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

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

[Out]

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

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Rubi [A]  time = 0.193491, 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} \[ -\frac{i 2^{n-\frac{5}{2}} \cos ^5(c+d x) (1+i \tan (c+d x))^{\frac{1}{2}-n} (a+i a \tan (c+d x))^{n+2} \text{Hypergeometric2F1}\left (-\frac{5}{2},\frac{7}{2}-n,-\frac{3}{2},\frac{1}{2} (1-i \tan (c+d x))\right )}{5 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/5)*2^(-5/2 + n)*Cos[c + d*x]^5*Hypergeometric2F1[-5/2, 7/2 - n, -3/2, (1 - I*Tan[c + d*x])/2]*(1 + I*Tan[
c + d*x])^(1/2 - n)*(a + I*a*Tan[c + d*x])^(2 + n))/(a^2*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 \cos ^5(c+d x) (a+i a \tan (c+d x))^n \, dx &=\left (\cos ^5(c+d x) (a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{5/2}\right ) \int \frac{(a+i a \tan (c+d x))^{-\frac{5}{2}+n}}{(a-i a \tan (c+d x))^{5/2}} \, dx\\ &=\frac{\left (a^2 \cos ^5(c+d x) (a-i a \tan (c+d x))^{5/2} (a+i a \tan (c+d x))^{5/2}\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{-\frac{7}{2}+n}}{(a-i a x)^{7/2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (2^{-\frac{7}{2}+n} \cos ^5(c+d x) (a-i a \tan (c+d x))^{5/2} (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 \frac{\left (\frac{1}{2}+\frac{i x}{2}\right )^{-\frac{7}{2}+n}}{(a-i a x)^{7/2}} \, dx,x,\tan (c+d x)\right )}{a d}\\ &=-\frac{i 2^{-\frac{5}{2}+n} \cos ^5(c+d x) \, _2F_1\left (-\frac{5}{2},\frac{7}{2}-n;-\frac{3}{2};\frac{1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac{1}{2}-n} (a+i a \tan (c+d x))^{2+n}}{5 a^2 d}\\ \end{align*}

Mathematica [A]  time = 6.01809, size = 149, normalized size = 1.59 \[ -\frac{i 2^{n-5} e^{-5 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^6 \left (e^{i d x}\right )^n \left (\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \sec ^{-n}(c+d x) (\cos (d x)+i \sin (d x))^{-n} \text{Hypergeometric2F1}\left (1,\frac{7}{2},n-\frac{3}{2},-e^{2 i (c+d x)}\right ) (a+i a \tan (c+d x))^n}{d (2 n-5)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*2^(-5 + n)*(E^(I*d*x))^n*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^n*(1 + E^((2*I)*(c + d*x)))^6*Hyper
geometric2F1[1, 7/2, -3/2 + n, -E^((2*I)*(c + d*x))]*(a + I*a*Tan[c + d*x])^n)/(d*E^((5*I)*(c + d*x))*(-5 + 2*
n)*Sec[c + d*x]^n*(Cos[d*x] + I*Sin[d*x])^n)

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Maple [F]  time = 0.734, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \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(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^n,x)

[Out]

int(cos(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} \cos \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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*cos(d*x + c)^5, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{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}{\left (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\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/32*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*(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)*e^(-5*I*d*x - 5*I*c), 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(cos(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} \cos \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(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*cos(d*x + c)^5, x)