3.10.33 \(\int \frac {(a+i a \tan (e+f x))^2}{(c-i c \tan (e+f x))^2} \, dx\) [933]

Optimal. Leaf size=28 \[ \frac {a^2 \tan (e+f x)}{f (c-i c \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3603, 3568, 34} \begin {gather*} \frac {a^2 \tan (e+f x)}{f (c-i c \tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 34

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_)), x_Symbol] :> Simp[d*x*((a + b*x)^(m + 1)/(b*(m + 2))), x] /
; FreeQ[{a, b, c, d, m}, x] && EqQ[a*d - b*c*(m + 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*} \int \frac {(a+i a \tan (e+f x))^2}{(c-i c \tan (e+f x))^2} \, dx &=\left (a^2 c^2\right ) \int \frac {\sec ^4(e+f x)}{(c-i c \tan (e+f x))^4} \, dx\\ &=\frac {\left (i a^2\right ) \text {Subst}\left (\int \frac {c-x}{(c+x)^3} \, dx,x,-i c \tan (e+f x)\right )}{c f}\\ &=\frac {a^2 \tan (e+f x)}{f (c-i c \tan (e+f x))^2}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 34, normalized size = 1.21 \begin {gather*} \frac {a^2 (-i \cos (4 (e+f x))+\sin (4 (e+f x)))}{4 c^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*((-I)*Cos[4*(e + f*x)] + Sin[4*(e + f*x)]))/(4*c^2*f)

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Maple [A]
time = 0.18, size = 39, normalized size = 1.39

method	result	size
risch	\(-\frac {i a^{2} {\mathrm e}^{4 i \left (f x +e \right )}}{4 c^{2} f}\)	\(22\)
derivativedivides	\(\frac {a^{2} \left (\frac {i}{\left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {1}{\tan \left (f x +e \right )+i}\right )}{f \,c^{2}}\)	\(39\)
default	\(\frac {a^{2} \left (\frac {i}{\left (\tan \left (f x +e \right )+i\right )^{2}}-\frac {1}{\tan \left (f x +e \right )+i}\right )}{f \,c^{2}}\)	\(39\)
norman	\(\frac {\frac {a^{2} \tan \left (f x +e \right )}{c f}-\frac {a^{2} \left (\tan ^{3}\left (f x +e \right )\right )}{c f}+\frac {2 i a^{2} \left (\tan ^{2}\left (f x +e \right )\right )}{c f}}{c \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [A]
time = 0.92, size = 21, normalized size = 0.75 \begin {gather*} -\frac {i \, a^{2} e^{\left (4 i \, f x + 4 i \, e\right )}}{4 \, c^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.13, size = 46, normalized size = 1.64 \begin {gather*} \begin {cases} - \frac {i a^{2} e^{4 i e} e^{4 i f x}}{4 c^{2} f} & \text {for}\: c^{2} f \neq 0 \\\frac {a^{2} x e^{4 i e}}{c^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.59, size = 54, normalized size = 1.93 \begin {gather*} -\frac {2 \, {\left (a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{c^{2} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.74, size = 28, normalized size = 1.00 \begin {gather*} -\frac {a^2\,\mathrm {tan}\left (e+f\,x\right )}{c^2\,f\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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