\(\int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))} \, dx\) [1086]

Optimal result

Integrand size = 28, antiderivative size = 128 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))} \, dx=\frac {x}{2 a (c+i d)}-\frac {c d x}{a (i c-d) \left (c^2+d^2\right )}-\frac {d^2 \log (c \cos (e+f x)+d \sin (e+f x))}{a (i c-d) \left (c^2+d^2\right ) f}-\frac {1}{2 (i c-d) f (a+i a \tan (e+f x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))} \, dx=-\frac {i \left (\frac {(c+3 i d) \log (i-\tan (e+f x))}{c+i d}-\frac {(c+i d) \log (i+\tan (e+f x))}{c-i d}-\frac {4 d^2 \log (c+d \tan (e+f x))}{c^2+d^2}+\frac {2 i}{-i+\tan (e+f x)}\right )}{4 a (c+i d) f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.83, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4034, 3042, 3960, 24, 3965, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3960

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 3965

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4013

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

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

rule 3960

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + 
b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[1/(2*a)   Int[(a + b*Tan[c + d*x])^ 
(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, 0]
 

rule 3965

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[a*(x/(a^ 
2 + b^2)), x] + Simp[b/(a^2 + b^2)   Int[(b - a*Tan[c + d*x])/(a + b*Tan[c 
+ d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0]
 

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 4034

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.39

method	result	size
risch	\(-\frac {x}{2 a \left (i d -c \right )}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )}}{4 a f \left (i d +c \right )}+\frac {2 d^{2} x}{a \left (i c^{2} d +i d^{3}+c^{3}+c \,d^{2}\right )}+\frac {2 d^{2} e}{a f \left (i c^{2} d +i d^{3}+c^{3}+c \,d^{2}\right )}+\frac {i d^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right )}{a f \left (i c^{2} d +i d^{3}+c^{3}+c \,d^{2}\right )}\)	\(178\)
norman	\(\frac {\frac {\tan \left (f x +e \right )}{2 a f \left (i d +c \right )}+\frac {i}{2 a f \left (i d +c \right )}+\frac {\left (2 i c d +c^{2}+d^{2}\right ) x}{2 \left (c^{2}+d^{2}\right ) a \left (i d +c \right )}+\frac {\left (2 i c d +c^{2}+d^{2}\right ) x \tan \left (f x +e \right )^{2}}{2 \left (c^{2}+d^{2}\right ) a \left (i d +c \right )}}{1+\tan \left (f x +e \right )^{2}}+\frac {d^{2} \ln \left (c +d \tan \left (f x +e \right )\right )}{a f \left (-i c^{3}-i c \,d^{2}+c^{2} d +d^{3}\right )}-\frac {d^{2} \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 a f \left (-i c^{3}-i c \,d^{2}+c^{2} d +d^{3}\right )}\)	\(224\)
derivativedivides	\(\frac {1}{f a \left (2 i d +2 c \right ) \left (-i+\tan \left (f x +e \right )\right )}-\frac {i \ln \left (1+\tan \left (f x +e \right )^{2}\right ) c}{8 f a \left (i d +c \right )^{2}}+\frac {3 \ln \left (1+\tan \left (f x +e \right )^{2}\right ) d}{8 f a \left (i d +c \right )^{2}}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) c}{4 f a \left (i d +c \right )^{2}}+\frac {3 i \arctan \left (\tan \left (f x +e \right )\right ) d}{4 f a \left (i d +c \right )^{2}}-\frac {i \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f a \left (4 i d -4 c \right )}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f a \left (4 i d -4 c \right )}-\frac {i d^{2} \ln \left (c +d \tan \left (f x +e \right )\right )}{f a \left (i d -c \right ) \left (i d +c \right )^{2}}\)	\(232\)
default	\(\frac {1}{f a \left (2 i d +2 c \right ) \left (-i+\tan \left (f x +e \right )\right )}-\frac {i \ln \left (1+\tan \left (f x +e \right )^{2}\right ) c}{8 f a \left (i d +c \right )^{2}}+\frac {3 \ln \left (1+\tan \left (f x +e \right )^{2}\right ) d}{8 f a \left (i d +c \right )^{2}}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) c}{4 f a \left (i d +c \right )^{2}}+\frac {3 i \arctan \left (\tan \left (f x +e \right )\right ) d}{4 f a \left (i d +c \right )^{2}}-\frac {i \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f a \left (4 i d -4 c \right )}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f a \left (4 i d -4 c \right )}-\frac {i d^{2} \ln \left (c +d \tan \left (f x +e \right )\right )}{f a \left (i d -c \right ) \left (i d +c \right )^{2}}\)	\(232\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))} \, dx=\frac {{\left (2 \, {\left (i \, c^{2} - 2 \, c d + 3 i \, d^{2}\right )} f x e^{\left (2 i \, f x + 2 i \, e\right )} - 4 \, d^{2} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d}{i \, c + d}\right ) - c^{2} - d^{2}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, {\left (i \, a c^{3} - a c^{2} d + i \, a c d^{2} - a d^{3}\right )} f} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 2.46 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.94 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))} \, dx=\frac {x \left (c + 3 i d\right )}{2 a c^{2} + 4 i a c d - 2 a d^{2}} + \begin {cases} \frac {i e^{- 2 i f x}}{4 a c f e^{2 i e} + 4 i a d f e^{2 i e}} & \text {for}\: 4 a c f e^{2 i e} + 4 i a d f e^{2 i e} \neq 0 \\x \left (- \frac {c + 3 i d}{2 a c^{2} + 4 i a c d - 2 a d^{2}} + \frac {c e^{2 i e} + c + 3 i d e^{2 i e} + i d}{2 a c^{2} e^{2 i e} + 4 i a c d e^{2 i e} - 2 a d^{2} e^{2 i e}}\right ) & \text {otherwise} \end {cases} + \frac {i d^{2} \log {\left (\frac {c + i d}{c e^{2 i e} - i d e^{2 i e}} + e^{2 i f x} \right )}}{a f \left (c - i d\right ) \left (c + i d\right )^{2}} \] Input:

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

Output:

x*(c + 3*I*d)/(2*a*c**2 + 4*I*a*c*d - 2*a*d**2) + Piecewise((I*exp(-2*I*f* 
x)/(4*a*c*f*exp(2*I*e) + 4*I*a*d*f*exp(2*I*e)), Ne(4*a*c*f*exp(2*I*e) + 4* 
I*a*d*f*exp(2*I*e), 0)), (x*(-(c + 3*I*d)/(2*a*c**2 + 4*I*a*c*d - 2*a*d**2 
) + (c*exp(2*I*e) + c + 3*I*d*exp(2*I*e) + I*d)/(2*a*c**2*exp(2*I*e) + 4*I 
*a*c*d*exp(2*I*e) - 2*a*d**2*exp(2*I*e))), True)) + I*d**2*log((c + I*d)/( 
c*exp(2*I*e) - I*d*exp(2*I*e)) + exp(2*I*f*x))/(a*f*(c - I*d)*(c + I*d)**2 
)
 

Maxima [F(-2)]

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

integrate(1/(a+I*a*tan(f*x+e))/(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 = 0.47 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))} \, dx=\frac {i \, d^{3} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{a c^{3} d f + i \, a c^{2} d^{2} f + a c d^{3} f + i \, a d^{4} f} - \frac {i \, {\left (c + 3 i \, d\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{4 \, {\left (a c^{2} f + 2 i \, a c d f - a d^{2} f\right )}} + \frac {i \, \log \left (\tan \left (f x + e\right ) + i\right )}{4 \, {\left (a c f - i \, a d f\right )}} + \frac {i \, {\left (-i \, c + d\right )}}{2 \, a {\left (c + i \, d\right )}^{2} f {\left (\tan \left (f x + e\right ) - i\right )}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 6.16 (sec) , antiderivative size = 2639, normalized size of antiderivative = 20.62 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))} \, dx=\text {Too large to display} \] Input:

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

Output:

symsum(log(-a*d^2*(c*1i + d)^2*(c*d + d^2*tan(e + f*x) + d^2*2i + 2*root(a 
^3*c^3*d^3*e^3*64i + 16*a^3*c^4*d^2*e^3 - 16*a^3*c^2*d^4*e^3 + a^3*c^5*d*e 
^3*32i + a^3*c*d^5*e^3*32i - 16*a^3*d^6*e^3 + 16*a^3*c^6*e^3 - 2*a*c^2*d^2 
*e + a*c^3*d*e*4i + a*c*d^3*e*4i + 13*a*d^4*e + a*c^4*e - c*d^2*1i + 3*d^3 
, e, k)*a*c^3 - root(a^3*c^3*d^3*e^3*64i + 16*a^3*c^4*d^2*e^3 - 16*a^3*c^2 
*d^4*e^3 + a^3*c^5*d*e^3*32i + a^3*c*d^5*e^3*32i - 16*a^3*d^6*e^3 + 16*a^3 
*c^6*e^3 - 2*a*c^2*d^2*e + a*c^3*d*e*4i + a*c*d^3*e*4i + 13*a*d^4*e + a*c^ 
4*e - c*d^2*1i + 3*d^3, e, k)*a*d^3*2i - 8*root(a^3*c^3*d^3*e^3*64i + 16*a 
^3*c^4*d^2*e^3 - 16*a^3*c^2*d^4*e^3 + a^3*c^5*d*e^3*32i + a^3*c*d^5*e^3*32 
i - 16*a^3*d^6*e^3 + 16*a^3*c^6*e^3 - 2*a*c^2*d^2*e + a*c^3*d*e*4i + a*c*d 
^3*e*4i + 13*a*d^4*e + a*c^4*e - c*d^2*1i + 3*d^3, e, k)^2*a^2*c^4*tan(e + 
 f*x) - 24*root(a^3*c^3*d^3*e^3*64i + 16*a^3*c^4*d^2*e^3 - 16*a^3*c^2*d^4* 
e^3 + a^3*c^5*d*e^3*32i + a^3*c*d^5*e^3*32i - 16*a^3*d^6*e^3 + 16*a^3*c^6* 
e^3 - 2*a*c^2*d^2*e + a*c^3*d*e*4i + a*c*d^3*e*4i + 13*a*d^4*e + a*c^4*e - 
 c*d^2*1i + 3*d^3, e, k)^2*a^2*d^4*tan(e + f*x) - 6*root(a^3*c^3*d^3*e^3*6 
4i + 16*a^3*c^4*d^2*e^3 - 16*a^3*c^2*d^4*e^3 + a^3*c^5*d*e^3*32i + a^3*c*d 
^5*e^3*32i - 16*a^3*d^6*e^3 + 16*a^3*c^6*e^3 - 2*a*c^2*d^2*e + a*c^3*d*e*4 
i + a*c*d^3*e*4i + 13*a*d^4*e + a*c^4*e - c*d^2*1i + 3*d^3, e, k)*a*c*d^2 
+ root(a^3*c^3*d^3*e^3*64i + 16*a^3*c^4*d^2*e^3 - 16*a^3*c^2*d^4*e^3 + a^3 
*c^5*d*e^3*32i + a^3*c*d^5*e^3*32i - 16*a^3*d^6*e^3 + 16*a^3*c^6*e^3 - ...
 

Reduce [F]

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

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

Output:

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