{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}+\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {x\,\left (-B+A\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a\,c^2} \]