\(\int (e \sec (c+d x))^{1-n} (a+i a \tan (c+d x))^n \, dx\) [488]

Optimal result

Integrand size = 30, antiderivative size = 118 \[ \int (e \sec (c+d x))^{1-n} (a+i a \tan (c+d x))^n \, dx=\frac {i 2^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-n}{2},\frac {1-n}{2},\frac {3-n}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{1-n} (1+i \tan (c+d x))^{\frac {1}{2} (-1-n)} (a+i a \tan (c+d x))^n}{d (1-n)} \]

[Out]

I*2^(1/2+1/2*n)*hypergeom([1/2-1/2*n, 1/2-1/2*n],[3/2-1/2*n],1/2-1/2*I*tan(d*x+c))*(e*sec(d*x+c))^(1-n)*(1+I*t
an(d*x+c))^(-1/2-1/2*n)*(a+I*a*tan(d*x+c))^n/d/(1-n)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 118, 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} \[ \int (e \sec (c+d x))^{1-n} (a+i a \tan (c+d x))^n \, dx=\frac {i 2^{\frac {n+1}{2}} (1+i \tan (c+d x))^{\frac {1}{2} (-n-1)} (a+i a \tan (c+d x))^n (e \sec (c+d x))^{1-n} \operatorname {Hypergeometric2F1}\left (\frac {1-n}{2},\frac {1-n}{2},\frac {3-n}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{d (1-n)} \]

[In]

Int[(e*Sec[c + d*x])^(1 - n)*(a + I*a*Tan[c + d*x])^n,x]

[Out]

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

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))^{1-n} (a-i a \tan (c+d x))^{\frac {1}{2} (-1+n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-1+n)}\right ) \int (a-i a \tan (c+d x))^{\frac {1-n}{2}} (a+i a \tan (c+d x))^{\frac {1-n}{2}+n} \, dx \\ & = \frac {\left (a^2 (e \sec (c+d x))^{1-n} (a-i a \tan (c+d x))^{\frac {1}{2} (-1+n)} (a+i a \tan (c+d x))^{\frac {1}{2} (-1+n)}\right ) \text {Subst}\left (\int (a-i a x)^{-1+\frac {1-n}{2}} (a+i a x)^{-1+\frac {1-n}{2}+n} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\left (2^{-\frac {1}{2}+\frac {n}{2}} a (e \sec (c+d x))^{1-n} (a-i a \tan (c+d x))^{\frac {1}{2} (-1+n)} (a+i a \tan (c+d x))^{\frac {1}{2}+\frac {1}{2} (-1+n)+\frac {n}{2}} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{-\frac {1}{2}-\frac {n}{2}}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{-1+\frac {1-n}{2}+n} (a-i a x)^{-1+\frac {1-n}{2}} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {i 2^{\frac {1+n}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1-n}{2},\frac {1-n}{2},\frac {3-n}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^{1-n} (1+i \tan (c+d x))^{\frac {1}{2} (-1-n)} (a+i a \tan (c+d x))^n}{d (1-n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.84 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.74 \[ \int (e \sec (c+d x))^{1-n} (a+i a \tan (c+d x))^n \, dx=-\frac {e (\operatorname {Hypergeometric2F1}(1,n,1+n,i \cos (c+d x)-\sin (c+d x))-\operatorname {Hypergeometric2F1}(1,n,1+n,-i \cos (c+d x)+\sin (c+d x))) (e \sec (c+d x))^{-n} (a+i a \tan (c+d x))^n}{d n} \]

[In]

Integrate[(e*Sec[c + d*x])^(1 - n)*(a + I*a*Tan[c + d*x])^n,x]

[Out]

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

Maple [F]

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

[In]

int((e*sec(d*x+c))^(1-n)*(a+I*a*tan(d*x+c))^n,x)

[Out]

int((e*sec(d*x+c))^(1-n)*(a+I*a*tan(d*x+c))^n,x)

Fricas [F]

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

[In]

integrate((e*sec(d*x+c))^(1-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))^(-n + 1)*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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate((e*sec(d*x+c))^(1-n)*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

-2*(a^n*e*cos(c*n + (d*n + d)*x + c) + I*a^n*e*sin(c*n + (d*n + d)*x + c) - 2*(I*a^n*d*e^(n + 1)*n - I*a^n*d*e
^(n + 1) + (I*a^n*d*e^(n + 1)*n - I*a^n*d*e^(n + 1))*cos(2*d*x + 2*c) - (a^n*d*e^(n + 1)*n - a^n*d*e^(n + 1))*
sin(2*d*x + 2*c))*integrate(((cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) + 1)*cos(c*n + (d*n + d)*x + c) + (sin(4*d
*x + 4*c) + 2*sin(2*d*x + 2*c))*sin(c*n + (d*n + d)*x + c))/((e^n*n - e^n)*cos(4*d*x + 4*c)^2 + 4*(e^n*n - e^n
)*cos(2*d*x + 2*c)^2 + (e^n*n - e^n)*sin(4*d*x + 4*c)^2 + 4*(e^n*n - e^n)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) +
4*(e^n*n - e^n)*sin(2*d*x + 2*c)^2 + e^n*n + 2*(e^n*n + 2*(e^n*n - e^n)*cos(2*d*x + 2*c) - e^n)*cos(4*d*x + 4*
c) + 4*(e^n*n - e^n)*cos(2*d*x + 2*c) - e^n), x) + 2*(a^n*d*e^(n + 1)*n - a^n*d*e^(n + 1) + (a^n*d*e^(n + 1)*n
- a^n*d*e^(n + 1))*cos(2*d*x + 2*c) - (-I*a^n*d*e^(n + 1)*n + I*a^n*d*e^(n + 1))*sin(2*d*x + 2*c))*integrate(
-((sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(c*n + (d*n + d)*x + c) - (cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c)
+ 1)*sin(c*n + (d*n + d)*x + c))/((e^n*n - e^n)*cos(4*d*x + 4*c)^2 + 4*(e^n*n - e^n)*cos(2*d*x + 2*c)^2 + (e^n
*n - e^n)*sin(4*d*x + 4*c)^2 + 4*(e^n*n - e^n)*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*(e^n*n - e^n)*sin(2*d*x +
2*c)^2 + e^n*n + 2*(e^n*n + 2*(e^n*n - e^n)*cos(2*d*x + 2*c) - e^n)*cos(4*d*x + 4*c) + 4*(e^n*n - e^n)*cos(2*
d*x + 2*c) - e^n), x))/(-I*d*e^n*n + I*d*e^n + (-I*d*e^n*n + I*d*e^n)*cos(2*d*x + 2*c) + (d*e^n*n - d*e^n)*sin
(2*d*x + 2*c))

Giac [F]

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

[In]

integrate((e*sec(d*x+c))^(1-n)*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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