\(\int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \, dx\) [670]

Optimal result

Integrand size = 24, antiderivative size = 46 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \, dx=a (A-i B) x-\frac {a (i A+B) \log (\cos (e+f x))}{f}+\frac {i a B \tan (e+f x)}{f} \]

[Out]

a*(A-I*B)*x-a*(I*A+B)*ln(cos(f*x+e))/f+I*a*B*tan(f*x+e)/f

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3606, 3556} \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \, dx=-\frac {a (B+i A) \log (\cos (e+f x))}{f}+a x (A-i B)+\frac {i a B \tan (e+f x)}{f} \]

[In]

Int[(a + I*a*Tan[e + f*x])*(A + B*Tan[e + f*x]),x]

[Out]

a*(A - I*B)*x - (a*(I*A + B)*Log[Cos[e + f*x]])/f + (I*a*B*Tan[e + f*x])/f

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3606

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

Rubi steps \begin{align*} \text {integral}& = a (A-i B) x+\frac {i a B \tan (e+f x)}{f}+(a (i A+B)) \int \tan (e+f x) \, dx \\ & = a (A-i B) x-\frac {a (i A+B) \log (\cos (e+f x))}{f}+\frac {i a B \tan (e+f x)}{f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.43 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \, dx=a A x-\frac {i a B \arctan (\tan (e+f x))}{f}-\frac {i a A \log (\cos (e+f x))}{f}-\frac {a B \log (\cos (e+f x))}{f}+\frac {i a B \tan (e+f x)}{f} \]

[In]

Integrate[(a + I*a*Tan[e + f*x])*(A + B*Tan[e + f*x]),x]

[Out]

a*A*x - (I*a*B*ArcTan[Tan[e + f*x]])/f - (I*a*A*Log[Cos[e + f*x]])/f - (a*B*Log[Cos[e + f*x]])/f + (I*a*B*Tan[
e + f*x])/f

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {a \left (i B \tan \left (f x +e \right )+\frac {\left (i A +B \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}+\left (-i B +A \right ) \arctan \left (\tan \left (f x +e \right )\right )\right )}{f}\)	\(50\)
default	\(\frac {a \left (i B \tan \left (f x +e \right )+\frac {\left (i A +B \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}+\left (-i B +A \right ) \arctan \left (\tan \left (f x +e \right )\right )\right )}{f}\)	\(50\)
norman	\(\left (-i a B +A a \right ) x +\frac {i a B \tan \left (f x +e \right )}{f}+\frac {\left (i A a +B a \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f}\)	\(52\)
parts	\(A a x +\frac {\left (i A a +B a \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f}+\frac {i B a \left (\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}\)	\(55\)
parallelrisch	\(\frac {-2 i B x a f +i A \ln \left (1+\tan \left (f x +e \right )^{2}\right ) a +2 A x a f +2 i a B \tan \left (f x +e \right )+B \ln \left (1+\tan \left (f x +e \right )^{2}\right ) a}{2 f}\)	\(61\)
risch	\(\frac {2 i a B e}{f}-\frac {2 a A e}{f}-\frac {2 a B}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) B}{f}-\frac {i a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) A}{f}\)	\(78\)

[In]

int((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/f*a*(I*B*tan(f*x+e)+1/2*(I*A+B)*ln(1+tan(f*x+e)^2)+(A-I*B)*arctan(tan(f*x+e)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.39 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \, dx=-\frac {2 \, B a - {\left ({\left (-i \, A - B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A - B\right )} a\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{f e^{\left (2 i \, f x + 2 i \, e\right )} + f} \]

[In]

integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e)),x, algorithm="fricas")

[Out]

-(2*B*a - ((-I*A - B)*a*e^(2*I*f*x + 2*I*e) + (-I*A - B)*a)*log(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(2*I*f*x + 2*I*
e) + f)

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.15 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \, dx=- \frac {2 B a}{f e^{2 i e} e^{2 i f x} + f} - \frac {i a \left (A - i B\right ) \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{f} \]

[In]

integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e)),x)

[Out]

-2*B*a/(f*exp(2*I*e)*exp(2*I*f*x) + f) - I*a*(A - I*B)*log(exp(2*I*f*x) + exp(-2*I*e))/f

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.09 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \, dx=\frac {2 \, {\left (f x + e\right )} {\left (A - i \, B\right )} a - {\left (-i \, A - B\right )} a \log \left (\tan \left (f x + e\right )^{2} + 1\right ) + 2 i \, B a \tan \left (f x + e\right )}{2 \, f} \]

[In]

integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e)),x, algorithm="maxima")

[Out]

1/2*(2*(f*x + e)*(A - I*B)*a - (-I*A - B)*a*log(tan(f*x + e)^2 + 1) + 2*I*B*a*tan(f*x + e))/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. 103 vs. \(2 (40) = 80\).

Time = 0.34 (sec) , antiderivative size = 103, normalized size of antiderivative = 2.24 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \, dx=\frac {-i \, A a e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - B a e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - i \, A a \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - B a \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 2 \, B a}{f e^{\left (2 i \, f x + 2 i \, e\right )} + f} \]

[In]

integrate((a+I*a*tan(f*x+e))*(A+B*tan(f*x+e)),x, algorithm="giac")

[Out]

(-I*A*a*e^(2*I*f*x + 2*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - B*a*e^(2*I*f*x + 2*I*e)*log(e^(2*I*f*x + 2*I*e) + 1
) - I*A*a*log(e^(2*I*f*x + 2*I*e) + 1) - B*a*log(e^(2*I*f*x + 2*I*e) + 1) - 2*B*a)/(f*e^(2*I*f*x + 2*I*e) + f)

Mupad [B] (verification not implemented)

Time = 8.88 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \, dx=\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\left (B\,a+A\,a\,1{}\mathrm {i}\right )}{f}+\frac {B\,a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}{f} \]

[In]

int((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i),x)

[Out]

(log(tan(e + f*x) + 1i)*(A*a*1i + B*a))/f + (B*a*tan(e + f*x)*1i)/f