\(\int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx\) [472]

Optimal result

Integrand size = 24, antiderivative size = 92 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx=\frac {i 2^{\frac {3}{2}+n} a \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {1}{2}-n,\frac {5}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) \sec ^3(c+d x) (1+i \tan (c+d x))^{-\frac {1}{2}-n} (a+i a \tan (c+d x))^{-1+n}}{3 d} \]

[Out]

1/3*I*2^(3/2+n)*a*hypergeom([3/2, -1/2-n],[5/2],1/2-1/2*I*tan(d*x+c))*sec(d*x+c)^3*(1+I*tan(d*x+c))^(-1/2-n)*(
a+I*a*tan(d*x+c))^(-1+n)/d

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3586, 3604, 72, 71} \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx=\frac {i a 2^{n+\frac {3}{2}} \sec ^3(c+d x) (1+i \tan (c+d x))^{-n-\frac {1}{2}} (a+i a \tan (c+d x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-n-\frac {1}{2},\frac {5}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{3 d} \]

[In]

Int[Sec[c + d*x]^3*(a + I*a*Tan[c + d*x])^n,x]

[Out]

((I/3)*2^(3/2 + n)*a*Hypergeometric2F1[3/2, -1/2 - n, 5/2, (1 - I*Tan[c + d*x])/2]*Sec[c + d*x]^3*(1 + I*Tan[c
+ d*x])^(-1/2 - n)*(a + I*a*Tan[c + d*x])^(-1 + n))/d

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}& = \frac {\sec ^3(c+d x) \int (a-i a \tan (c+d x))^{3/2} (a+i a \tan (c+d x))^{\frac {3}{2}+n} \, dx}{(a-i a \tan (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}} \\ & = \frac {\left (a^2 \sec ^3(c+d x)\right ) \text {Subst}\left (\int \sqrt {a-i a x} (a+i a x)^{\frac {1}{2}+n} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}} \\ & = \frac {\left (2^{\frac {1}{2}+n} a^2 \sec ^3(c+d x) (a+i a \tan (c+d x))^{-1+n} \left (\frac {a+i a \tan (c+d x)}{a}\right )^{-\frac {1}{2}-n}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{\frac {1}{2}+n} \sqrt {a-i a x} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{3/2}} \\ & = \frac {i 2^{\frac {3}{2}+n} a \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {1}{2}-n,\frac {5}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) \sec ^3(c+d x) (1+i \tan (c+d x))^{-\frac {1}{2}-n} (a+i a \tan (c+d x))^{-1+n}}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 13.91 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.62 \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx=-\frac {i 2^{3+n} e^{3 i (c+d x)} \left (e^{i d x}\right )^n \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {5}{2}+n,-e^{2 i (c+d x)}\right ) \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} (a+i a \tan (c+d x))^n}{d \left (1+e^{2 i (c+d x)}\right )^2 (3+2 n)} \]

[In]

Integrate[Sec[c + d*x]^3*(a + I*a*Tan[c + d*x])^n,x]

[Out]

((-I)*2^(3 + n)*E^((3*I)*(c + d*x))*(E^(I*d*x))^n*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^n*Hypergeometric
2F1[-1/2, 1, 5/2 + n, -E^((2*I)*(c + d*x))]*(a + I*a*Tan[c + d*x])^n)/(d*(1 + E^((2*I)*(c + d*x)))^2*(3 + 2*n)
*Sec[c + d*x]^n*(Cos[d*x] + I*Sin[d*x])^n)

Maple [F]

\[\int \left (\sec ^{3}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]

[In]

int(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^n,x)

[Out]

int(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^n,x)

Fricas [F]

\[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

integral(8*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*e^(3*I*d*x + 3*I*c)/(e^(6*I*d*x + 6*I*c) + 3*
e^(4*I*d*x + 4*I*c) + 3*e^(2*I*d*x + 2*I*c) + 1), x)

Sympy [F]

\[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \sec ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**3*(a+I*a*tan(d*x+c))**n,x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**n*sec(c + d*x)**3, x)

Maxima [F]

\[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((I*a*tan(d*x + c) + a)^n*sec(d*x + c)^3, x)

Giac [F]

\[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(sec(d*x+c)^3*(a+I*a*tan(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^n*sec(d*x + c)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \sec ^3(c+d x) (a+i a \tan (c+d x))^n \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\cos \left (c+d\,x\right )}^3} \,d x \]

[In]

int((a + a*tan(c + d*x)*1i)^n/cos(c + d*x)^3,x)

[Out]

int((a + a*tan(c + d*x)*1i)^n/cos(c + d*x)^3, x)