3.502 \(\int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-1-n} \, dx\)

Optimal. Leaf size=66 \[ \frac{i (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n} \text{Hypergeometric2F1}\left (2,n,n+1,\frac{1}{2} (1-i \tan (e+f x))\right )}{4 a f n} \]

[Out]

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

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Rubi [A]  time = 0.182586, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3505, 3522, 3487, 68} \[ \frac{i (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n} \text{Hypergeometric2F1}\left (2,n,n+1,\frac{1}{2} (1-i \tan (e+f x))\right )}{4 a f n} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(2*n)*(a + I*a*Tan[e + f*x])^(-1 - n),x]

[Out]

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

Rule 3505

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*S
ec[e + f*x])^m/((a + b*Tan[e + f*x])^(m/2)*(a - b*Tan[e + f*x])^(m/2)), Int[(a + b*Tan[e + f*x])^(m/2 + n)*(a
- b*Tan[e + f*x])^(m/2), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]

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

Mathematica [B]  time = 13.9642, size = 165, normalized size = 2.5 \[ \frac{i e^{i e} 2^{n-2} \left (1+e^{2 i (e+f x)}\right )^2 \left (e^{i f x}\right )^{-n} \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^n \sec ^{1-n}(e+f x) (\cos (f x)+i \sin (f x))^{n+1} \text{Hypergeometric2F1}\left (2,2-n,3-n,1+e^{2 i (e+f x)}\right ) (a+i a \tan (e+f x))^{-n-1} (d \sec (e+f x))^{2 n}}{f (n-2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Sec[e + f*x])^(2*n)*(a + I*a*Tan[e + f*x])^(-1 - n),x]

[Out]

(I*2^(-2 + n)*E^(I*e)*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^n*(1 + E^((2*I)*(e + f*x)))^2*Hypergeometric
2F1[2, 2 - n, 3 - n, 1 + E^((2*I)*(e + f*x))]*Sec[e + f*x]^(1 - n)*(d*Sec[e + f*x])^(2*n)*(Cos[f*x] + I*Sin[f*
x])^(1 + n)*(a + I*a*Tan[e + f*x])^(-1 - n))/((E^(I*f*x))^n*f*(-2 + n))

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Maple [F]  time = 1.344, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{2\,n} \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{-1-n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(2*n)*(a+I*a*tan(f*x+e))^(-1-n),x)

[Out]

int((d*sec(f*x+e))^(2*n)*(a+I*a*tan(f*x+e))^(-1-n),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((d*sec(f*x+e))^(2*n)*(a+I*a*tan(f*x+e))^(-1-n),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 (\left (\frac{2 \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{-n - 1} \left (\frac{2 \, d e^{\left (i \, f x + i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{2 \, n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(2*n)*(a+I*a*tan(f*x+e))^(-1-n),x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{2 \, n}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{-n - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(2*n)*(a+I*a*tan(f*x+e))^(-1-n),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(2*n)*(I*a*tan(f*x + e) + a)^(-n - 1), x)