\(\int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-n} \, dx\) [503]

Optimal result

Integrand size = 30, antiderivative size = 63 \[ \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-n} \, dx=\frac {i \operatorname {Hypergeometric2F1}\left (1,n,1+n,\frac {1}{2} (1-i \tan (e+f x))\right ) (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-n}}{2 f n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3573, 3562, 70} \[ \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-n} \, dx=\frac {i (a+i a \tan (e+f x))^{-n} (d \sec (e+f x))^{2 n} \operatorname {Hypergeometric2F1}\left (1,n,n+1,\frac {1}{2} (1-i \tan (e+f x))\right )}{2 f n} \]

[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 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 3562

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 3573

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*FracPart
[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*} \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 (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 ) \text {Subst}\left (\int \frac {(a+x)^{-1+n}}{a-x} \, dx,x,-i a \tan (e+f x)\right )}{f} \\ & = \frac {i \operatorname {Hypergeometric2F1}\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] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(150\) vs. \(2(63)=126\).

Time = 2.49 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.38 \[ \int (d \sec (e+f x))^{2 n} (a+i a \tan (e+f x))^{-n} \, dx=-\frac {i 2^{-1+n} \left (e^{i f x}\right )^{-n} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^n \left (1+e^{2 i (e+f x)}\right ) \operatorname {Hypergeometric2F1}\left (1,1-n,2-n,1+e^{2 i (e+f x)}\right ) \sec ^{-n}(e+f x) (d \sec (e+f x))^{2 n} (\cos (f x)+i \sin (f x))^n (a+i a \tan (e+f x))^{-n}}{f (-1+n)} \]

[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)

Maple [F]

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

[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)

Fricas [F]

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

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[In]

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

[Out]

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