3.1046 \(\int (a+i a \tan (e+f x)) (c-i c \tan (e+f x))^n \, dx\)

Optimal. Leaf size=26 \[ \frac{i a (c-i c \tan (e+f x))^n}{f n} \]

[Out]

(I*a*(c - I*c*Tan[e + f*x])^n)/(f*n)

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Rubi [A]  time = 0.082872, antiderivative size = 26, normalized size of antiderivative = 1., 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 = {3522, 3487, 32} \[ \frac{i a (c-i c \tan (e+f x))^n}{f n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*a*(c - I*c*Tan[e + f*x])^n)/(f*n)

Rule 3522

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]))

Rule 3487

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 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]

Rubi steps

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

Mathematica [A]  time = 0.730002, size = 51, normalized size = 1.96 \[ \frac{i a (c \sec (e+f x))^n \exp (n (-\log (c \sec (e+f x))+\log (c-i c \tan (e+f x))))}{f n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 25, normalized size = 1. \begin{align*}{\frac{ia \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{n}}{fn}} \end{align*}

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))^n,x)

[Out]

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

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Maxima [A]  time = 1.67797, size = 32, normalized size = 1.23 \begin{align*} \frac{i \, a c^{n}{\left (-i \, \tan \left (f x + e\right ) + 1\right )}^{n}}{f n} \end{align*}

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))^n,x, algorithm="maxima")

[Out]

I*a*c^n*(-I*tan(f*x + e) + 1)^n/(f*n)

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Fricas [A]  time = 1.67238, size = 61, normalized size = 2.35 \begin{align*} \frac{i \, a \left (\frac{2 \, c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n}}{f n} \end{align*}

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))^n,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 17.0957, size = 119, normalized size = 4.58 \begin{align*} \begin{cases} x \left (i a \tan{\left (e \right )} + a\right ) & \text{for}\: f = 0 \wedge n = 0 \\x \left (i a \tan{\left (e \right )} + a\right ) \left (- i c \tan{\left (e \right )} + c\right )^{n} & \text{for}\: f = 0 \\a x + \frac{i a \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} & \text{for}\: n = 0 \\\frac{i a \left (- i c \tan{\left (e + f x \right )} + c\right )^{n} \tan{\left (e + f x \right )}}{f n \tan{\left (e + f x \right )} + i f n} - \frac{a \left (- i c \tan{\left (e + f x \right )} + c\right )^{n}}{f n \tan{\left (e + f x \right )} + i f n} & \text{otherwise} \end{cases} \end{align*}

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))**n,x)

[Out]

Piecewise((x*(I*a*tan(e) + a), Eq(f, 0) & Eq(n, 0)), (x*(I*a*tan(e) + a)*(-I*c*tan(e) + c)**n, Eq(f, 0)), (a*x
+ I*a*log(tan(e + f*x)**2 + 1)/(2*f), Eq(n, 0)), (I*a*(-I*c*tan(e + f*x) + c)**n*tan(e + f*x)/(f*n*tan(e + f*
x) + I*f*n) - a*(-I*c*tan(e + f*x) + c)**n/(f*n*tan(e + f*x) + I*f*n), True))

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Giac [A]  time = 1.67333, size = 31, normalized size = 1.19 \begin{align*} \frac{i \,{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{n} a}{f n} \end{align*}

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))^n,x, algorithm="giac")

[Out]

I*(-I*c*tan(f*x + e) + c)^n*a/(f*n)