\(\int \frac {1}{(c+d x)^2 (a+i a \tan (e+f x))^2} \, dx\) [28]

Optimal result

Integrand size = 23, antiderivative size = 436 \[ \int \frac {1}{(c+d x)^2 (a+i a \tan (e+f x))^2} \, dx=-\frac {1}{4 a^2 d (c+d x)}-\frac {\cos (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac {\cos ^2(2 e+2 f x)}{4 a^2 d (c+d x)}-\frac {i f \cos \left (2 e-\frac {2 c f}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right )}{a^2 d^2}-\frac {i f \cos \left (4 e-\frac {4 c f}{d}\right ) \operatorname {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right )}{a^2 d^2}-\frac {f \operatorname {CosIntegral}\left (\frac {4 c f}{d}+4 f x\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{a^2 d^2}-\frac {f \operatorname {CosIntegral}\left (\frac {2 c f}{d}+2 f x\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}+\frac {i \sin (2 e+2 f x)}{2 a^2 d (c+d x)}+\frac {\sin ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {i \sin (4 e+4 f x)}{4 a^2 d (c+d x)}-\frac {f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{a^2 d^2}+\frac {i f \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (\frac {2 c f}{d}+2 f x\right )}{a^2 d^2}-\frac {f \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{a^2 d^2}+\frac {i f \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (\frac {4 c f}{d}+4 f x\right )}{a^2 d^2} \] Output:

-1/4/a^2/d/(d*x+c)-1/2*cos(2*f*x+2*e)/a^2/d/(d*x+c)-1/4*cos(2*f*x+2*e)^2/a 
^2/d/(d*x+c)-I*f*cos(-2*e+2*c*f/d)*Ci(2*c*f/d+2*f*x)/a^2/d^2-I*f*cos(-4*e+ 
4*c*f/d)*Ci(4*c*f/d+4*f*x)/a^2/d^2+f*Ci(4*c*f/d+4*f*x)*sin(-4*e+4*c*f/d)/a 
^2/d^2+f*Ci(2*c*f/d+2*f*x)*sin(-2*e+2*c*f/d)/a^2/d^2+1/2*I*sin(2*f*x+2*e)/ 
a^2/d/(d*x+c)+1/4*sin(2*f*x+2*e)^2/a^2/d/(d*x+c)+1/4*I*sin(4*f*x+4*e)/a^2/ 
d/(d*x+c)-f*cos(-2*e+2*c*f/d)*Si(2*c*f/d+2*f*x)/a^2/d^2-I*f*sin(-2*e+2*c*f 
/d)*Si(2*c*f/d+2*f*x)/a^2/d^2-f*cos(-4*e+4*c*f/d)*Si(4*c*f/d+4*f*x)/a^2/d^ 
2-I*f*sin(-4*e+4*c*f/d)*Si(4*c*f/d+4*f*x)/a^2/d^2
 

Mathematica [A] (verified)

Time = 1.55 (sec) , antiderivative size = 467, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(c+d x)^2 (a+i a \tan (e+f x))^2} \, dx=-\frac {\left (\cos \left (2 \left (e+f \left (-\frac {c}{d}+x\right )\right )\right )-i \sin \left (2 \left (e+f \left (-\frac {c}{d}+x\right )\right )\right )\right ) \left (2 d \cos \left (\frac {2 c f}{d}\right )+d \cos \left (2 \left (e+f \left (-\frac {c}{d}+x\right )\right )\right )+d \cos \left (2 \left (e+f \left (\frac {c}{d}+x\right )\right )\right )-2 i d \sin \left (\frac {2 c f}{d}\right )+4 i f (c+d x) \operatorname {CosIntegral}\left (\frac {2 f (c+d x)}{d}\right ) (\cos (2 f x)+i \sin (2 f x))+i d \sin \left (2 \left (e+f \left (-\frac {c}{d}+x\right )\right )\right )-i d \sin \left (2 \left (e+f \left (\frac {c}{d}+x\right )\right )\right )+4 f (c+d x) \operatorname {CosIntegral}\left (\frac {4 f (c+d x)}{d}\right ) \left (i \cos \left (2 e-\frac {2 f (c+d x)}{d}\right )+\sin \left (2 e-\frac {2 f (c+d x)}{d}\right )\right )+4 c f \cos (2 f x) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 d f x \cos (2 f x) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 i c f \sin (2 f x) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 i d f x \sin (2 f x) \text {Si}\left (\frac {2 f (c+d x)}{d}\right )+4 c f \cos \left (2 e-\frac {2 f (c+d x)}{d}\right ) \text {Si}\left (\frac {4 f (c+d x)}{d}\right )+4 d f x \cos \left (2 e-\frac {2 f (c+d x)}{d}\right ) \text {Si}\left (\frac {4 f (c+d x)}{d}\right )-4 i c f \sin \left (2 e-\frac {2 f (c+d x)}{d}\right ) \text {Si}\left (\frac {4 f (c+d x)}{d}\right )-4 i d f x \sin \left (2 e-\frac {2 f (c+d x)}{d}\right ) \text {Si}\left (\frac {4 f (c+d x)}{d}\right )\right )}{4 a^2 d^2 (c+d x)} \] Input:

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

Output:

-1/4*((Cos[2*(e + f*(-(c/d) + x))] - I*Sin[2*(e + f*(-(c/d) + x))])*(2*d*C 
os[(2*c*f)/d] + d*Cos[2*(e + f*(-(c/d) + x))] + d*Cos[2*(e + f*(c/d + x))] 
 - (2*I)*d*Sin[(2*c*f)/d] + (4*I)*f*(c + d*x)*CosIntegral[(2*f*(c + d*x))/ 
d]*(Cos[2*f*x] + I*Sin[2*f*x]) + I*d*Sin[2*(e + f*(-(c/d) + x))] - I*d*Sin 
[2*(e + f*(c/d + x))] + 4*f*(c + d*x)*CosIntegral[(4*f*(c + d*x))/d]*(I*Co 
s[2*e - (2*f*(c + d*x))/d] + Sin[2*e - (2*f*(c + d*x))/d]) + 4*c*f*Cos[2*f 
*x]*SinIntegral[(2*f*(c + d*x))/d] + 4*d*f*x*Cos[2*f*x]*SinIntegral[(2*f*( 
c + d*x))/d] + (4*I)*c*f*Sin[2*f*x]*SinIntegral[(2*f*(c + d*x))/d] + (4*I) 
*d*f*x*Sin[2*f*x]*SinIntegral[(2*f*(c + d*x))/d] + 4*c*f*Cos[2*e - (2*f*(c 
 + d*x))/d]*SinIntegral[(4*f*(c + d*x))/d] + 4*d*f*x*Cos[2*e - (2*f*(c + d 
*x))/d]*SinIntegral[(4*f*(c + d*x))/d] - (4*I)*c*f*Sin[2*e - (2*f*(c + d*x 
))/d]*SinIntegral[(4*f*(c + d*x))/d] - (4*I)*d*f*x*Sin[2*e - (2*f*(c + d*x 
))/d]*SinIntegral[(4*f*(c + d*x))/d]))/(a^2*d^2*(c + d*x))
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 436, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 4211, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4211

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {f \operatorname {CosIntegral}\left (4 x f+\frac {4 c f}{d}\right ) \sin \left (4 e-\frac {4 c f}{d}\right )}{a^2 d^2}-\frac {f \operatorname {CosIntegral}\left (2 x f+\frac {2 c f}{d}\right ) \sin \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {i f \operatorname {CosIntegral}\left (2 x f+\frac {2 c f}{d}\right ) \cos \left (2 e-\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {i f \operatorname {CosIntegral}\left (4 x f+\frac {4 c f}{d}\right ) \cos \left (4 e-\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {i f \sin \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a^2 d^2}+\frac {i f \sin \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{a^2 d^2}-\frac {f \cos \left (2 e-\frac {2 c f}{d}\right ) \text {Si}\left (2 x f+\frac {2 c f}{d}\right )}{a^2 d^2}-\frac {f \cos \left (4 e-\frac {4 c f}{d}\right ) \text {Si}\left (4 x f+\frac {4 c f}{d}\right )}{a^2 d^2}+\frac {\sin ^2(2 e+2 f x)}{4 a^2 d (c+d x)}+\frac {i \sin (2 e+2 f x)}{2 a^2 d (c+d x)}+\frac {i \sin (4 e+4 f x)}{4 a^2 d (c+d x)}-\frac {\cos ^2(2 e+2 f x)}{4 a^2 d (c+d x)}-\frac {\cos (2 e+2 f x)}{2 a^2 d (c+d x)}-\frac {1}{4 a^2 d (c+d x)}\)

Input:

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

Output:

-1/4*1/(a^2*d*(c + d*x)) - Cos[2*e + 2*f*x]/(2*a^2*d*(c + d*x)) - Cos[2*e 
+ 2*f*x]^2/(4*a^2*d*(c + d*x)) - (I*f*Cos[2*e - (2*c*f)/d]*CosIntegral[(2* 
c*f)/d + 2*f*x])/(a^2*d^2) - (I*f*Cos[4*e - (4*c*f)/d]*CosIntegral[(4*c*f) 
/d + 4*f*x])/(a^2*d^2) - (f*CosIntegral[(4*c*f)/d + 4*f*x]*Sin[4*e - (4*c* 
f)/d])/(a^2*d^2) - (f*CosIntegral[(2*c*f)/d + 2*f*x]*Sin[2*e - (2*c*f)/d]) 
/(a^2*d^2) + ((I/2)*Sin[2*e + 2*f*x])/(a^2*d*(c + d*x)) + Sin[2*e + 2*f*x] 
^2/(4*a^2*d*(c + d*x)) + ((I/4)*Sin[4*e + 4*f*x])/(a^2*d*(c + d*x)) - (f*C 
os[2*e - (2*c*f)/d]*SinIntegral[(2*c*f)/d + 2*f*x])/(a^2*d^2) + (I*f*Sin[2 
*e - (2*c*f)/d]*SinIntegral[(2*c*f)/d + 2*f*x])/(a^2*d^2) - (f*Cos[4*e - ( 
4*c*f)/d]*SinIntegral[(4*c*f)/d + 4*f*x])/(a^2*d^2) + (I*f*Sin[4*e - (4*c* 
f)/d]*SinIntegral[(4*c*f)/d + 4*f*x])/(a^2*d^2)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4211

Int[((c_.) + (d_.)*(x_))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), 
x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (1/(2*a) + Cos[2*e + 2*f*x]/( 
2*a) + Sin[2*e + 2*f*x]/(2*b))^(-n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] 
 && EqQ[a^2 + b^2, 0] && ILtQ[m, 0] && ILtQ[n, 0]
 

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.40

method	result	size
risch	\(-\frac {1}{4 a^{2} d \left (d x +c \right )}-\frac {f \,{\mathrm e}^{-4 i \left (f x +e \right )}}{4 a^{2} \left (d f x +c f \right ) d}+\frac {i f \,{\mathrm e}^{\frac {4 i \left (c f -d e \right )}{d}} \operatorname {expIntegral}_{1}\left (4 i f x +4 i e +\frac {4 i \left (c f -d e \right )}{d}\right )}{a^{2} d^{2}}-\frac {f \,{\mathrm e}^{-2 i \left (f x +e \right )}}{2 a^{2} \left (d f x +c f \right ) d}+\frac {i f \,{\mathrm e}^{\frac {2 i \left (c f -d e \right )}{d}} \operatorname {expIntegral}_{1}\left (2 i f x +2 i e +\frac {2 i \left (c f -d e \right )}{d}\right )}{a^{2} d^{2}}\)	\(175\)

Input:

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

Output:

-1/4/a^2/d/(d*x+c)-1/4/a^2*f*exp(-4*I*(f*x+e))/(d*f*x+c*f)/d+I/a^2*f/d^2*e 
xp(4*I*(c*f-d*e)/d)*Ei(1,4*I*f*x+4*I*e+4*I*(c*f-d*e)/d)-1/2/a^2*f*exp(-2*I 
*(f*x+e))/(d*f*x+c*f)/d+I/a^2*f/d^2*exp(2*I*(c*f-d*e)/d)*Ei(1,2*I*f*x+2*I* 
e+2*I*(c*f-d*e)/d)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.33 \[ \int \frac {1}{(c+d x)^2 (a+i a \tan (e+f x))^2} \, dx=-\frac {{\left ({\left (4 \, {\left (i \, d f x + i \, c f\right )} {\rm Ei}\left (-\frac {2 \, {\left (i \, d f x + i \, c f\right )}}{d}\right ) e^{\left (-\frac {2 \, {\left (i \, d e - i \, c f\right )}}{d}\right )} + 4 \, {\left (i \, d f x + i \, c f\right )} {\rm Ei}\left (-\frac {4 \, {\left (i \, d f x + i \, c f\right )}}{d}\right ) e^{\left (-\frac {4 \, {\left (i \, d e - i \, c f\right )}}{d}\right )} + d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d e^{\left (2 i \, f x + 2 i \, e\right )} + d\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{4 \, {\left (a^{2} d^{3} x + a^{2} c d^{2}\right )}} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {1}{(c+d x)^2 (a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {1}{c^{2} \tan ^{2}{\left (e + f x \right )} - 2 i c^{2} \tan {\left (e + f x \right )} - c^{2} + 2 c d x \tan ^{2}{\left (e + f x \right )} - 4 i c d x \tan {\left (e + f x \right )} - 2 c d x + d^{2} x^{2} \tan ^{2}{\left (e + f x \right )} - 2 i d^{2} x^{2} \tan {\left (e + f x \right )} - d^{2} x^{2}}\, dx}{a^{2}} \] Input:

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

Output:

-Integral(1/(c**2*tan(e + f*x)**2 - 2*I*c**2*tan(e + f*x) - c**2 + 2*c*d*x 
*tan(e + f*x)**2 - 4*I*c*d*x*tan(e + f*x) - 2*c*d*x + d**2*x**2*tan(e + f* 
x)**2 - 2*I*d**2*x**2*tan(e + f*x) - d**2*x**2), x)/a**2
 

Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.48 \[ \int \frac {1}{(c+d x)^2 (a+i a \tan (e+f x))^2} \, dx=-\frac {2 \, f^{2} \cos \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) E_{2}\left (-\frac {2 \, {\left (-i \, {\left (f x + e\right )} d + i \, d e - i \, c f\right )}}{d}\right ) + f^{2} \cos \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) E_{2}\left (-\frac {4 \, {\left (-i \, {\left (f x + e\right )} d + i \, d e - i \, c f\right )}}{d}\right ) + 2 i \, f^{2} E_{2}\left (-\frac {2 \, {\left (-i \, {\left (f x + e\right )} d + i \, d e - i \, c f\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (d e - c f\right )}}{d}\right ) + i \, f^{2} E_{2}\left (-\frac {4 \, {\left (-i \, {\left (f x + e\right )} d + i \, d e - i \, c f\right )}}{d}\right ) \sin \left (-\frac {4 \, {\left (d e - c f\right )}}{d}\right ) + f^{2}}{4 \, {\left ({\left (f x + e\right )} a^{2} d^{2} - a^{2} d^{2} e + a^{2} c d f\right )} f} \] Input:

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

Output:

-1/4*(2*f^2*cos(-2*(d*e - c*f)/d)*exp_integral_e(2, -2*(-I*(f*x + e)*d + I 
*d*e - I*c*f)/d) + f^2*cos(-4*(d*e - c*f)/d)*exp_integral_e(2, -4*(-I*(f*x 
 + e)*d + I*d*e - I*c*f)/d) + 2*I*f^2*exp_integral_e(2, -2*(-I*(f*x + e)*d 
 + I*d*e - I*c*f)/d)*sin(-2*(d*e - c*f)/d) + I*f^2*exp_integral_e(2, -4*(- 
I*(f*x + e)*d + I*d*e - I*c*f)/d)*sin(-4*(d*e - c*f)/d) + f^2)/(((f*x + e) 
*a^2*d^2 - a^2*d^2*e + a^2*c*d*f)*f)
 

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1967 vs. \(2 (410) = 820\).

Time = 11.12 (sec) , antiderivative size = 1967, normalized size of antiderivative = 4.51 \[ \int \frac {1}{(c+d x)^2 (a+i a \tan (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(-4*I*(d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f)*f^2*cos(-4*(d*e - 
c*f)/d)*cos_integral(4*((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f) - d* 
e + c*f)/d) + 4*I*d*e*f^2*cos(-4*(d*e - c*f)/d)*cos_integral(4*((d*x + c)* 
(d*e/(d*x + c) - c*f/(d*x + c) + f) - d*e + c*f)/d) - 4*I*c*f^3*cos(-4*(d* 
e - c*f)/d)*cos_integral(4*((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f) 
- d*e + c*f)/d) - 4*I*(d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f)*f^2*co 
s(-2*(d*e - c*f)/d)*cos_integral(2*((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + 
c) + f) - d*e + c*f)/d) + 4*I*d*e*f^2*cos(-2*(d*e - c*f)/d)*cos_integral(2 
*((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f) - d*e + c*f)/d) - 4*I*c*f^ 
3*cos(-2*(d*e - c*f)/d)*cos_integral(2*((d*x + c)*(d*e/(d*x + c) - c*f/(d* 
x + c) + f) - d*e + c*f)/d) + 4*(d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + 
 f)*f^2*cos_integral(2*((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f) - d* 
e + c*f)/d)*sin(-2*(d*e - c*f)/d) - 4*d*e*f^2*cos_integral(2*((d*x + c)*(d 
*e/(d*x + c) - c*f/(d*x + c) + f) - d*e + c*f)/d)*sin(-2*(d*e - c*f)/d) + 
4*c*f^3*cos_integral(2*((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f) - d* 
e + c*f)/d)*sin(-2*(d*e - c*f)/d) + 4*(d*x + c)*(d*e/(d*x + c) - c*f/(d*x 
+ c) + f)*f^2*cos_integral(4*((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) + f 
) - d*e + c*f)/d)*sin(-4*(d*e - c*f)/d) - 4*d*e*f^2*cos_integral(4*((d*x + 
 c)*(d*e/(d*x + c) - c*f/(d*x + c) + f) - d*e + c*f)/d)*sin(-4*(d*e - c*f) 
/d) + 4*c*f^3*cos_integral(4*((d*x + c)*(d*e/(d*x + c) - c*f/(d*x + c) ...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{(c+d x)^2 (a+i a \tan (e+f x))^2} \, dx=-\frac {\int \frac {1}{\tan \left (f x +e \right )^{2} c^{2}+2 \tan \left (f x +e \right )^{2} c d x +\tan \left (f x +e \right )^{2} d^{2} x^{2}-2 \tan \left (f x +e \right ) c^{2} i -4 \tan \left (f x +e \right ) c d i x -2 \tan \left (f x +e \right ) d^{2} i \,x^{2}-c^{2}-2 c d x -d^{2} x^{2}}d x}{a^{2}} \] Input:

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

Output:

( - int(1/(tan(e + f*x)**2*c**2 + 2*tan(e + f*x)**2*c*d*x + tan(e + f*x)** 
2*d**2*x**2 - 2*tan(e + f*x)*c**2*i - 4*tan(e + f*x)*c*d*i*x - 2*tan(e + f 
*x)*d**2*i*x**2 - c**2 - 2*c*d*x - d**2*x**2),x))/a**2