\(\int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))} \, dx\) [1087]

Optimal result

Integrand size = 28, antiderivative size = 174 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))} \, dx=\frac {\left (c^3+3 i c^2 d-3 c d^2+3 i d^3\right ) x}{4 a^2 (c-i d) (c+i d)^3}-\frac {d^3 \log (c \cos (e+f x)+d \sin (e+f x))}{a^2 (c-i d) (c+i d)^3 f}+\frac {i c-3 d}{4 a^2 (c+i d)^2 f (1+i \tan (e+f x))}-\frac {1}{4 (i c-d) f (a+i a \tan (e+f x))^2} \] Output:

1/4*(c^3+3*I*c^2*d-3*c*d^2+3*I*d^3)*x/a^2/(c-I*d)/(c+I*d)^3-d^3*ln(c*cos(f 
*x+e)+d*sin(f*x+e))/a^2/(c-I*d)/(c+I*d)^3/f+1/4*(I*c-3*d)/a^2/(c+I*d)^2/f/ 
(1+I*tan(f*x+e))-1/4/(I*c-d)/f/(a+I*a*tan(f*x+e))^2
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))} \, dx=-\frac {\frac {i \left (c^2+4 i c d-7 d^2\right ) \log (i-\tan (e+f x))}{c+i d}+\frac {(c+i d)^2 \log (i+\tan (e+f x))}{i c+d}+\frac {8 d^3 \log (c+d \tan (e+f x))}{c^2+d^2}+\frac {2 i (c+i d)}{(-i+\tan (e+f x))^2}-\frac {2 (c+3 i d)}{-i+\tan (e+f x)}}{8 a^2 (c+i d)^2 f} \] Input:

Integrate[1/((a + I*a*Tan[e + f*x])^2*(c + d*Tan[e + f*x])),x]
 

Output:

-1/8*((I*(c^2 + (4*I)*c*d - 7*d^2)*Log[I - Tan[e + f*x]])/(c + I*d) + ((c 
+ I*d)^2*Log[I + Tan[e + f*x]])/(I*c + d) + (8*d^3*Log[c + d*Tan[e + f*x]] 
)/(c^2 + d^2) + ((2*I)*(c + I*d))/(-I + Tan[e + f*x])^2 - (2*(c + (3*I)*d) 
)/(-I + Tan[e + f*x]))/(a^2*(c + I*d)^2*f)
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4042, 27, 3042, 4079, 3042, 4014, 3042, 4013}

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.

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))}dx\)

\(\Big \downarrow \) 4042

\(\displaystyle -\frac {\int -\frac {2 (a (i c-2 d)+i a d \tan (e+f x))}{(i \tan (e+f x) a+a) (c+d \tan (e+f x))}dx}{4 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (i c-2 d)+i a d \tan (e+f x)}{(i \tan (e+f x) a+a) (c+d \tan (e+f x))}dx}{2 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (i c-2 d)+i a d \tan (e+f x)}{(i \tan (e+f x) a+a) (c+d \tan (e+f x))}dx}{2 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {-\frac {\int \frac {\left (c^2+3 i d c-4 d^2\right ) a^2+(c+3 i d) d \tan (e+f x) a^2}{c+d \tan (e+f x)}dx}{2 a^2 (-d+i c)}-\frac {c+3 i d}{2 f (c+i d) (1+i \tan (e+f x))}}{2 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\left (c^2+3 i d c-4 d^2\right ) a^2+(c+3 i d) d \tan (e+f x) a^2}{c+d \tan (e+f x)}dx}{2 a^2 (-d+i c)}-\frac {c+3 i d}{2 f (c+i d) (1+i \tan (e+f x))}}{2 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 4014

\(\displaystyle \frac {-\frac {-\frac {4 a^2 d^3 \int \frac {d-c \tan (e+f x)}{c+d \tan (e+f x)}dx}{c^2+d^2}+\frac {a^2 x \left (c^3+3 i c^2 d-3 c d^2+3 i d^3\right )}{c^2+d^2}}{2 a^2 (-d+i c)}-\frac {c+3 i d}{2 f (c+i d) (1+i \tan (e+f x))}}{2 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {4 a^2 d^3 \int \frac {d-c \tan (e+f x)}{c+d \tan (e+f x)}dx}{c^2+d^2}+\frac {a^2 x \left (c^3+3 i c^2 d-3 c d^2+3 i d^3\right )}{c^2+d^2}}{2 a^2 (-d+i c)}-\frac {c+3 i d}{2 f (c+i d) (1+i \tan (e+f x))}}{2 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 4013

\(\displaystyle \frac {-\frac {-\frac {4 a^2 d^3 \log (c \cos (e+f x)+d \sin (e+f x))}{f \left (c^2+d^2\right )}+\frac {a^2 x \left (c^3+3 i c^2 d-3 c d^2+3 i d^3\right )}{c^2+d^2}}{2 a^2 (-d+i c)}-\frac {c+3 i d}{2 f (c+i d) (1+i \tan (e+f x))}}{2 a^2 (-d+i c)}-\frac {1}{4 f (-d+i c) (a+i a \tan (e+f x))^2}\)

Input:

Int[1/((a + I*a*Tan[e + f*x])^2*(c + d*Tan[e + f*x])),x]
 

Output:

(-1/2*((a^2*(c^3 + (3*I)*c^2*d - 3*c*d^2 + (3*I)*d^3)*x)/(c^2 + d^2) - (4* 
a^2*d^3*Log[c*Cos[e + f*x] + d*Sin[e + f*x]])/((c^2 + d^2)*f))/(a^2*(I*c - 
 d)) - (c + (3*I)*d)/(2*(c + I*d)*f*(1 + I*Tan[e + f*x])))/(2*a^2*(I*c - d 
)) - 1/(4*(I*c - d)*f*(a + I*a*Tan[e + f*x])^2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4013

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)* 
(x_)]), x_Symbol] :> Simp[(c/(b*f))*Log[RemoveContent[a*Cos[e + f*x] + b*Si 
n[e + f*x], x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 
 NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]
 

rule 4014

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. 
)*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a 
*d)/(a^2 + b^2)   Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] /; 
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && N 
eQ[a*c + b*d, 0]
 

rule 4042

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e 
 + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   In 
t[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m 
 + n + 1) + b*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, 
e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] 
 && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4079

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( 
b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m 
 + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m 
- b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free 
Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] 
 && LtQ[m, 0] &&  !GtQ[n, 0]
 

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.35

method	result	size
risch	\(-\frac {x}{4 a^{2} \left (i d -c \right )}-\frac {{\mathrm e}^{-2 i \left (f x +e \right )} d}{2 a^{2} \left (i d +c \right )^{2} f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} c}{4 a^{2} \left (i d +c \right )^{2} f}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )}}{16 a^{2} \left (i d +c \right ) f}+\frac {2 i d^{3} x}{a^{2} \left (2 i c^{3} d +2 i c \,d^{3}+c^{4}-d^{4}\right )}+\frac {2 i d^{3} e}{a^{2} f \left (2 i c^{3} d +2 i c \,d^{3}+c^{4}-d^{4}\right )}-\frac {d^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right )}{a^{2} f \left (2 i c^{3} d +2 i c \,d^{3}+c^{4}-d^{4}\right )}\)	\(235\)
norman	\(\frac {\frac {-i c +2 d}{2 a f \left (-2 i c d -c^{2}+d^{2}\right )}+\frac {\left (3 i d +c \right ) \tan \left (f x +e \right )^{3}}{4 a f \left (2 i c d +c^{2}-d^{2}\right )}+\frac {\left (5 i d +3 c \right ) \tan \left (f x +e \right )}{4 a f \left (2 i c d +c^{2}-d^{2}\right )}+\frac {\left (3 i c^{2} d +3 i d^{3}+c^{3}-3 c \,d^{2}\right ) x}{4 \left (c^{2}+d^{2}\right ) a \left (2 i c d +c^{2}-d^{2}\right )}-\frac {d \tan \left (f x +e \right )^{2}}{2 a f \left (2 i c d +c^{2}-d^{2}\right )}+\frac {\left (3 i c^{2} d +3 i d^{3}+c^{3}-3 c \,d^{2}\right ) x \tan \left (f x +e \right )^{2}}{2 \left (c^{2}+d^{2}\right ) a \left (2 i c d +c^{2}-d^{2}\right )}+\frac {\left (3 i c^{2} d +3 i d^{3}+c^{3}-3 c \,d^{2}\right ) x \tan \left (f x +e \right )^{4}}{4 \left (c^{2}+d^{2}\right ) a \left (2 i c d +c^{2}-d^{2}\right )}}{a \left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {d^{3} \ln \left (c +d \tan \left (f x +e \right )\right )}{a^{2} f \left (-2 i c^{3} d -2 i c \,d^{3}-c^{4}+d^{4}\right )}-\frac {d^{3} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 a^{2} f \left (-2 i c^{3} d -2 i c \,d^{3}-c^{4}+d^{4}\right )}\)	\(431\)
derivativedivides	\(\frac {d^{3} \ln \left (c +d \tan \left (f x +e \right )\right )}{f \,a^{2} \left (i d -c \right ) \left (i d +c \right )^{3}}-\frac {i \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,a^{2} \left (8 i d -8 c \right )}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2} \left (8 i d -8 c \right )}-\frac {i \ln \left (1+\tan \left (f x +e \right )^{2}\right ) c^{2}}{16 f \,a^{2} \left (i d +c \right )^{3}}+\frac {7 i \ln \left (1+\tan \left (f x +e \right )^{2}\right ) d^{2}}{16 f \,a^{2} \left (i d +c \right )^{3}}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right ) c d}{4 f \,a^{2} \left (i d +c \right )^{3}}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) c^{2}}{8 f \,a^{2} \left (i d +c \right )^{3}}-\frac {7 \arctan \left (\tan \left (f x +e \right )\right ) d^{2}}{8 f \,a^{2} \left (i d +c \right )^{3}}+\frac {i \arctan \left (\tan \left (f x +e \right )\right ) c d}{2 f \,a^{2} \left (i d +c \right )^{3}}+\frac {i c d}{f \,a^{2} \left (i d +c \right )^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {c^{2}}{4 f \,a^{2} \left (i d +c \right )^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {3 d^{2}}{4 f \,a^{2} \left (i d +c \right )^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {i c^{2}}{4 f \,a^{2} \left (i d +c \right )^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {i d^{2}}{4 f \,a^{2} \left (i d +c \right )^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {c d}{2 f \,a^{2} \left (i d +c \right )^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}\)	\(446\)
default	\(\frac {d^{3} \ln \left (c +d \tan \left (f x +e \right )\right )}{f \,a^{2} \left (i d -c \right ) \left (i d +c \right )^{3}}-\frac {i \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \,a^{2} \left (8 i d -8 c \right )}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2} \left (8 i d -8 c \right )}-\frac {i \ln \left (1+\tan \left (f x +e \right )^{2}\right ) c^{2}}{16 f \,a^{2} \left (i d +c \right )^{3}}+\frac {7 i \ln \left (1+\tan \left (f x +e \right )^{2}\right ) d^{2}}{16 f \,a^{2} \left (i d +c \right )^{3}}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right ) c d}{4 f \,a^{2} \left (i d +c \right )^{3}}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) c^{2}}{8 f \,a^{2} \left (i d +c \right )^{3}}-\frac {7 \arctan \left (\tan \left (f x +e \right )\right ) d^{2}}{8 f \,a^{2} \left (i d +c \right )^{3}}+\frac {i \arctan \left (\tan \left (f x +e \right )\right ) c d}{2 f \,a^{2} \left (i d +c \right )^{3}}+\frac {i c d}{f \,a^{2} \left (i d +c \right )^{3} \left (-i+\tan \left (f x +e \right )\right )}+\frac {c^{2}}{4 f \,a^{2} \left (i d +c \right )^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {3 d^{2}}{4 f \,a^{2} \left (i d +c \right )^{3} \left (-i+\tan \left (f x +e \right )\right )}-\frac {i c^{2}}{4 f \,a^{2} \left (i d +c \right )^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {i d^{2}}{4 f \,a^{2} \left (i d +c \right )^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}+\frac {c d}{2 f \,a^{2} \left (i d +c \right )^{3} \left (-i+\tan \left (f x +e \right )\right )^{2}}\)	\(446\)

Input:

int(1/(a+I*a*tan(f*x+e))^2/(c+d*tan(f*x+e)),x,method=_RETURNVERBOSE)
 

Output:

-1/4*x/a^2/(I*d-c)-1/2/a^2/(c+I*d)^2/f*exp(-2*I*(f*x+e))*d+1/4*I/a^2/(c+I* 
d)^2/f*exp(-2*I*(f*x+e))*c+1/16*I/a^2/(c+I*d)/f*exp(-4*I*(f*x+e))+2*I*d^3/ 
a^2/(2*I*c^3*d+2*I*c*d^3+c^4-d^4)*x+2*I*d^3/a^2/f/(2*I*c^3*d+2*I*c*d^3+c^4 
-d^4)*e-d^3/a^2/f/(2*I*c^3*d+2*I*c*d^3+c^4-d^4)*ln(exp(2*I*(f*x+e))-(c+I*d 
)/(I*d-c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))} \, dx=-\frac {{\left (16 \, d^{3} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d}{i \, c + d}\right ) - 4 \, {\left (c^{3} + 3 i \, c^{2} d - 3 \, c d^{2} + 7 i \, d^{3}\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} - i \, c^{3} + c^{2} d - i \, c d^{2} + d^{3} - 4 \, {\left (i \, c^{3} - 2 \, c^{2} d + i \, c d^{2} - 2 \, d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{16 \, {\left (a^{2} c^{4} + 2 i \, a^{2} c^{3} d + 2 i \, a^{2} c d^{3} - a^{2} d^{4}\right )} f} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^2/(c+d*tan(f*x+e)),x, algorithm="fricas")
 

Output:

-1/16*(16*d^3*e^(4*I*f*x + 4*I*e)*log(((I*c + d)*e^(2*I*f*x + 2*I*e) + I*c 
 - d)/(I*c + d)) - 4*(c^3 + 3*I*c^2*d - 3*c*d^2 + 7*I*d^3)*f*x*e^(4*I*f*x 
+ 4*I*e) - I*c^3 + c^2*d - I*c*d^2 + d^3 - 4*(I*c^3 - 2*c^2*d + I*c*d^2 - 
2*d^3)*e^(2*I*f*x + 2*I*e))*e^(-4*I*f*x - 4*I*e)/((a^2*c^4 + 2*I*a^2*c^3*d 
 + 2*I*a^2*c*d^3 - a^2*d^4)*f)
 

Sympy [A] (verification not implemented)

Time = 5.50 (sec) , antiderivative size = 610, normalized size of antiderivative = 3.51 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))} \, dx=\frac {x \left (c^{2} + 4 i c d - 7 d^{2}\right )}{4 a^{2} c^{3} + 12 i a^{2} c^{2} d - 12 a^{2} c d^{2} - 4 i a^{2} d^{3}} + \begin {cases} \frac {\left (4 i a^{2} c^{2} f e^{2 i e} - 8 a^{2} c d f e^{2 i e} - 4 i a^{2} d^{2} f e^{2 i e}\right ) e^{- 4 i f x} + \left (16 i a^{2} c^{2} f e^{4 i e} - 48 a^{2} c d f e^{4 i e} - 32 i a^{2} d^{2} f e^{4 i e}\right ) e^{- 2 i f x}}{64 a^{4} c^{3} f^{2} e^{6 i e} + 192 i a^{4} c^{2} d f^{2} e^{6 i e} - 192 a^{4} c d^{2} f^{2} e^{6 i e} - 64 i a^{4} d^{3} f^{2} e^{6 i e}} & \text {for}\: 64 a^{4} c^{3} f^{2} e^{6 i e} + 192 i a^{4} c^{2} d f^{2} e^{6 i e} - 192 a^{4} c d^{2} f^{2} e^{6 i e} - 64 i a^{4} d^{3} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac {c^{2} + 4 i c d - 7 d^{2}}{4 a^{2} c^{3} + 12 i a^{2} c^{2} d - 12 a^{2} c d^{2} - 4 i a^{2} d^{3}} + \frac {c^{2} e^{4 i e} + 2 c^{2} e^{2 i e} + c^{2} + 4 i c d e^{4 i e} + 6 i c d e^{2 i e} + 2 i c d - 7 d^{2} e^{4 i e} - 4 d^{2} e^{2 i e} - d^{2}}{4 a^{2} c^{3} e^{4 i e} + 12 i a^{2} c^{2} d e^{4 i e} - 12 a^{2} c d^{2} e^{4 i e} - 4 i a^{2} d^{3} e^{4 i e}}\right ) & \text {otherwise} \end {cases} - \frac {d^{3} \log {\left (\frac {c + i d}{c e^{2 i e} - i d e^{2 i e}} + e^{2 i f x} \right )}}{a^{2} f \left (c - i d\right ) \left (c + i d\right )^{3}} \] Input:

integrate(1/(a+I*a*tan(f*x+e))**2/(c+d*tan(f*x+e)),x)
 

Output:

x*(c**2 + 4*I*c*d - 7*d**2)/(4*a**2*c**3 + 12*I*a**2*c**2*d - 12*a**2*c*d* 
*2 - 4*I*a**2*d**3) + Piecewise((((4*I*a**2*c**2*f*exp(2*I*e) - 8*a**2*c*d 
*f*exp(2*I*e) - 4*I*a**2*d**2*f*exp(2*I*e))*exp(-4*I*f*x) + (16*I*a**2*c** 
2*f*exp(4*I*e) - 48*a**2*c*d*f*exp(4*I*e) - 32*I*a**2*d**2*f*exp(4*I*e))*e 
xp(-2*I*f*x))/(64*a**4*c**3*f**2*exp(6*I*e) + 192*I*a**4*c**2*d*f**2*exp(6 
*I*e) - 192*a**4*c*d**2*f**2*exp(6*I*e) - 64*I*a**4*d**3*f**2*exp(6*I*e)), 
 Ne(64*a**4*c**3*f**2*exp(6*I*e) + 192*I*a**4*c**2*d*f**2*exp(6*I*e) - 192 
*a**4*c*d**2*f**2*exp(6*I*e) - 64*I*a**4*d**3*f**2*exp(6*I*e), 0)), (x*(-( 
c**2 + 4*I*c*d - 7*d**2)/(4*a**2*c**3 + 12*I*a**2*c**2*d - 12*a**2*c*d**2 
- 4*I*a**2*d**3) + (c**2*exp(4*I*e) + 2*c**2*exp(2*I*e) + c**2 + 4*I*c*d*e 
xp(4*I*e) + 6*I*c*d*exp(2*I*e) + 2*I*c*d - 7*d**2*exp(4*I*e) - 4*d**2*exp( 
2*I*e) - d**2)/(4*a**2*c**3*exp(4*I*e) + 12*I*a**2*c**2*d*exp(4*I*e) - 12* 
a**2*c*d**2*exp(4*I*e) - 4*I*a**2*d**3*exp(4*I*e))), True)) - d**3*log((c 
+ I*d)/(c*exp(2*I*e) - I*d*exp(2*I*e)) + exp(2*I*f*x))/(a**2*f*(c - I*d)*( 
c + I*d)**3)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^2/(c+d*tan(f*x+e)),x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.25 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))} \, dx=-\frac {i \, d^{4} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{i \, a^{2} c^{4} d f - 2 \, a^{2} c^{3} d^{2} f - 2 \, a^{2} c d^{4} f - i \, a^{2} d^{5} f} - \frac {{\left (c^{2} + 4 i \, c d - 7 \, d^{2}\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{-8 i \, a^{2} c^{3} f + 24 \, a^{2} c^{2} d f + 24 i \, a^{2} c d^{2} f - 8 \, a^{2} d^{3} f} + \frac {\log \left (\tan \left (f x + e\right ) + i\right )}{-8 i \, a^{2} c f - 8 \, a^{2} d f} + \frac {-2 i \, c^{2} + 6 \, c d + 4 i \, d^{2} + i \, {\left (-i \, c^{2} + 4 \, c d + 3 i \, d^{2}\right )} \tan \left (f x + e\right )}{4 \, a^{2} {\left (c + i \, d\right )}^{3} f {\left (\tan \left (f x + e\right ) - i\right )}^{2}} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^2/(c+d*tan(f*x+e)),x, algorithm="giac")
 

Output:

-I*d^4*log(abs(d*tan(f*x + e) + c))/(I*a^2*c^4*d*f - 2*a^2*c^3*d^2*f - 2*a 
^2*c*d^4*f - I*a^2*d^5*f) - (c^2 + 4*I*c*d - 7*d^2)*log(tan(f*x + e) - I)/ 
(-8*I*a^2*c^3*f + 24*a^2*c^2*d*f + 24*I*a^2*c*d^2*f - 8*a^2*d^3*f) + log(t 
an(f*x + e) + I)/(-8*I*a^2*c*f - 8*a^2*d*f) + 1/4*(-2*I*c^2 + 6*c*d + 4*I* 
d^2 + I*(-I*c^2 + 4*c*d + 3*I*d^2)*tan(f*x + e))/(a^2*(c + I*d)^3*f*(tan(f 
*x + e) - I)^2)
 

Mupad [B] (verification not implemented)

Time = 6.22 (sec) , antiderivative size = 1384, normalized size of antiderivative = 7.95 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))} \, dx=\text {Too large to display} \] Input:

int(1/((a + a*tan(e + f*x)*1i)^2*(c + d*tan(e + f*x))),x)
 

Output:

symsum(log(root(640*a^6*c^4*d^4*e^3 - a^6*c^5*d^3*e^3*256i + a^6*c^3*d^5*e 
^3*256i + 256*a^6*c^6*d^2*e^3 + 256*a^6*c^2*d^6*e^3 - a^6*c^7*d*e^3*256i + 
 a^6*c*d^7*e^3*256i - 64*a^6*d^8*e^3 - 64*a^6*c^8*e^3 + a^2*c*d^5*e*18i - 
a^2*c^5*d*e*6i + a^2*c^3*d^3*e*12i + 15*a^2*c^4*d^2*e + 9*a^2*c^2*d^4*e + 
57*a^2*d^6*e - a^2*c^6*e - c^2*d^3 - c*d^4*4i + 7*d^5, e, k)*(root(640*a^6 
*c^4*d^4*e^3 - a^6*c^5*d^3*e^3*256i + a^6*c^3*d^5*e^3*256i + 256*a^6*c^6*d 
^2*e^3 + 256*a^6*c^2*d^6*e^3 - a^6*c^7*d*e^3*256i + a^6*c*d^7*e^3*256i - 6 
4*a^6*d^8*e^3 - 64*a^6*c^8*e^3 + a^2*c*d^5*e*18i - a^2*c^5*d*e*6i + a^2*c^ 
3*d^3*e*12i + 15*a^2*c^4*d^2*e + 9*a^2*c^2*d^4*e + 57*a^2*d^6*e - a^2*c^6* 
e - c^2*d^3 - c*d^4*4i + 7*d^5, e, k)*((a^2*d^6 + a^2*c*d^5*4i - 6*a^2*c^2 
*d^4 - a^2*c^3*d^3*4i + a^2*c^4*d^2)*(128*a^4*c*d^5 + 128*a^4*c^5*d - a^4* 
c^2*d^4*512i - 768*a^4*c^3*d^3 + a^4*c^4*d^2*512i) - tan(e + f*x)*(a^2*d^6 
 + a^2*c*d^5*4i - 6*a^2*c^2*d^4 - a^2*c^3*d^3*4i + a^2*c^4*d^2)*(32*a^4*c^ 
6 - 96*a^4*d^6 + a^4*c*d^5*384i + a^4*c^5*d*128i + 608*a^4*c^2*d^4 - a^4*c 
^3*d^3*512i - 288*a^4*c^4*d^2)) + (a^2*d^6 + a^2*c*d^5*4i - 6*a^2*c^2*d^4 
- a^2*c^3*d^3*4i + a^2*c^4*d^2)*(4*a^2*c^5 + a^2*d^5*12i + 44*a^2*c*d^4 + 
a^2*c^4*d*20i - a^2*c^2*d^3*64i - 48*a^2*c^3*d^2) + tan(e + f*x)*(48*a^2*d 
^5 - a^2*c*d^4*120i + 8*a^2*c^4*d - 104*a^2*c^2*d^3 + a^2*c^3*d^2*40i)*(a^ 
2*d^6 + a^2*c*d^5*4i - 6*a^2*c^2*d^4 - a^2*c^3*d^3*4i + a^2*c^4*d^2)) - (1 
3*c*d^3 - c^3*d + d^4*12i - c^2*d^2*6i)*(a^2*d^6 + a^2*c*d^5*4i - 6*a^2...
 

Reduce [F]

\[ \int \frac {1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))} \, dx=-\frac {\int \frac {1}{\tan \left (f x +e \right )^{3} d +\tan \left (f x +e \right )^{2} c -2 \tan \left (f x +e \right )^{2} d i -2 \tan \left (f x +e \right ) c i -d \tan \left (f x +e \right )-c}d x}{a^{2}} \] Input:

int(1/(a+I*a*tan(f*x+e))^2/(c+d*tan(f*x+e)),x)
 

Output:

( - int(1/(tan(e + f*x)**3*d + tan(e + f*x)**2*c - 2*tan(e + f*x)**2*d*i - 
 2*tan(e + f*x)*c*i - tan(e + f*x)*d - c),x))/a**2