\(\int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-1-n} \, dx\) [502]

Optimal result

Integrand size = 32, antiderivative size = 66 \[ \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-1-n} \, dx=\frac {i \operatorname {Hypergeometric2F1}\left (2,n,1+n,\frac {1}{2} (1-i \tan (e+f x))\right ) (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-n}}{4 a f n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3586, 3603, 3568, 70} \[ \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-1-n} \, dx=\frac {i (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n} \operatorname {Hypergeometric2F1}\left (2,n,n+1,\frac {1}{2} (1-i \tan (e+f x))\right )}{4 a f n} \]

[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 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 3568

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 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 3603

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]))

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

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(165\) vs. \(2(66)=132\).

Time = 12.19 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.50 \[ \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-1-n} \, dx=\frac {i 2^{-2+n} e^{i e} \left (e^{i f x}\right )^{-n} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^n \left (1+e^{2 i (e+f x)}\right )^2 \operatorname {Hypergeometric2F1}\left (2,2-n,3-n,1+e^{2 i (e+f x)}\right ) \sec ^{1-n}(e+f x) (d \sec (e+f x))^{2 n} (\cos (f x)+i \sin (f x))^{1+n} (a+i a \tan (e+f x))^{-1-n}}{f (-2+n)} \]

[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))

Maple [F]

\[\int \left (d \sec \left (f x +e \right )\right )^{2 n} \left (a +i a \tan \left (f x +e \right )\right )^{-1-n}d x\]

[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)

Fricas [F]

\[ \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-1-n} \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{2 \, n} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{-n - 1} \,d x } \]

[In]

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

Sympy [F]

\[ \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-1-n} \, dx=\int \left (d \sec {\left (e + f x \right )}\right )^{2 n} \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{- n - 1}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-1-n} \, dx=\text {Exception raised: RuntimeError} \]

[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 >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

\[ \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-1-n} \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{2 \, n} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{-n - 1} \,d x } \]

[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)

Mupad [F(-1)]

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

[In]

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

[Out]

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