3.1047 \(\int \frac{(c-i c \tan (e+f x))^n}{a+i a \tan (e+f x)} \, dx\)

Optimal. Leaf size=52 \[ \frac{i (c-i c \tan (e+f x))^n \, _2F_1\left (2,n;n+1;\frac{1}{2} (1-i \tan (e+f x))\right )}{4 a f n} \]

[Out]

((I/4)*Hypergeometric2F1[2, n, 1 + n, (1 - I*Tan[e + f*x])/2]*(c - I*c*Tan[e + f*x])^n)/(a*f*n)

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Rubi [A]  time = 0.124293, antiderivative size = 52, normalized size of antiderivative = 1., 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 = {3522, 3487, 68} \[ \frac{i (c-i c \tan (e+f x))^n \, _2F_1\left (2,n;n+1;\frac{1}{2} (1-i \tan (e+f x))\right )}{4 a f n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/4)*Hypergeometric2F1[2, n, 1 + n, (1 - I*Tan[e + f*x])/2]*(c - I*c*Tan[e + f*x])^n)/(a*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 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{(c-i c \tan (e+f x))^n}{a+i a \tan (e+f x)} \, dx &=\frac{\int \cos ^2(e+f x) (c-i c \tan (e+f x))^{1+n} \, dx}{a c}\\ &=\frac{\left (i c^2\right ) \operatorname{Subst}\left (\int \frac{(c+x)^{-1+n}}{(c-x)^2} \, dx,x,-i c \tan (e+f x)\right )}{a f}\\ &=\frac{i \, _2F_1\left (2,n;1+n;\frac{1}{2} (1-i \tan (e+f x))\right ) (c-i c \tan (e+f x))^n}{4 a f n}\\ \end{align*}

Mathematica [A]  time = 57.6802, size = 79, normalized size = 1.52 \[ \frac{i 2^{n-2} \left (1+e^{2 i (e+f x)}\right )^2 \left (\frac{c}{1+e^{2 i (e+f x)}}\right )^n \, _2F_1\left (2,2-n;3-n;1+e^{2 i (e+f x)}\right )}{a f (n-2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*2^(-2 + n)*(c/(1 + E^((2*I)*(e + f*x))))^n*(1 + E^((2*I)*(e + f*x)))^2*Hypergeometric2F1[2, 2 - n, 3 - n, 1
+ E^((2*I)*(e + f*x))])/(a*f*(-2 + n))

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Maple [F]  time = 0.527, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{n}}{a+ia\tan \left ( fx+e \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (\frac{2 \, c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n}{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{n}}{i \, a \tan \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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