\(\int (e \sec (c+d x))^{-2-2 n} (a+i a \tan (c+d x))^n \, dx\) [499]

Optimal result

Integrand size = 30, antiderivative size = 74 \[ \int (e \sec (c+d x))^{-2-2 n} (a+i a \tan (c+d x))^n \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (2,-1-n,-n,\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{-2 (1+n)} (a+i a \tan (c+d x))^{1+n}}{4 a d (1+n)} \]

[Out]

-1/4*I*hypergeom([2, -1-n],[-n],1/2-1/2*I*tan(d*x+c))*(a+I*a*tan(d*x+c))^(1+n)/a/d/(1+n)/((e*sec(d*x+c))^(2+2*
n))

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3586, 3604, 7, 70} \[ \int (e \sec (c+d x))^{-2-2 n} (a+i a \tan (c+d x))^n \, dx=-\frac {i (a+i a \tan (c+d x))^{n+1} (e \sec (c+d x))^{-2 (n+1)} \operatorname {Hypergeometric2F1}\left (2,-n-1,-n,\frac {1}{2} (1-i \tan (c+d x))\right )}{4 a d (n+1)} \]

[In]

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

[Out]

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

Rule 7

Int[(u_.)*(Px_)^(p_), x_Symbol] :> Int[u*Px^Simplify[p], x] /; PolyQ[Px, x] &&  !RationalQ[p] && FreeQ[p, x] &
& RationalQ[Simplify[p]]

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3586

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 3604

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

Rubi steps \begin{align*} \text {integral}& = \left ((e \sec (c+d x))^{-2-2 n} (a-i a \tan (c+d x))^{\frac {1}{2} (2+2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (2+2 n)}\right ) \int (a-i a \tan (c+d x))^{\frac {1}{2} (-2-2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-2-2 n)+n} \, dx \\ & = \frac {\left (a^2 (e \sec (c+d x))^{-2-2 n} (a-i a \tan (c+d x))^{\frac {1}{2} (2+2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (2+2 n)}\right ) \text {Subst}\left (\int (a-i a x)^{-1+\frac {1}{2} (-2-2 n)} (a+i a x)^{-1+\frac {1}{2} (-2-2 n)+n} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\left (a^2 (e \sec (c+d x))^{-2-2 n} (a-i a \tan (c+d x))^{\frac {1}{2} (2+2 n)} (a+i a \tan (c+d x))^{\frac {1}{2} (2+2 n)}\right ) \text {Subst}\left (\int \frac {(a-i a x)^{-1+\frac {1}{2} (-2-2 n)}}{(a+i a x)^2} \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {i \operatorname {Hypergeometric2F1}\left (2,-1-n,-n,\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{-2 (1+n)} (a+i a \tan (c+d x))^{1+n}}{4 a d (1+n)} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(151\) vs. \(2(74)=148\).

Time = 14.69 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.04 \[ \int (e \sec (c+d x))^{-2-2 n} (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{-3-n} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-n} \left (1+e^{2 i (c+d x)}\right )^3 \operatorname {Hypergeometric2F1}\left (2,3+n,4+n,1+e^{2 i (c+d x)}\right ) \sec ^n(c+d x) (e \sec (c+d x))^{-2 n} (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d e^2 (3+n)} \]

[In]

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

[Out]

((-I)*2^(-3 - n)*(E^(I*d*x))^n*(1 + E^((2*I)*(c + d*x)))^3*Hypergeometric2F1[2, 3 + n, 4 + n, 1 + E^((2*I)*(c
+ d*x))]*Sec[c + d*x]^n*(a + I*a*Tan[c + d*x])^n)/(d*e^2*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^n*(3 + n)
*(e*Sec[c + d*x])^(2*n)*(Cos[d*x] + I*Sin[d*x])^n)

Maple [F]

\[\int \left (e \sec \left (d x +c \right )\right )^{-2-2 n} \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (e \sec (c+d x))^{-2-2 n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-2 \, n - 2} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int (e \sec (c+d x))^{-2-2 n} (a+i a \tan (c+d x))^n \, dx=\int \left (e \sec {\left (c + d x \right )}\right )^{- 2 n - 2} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int (e \sec (c+d x))^{-2-2 n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-2 \, n - 2} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (e \sec (c+d x))^{-2-2 n} (a+i a \tan (c+d x))^n \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{-2 \, n - 2} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (e \sec (c+d x))^{-2-2 n} (a+i a \tan (c+d x))^n \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{2\,n+2}} \,d x \]

[In]

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

[Out]

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