3.10.34 \(\int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^2} \, dx\) [934]

Optimal. Leaf size=25 \[ -\frac {i a}{2 f (c-i c \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 25, 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}{2 f (c-i c \tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(51\) vs. \(2(25)=50\).
time = 0.40, size = 51, normalized size = 2.04 \begin {gather*} \frac {a (3 \cos (e+f x)-i \sin (e+f x)) (-i \cos (3 (e+f x))+\sin (3 (e+f x)))}{8 c^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(3*Cos[e + f*x] - I*Sin[e + f*x])*((-I)*Cos[3*(e + f*x)] + Sin[3*(e + f*x)]))/(8*c^2*f)

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Maple [A]
time = 0.17, size = 22, normalized size = 0.88

method	result	size
derivativedivides	\(\frac {i a}{2 f \,c^{2} \left (\tan \left (f x +e \right )+i\right )^{2}}\)	\(22\)
default	\(\frac {i a}{2 f \,c^{2} \left (\tan \left (f x +e \right )+i\right )^{2}}\)	\(22\)
risch	\(-\frac {i a \,{\mathrm e}^{4 i \left (f x +e \right )}}{8 c^{2} f}-\frac {i a \,{\mathrm e}^{2 i \left (f x +e \right )}}{4 c^{2} f}\)	\(40\)
norman	\(\frac {\frac {a \tan \left (f x +e \right )}{c f}-\frac {i a}{2 c f}+\frac {i a \left (\tan ^{2}\left (f x +e \right )\right )}{2 c f}}{c \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}\)	\(60\)

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))^2,x,method=_RETURNVERBOSE)

[Out]

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

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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))^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 = 1.04, size = 35, normalized size = 1.40 \begin {gather*} \frac {-i \, a e^{\left (4 i \, f x + 4 i \, e\right )} - 2 i \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{8 \, c^{2} 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))^2,x, algorithm="fricas")

[Out]

1/8*(-I*a*e^(4*I*f*x + 4*I*e) - 2*I*a*e^(2*I*f*x + 2*I*e))/(c^2*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. 88 vs. \(2 (20) = 40\).
time = 0.14, size = 88, normalized size = 3.52 \begin {gather*} \begin {cases} \frac {- 4 i a c^{2} f e^{4 i e} e^{4 i f x} - 8 i a c^{2} f e^{2 i e} e^{2 i f x}}{32 c^{4} f^{2}} & \text {for}\: c^{4} f^{2} \neq 0 \\\frac {x \left (a e^{4 i e} + a e^{2 i e}\right )}{2 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))/(c-I*c*tan(f*x+e))**2,x)

[Out]

Piecewise(((-4*I*a*c**2*f*exp(4*I*e)*exp(4*I*f*x) - 8*I*a*c**2*f*exp(2*I*e)*exp(2*I*f*x))/(32*c**4*f**2), Ne(c
**4*f**2, 0)), (x*(a*exp(4*I*e) + a*exp(2*I*e))/(2*c**2), True))

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (20) = 40\).
time = 0.51, size = 65, normalized size = 2.60 \begin {gather*} -\frac {2 \, {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + i \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a \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))/(c-I*c*tan(f*x+e))^2,x, algorithm="giac")

[Out]

-2*(a*tan(1/2*f*x + 1/2*e)^3 + I*a*tan(1/2*f*x + 1/2*e)^2 - a*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 = 4.61, size = 21, normalized size = 0.84 \begin {gather*} \frac {a\,1{}\mathrm {i}}{2\,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)/(c - c*tan(e + f*x)*1i)^2,x)

[Out]

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

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