\(\int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^2} \, dx\) [1092]

Optimal result

Integrand size = 28, antiderivative size = 202 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^2} \, dx=\frac {\left (c^3+3 i c^2 d+3 c d^2-3 i d^3\right ) x}{2 a (c-i d)^2 (c+i d)^3}+\frac {(3 c-i d) d^2 \log (c \cos (e+f x)+d \sin (e+f x))}{a (i c-d)^3 (c-i d)^2 f}+\frac {(c-3 i d) d}{2 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))}-\frac {1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3633, 3610, 3612, 3611} \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^2} \, dx=\frac {x \left (c^3+3 i c^2 d+3 c d^2-3 i d^3\right )}{2 a (c-i d)^2 (c+i d)^3}+\frac {d^2 (3 c-i d) \log (c \cos (e+f x)+d \sin (e+f x))}{a f (-d+i c)^3 (c-i d)^2}+\frac {d (c-3 i d)}{2 a f (c-i d) (c+i d)^2 (c+d \tan (e+f x))}-\frac {1}{2 f (-d+i c) (a+i a \tan (e+f x)) (c+d \tan (e+f x))} \]

[In]

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

[Out]

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

Rule 3610

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

Rule 3611

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*Sin[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 3612

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] + Dist[(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] && NeQ[a*c + b*d, 0]

Rule 3633

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-a
)*((c + d*Tan[e + f*x])^(n + 1)/(2*f*(b*c - a*d)*(a + b*Tan[e + f*x]))), x] + Dist[1/(2*a*(b*c - a*d)), Int[(c
+ d*Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*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] &&  !GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))}+\frac {\int \frac {a (i c-3 d)+2 i a d \tan (e+f x)}{(c+d \tan (e+f x))^2} \, dx}{2 a^2 (i c-d)} \\ & = \frac {(c-3 i d) d}{2 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))}-\frac {1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))}+\frac {\int \frac {-a \left (3 c d-i \left (c^2+2 d^2\right )\right )+a d (i c+3 d) \tan (e+f x)}{c+d \tan (e+f x)} \, dx}{2 a^2 (i c-d) \left (c^2+d^2\right )} \\ & = \frac {\left (c^3+3 i c^2 d+3 c d^2-3 i d^3\right ) x}{2 a (c-i d)^2 (c+i d)^3}+\frac {(c-3 i d) d}{2 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))}-\frac {1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))}-\frac {\left ((3 c-i d) d^2\right ) \int \frac {d-c \tan (e+f x)}{c+d \tan (e+f x)} \, dx}{a (i c-d) \left (c^2+d^2\right )^2} \\ & = \frac {\left (c^3+3 i c^2 d+3 c d^2-3 i d^3\right ) x}{2 a (c-i d)^2 (c+i d)^3}+\frac {(3 c-i d) d^2 \log (c \cos (e+f x)+d \sin (e+f x))}{a (i c-d)^3 (c-i d)^2 f}+\frac {(c-3 i d) d}{2 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))}-\frac {1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.17 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^2} \, dx=\frac {\frac {2 \log (i-\tan (e+f x))}{i c-d}-\frac {2 \log (i+\tan (e+f x))}{i c+d}+\frac {4 d \log (c+d \tan (e+f x))}{c^2+d^2}+\frac {2}{(-i+\tan (e+f x)) (c+d \tan (e+f x))}+(c-3 i d) \left (\frac {i \log (i-\tan (e+f x))}{(c+i d)^2}-\frac {i \log (i+\tan (e+f x))}{(c-i d)^2}+\frac {2 d \left (-2 c \log (c+d \tan (e+f x))+\frac {c^2+d^2}{c+d \tan (e+f x)}\right )}{\left (c^2+d^2\right )^2}\right )}{4 a (c+i d) f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.77

method	result	size
derivativedivides	\(-\frac {i d^{2} c^{2}}{f a \left (i d -c \right )^{2} \left (i d +c \right )^{3} \left (c +d \tan \left (f x +e \right )\right )}-\frac {i d^{4}}{f a \left (i d -c \right )^{2} \left (i d +c \right )^{3} \left (c +d \tan \left (f x +e \right )\right )}+\frac {3 i d^{2} \ln \left (c +d \tan \left (f x +e \right )\right ) c}{f a \left (i d -c \right )^{2} \left (i d +c \right )^{3}}+\frac {d^{3} \ln \left (c +d \tan \left (f x +e \right )\right )}{f a \left (i d -c \right )^{2} \left (i d +c \right )^{3}}-\frac {i \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) c}{8 f a \left (i d +c \right )^{3}}+\frac {5 \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) d}{8 f a \left (i d +c \right )^{3}}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) c}{4 f a \left (i d +c \right )^{3}}+\frac {5 i \arctan \left (\tan \left (f x +e \right )\right ) d}{4 f a \left (i d +c \right )^{3}}+\frac {1}{2 f a \left (i d +c \right )^{2} \left (\tan \left (f x +e \right )-i\right )}+\frac {i \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{8 f a \left (i d -c \right )^{2}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{4 f a \left (i d -c \right )^{2}}\)	\(358\)
default	\(-\frac {i d^{2} c^{2}}{f a \left (i d -c \right )^{2} \left (i d +c \right )^{3} \left (c +d \tan \left (f x +e \right )\right )}-\frac {i d^{4}}{f a \left (i d -c \right )^{2} \left (i d +c \right )^{3} \left (c +d \tan \left (f x +e \right )\right )}+\frac {3 i d^{2} \ln \left (c +d \tan \left (f x +e \right )\right ) c}{f a \left (i d -c \right )^{2} \left (i d +c \right )^{3}}+\frac {d^{3} \ln \left (c +d \tan \left (f x +e \right )\right )}{f a \left (i d -c \right )^{2} \left (i d +c \right )^{3}}-\frac {i \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) c}{8 f a \left (i d +c \right )^{3}}+\frac {5 \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) d}{8 f a \left (i d +c \right )^{3}}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) c}{4 f a \left (i d +c \right )^{3}}+\frac {5 i \arctan \left (\tan \left (f x +e \right )\right ) d}{4 f a \left (i d +c \right )^{3}}+\frac {1}{2 f a \left (i d +c \right )^{2} \left (\tan \left (f x +e \right )-i\right )}+\frac {i \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{8 f a \left (i d -c \right )^{2}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{4 f a \left (i d -c \right )^{2}}\)	\(358\)
norman	\(\frac {\frac {1}{2 a f \left (-i c +d \right )}+\frac {c \left (3 i c^{2} d -3 i d^{3}+c^{3}+3 c \,d^{2}\right ) x}{2 \left (-i d +c \right )^{2} \left (i d +c \right )^{3} a}+\frac {c \left (3 i c^{2} d -3 i d^{3}+c^{3}+3 c \,d^{2}\right ) x \left (\tan ^{2}\left (f x +e \right )\right )}{2 \left (-i d +c \right )^{2} \left (i d +c \right )^{3} a}+\frac {d \left (3 i c^{2} d -3 i d^{3}+c^{3}+3 c \,d^{2}\right ) x \tan \left (f x +e \right )}{2 \left (-i d +c \right )^{2} \left (i d +c \right )^{3} a}+\frac {d \left (3 i c^{2} d -3 i d^{3}+c^{3}+3 c \,d^{2}\right ) x \left (\tan ^{3}\left (f x +e \right )\right )}{2 \left (-i d +c \right )^{2} \left (i d +c \right )^{3} a}+\frac {\left (3 i d^{3}+c^{3}\right ) \tan \left (f x +e \right )}{2 a f \left (-i d +c \right ) \left (i d +c \right )^{2} c}+\frac {d \left (i c d +3 d^{2}\right ) \left (\tan ^{3}\left (f x +e \right )\right )}{2 a \left (i c +d \right ) \left (-i c +d \right )^{2} f c}}{\left (c +d \tan \left (f x +e \right )\right ) \left (1+\tan ^{2}\left (f x +e \right )\right )}+\frac {i d^{2} \left (3 i c +d \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 a f \left (-i c^{5}-2 i c^{3} d^{2}-i c \,d^{4}+c^{4} d +2 c^{2} d^{3}+d^{5}\right )}-\frac {i d^{2} \left (3 i c +d \right ) \ln \left (c +d \tan \left (f x +e \right )\right )}{a f \left (-i c^{5}-2 i c^{3} d^{2}-i c \,d^{4}+c^{4} d +2 c^{2} d^{3}+d^{5}\right )}\)	\(480\)
risch	\(-\frac {x}{2 a \left (2 i c d -c^{2}+d^{2}\right )}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )}}{4 a \left (2 i c d +c^{2}-d^{2}\right ) f}+\frac {6 d^{2} c x}{a \left (i c^{4} d +2 i c^{2} d^{3}+i d^{5}+c^{5}+2 c^{3} d^{2}+c \,d^{4}\right )}+\frac {6 d^{2} c e}{a f \left (i c^{4} d +2 i c^{2} d^{3}+i d^{5}+c^{5}+2 c^{3} d^{2}+c \,d^{4}\right )}-\frac {2 i d^{3} x}{a \left (i c^{4} d +2 i c^{2} d^{3}+i d^{5}+c^{5}+2 c^{3} d^{2}+c \,d^{4}\right )}-\frac {2 i d^{3} e}{a f \left (i c^{4} d +2 i c^{2} d^{3}+i d^{5}+c^{5}+2 c^{3} d^{2}+c \,d^{4}\right )}-\frac {2 i d^{3}}{\left (-i c +d \right )^{2} f a \left (i c +d \right )^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )} d +i c \,{\mathrm e}^{2 i \left (f x +e \right )}-d +i c \right )}+\frac {3 i d^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right ) c}{a f \left (i c^{4} d +2 i c^{2} d^{3}+i d^{5}+c^{5}+2 c^{3} d^{2}+c \,d^{4}\right )}+\frac {d^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {i d +c}{i d -c}\right )}{a f \left (i c^{4} d +2 i c^{2} d^{3}+i d^{5}+c^{5}+2 c^{3} d^{2}+c \,d^{4}\right )}\)	\(493\)

[In]

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

[Out]

-I/f/a*d^2/(I*d-c)^2/(c+I*d)^3/(c+d*tan(f*x+e))*c^2-I/f/a*d^4/(I*d-c)^2/(c+I*d)^3/(c+d*tan(f*x+e))+3*I/f/a*d^2
/(I*d-c)^2/(c+I*d)^3*ln(c+d*tan(f*x+e))*c+1/f/a*d^3/(I*d-c)^2/(c+I*d)^3*ln(c+d*tan(f*x+e))-1/8*I/f/a/(c+I*d)^3
*ln(1+tan(f*x+e)^2)*c+5/8/f/a/(c+I*d)^3*ln(1+tan(f*x+e)^2)*d+1/4/f/a/(c+I*d)^3*arctan(tan(f*x+e))*c+5/4*I/f/a/
(c+I*d)^3*arctan(tan(f*x+e))*d+1/2/f/a/(c+I*d)^2/(tan(f*x+e)-I)+1/8*I/f/a/(I*d-c)^2*ln(1+tan(f*x+e)^2)+1/4/f/a
/(I*d-c)^2*arctan(tan(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.63 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^2} \, dx=\frac {i \, c^{4} + 2 i \, c^{2} d^{2} + i \, d^{4} + 2 \, {\left (c^{4} + 2 i \, c^{3} d + 12 \, c^{2} d^{2} - 14 i \, c d^{3} - 5 \, d^{4}\right )} f x e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (i \, c^{4} + 2 \, c^{3} d - 6 \, c d^{3} - 9 i \, d^{4} + 2 \, {\left (c^{4} + 4 i \, c^{3} d + 6 \, c^{2} d^{2} + 4 i \, c d^{3} + 5 \, d^{4}\right )} f x\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 \, {\left ({\left (-3 i \, c^{2} d^{2} - 4 \, c d^{3} + i \, d^{4}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-3 i \, c^{2} d^{2} + 2 \, c d^{3} - i \, d^{4}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (\frac {{\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c - d}{i \, c + d}\right )}{4 \, {\left ({\left (a c^{6} + 3 \, a c^{4} d^{2} + 3 \, a c^{2} d^{4} + a d^{6}\right )} f e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (a c^{6} + 2 i \, a c^{5} d + a c^{4} d^{2} + 4 i \, a c^{3} d^{3} - a c^{2} d^{4} + 2 i \, a c d^{5} - a d^{6}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \]

[In]

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

[Out]

1/4*(I*c^4 + 2*I*c^2*d^2 + I*d^4 + 2*(c^4 + 2*I*c^3*d + 12*c^2*d^2 - 14*I*c*d^3 - 5*d^4)*f*x*e^(4*I*f*x + 4*I*
e) + (I*c^4 + 2*c^3*d - 6*c*d^3 - 9*I*d^4 + 2*(c^4 + 4*I*c^3*d + 6*c^2*d^2 + 4*I*c*d^3 + 5*d^4)*f*x)*e^(2*I*f*
x + 2*I*e) - 4*((-3*I*c^2*d^2 - 4*c*d^3 + I*d^4)*e^(4*I*f*x + 4*I*e) + (-3*I*c^2*d^2 + 2*c*d^3 - I*d^4)*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)))/((a*c^6 + 3*a*c^4*d^2 + 3*a*c^2*d^4 +
a*d^6)*f*e^(4*I*f*x + 4*I*e) + (a*c^6 + 2*I*a*c^5*d + a*c^4*d^2 + 4*I*a*c^3*d^3 - a*c^2*d^4 + 2*I*a*c*d^5 - a
*d^6)*f*e^(2*I*f*x + 2*I*e))

Sympy [A] (verification not implemented)

Time = 12.98 (sec) , antiderivative size = 510, normalized size of antiderivative = 2.52 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^2} \, dx=- \frac {2 d^{3}}{a c^{5} f + i a c^{4} d f + 2 a c^{3} d^{2} f + 2 i a c^{2} d^{3} f + a c d^{4} f + i a d^{5} f + \left (a c^{5} f e^{2 i e} - i a c^{4} d f e^{2 i e} + 2 a c^{3} d^{2} f e^{2 i e} - 2 i a c^{2} d^{3} f e^{2 i e} + a c d^{4} f e^{2 i e} - i a d^{5} f e^{2 i e}\right ) e^{2 i f x}} + \frac {x \left (c + 5 i d\right )}{2 a c^{3} + 6 i a c^{2} d - 6 a c d^{2} - 2 i a d^{3}} + \begin {cases} \frac {i e^{- 2 i f x}}{4 a c^{2} f e^{2 i e} + 8 i a c d f e^{2 i e} - 4 a d^{2} f e^{2 i e}} & \text {for}\: 4 a c^{2} f e^{2 i e} + 8 i a c d f e^{2 i e} - 4 a d^{2} f e^{2 i e} \neq 0 \\x \left (- \frac {c + 5 i d}{2 a c^{3} + 6 i a c^{2} d - 6 a c d^{2} - 2 i a d^{3}} + \frac {c e^{2 i e} + c + 5 i d e^{2 i e} + i d}{2 a c^{3} e^{2 i e} + 6 i a c^{2} d e^{2 i e} - 6 a c d^{2} e^{2 i e} - 2 i a d^{3} e^{2 i e}}\right ) & \text {otherwise} \end {cases} + \frac {i d^{2} \cdot \left (3 c - i d\right ) \log {\left (\frac {c + i d}{c e^{2 i e} - i d e^{2 i e}} + e^{2 i f x} \right )}}{a f \left (c - i d\right )^{2} \left (c + i d\right )^{3}} \]

[In]

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

[Out]

-2*d**3/(a*c**5*f + I*a*c**4*d*f + 2*a*c**3*d**2*f + 2*I*a*c**2*d**3*f + a*c*d**4*f + I*a*d**5*f + (a*c**5*f*e
xp(2*I*e) - I*a*c**4*d*f*exp(2*I*e) + 2*a*c**3*d**2*f*exp(2*I*e) - 2*I*a*c**2*d**3*f*exp(2*I*e) + a*c*d**4*f*e
xp(2*I*e) - I*a*d**5*f*exp(2*I*e))*exp(2*I*f*x)) + x*(c + 5*I*d)/(2*a*c**3 + 6*I*a*c**2*d - 6*a*c*d**2 - 2*I*a
*d**3) + Piecewise((I*exp(-2*I*f*x)/(4*a*c**2*f*exp(2*I*e) + 8*I*a*c*d*f*exp(2*I*e) - 4*a*d**2*f*exp(2*I*e)),
Ne(4*a*c**2*f*exp(2*I*e) + 8*I*a*c*d*f*exp(2*I*e) - 4*a*d**2*f*exp(2*I*e), 0)), (x*(-(c + 5*I*d)/(2*a*c**3 + 6
*I*a*c**2*d - 6*a*c*d**2 - 2*I*a*d**3) + (c*exp(2*I*e) + c + 5*I*d*exp(2*I*e) + I*d)/(2*a*c**3*exp(2*I*e) + 6*
I*a*c**2*d*exp(2*I*e) - 6*a*c*d**2*exp(2*I*e) - 2*I*a*d**3*exp(2*I*e))), True)) + I*d**2*(3*c - I*d)*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)**2*(c + I*d)**3)

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.60 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^2} \, dx=-\frac {\frac {8 \, {\left (-3 i \, c d^{3} - d^{4}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{a c^{5} d + i \, a c^{4} d^{2} + 2 \, a c^{3} d^{3} + 2 i \, a c^{2} d^{4} + a c d^{5} + i \, a d^{6}} + \frac {2 \, {\left (i \, c - 5 \, d\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a c^{3} + 3 i \, a c^{2} d - 3 \, a c d^{2} - i \, a d^{3}} + \frac {16 \, \log \left (\tan \left (f x + e\right ) + i\right )}{8 i \, a c^{2} + 16 \, a c d - 8 i \, a d^{2}} - \frac {i \, c^{2} d \tan \left (f x + e\right )^{2} - 2 \, c d^{2} \tan \left (f x + e\right )^{2} - i \, d^{3} \tan \left (f x + e\right )^{2} + i \, c^{3} \tan \left (f x + e\right ) + 3 \, c^{2} d \tan \left (f x + e\right ) - 15 i \, c d^{2} \tan \left (f x + e\right ) - 13 \, d^{3} \tan \left (f x + e\right ) + 5 \, c^{3} - 6 i \, c^{2} d - 13 \, c d^{2} + 8 i \, d^{3}}{{\left (a c^{4} + 2 \, a c^{2} d^{2} + a d^{4}\right )} {\left (d \tan \left (f x + e\right )^{2} + c \tan \left (f x + e\right ) - i \, d \tan \left (f x + e\right ) - i \, c\right )}}}{8 \, f} \]

[In]

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

[Out]

-1/8*(8*(-3*I*c*d^3 - d^4)*log(d*tan(f*x + e) + c)/(a*c^5*d + I*a*c^4*d^2 + 2*a*c^3*d^3 + 2*I*a*c^2*d^4 + a*c*
d^5 + I*a*d^6) + 2*(I*c - 5*d)*log(tan(f*x + e) - I)/(a*c^3 + 3*I*a*c^2*d - 3*a*c*d^2 - I*a*d^3) + 16*log(tan(
f*x + e) + I)/(8*I*a*c^2 + 16*a*c*d - 8*I*a*d^2) - (I*c^2*d*tan(f*x + e)^2 - 2*c*d^2*tan(f*x + e)^2 - I*d^3*ta
n(f*x + e)^2 + I*c^3*tan(f*x + e) + 3*c^2*d*tan(f*x + e) - 15*I*c*d^2*tan(f*x + e) - 13*d^3*tan(f*x + e) + 5*c
^3 - 6*I*c^2*d - 13*c*d^2 + 8*I*d^3)/((a*c^4 + 2*a*c^2*d^2 + a*d^4)*(d*tan(f*x + e)^2 + c*tan(f*x + e) - I*d*t
an(f*x + e) - I*c)))/f

Mupad [B] (verification not implemented)

Time = 10.50 (sec) , antiderivative size = 1334, normalized size of antiderivative = 6.60 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

symsum(log(tan(e + f*x)*(a*d^4 - a*c^2*d^2 + a*c*d^3*2i)*(c*d^3*6i + 9*d^4 - c^2*d^2) - (a*d^4 - a*c^2*d^2 + a
*c*d^3*2i)*(11*c*d^3 + c^3*d - d^4*6i) - root(a^3*c^5*d^5*e^3*192i + a^3*c^7*d^3*e^3*128i + a^3*c^3*d^7*e^3*12
8i + 48*a^3*c^8*d^2*e^3 - 48*a^3*c^2*d^8*e^3 + 32*a^3*c^6*d^4*e^3 - 32*a^3*c^4*d^6*e^3 + a^3*c^9*d*e^3*32i + a
^3*c*d^9*e^3*32i - 16*a^3*d^10*e^3 + 16*a^3*c^10*e^3 + 135*a*c^2*d^4*e + a*c^3*d^3*e*12i - 3*a*c^4*d^2*e - a*c
*d^5*e*90i + a*c^5*d*e*6i - 21*a*d^6*e + a*c^6*e - c^2*d^2*3i + 14*c*d^3 - d^4*5i, e, k)*((a*d^4 - a*c^2*d^2 +
a*c*d^3*2i)*(2*a*c^6 - 6*a*d^6 - 10*a*c^2*d^4 + a*c^3*d^3*16i - 2*a*c^4*d^2 + a*c*d^5*8i + a*c^5*d*8i) + root
(a^3*c^5*d^5*e^3*192i + a^3*c^7*d^3*e^3*128i + a^3*c^3*d^7*e^3*128i + 48*a^3*c^8*d^2*e^3 - 48*a^3*c^2*d^8*e^3
+ 32*a^3*c^6*d^4*e^3 - 32*a^3*c^4*d^6*e^3 + a^3*c^9*d*e^3*32i + a^3*c*d^9*e^3*32i - 16*a^3*d^10*e^3 + 16*a^3*c
^10*e^3 + 135*a*c^2*d^4*e + a*c^3*d^3*e*12i - 3*a*c^4*d^2*e - a*c*d^5*e*90i + a*c^5*d*e*6i - 21*a*d^6*e + a*c^
6*e - c^2*d^2*3i + 14*c*d^3 - d^4*5i, e, k)*((a*d^4 - a*c^2*d^2 + a*c*d^3*2i)*(32*a^2*c^7*d - 32*a^2*c*d^7 + a
^2*c^2*d^6*64i - 32*a^2*c^3*d^5 + a^2*c^4*d^4*128i + 32*a^2*c^5*d^3 + a^2*c^6*d^2*64i) + tan(e + f*x)*(a*d^4 -
a*c^2*d^2 + a*c*d^3*2i)*(a^2*c*d^7*48i - 24*a^2*d^8 - 8*a^2*c^8 - a^2*c^7*d*16i - 16*a^2*c^2*d^6 + a^2*c^3*d^
5*80i + 32*a^2*c^4*d^4 + a^2*c^5*d^3*16i + 16*a^2*c^6*d^2)) + tan(e + f*x)*(a*d^4 - a*c^2*d^2 + a*c*d^3*2i)*(a
*d^6*12i + a*c^2*d^4*40i - 8*a*c^3*d^3 + a*c^4*d^2*28i - 12*a*c*d^5 + 4*a*c^5*d)))*root(a^3*c^5*d^5*e^3*192i +
a^3*c^7*d^3*e^3*128i + a^3*c^3*d^7*e^3*128i + 48*a^3*c^8*d^2*e^3 - 48*a^3*c^2*d^8*e^3 + 32*a^3*c^6*d^4*e^3 -
32*a^3*c^4*d^6*e^3 + a^3*c^9*d*e^3*32i + a^3*c*d^9*e^3*32i - 16*a^3*d^10*e^3 + 16*a^3*c^10*e^3 + 135*a*c^2*d^4
*e + a*c^3*d^3*e*12i - 3*a*c^4*d^2*e - a*c*d^5*e*90i + a*c^5*d*e*6i - 21*a*d^6*e + a*c^6*e - c^2*d^2*3i + 14*c
*d^3 - d^4*5i, e, k), k, 1, 3)/f - ((c*d*2i - 2*c^2 + 4*d^2)/(4*a*d*(c*d^2 + c^2*d*1i + c^3 + d^3*1i)) - (tan(
e + f*x)*(c - d*3i))/(2*a*(c*d^2 + c^2*d*1i + c^3 + d^3*1i)))/(f*(tan(e + f*x)*(c/d - 1i) - (c*1i)/d + tan(e +
f*x)^2))