\(\int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx\) [571]

Optimal result

Integrand size = 29, antiderivative size = 88 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\sin ^{1+n}(c+d x)}{a^4 d (1+n)}-\frac {4 \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{a^4 d}+\frac {4 \sin ^{1+n}(c+d x)}{d \left (a^4+a^4 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2915, 91, 81, 66} \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {4 \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}(1,n+1,n+2,-\sin (c+d x))}{a^4 d}+\frac {\sin ^{n+1}(c+d x)}{a^4 d (n+1)}+\frac {4 \sin ^{n+1}(c+d x)}{d \left (a^4 \sin (c+d x)+a^4\right )} \]

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^4,x]

[Out]

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

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 81

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

Rule 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 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)^2} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {4 \sin ^{1+n}(c+d x)}{d \left (a^4+a^4 \sin (c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {(a (3+4 n)-x) \left (\frac {x}{a}\right )^n}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\sin ^{1+n}(c+d x)}{a^4 d (1+n)}+\frac {4 \sin ^{1+n}(c+d x)}{d \left (a^4+a^4 \sin (c+d x)\right )}-\frac {(4 (1+n)) \text {Subst}\left (\int \frac {\left (\frac {x}{a}\right )^n}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^4 d} \\ & = \frac {\sin ^{1+n}(c+d x)}{a^4 d (1+n)}-\frac {4 \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) \sin ^{1+n}(c+d x)}{a^4 d}+\frac {4 \sin ^{1+n}(c+d x)}{d \left (a^4+a^4 \sin (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\sin ^{1+n}(c+d x) (5+4 n+\sin (c+d x)-4 (1+n) \operatorname {Hypergeometric2F1}(1,1+n,2+n,-\sin (c+d x)) (1+\sin (c+d x)))}{a^4 d (1+n) (1+\sin (c+d x))} \]

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^4,x]

[Out]

(Sin[c + d*x]^(1 + n)*(5 + 4*n + Sin[c + d*x] - 4*(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, -Sin[c + d*x]]*(1
+ Sin[c + d*x])))/(a^4*d*(1 + n)*(1 + Sin[c + d*x]))

Maple [F]

\[\int \frac {\left (\cos ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{\left (a +a \sin \left (d x +c \right )\right )^{4}}d x\]

[In]

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x)

[Out]

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x)

Fricas [F]

\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

integral(sin(d*x + c)^n*cos(d*x + c)^5/(a^4*cos(d*x + c)^4 - 8*a^4*cos(d*x + c)^2 + 8*a^4 - 4*(a^4*cos(d*x + c
)^2 - 2*a^4)*sin(d*x + c)), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**5*sin(d*x+c)**n/(a+a*sin(d*x+c))**4,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)^n*cos(d*x + c)^5/(a*sin(d*x + c) + a)^4, x)

Giac [F]

\[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int { \frac {\sin \left (d x + c\right )^{n} \cos \left (d x + c\right )^{5}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{4}} \,d x } \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

integrate(sin(d*x + c)^n*cos(d*x + c)^5/(a*sin(d*x + c) + a)^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^n}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^4} \,d x \]

[In]

int((cos(c + d*x)^5*sin(c + d*x)^n)/(a + a*sin(c + d*x))^4,x)

[Out]

int((cos(c + d*x)^5*sin(c + d*x)^n)/(a + a*sin(c + d*x))^4, x)