3.471 \(\int \sec ^5(c+d x) (a+i a \tan (c+d x))^n \, dx\)

Optimal. Leaf size=94 \[ \frac{i a^2 2^{n+\frac{5}{2}} \sec ^5(c+d x) (1+i \tan (c+d x))^{-n-\frac{1}{2}} (a+i a \tan (c+d x))^{n-2} \text{Hypergeometric2F1}\left (\frac{5}{2},-n-\frac{3}{2},\frac{7}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{5 d} \]

[Out]

((I/5)*2^(5/2 + n)*a^2*Hypergeometric2F1[5/2, -3/2 - n, 7/2, (1 - I*Tan[c + d*x])/2]*Sec[c + d*x]^5*(1 + I*Tan
[c + d*x])^(-1/2 - n)*(a + I*a*Tan[c + d*x])^(-2 + n))/d

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Rubi [A]  time = 0.187376, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3505, 3523, 70, 69} \[ \frac{i a^2 2^{n+\frac{5}{2}} \sec ^5(c+d x) (1+i \tan (c+d x))^{-n-\frac{1}{2}} (a+i a \tan (c+d x))^{n-2} \text{Hypergeometric2F1}\left (\frac{5}{2},-n-\frac{3}{2},\frac{7}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + I*a*Tan[c + d*x])^n,x]

[Out]

((I/5)*2^(5/2 + n)*a^2*Hypergeometric2F1[5/2, -3/2 - n, 7/2, (1 - I*Tan[c + d*x])/2]*Sec[c + d*x]^5*(1 + I*Tan
[c + d*x])^(-1/2 - n)*(a + I*a*Tan[c + d*x])^(-2 + n))/d

Rule 3505

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 3523

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]

Rule 70

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 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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]))

Rubi steps

\begin{align*} \int \sec ^5(c+d x) (a+i a \tan (c+d x))^n \, dx &=\frac{\sec ^5(c+d x) \int (a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{\frac{5}{2}+n} \, dx}{(a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{5/2}}\\ &=\frac{\left (a^2 \sec ^5(c+d x)\right ) \operatorname{Subst}\left (\int (a-i a x)^{3/2} (a+i a x)^{\frac{3}{2}+n} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{5/2}}\\ &=\frac{\left (2^{\frac{3}{2}+n} a^3 \sec ^5(c+d x) (a+i a \tan (c+d x))^{-2+n} \left (\frac{a+i a \tan (c+d x)}{a}\right )^{-\frac{1}{2}-n}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{i x}{2}\right )^{\frac{3}{2}+n} (a-i a x)^{3/2} \, dx,x,\tan (c+d x)\right )}{d (a-i a \tan (c+d x))^{5/2}}\\ &=\frac{i 2^{\frac{5}{2}+n} a^2 \, _2F_1\left (\frac{5}{2},-\frac{3}{2}-n;\frac{7}{2};\frac{1}{2} (1-i \tan (c+d x))\right ) \sec ^5(c+d x) (1+i \tan (c+d x))^{-\frac{1}{2}-n} (a+i a \tan (c+d x))^{-2+n}}{5 d}\\ \end{align*}

Mathematica [A]  time = 13.9095, size = 149, normalized size = 1.59 \[ -\frac{i 2^{n+5} e^{5 i (c+d x)} \left (e^{i d x}\right )^n \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} \text{Hypergeometric2F1}\left (-\frac{3}{2},1,n+\frac{7}{2},-e^{2 i (c+d x)}\right ) (a+i a \tan (c+d x))^n}{d (2 n+5) \left (1+e^{2 i (c+d x)}\right )^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^5*(a + I*a*Tan[c + d*x])^n,x]

[Out]

((-I)*2^(5 + n)*E^((5*I)*(c + d*x))*(E^(I*d*x))^n*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^n*Hypergeometric
2F1[-3/2, 1, 7/2 + n, -E^((2*I)*(c + d*x))]*(a + I*a*Tan[c + d*x])^n)/(d*(1 + E^((2*I)*(c + d*x)))^4*(5 + 2*n)
*Sec[c + d*x]^n*(Cos[d*x] + I*Sin[d*x])^n)

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Maple [F]  time = 0.345, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{5} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^n,x)

[Out]

int(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^n,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(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)^5, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{32 \, \left (\frac{2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n} e^{\left (5 i \, d x + 5 i \, c\right )}}{e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

integral(32*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*e^(5*I*d*x + 5*I*c)/(e^(10*I*d*x + 10*I*c) +
5*e^(8*I*d*x + 8*I*c) + 10*e^(6*I*d*x + 6*I*c) + 10*e^(4*I*d*x + 4*I*c) + 5*e^(2*I*d*x + 2*I*c) + 1), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*(a+I*a*tan(d*x+c))**n,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \sec \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(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)^5, x)