\(\int (e \sec (c+d x))^{2-n} (a+i a \tan (c+d x))^n \, dx\) [489]

Optimal result

Integrand size = 30, antiderivative size = 113 \[ \int (e \sec (c+d x))^{2-n} (a+i a \tan (c+d x))^n \, dx=\frac {i 2^{1+\frac {n}{2}} a \operatorname {Hypergeometric2F1}\left (\frac {2-n}{2},-\frac {n}{2},\frac {4-n}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{2-n} (1+i \tan (c+d x))^{-n/2} (a+i a \tan (c+d x))^{-1+n}}{d (2-n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 113, 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, 72, 71} \[ \int (e \sec (c+d x))^{2-n} (a+i a \tan (c+d x))^n \, dx=\frac {i a 2^{\frac {n}{2}+1} (1+i \tan (c+d x))^{-n/2} (a+i a \tan (c+d x))^{n-1} (e \sec (c+d x))^{2-n} \operatorname {Hypergeometric2F1}\left (\frac {2-n}{2},-\frac {n}{2},\frac {4-n}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d (2-n)} \]

[In]

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

[Out]

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

Rule 71

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

Rule 72

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

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

Mathematica [A] (verified)

Time = 14.18 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.99 \[ \int (e \sec (c+d x))^{2-n} (a+i a \tan (c+d x))^n \, dx=\frac {4 e^2 \operatorname {Hypergeometric2F1}\left (2,1-\frac {n}{2},2-\frac {n}{2},-\cos (2 (c+d x))+i \sin (2 (c+d x))\right ) (e \sec (c+d x))^{-n} (\cos (2 c)-i \sin (2 c)) (i+\tan (d x)) (a+i a \tan (c+d x))^n}{d (-2+n) (-1-i \tan (d x))} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

4*(4*a^n*e^2*cos(d*n*x + c*n) + 4*I*a^n*e^2*sin(d*n*x + c*n) - (a^n*e^2*n - 4*a^n*e^2)*cos(c*n + (d*n + 2*d)*x
+ 2*c) - 4*(I*a^n*d*e^(n + 2)*n^3 - 6*I*a^n*d*e^(n + 2)*n^2 + 8*I*a^n*d*e^(n + 2)*n + (I*a^n*d*e^(n + 2)*n^3
- 6*I*a^n*d*e^(n + 2)*n^2 + 8*I*a^n*d*e^(n + 2)*n)*cos(4*d*x + 4*c) + 2*(I*a^n*d*e^(n + 2)*n^3 - 6*I*a^n*d*e^(
n + 2)*n^2 + 8*I*a^n*d*e^(n + 2)*n)*cos(2*d*x + 2*c) - (a^n*d*e^(n + 2)*n^3 - 6*a^n*d*e^(n + 2)*n^2 + 8*a^n*d*
e^(n + 2)*n)*sin(4*d*x + 4*c) - 2*(a^n*d*e^(n + 2)*n^3 - 6*a^n*d*e^(n + 2)*n^2 + 8*a^n*d*e^(n + 2)*n)*sin(2*d*
x + 2*c))*integrate(((cos(6*d*x + 6*c) + 3*cos(4*d*x + 4*c) + 3*cos(2*d*x + 2*c) + 1)*cos(d*n*x + c*n) + (sin(
6*d*x + 6*c) + 3*sin(4*d*x + 4*c) + 3*sin(2*d*x + 2*c))*sin(d*n*x + c*n))/(e^n*n^2 + (e^n*n^2 - 6*e^n*n + 8*e^
n)*cos(6*d*x + 6*c)^2 + 9*(e^n*n^2 - 6*e^n*n + 8*e^n)*cos(4*d*x + 4*c)^2 + 9*(e^n*n^2 - 6*e^n*n + 8*e^n)*cos(2
*d*x + 2*c)^2 + (e^n*n^2 - 6*e^n*n + 8*e^n)*sin(6*d*x + 6*c)^2 + 9*(e^n*n^2 - 6*e^n*n + 8*e^n)*sin(4*d*x + 4*c
)^2 + 18*(e^n*n^2 - 6*e^n*n + 8*e^n)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*(e^n*n^2 - 6*e^n*n + 8*e^n)*sin(2*d
*x + 2*c)^2 - 6*e^n*n + 2*(e^n*n^2 - 6*e^n*n + 3*(e^n*n^2 - 6*e^n*n + 8*e^n)*cos(4*d*x + 4*c) + 3*(e^n*n^2 - 6
*e^n*n + 8*e^n)*cos(2*d*x + 2*c) + 8*e^n)*cos(6*d*x + 6*c) + 6*(e^n*n^2 - 6*e^n*n + 3*(e^n*n^2 - 6*e^n*n + 8*e
^n)*cos(2*d*x + 2*c) + 8*e^n)*cos(4*d*x + 4*c) + 6*(e^n*n^2 - 6*e^n*n + 8*e^n)*cos(2*d*x + 2*c) + 6*((e^n*n^2
- 6*e^n*n + 8*e^n)*sin(4*d*x + 4*c) + (e^n*n^2 - 6*e^n*n + 8*e^n)*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 8*e^n),
x) + 4*(a^n*d*e^(n + 2)*n^3 - 6*a^n*d*e^(n + 2)*n^2 + 8*a^n*d*e^(n + 2)*n + (a^n*d*e^(n + 2)*n^3 - 6*a^n*d*e^
(n + 2)*n^2 + 8*a^n*d*e^(n + 2)*n)*cos(4*d*x + 4*c) + 2*(a^n*d*e^(n + 2)*n^3 - 6*a^n*d*e^(n + 2)*n^2 + 8*a^n*d
*e^(n + 2)*n)*cos(2*d*x + 2*c) - (-I*a^n*d*e^(n + 2)*n^3 + 6*I*a^n*d*e^(n + 2)*n^2 - 8*I*a^n*d*e^(n + 2)*n)*si
n(4*d*x + 4*c) - 2*(-I*a^n*d*e^(n + 2)*n^3 + 6*I*a^n*d*e^(n + 2)*n^2 - 8*I*a^n*d*e^(n + 2)*n)*sin(2*d*x + 2*c)
)*integrate(-((sin(6*d*x + 6*c) + 3*sin(4*d*x + 4*c) + 3*sin(2*d*x + 2*c))*cos(d*n*x + c*n) - (cos(6*d*x + 6*c
) + 3*cos(4*d*x + 4*c) + 3*cos(2*d*x + 2*c) + 1)*sin(d*n*x + c*n))/(e^n*n^2 + (e^n*n^2 - 6*e^n*n + 8*e^n)*cos(
6*d*x + 6*c)^2 + 9*(e^n*n^2 - 6*e^n*n + 8*e^n)*cos(4*d*x + 4*c)^2 + 9*(e^n*n^2 - 6*e^n*n + 8*e^n)*cos(2*d*x +
2*c)^2 + (e^n*n^2 - 6*e^n*n + 8*e^n)*sin(6*d*x + 6*c)^2 + 9*(e^n*n^2 - 6*e^n*n + 8*e^n)*sin(4*d*x + 4*c)^2 + 1
8*(e^n*n^2 - 6*e^n*n + 8*e^n)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 9*(e^n*n^2 - 6*e^n*n + 8*e^n)*sin(2*d*x + 2*
c)^2 - 6*e^n*n + 2*(e^n*n^2 - 6*e^n*n + 3*(e^n*n^2 - 6*e^n*n + 8*e^n)*cos(4*d*x + 4*c) + 3*(e^n*n^2 - 6*e^n*n
+ 8*e^n)*cos(2*d*x + 2*c) + 8*e^n)*cos(6*d*x + 6*c) + 6*(e^n*n^2 - 6*e^n*n + 3*(e^n*n^2 - 6*e^n*n + 8*e^n)*cos
(2*d*x + 2*c) + 8*e^n)*cos(4*d*x + 4*c) + 6*(e^n*n^2 - 6*e^n*n + 8*e^n)*cos(2*d*x + 2*c) + 6*((e^n*n^2 - 6*e^n
*n + 8*e^n)*sin(4*d*x + 4*c) + (e^n*n^2 - 6*e^n*n + 8*e^n)*sin(2*d*x + 2*c))*sin(6*d*x + 6*c) + 8*e^n), x) - (
I*a^n*e^2*n - 4*I*a^n*e^2)*sin(c*n + (d*n + 2*d)*x + 2*c))/(-I*d*e^n*n^2 + 6*I*d*e^n*n - 8*I*d*e^n + (-I*d*e^n
*n^2 + 6*I*d*e^n*n - 8*I*d*e^n)*cos(4*d*x + 4*c) - 2*(I*d*e^n*n^2 - 6*I*d*e^n*n + 8*I*d*e^n)*cos(2*d*x + 2*c)
+ (d*e^n*n^2 - 6*d*e^n*n + 8*d*e^n)*sin(4*d*x + 4*c) + 2*(d*e^n*n^2 - 6*d*e^n*n + 8*d*e^n)*sin(2*d*x + 2*c))

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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