\(\int \frac {(a+i a \tan (e+f x))^2}{c-i c \tan (e+f x)} \, dx\) [926]

Optimal result

Integrand size = 31, antiderivative size = 55 \[ \int \frac {(a+i a \tan (e+f x))^2}{c-i c \tan (e+f x)} \, dx=-\frac {a^2 x}{c}+\frac {i a^2 \log (\cos (e+f x))}{c f}-\frac {2 i a^2}{f (c-i c \tan (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3603, 3568, 45} \[ \int \frac {(a+i a \tan (e+f x))^2}{c-i c \tan (e+f x)} \, dx=-\frac {2 i a^2}{f (c-i c \tan (e+f x))}+\frac {i a^2 \log (\cos (e+f x))}{c f}-\frac {a^2 x}{c} \]

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps \begin{align*} \text {integral}& = \left (a^2 c^2\right ) \int \frac {\sec ^4(e+f x)}{(c-i c \tan (e+f x))^3} \, dx \\ & = \frac {\left (i a^2\right ) \text {Subst}\left (\int \frac {c-x}{(c+x)^2} \, dx,x,-i c \tan (e+f x)\right )}{c f} \\ & = \frac {\left (i a^2\right ) \text {Subst}\left (\int \left (\frac {1}{-c-x}+\frac {2 c}{(c+x)^2}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c f} \\ & = -\frac {a^2 x}{c}+\frac {i a^2 \log (\cos (e+f x))}{c f}-\frac {2 i a^2}{f (c-i c \tan (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.54 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82 \[ \int \frac {(a+i a \tan (e+f x))^2}{c-i c \tan (e+f x)} \, dx=\frac {i a^2 \left (-\log (i+\tan (e+f x))-\frac {2 c}{c-i c \tan (e+f x)}\right )}{c f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.07

method	result	size
risch	\(-\frac {i a^{2} {\mathrm e}^{2 i \left (f x +e \right )}}{c f}+\frac {2 a^{2} e}{c f}+\frac {i a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{c f}\)	\(59\)
derivativedivides	\(-\frac {i a^{2} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f c}-\frac {a^{2} \arctan \left (\tan \left (f x +e \right )\right )}{f c}+\frac {2 a^{2}}{f c \left (\tan \left (f x +e \right )+i\right )}\)	\(65\)
default	\(-\frac {i a^{2} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f c}-\frac {a^{2} \arctan \left (\tan \left (f x +e \right )\right )}{f c}+\frac {2 a^{2}}{f c \left (\tan \left (f x +e \right )+i\right )}\)	\(65\)
norman	\(\frac {-\frac {2 i a^{2}}{c f}-\frac {a^{2} x}{c}-\frac {a^{2} x \left (\tan ^{2}\left (f x +e \right )\right )}{c}+\frac {2 a^{2} \tan \left (f x +e \right )}{c f}}{1+\tan ^{2}\left (f x +e \right )}-\frac {i a^{2} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f c}\)	\(94\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int \frac {(a+i a \tan (e+f x))^2}{c-i c \tan (e+f x)} \, dx=\frac {-i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{2} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{c f} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.24 \[ \int \frac {(a+i a \tan (e+f x))^2}{c-i c \tan (e+f x)} \, dx=\frac {i a^{2} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c f} + \begin {cases} - \frac {i a^{2} e^{2 i e} e^{2 i f x}}{c f} & \text {for}\: c f \neq 0 \\\frac {2 a^{2} x e^{2 i e}}{c} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

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

[Out]

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

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. 119 vs. \(2 (49) = 98\).

Time = 0.42 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.16 \[ \int \frac {(a+i a \tan (e+f x))^2}{c-i c \tan (e+f x)} \, dx=-\frac {-\frac {i \, a^{2} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c} + \frac {2 i \, a^{2} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c} - \frac {i \, a^{2} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c} + \frac {-3 i \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 10 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 i \, a^{2}}{c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{2}}}{f} \]

[In]

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

[Out]

-(-I*a^2*log(tan(1/2*f*x + 1/2*e) + 1)/c + 2*I*a^2*log(tan(1/2*f*x + 1/2*e) + I)/c - I*a^2*log(tan(1/2*f*x + 1
/2*e) - 1)/c + (-3*I*a^2*tan(1/2*f*x + 1/2*e)^2 + 10*a^2*tan(1/2*f*x + 1/2*e) + 3*I*a^2)/(c*(tan(1/2*f*x + 1/2
*e) + I)^2))/f

Mupad [B] (verification not implemented)

Time = 5.63 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82 \[ \int \frac {(a+i a \tan (e+f x))^2}{c-i c \tan (e+f x)} \, dx=\frac {2\,a^2}{c\,f\,\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {a^2\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{c\,f} \]

[In]

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

[Out]

(2*a^2)/(c*f*(tan(e + f*x) + 1i)) - (a^2*log(tan(e + f*x) + 1i)*1i)/(c*f)