\(\int \cos ^4(c+d x) (a+i a \tan (c+d x))^n \, dx\) [469]

Optimal result

Integrand size = 24, antiderivative size = 60 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^n \, dx=\frac {i a^2 \operatorname {Hypergeometric2F1}\left (3,-2+n,-1+n,\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^{-2+n}}{8 d (2-n)} \]

[Out]

1/8*I*a^2*hypergeom([3, -2+n],[-1+n],1/2+1/2*I*tan(d*x+c))*(a+I*a*tan(d*x+c))^(-2+n)/d/(2-n)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 70} \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^n \, dx=\frac {i a^2 (a+i a \tan (c+d x))^{n-2} \operatorname {Hypergeometric2F1}\left (3,n-2,n-1,\frac {1}{2} (i \tan (c+d x)+1)\right )}{8 d (2-n)} \]

[In]

Int[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^n,x]

[Out]

((I/8)*a^2*Hypergeometric2F1[3, -2 + n, -1 + n, (1 + I*Tan[c + d*x])/2]*(a + I*a*Tan[c + d*x])^(-2 + n))/(d*(2
- n))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (i a^5\right ) \text {Subst}\left (\int \frac {(a+x)^{-3+n}}{(a-x)^3} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = \frac {i a^2 \operatorname {Hypergeometric2F1}\left (3,-2+n,-1+n,\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^{-2+n}}{8 d (2-n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.97 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^n \, dx=-\frac {i a^2 \operatorname {Hypergeometric2F1}\left (3,-2+n,-1+n,\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^{-2+n}}{8 d (-2+n)} \]

[In]

Integrate[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^n,x]

[Out]

((-1/8*I)*a^2*Hypergeometric2F1[3, -2 + n, -1 + n, (1 + I*Tan[c + d*x])/2]*(a + I*a*Tan[c + d*x])^(-2 + n))/(d
*(-2 + n))

Maple [F]

\[\int \left (\cos ^{4}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{n}d x\]

[In]

int(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^n,x)

[Out]

int(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^n,x)

Fricas [F]

\[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

integral(1/16*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*(e^(8*I*d*x + 8*I*c) + 4*e^(6*I*d*x + 6*I*
c) + 6*e^(4*I*d*x + 4*I*c) + 4*e^(2*I*d*x + 2*I*c) + 1)*e^(-4*I*d*x - 4*I*c), x)

Sympy [F]

\[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^n \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \cos ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(cos(d*x+c)**4*(a+I*a*tan(d*x+c))**n,x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**n*cos(c + d*x)**4, x)

Maxima [F]

\[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cos(d*x+c)^4*(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)^4, x)

Giac [F]

\[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^n \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(cos(d*x+c)^4*(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)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^n \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]

[In]

int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^n,x)

[Out]

int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^n, x)