3.10.27 \(\int \frac {a+i a \tan (e+f x)}{c-i c \tan (e+f x)} \, dx\) [927]

Optimal. Leaf size=23 \[ -\frac {i a}{f (c-i c \tan (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3603, 3568, 32} \begin {gather*} -\frac {i a}{f (c-i c \tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*a)/(f*(c - I*c*Tan[e + f*x]))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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)}{c-i c \tan (e+f x)} \, dx &=(a c) \int \frac {\sec ^2(e+f x)}{(c-i c \tan (e+f x))^2} \, dx\\ &=\frac {(i a) \text {Subst}\left (\int \frac {1}{(c+x)^2} \, dx,x,-i c \tan (e+f x)\right )}{f}\\ &=-\frac {i a}{f (c-i c \tan (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 32, normalized size = 1.39 \begin {gather*} \frac {a (-i \cos (2 (e+f x))+\sin (2 (e+f x)))}{2 c f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.15, size = 20, normalized size = 0.87

method	result	size
derivativedivides	\(\frac {a}{f c \left (\tan \left (f x +e \right )+i\right )}\)	\(20\)
default	\(\frac {a}{f c \left (\tan \left (f x +e \right )+i\right )}\)	\(20\)
risch	\(-\frac {i a \,{\mathrm e}^{2 i \left (f x +e \right )}}{2 c f}\)	\(20\)
norman	\(\frac {-\frac {i a}{c f}+\frac {a \tan \left (f x +e \right )}{c f}}{1+\tan ^{2}\left (f x +e \right )}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*a/c/(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))/(c-I*c*tan(f*x+e)),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 = 1.05, size = 19, normalized size = 0.83 \begin {gather*} -\frac {i \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{2 \, c f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*a*e^(2*I*f*x + 2*I*e)/(c*f)

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Sympy [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (17) = 34\).
time = 0.08, size = 37, normalized size = 1.61 \begin {gather*} \begin {cases} - \frac {i a e^{2 i e} e^{2 i f x}}{2 c f} & \text {for}\: c f \neq 0 \\\frac {a x e^{2 i e}}{c} & \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))/(c-I*c*tan(f*x+e)),x)

[Out]

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

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Giac [A]
time = 0.45, size = 33, normalized size = 1.43 \begin {gather*} -\frac {2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{c f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.70, size = 19, normalized size = 0.83 \begin {gather*} \frac {a}{c\,f\,\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/(c*f*(tan(e + f*x) + 1i))

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