3.5.76 \(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^n \, dx\) [476]

Optimal. Leaf size=94 \[ -\frac {i 2^{-\frac {5}{2}+n} \cos ^5(c+d x) \, _2F_1\left (-\frac {5}{2},\frac {7}{2}-n;-\frac {3}{2};\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{\frac {1}{2}-n} (a+i a \tan (c+d x))^{2+n}}{5 a^2 d} \]

[Out]

-1/5*I*2^(-5/2+n)*cos(d*x+c)^5*hypergeom([-5/2, 7/2-n],[-3/2],1/2-1/2*I*tan(d*x+c))*(1+I*tan(d*x+c))^(1/2-n)*(
a+I*a*tan(d*x+c))^(2+n)/a^2/d

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Rubi [A]
time = 0.14, antiderivative size = 94, normalized size of antiderivative = 1.00, 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 = {3586, 3604, 72, 71} \begin {gather*} -\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} \, _2F_1\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/5*I)*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*Ta
n[c + d*x])^(1/2 - n)*(a + I*a*Tan[c + d*x])^(2 + n))/(a^2*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*} \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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 10.80, size = 149, normalized size = 1.59 \begin {gather*} -\frac {i 2^{-5+n} 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 \left (1+e^{2 i (c+d x)}\right )^6 \, _2F_1\left (1,\frac {7}{2};-\frac {3}{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 (-5+2 n)} \end {gather*}

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 = 0.99, size = 0, normalized size = 0.00 \[\int \left (\cos ^{5}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^5\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^n,x)

[Out]

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^n, x)

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