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

Optimal. Leaf size=65 \[ -\frac {i (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-2 n} \, _2F_1\left (1,-n;1-n;\frac {1}{2} (1-i \tan (c+d x))\right )}{2 d n} \]

[Out]

-1/2*I*hypergeom([1, -n],[1-n],1/2-1/2*I*tan(d*x+c))*(a+I*a*tan(d*x+c))^n/d/n/((e*sec(d*x+c))^(2*n))

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Rubi [A]  time = 0.08, antiderivative size = 65, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [B]  time = 1.78, size = 146, normalized size = 2.25 \[ \frac {i 2^{-n-1} \left (1+e^{2 i (c+d x)}\right ) \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-n} \, _2F_1\left (1,n+1;n+2;1+e^{2 i (c+d x)}\right ) \sec ^n(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n (e \sec (c+d x))^{-2 n}}{d (n+1)} \]

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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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e^{\left (i \, d n x + i \, c n + n \log \left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right ) + n \log \left (\frac {a}{e}\right )\right )}}{\left (\frac {2 \, e e^{\left (i \, d x + i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{2 \, n}}, x\right ) \]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\left (e \sec \left (d x + c\right )\right )^{2 \, n}}\,{d x} \]

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)

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maple [F]  time = 1.36, size = 0, normalized size = 0.00 \[ \int \left (a +i a \tan \left (d x +c \right )\right )^{n} \left (e \sec \left (d x +c \right )\right )^{-2 n}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\left (e \sec \left (d x + c\right )\right )^{2 \, n}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{2\,n}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^n/(e/cos(c + d*x))^(2*n),x)

[Out]

int((a + a*tan(c + d*x)*1i)^n/(e/cos(c + d*x))^(2*n), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \sec {\left (c + d x \right )}\right )^{- 2 n} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \]

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]

Integral((e*sec(c + d*x))**(-2*n)*(I*a*(tan(c + d*x) - I))**n, x)

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