Integrand size = 41, antiderivative size = 221 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=\frac {5 (3 A+i B) x}{64 a^2 c^4}-\frac {i A-B}{64 a^2 c^4 f (i-\tan (e+f x))^2}-\frac {5 A+3 i B}{64 a^2 c^4 f (i-\tan (e+f x))}-\frac {i A+B}{32 a^2 c^4 f (i+\tan (e+f x))^4}-\frac {3 A-i B}{48 a^2 c^4 f (i+\tan (e+f x))^3}+\frac {3 i A}{32 a^2 c^4 f (i+\tan (e+f x))^2}+\frac {5 A+i B}{32 a^2 c^4 f (i+\tan (e+f x))} \] Output:
5/64*(3*A+I*B)*x/a^2/c^4-1/64*(I*A-B)/a^2/c^4/f/(I-tan(f*x+e))^2-1/64*(5*A +3*I*B)/a^2/c^4/f/(I-tan(f*x+e))-1/32*(I*A+B)/a^2/c^4/f/(I+tan(f*x+e))^4-1 /48*(3*A-I*B)/a^2/c^4/f/(I+tan(f*x+e))^3+3/32*I*A/a^2/c^4/f/(I+tan(f*x+e)) ^2+1/32*(5*A+I*B)/a^2/c^4/f/(I+tan(f*x+e))
Time = 4.15 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.96 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=\frac {\sec ^5(e+f x) (240 i A \cos (e+f x)+5 (-9 i A+11 B) \cos (3 (e+f x))-3 i A \cos (5 (e+f x))+9 B \cos (5 (e+f x))+102 A \sin (e+f x)+34 i B \sin (e+f x)-60 (3 A+i B) \arctan (\tan (e+f x)) \sec (e+f x) (\cos (2 (e+f x))-i \sin (2 (e+f x)))-87 A \sin (3 (e+f x))-29 i B \sin (3 (e+f x))-9 A \sin (5 (e+f x))-3 i B \sin (5 (e+f x)))}{768 a^2 c^4 f (-i+\tan (e+f x))^2 (i+\tan (e+f x))^4} \] Input:
Integrate[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^4),x]
Output:
(Sec[e + f*x]^5*((240*I)*A*Cos[e + f*x] + 5*((-9*I)*A + 11*B)*Cos[3*(e + f *x)] - (3*I)*A*Cos[5*(e + f*x)] + 9*B*Cos[5*(e + f*x)] + 102*A*Sin[e + f*x ] + (34*I)*B*Sin[e + f*x] - 60*(3*A + I*B)*ArcTan[Tan[e + f*x]]*Sec[e + f* x]*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)]) - 87*A*Sin[3*(e + f*x)] - (29*I )*B*Sin[3*(e + f*x)] - 9*A*Sin[5*(e + f*x)] - (3*I)*B*Sin[5*(e + f*x)]))/( 768*a^2*c^4*f*(-I + Tan[e + f*x])^2*(I + Tan[e + f*x])^4)
Time = 0.72 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.80, 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.
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{a^3 c^5 (1-i \tan (e+f x))^5 (i \tan (e+f x)+1)^3}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {A+B \tan (e+f x)}{(1-i \tan (e+f x))^5 (i \tan (e+f x)+1)^3}d\tan (e+f x)}{a^2 c^4 f}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {\int \left (-\frac {3 i A}{16 (\tan (e+f x)+i)^3}+\frac {5 (3 A+i B)}{64 \left (\tan ^2(e+f x)+1\right )}+\frac {-5 A-3 i B}{64 (\tan (e+f x)-i)^2}+\frac {-5 A-i B}{32 (\tan (e+f x)+i)^2}+\frac {i (A+i B)}{32 (\tan (e+f x)-i)^3}+\frac {3 A-i B}{16 (\tan (e+f x)+i)^4}+\frac {i A+B}{8 (\tan (e+f x)+i)^5}\right )d\tan (e+f x)}{a^2 c^4 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {5}{64} (3 A+i B) \arctan (\tan (e+f x))-\frac {5 A+3 i B}{64 (-\tan (e+f x)+i)}+\frac {5 A+i B}{32 (\tan (e+f x)+i)}-\frac {-B+i A}{64 (-\tan (e+f x)+i)^2}-\frac {3 A-i B}{48 (\tan (e+f x)+i)^3}-\frac {B+i A}{32 (\tan (e+f x)+i)^4}+\frac {3 i A}{32 (\tan (e+f x)+i)^2}}{a^2 c^4 f}\) |
Input:
Int[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^ 4),x]
Output:
((5*(3*A + I*B)*ArcTan[Tan[e + f*x]])/64 - (I*A - B)/(64*(I - Tan[e + f*x] )^2) - (5*A + (3*I)*B)/(64*(I - Tan[e + f*x])) - (I*A + B)/(32*(I + Tan[e + f*x])^4) - (3*A - I*B)/(48*(I + Tan[e + f*x])^3) + (((3*I)/32)*A)/(I + T an[e + f*x])^2 + (5*A + I*B)/(32*(I + Tan[e + f*x])))/(a^2*c^4*f)
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 = 0.56 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.27
| method | result | size |
| norman | \(\frac {\frac {5 \left (i B +3 A \right ) x}{64 a c}-\frac {3 i A +B}{12 a c f}+\frac {B \tan \left (f x +e \right )^{2}}{6 a c f}+\frac {73 \left (i B +3 A \right ) \tan \left (f x +e \right )^{3}}{192 a c f}+\frac {55 \left (i B +3 A \right ) \tan \left (f x +e \right )^{5}}{192 a c f}+\frac {5 \left (i B +3 A \right ) \tan \left (f x +e \right )^{7}}{64 a c f}+\frac {5 \left (i B +3 A \right ) x \tan \left (f x +e \right )^{2}}{16 a c}+\frac {15 \left (i B +3 A \right ) x \tan \left (f x +e \right )^{4}}{32 a c}+\frac {5 \left (i B +3 A \right ) x \tan \left (f x +e \right )^{6}}{16 a c}+\frac {5 \left (i B +3 A \right ) x \tan \left (f x +e \right )^{8}}{64 a c}+\frac {\left (-5 i B +49 A \right ) \tan \left (f x +e \right )}{64 a c f}}{a \,c^{3} \left (1+\tan \left (f x +e \right )^{2}\right )^{4}}\) | \(281\) |
| risch | \(\frac {5 i x B}{64 a^{2} c^{4}}+\frac {15 x A}{64 a^{2} c^{4}}-\frac {{\mathrm e}^{8 i \left (f x +e \right )} B}{512 a^{2} c^{4} f}-\frac {i {\mathrm e}^{8 i \left (f x +e \right )} A}{512 a^{2} c^{4} f}-\frac {{\mathrm e}^{6 i \left (f x +e \right )} B}{96 a^{2} c^{4} f}-\frac {i {\mathrm e}^{6 i \left (f x +e \right )} A}{64 a^{2} c^{4} f}-\frac {3 \cos \left (4 f x +4 e \right ) B}{128 a^{2} c^{4} f}-\frac {7 i \cos \left (4 f x +4 e \right ) A}{128 a^{2} c^{4} f}-\frac {i \sin \left (4 f x +4 e \right ) B}{64 a^{2} c^{4} f}+\frac {\sin \left (4 f x +4 e \right ) A}{16 a^{2} c^{4} f}-\frac {\cos \left (2 f x +2 e \right ) B}{32 a^{2} c^{4} f}-\frac {7 i \cos \left (2 f x +2 e \right ) A}{64 a^{2} c^{4} f}+\frac {i \sin \left (2 f x +2 e \right ) B}{32 a^{2} c^{4} f}+\frac {13 \sin \left (2 f x +2 e \right ) A}{64 a^{2} c^{4} f}\) | \(281\) |
| derivativedivides | \(-\frac {i A}{32 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}+\frac {i B}{48 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {15 A \arctan \left (\tan \left (f x +e \right )\right )}{64 f \,a^{2} c^{4}}+\frac {B}{64 f \,a^{2} c^{4} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {i A}{64 f \,a^{2} c^{4} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {5 A}{64 f \,a^{2} c^{4} \left (-i+\tan \left (f x +e \right )\right )}+\frac {5 i B \arctan \left (\tan \left (f x +e \right )\right )}{64 f \,a^{2} c^{4}}-\frac {A}{16 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {3 i A}{32 a^{2} c^{4} f \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {i B}{32 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )}+\frac {3 i B}{64 f \,a^{2} c^{4} \left (-i+\tan \left (f x +e \right )\right )}-\frac {B}{32 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}+\frac {5 A}{32 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )}\) | \(300\) |
| default | \(-\frac {i A}{32 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}+\frac {i B}{48 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {15 A \arctan \left (\tan \left (f x +e \right )\right )}{64 f \,a^{2} c^{4}}+\frac {B}{64 f \,a^{2} c^{4} \left (-i+\tan \left (f x +e \right )\right )^{2}}-\frac {i A}{64 f \,a^{2} c^{4} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {5 A}{64 f \,a^{2} c^{4} \left (-i+\tan \left (f x +e \right )\right )}+\frac {5 i B \arctan \left (\tan \left (f x +e \right )\right )}{64 f \,a^{2} c^{4}}-\frac {A}{16 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {3 i A}{32 a^{2} c^{4} f \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {i B}{32 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )}+\frac {3 i B}{64 f \,a^{2} c^{4} \left (-i+\tan \left (f x +e \right )\right )}-\frac {B}{32 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )^{4}}+\frac {5 A}{32 f \,a^{2} c^{4} \left (i+\tan \left (f x +e \right )\right )}\) | \(300\) |
Input:
int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^4,x,method=_R ETURNVERBOSE)
Output:
(5/64*(3*A+I*B)/a/c*x-1/12*(3*I*A+B)/a/c/f+1/6/a/c/f*B*tan(f*x+e)^2+73/192 *(3*A+I*B)/a/c/f*tan(f*x+e)^3+55/192*(3*A+I*B)/a/c/f*tan(f*x+e)^5+5/64*(3* A+I*B)/a/c/f*tan(f*x+e)^7+5/16*(3*A+I*B)/a/c*x*tan(f*x+e)^2+15/32*(3*A+I*B )/a/c*x*tan(f*x+e)^4+5/16*(3*A+I*B)/a/c*x*tan(f*x+e)^6+5/64*(3*A+I*B)/a/c* x*tan(f*x+e)^8+1/64*(49*A-5*I*B)/a/c/f*tan(f*x+e))/a/c^3/(1+tan(f*x+e)^2)^ 4
Time = 0.07 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.57 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=\frac {{\left (120 \, {\left (3 \, A + i \, B\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} - 3 \, {\left (i \, A + B\right )} e^{\left (12 i \, f x + 12 i \, e\right )} - 8 \, {\left (3 i \, A + 2 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} - 30 \, {\left (3 i \, A + B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} - 240 i \, A e^{\left (6 i \, f x + 6 i \, e\right )} - 24 \, {\left (-3 i \, A + 2 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 i \, A - 6 \, B\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{1536 \, a^{2} c^{4} f} \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^4,x, al gorithm="fricas")
Output:
1/1536*(120*(3*A + I*B)*f*x*e^(4*I*f*x + 4*I*e) - 3*(I*A + B)*e^(12*I*f*x + 12*I*e) - 8*(3*I*A + 2*B)*e^(10*I*f*x + 10*I*e) - 30*(3*I*A + B)*e^(8*I* f*x + 8*I*e) - 240*I*A*e^(6*I*f*x + 6*I*e) - 24*(-3*I*A + 2*B)*e^(2*I*f*x + 2*I*e) + 6*I*A - 6*B)*e^(-4*I*f*x - 4*I*e)/(a^2*c^4*f)
Time = 0.53 (sec) , antiderivative size = 498, normalized size of antiderivative = 2.25 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=\begin {cases} \frac {\left (- 2061584302080 i A a^{10} c^{20} f^{5} e^{8 i e} e^{2 i f x} + \left (51539607552 i A a^{10} c^{20} f^{5} e^{2 i e} - 51539607552 B a^{10} c^{20} f^{5} e^{2 i e}\right ) e^{- 4 i f x} + \left (618475290624 i A a^{10} c^{20} f^{5} e^{4 i e} - 412316860416 B a^{10} c^{20} f^{5} e^{4 i e}\right ) e^{- 2 i f x} + \left (- 773094113280 i A a^{10} c^{20} f^{5} e^{10 i e} - 257698037760 B a^{10} c^{20} f^{5} e^{10 i e}\right ) e^{4 i f x} + \left (- 206158430208 i A a^{10} c^{20} f^{5} e^{12 i e} - 137438953472 B a^{10} c^{20} f^{5} e^{12 i e}\right ) e^{6 i f x} + \left (- 25769803776 i A a^{10} c^{20} f^{5} e^{14 i e} - 25769803776 B a^{10} c^{20} f^{5} e^{14 i e}\right ) e^{8 i f x}\right ) e^{- 6 i e}}{13194139533312 a^{12} c^{24} f^{6}} & \text {for}\: a^{12} c^{24} f^{6} e^{6 i e} \neq 0 \\x \left (- \frac {15 A + 5 i B}{64 a^{2} c^{4}} + \frac {\left (A e^{12 i e} + 6 A e^{10 i e} + 15 A e^{8 i e} + 20 A e^{6 i e} + 15 A e^{4 i e} + 6 A e^{2 i e} + A - i B e^{12 i e} - 4 i B e^{10 i e} - 5 i B e^{8 i e} + 5 i B e^{4 i e} + 4 i B e^{2 i e} + i B\right ) e^{- 4 i e}}{64 a^{2} c^{4}}\right ) & \text {otherwise} \end {cases} + \frac {x \left (15 A + 5 i B\right )}{64 a^{2} c^{4}} \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))**2/(c-I*c*tan(f*x+e))**4,x)
Output:
Piecewise(((-2061584302080*I*A*a**10*c**20*f**5*exp(8*I*e)*exp(2*I*f*x) + (51539607552*I*A*a**10*c**20*f**5*exp(2*I*e) - 51539607552*B*a**10*c**20*f **5*exp(2*I*e))*exp(-4*I*f*x) + (618475290624*I*A*a**10*c**20*f**5*exp(4*I *e) - 412316860416*B*a**10*c**20*f**5*exp(4*I*e))*exp(-2*I*f*x) + (-773094 113280*I*A*a**10*c**20*f**5*exp(10*I*e) - 257698037760*B*a**10*c**20*f**5* exp(10*I*e))*exp(4*I*f*x) + (-206158430208*I*A*a**10*c**20*f**5*exp(12*I*e ) - 137438953472*B*a**10*c**20*f**5*exp(12*I*e))*exp(6*I*f*x) + (-25769803 776*I*A*a**10*c**20*f**5*exp(14*I*e) - 25769803776*B*a**10*c**20*f**5*exp( 14*I*e))*exp(8*I*f*x))*exp(-6*I*e)/(13194139533312*a**12*c**24*f**6), Ne(a **12*c**24*f**6*exp(6*I*e), 0)), (x*(-(15*A + 5*I*B)/(64*a**2*c**4) + (A*e xp(12*I*e) + 6*A*exp(10*I*e) + 15*A*exp(8*I*e) + 20*A*exp(6*I*e) + 15*A*ex p(4*I*e) + 6*A*exp(2*I*e) + A - I*B*exp(12*I*e) - 4*I*B*exp(10*I*e) - 5*I* B*exp(8*I*e) + 5*I*B*exp(4*I*e) + 4*I*B*exp(2*I*e) + I*B)*exp(-4*I*e)/(64* a**2*c**4)), True)) + x*(15*A + 5*I*B)/(64*a**2*c**4)
Exception generated. \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^4,x, al gorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.60 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.78 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=\frac {5 \, {\left (3 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) + i\right )}{128 \, a^{2} c^{4} f} - \frac {5 \, {\left (3 i \, A - B\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{128 \, a^{2} c^{4} f} + \frac {15 \, {\left (3 \, A + i \, B\right )} \tan \left (f x + e\right )^{5} - 30 \, {\left (-3 i \, A + B\right )} \tan \left (f x + e\right )^{4} + 10 \, {\left (3 \, A + i \, B\right )} \tan \left (f x + e\right )^{3} - 50 \, {\left (-3 i \, A + B\right )} \tan \left (f x + e\right )^{2} - 17 \, {\left (3 \, A + i \, B\right )} \tan \left (f x + e\right ) + 48 i \, A + 16 \, B}{192 \, a^{2} c^{4} f {\left (\tan \left (f x + e\right ) + i\right )}^{4} {\left (\tan \left (f x + e\right ) - i\right )}^{2}} \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^4,x, al gorithm="giac")
Output:
5/128*(3*I*A - B)*log(tan(f*x + e) + I)/(a^2*c^4*f) - 5/128*(3*I*A - B)*lo g(tan(f*x + e) - I)/(a^2*c^4*f) + 1/192*(15*(3*A + I*B)*tan(f*x + e)^5 - 3 0*(-3*I*A + B)*tan(f*x + e)^4 + 10*(3*A + I*B)*tan(f*x + e)^3 - 50*(-3*I*A + B)*tan(f*x + e)^2 - 17*(3*A + I*B)*tan(f*x + e) + 48*I*A + 16*B)/(a^2*c ^4*f*(tan(f*x + e) + I)^4*(tan(f*x + e) - I)^2)
Time = 6.89 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.12 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=\frac {\frac {B}{12\,a^2\,c^4}+{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (-\frac {5\,B}{32\,a^2\,c^4}+\frac {A\,15{}\mathrm {i}}{32\,a^2\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {5\,A}{32\,a^2\,c^4}+\frac {B\,5{}\mathrm {i}}{96\,a^2\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (\frac {15\,A}{64\,a^2\,c^4}+\frac {B\,5{}\mathrm {i}}{64\,a^2\,c^4}\right )+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (-\frac {25\,B}{96\,a^2\,c^4}+\frac {A\,25{}\mathrm {i}}{32\,a^2\,c^4}\right )-\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {17\,A}{64\,a^2\,c^4}+\frac {B\,17{}\mathrm {i}}{192\,a^2\,c^4}\right )+\frac {A\,1{}\mathrm {i}}{4\,a^2\,c^4}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^6+{\mathrm {tan}\left (e+f\,x\right )}^5\,2{}\mathrm {i}+{\mathrm {tan}\left (e+f\,x\right )}^4+{\mathrm {tan}\left (e+f\,x\right )}^3\,4{}\mathrm {i}-{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}-1\right )}+\frac {5\,x\,\left (3\,A+B\,1{}\mathrm {i}\right )}{64\,a^2\,c^4} \] Input:
int((A + B*tan(e + f*x))/((a + a*tan(e + f*x)*1i)^2*(c - c*tan(e + f*x)*1i )^4),x)
Output:
(tan(e + f*x)^4*((A*15i)/(32*a^2*c^4) - (5*B)/(32*a^2*c^4)) - tan(e + f*x) *((17*A)/(64*a^2*c^4) + (B*17i)/(192*a^2*c^4)) + tan(e + f*x)^3*((5*A)/(32 *a^2*c^4) + (B*5i)/(96*a^2*c^4)) + tan(e + f*x)^5*((15*A)/(64*a^2*c^4) + ( B*5i)/(64*a^2*c^4)) + tan(e + f*x)^2*((A*25i)/(32*a^2*c^4) - (25*B)/(96*a^ 2*c^4)) + (A*1i)/(4*a^2*c^4) + B/(12*a^2*c^4))/(f*(tan(e + f*x)*2i - tan(e + f*x)^2 + tan(e + f*x)^3*4i + tan(e + f*x)^4 + tan(e + f*x)^5*2i + tan(e + f*x)^6 - 1)) + (5*x*(3*A + B*1i))/(64*a^2*c^4)
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^4} \, dx=\frac {-\left (\int \frac {\tan \left (f x +e \right )}{\tan \left (f x +e \right )^{6}+2 \tan \left (f x +e \right )^{5} i +\tan \left (f x +e \right )^{4}+4 \tan \left (f x +e \right )^{3} i -\tan \left (f x +e \right )^{2}+2 \tan \left (f x +e \right ) i -1}d x \right ) b -\left (\int \frac {1}{\tan \left (f x +e \right )^{6}+2 \tan \left (f x +e \right )^{5} i +\tan \left (f x +e \right )^{4}+4 \tan \left (f x +e \right )^{3} i -\tan \left (f x +e \right )^{2}+2 \tan \left (f x +e \right ) i -1}d x \right ) a}{a^{2} c^{4}} \] Input:
int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^4,x)
Output:
( - (int(tan(e + f*x)/(tan(e + f*x)**6 + 2*tan(e + f*x)**5*i + tan(e + f*x )**4 + 4*tan(e + f*x)**3*i - tan(e + f*x)**2 + 2*tan(e + f*x)*i - 1),x)*b + int(1/(tan(e + f*x)**6 + 2*tan(e + f*x)**5*i + tan(e + f*x)**4 + 4*tan(e + f*x)**3*i - tan(e + f*x)**2 + 2*tan(e + f*x)*i - 1),x)*a))/(a**2*c**4)