\(\int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [570]

Optimal result

Integrand size = 29, antiderivative size = 85 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {3 \sin ^{1+n}(c+d x)}{a^3 d (1+n)}+\frac {4 \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{a^3 d (1+n)}+\frac {\sin ^{2+n}(c+d x)}{a^3 d (2+n)} \]

[Out]

-3*sin(d*x+c)^(1+n)/a^3/d/(1+n)+4*hypergeom([1, 1+n],[2+n],-sin(d*x+c))*sin(d*x+c)^(1+n)/a^3/d/(1+n)+sin(d*x+c
)^(2+n)/a^3/d/(2+n)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 90, 66} \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(1,n+1,n+2,-\sin (c+d x))}{a^3 d (n+1)}-\frac {3 \sin ^{n+1}(c+d x)}{a^3 d (n+1)}+\frac {\sin ^{n+2}(c+d x)}{a^3 d (n+2)} \]

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^3,x]

[Out]

(-3*Sin[c + d*x]^(1 + n))/(a^3*d*(1 + n)) + (4*Hypergeometric2F1[1, 1 + n, 2 + n, -Sin[c + d*x]]*Sin[c + d*x]^
(1 + n))/(a^3*d*(1 + n)) + Sin[c + d*x]^(2 + n)/(a^3*d*(2 + n))

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x)^2 \left (\frac {x}{a}\right )^n}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \left (-3 a \left (\frac {x}{a}\right )^n+a \left (\frac {x}{a}\right )^{1+n}+\frac {4 a^2 \left (\frac {x}{a}\right )^n}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = -\frac {3 \sin ^{1+n}(c+d x)}{a^3 d (1+n)}+\frac {\sin ^{2+n}(c+d x)}{a^3 d (2+n)}+\frac {4 \text {Subst}\left (\int \frac {\left (\frac {x}{a}\right )^n}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = -\frac {3 \sin ^{1+n}(c+d x)}{a^3 d (1+n)}+\frac {4 \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{a^3 d (1+n)}+\frac {\sin ^{2+n}(c+d x)}{a^3 d (2+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.75 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sin ^{1+n}(c+d x) (-3 (2+n)+4 (2+n) \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x))+(1+n) \sin (c+d x))}{a^3 d (1+n) (2+n)} \]

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^3,x]

[Out]

(Sin[c + d*x]^(1 + n)*(-3*(2 + n) + 4*(2 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, -Sin[c + d*x]] + (1 + n)*Sin[
c + d*x]))/(a^3*d*(1 + n)*(2 + n))

Maple [F]

\[\int \frac {\left (\cos ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{\left (a +a \sin \left (d x +c \right )\right )^{3}}d x\]

[In]

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x)

[Out]

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x)

Fricas [F]

\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(-sin(d*x + c)^n*cos(d*x + c)^5/(3*a^3*cos(d*x + c)^2 - 4*a^3 + (a^3*cos(d*x + c)^2 - 4*a^3)*sin(d*x +
c)), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*sin(d*x+c)**n/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)^n*cos(d*x + c)^5/(a*sin(d*x + c) + a)^3, x)

Giac [F]

\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^n*cos(d*x + c)^5/(a*sin(d*x + c) + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^n}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^3} \,d x \]

[In]

int((cos(c + d*x)^5*sin(c + d*x)^n)/(a + a*sin(c + d*x))^3,x)

[Out]

int((cos(c + d*x)^5*sin(c + d*x)^n)/(a + a*sin(c + d*x))^3, x)