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

Optimal. Leaf size=65 \[ -\frac{i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-2 n} \text{Hypergeometric2F1}\left (1,-n,1-n,\frac{1}{2} (1-i \tan (c+d x))\right )}{2 d n} \]

[Out]

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

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Rubi [A]  time = 0.0781432, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3492, 3481, 68} \[ -\frac{i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-2 n} \text{Hypergeometric2F1}\left (1,-n,1-n,\frac{1}{2} (1-i \tan (c+d x))\right )}{2 d n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/2)*Hypergeometric2F1[1, -n, 1 - n, (1 - I*Tan[c + d*x])/2]*(a + I*a*Tan[c + d*x])^n)/(d*n*(e*Sec[c + d*x]
)^(2*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 (e \sec (c+d x))^{-2 n} (a+i a \tan (c+d x))^n \, dx &=\left ((e \sec (c+d x))^{-2 n} (a-i a \tan (c+d x))^n (a+i a \tan (c+d x))^n\right ) \int (a-i a \tan (c+d x))^{-n} \, dx\\ &=\frac{\left (i a (e \sec (c+d x))^{-2 n} (a-i a \tan (c+d x))^n (a+i a \tan (c+d x))^n\right ) \operatorname{Subst}\left (\int \frac{(a+x)^{-1-n}}{a-x} \, dx,x,-i a \tan (c+d x)\right )}{d}\\ &=-\frac{i \, _2F_1\left (1,-n;1-n;\frac{1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{-2 n} (a+i a \tan (c+d x))^n}{2 d n}\\ \end{align*}

Mathematica [B]  time = 1.65388, size = 146, normalized size = 2.25 \[ \frac{i 2^{-n-1} \left (1+e^{2 i (c+d x)}\right ) \left (e^{i d x}\right )^n \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-n} \sec ^n(c+d x) (\cos (d x)+i \sin (d x))^{-n} \text{Hypergeometric2F1}\left (1,n+1,n+2,1+e^{2 i (c+d x)}\right ) (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-2 n}}{d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.487, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{n}}{ \left ( e\sec \left ( dx+c \right ) \right ) ^{2\,n}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^n/((e*sec(d*x+c))^(2*n)),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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