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

Optimal. Leaf size=63 \[ \frac{i (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n} \text{Hypergeometric2F1}\left (1,n,n+1,\frac{1}{2} (1-i \tan (e+f x))\right )}{2 f n} \]

[Out]

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

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Rubi [A]  time = 0.0744578, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3492, 3481, 68} \[ \frac{i (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n} \text{Hypergeometric2F1}\left (1,n,n+1,\frac{1}{2} (1-i \tan (e+f x))\right )}{2 f n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3492

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[((a/d)
^(2*IntPart[n])*(a + b*Tan[e + f*x])^FracPart[n]*(a - b*Tan[e + f*x])^FracPart[n])/(d*Sec[e + f*x])^(2*FracPar
t[n]), Int[1/(a - b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] && EqQ[Sim
plify[m/2 + n], 0]

Rule 3481

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Dist[b/d, Subst[Int[(a + x)^(n - 1)/(a - x), x]
, x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

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))^{-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 (a-i a \tan (e+f x))^n \, dx\\ &=\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} \, dx,x,-i a \tan (e+f x)\right )}{f}\\ &=\frac{i \, _2F_1\left (1,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}}{2 f n}\\ \end{align*}

Mathematica [B]  time = 1.05455, size = 150, normalized size = 2.38 \[ -\frac{i 2^{n-1} \left (1+e^{2 i (e+f x)}\right ) \left (e^{i f x}\right )^{-n} \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^n \sec ^{-n}(e+f x) (\cos (f x)+i \sin (f x))^n \text{Hypergeometric2F1}\left (1,1-n,2-n,1+e^{2 i (e+f x)}\right ) (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n}}{f (n-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.486, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d\sec \left ( fx+e \right ) \right ) ^{2\,n}}{ \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{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))^n),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \sec \left (f x + e\right )\right )^{2 \, n}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{n}}\,{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))^n),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\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}}{\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}}, 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))^n),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**n),x)

[Out]

Timed out

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

[Out]

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