3.568 \(\int \frac{\cos ^5(c+d x) \sin ^n(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=91 \[ \frac{\sin ^{n+1}(c+d x)}{a d (n+1)}-\frac{\sin ^{n+2}(c+d x)}{a d (n+2)}-\frac{\sin ^{n+3}(c+d x)}{a d (n+3)}+\frac{\sin ^{n+4}(c+d x)}{a d (n+4)} \]

[Out]

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

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Rubi [A]  time = 0.139479, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {2836, 75} \[ \frac{\sin ^{n+1}(c+d x)}{a d (n+1)}-\frac{\sin ^{n+2}(c+d x)}{a d (n+2)}-\frac{\sin ^{n+3}(c+d x)}{a d (n+3)}+\frac{\sin ^{n+4}(c+d x)}{a d (n+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2836

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*x)/b
)^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]

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

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

Mathematica [A]  time = 0.689956, size = 74, normalized size = 0.81 \[ \frac{\sin ^{n+1}(c+d x) \left (-\frac{(n+4) \sin ^2(c+d x)}{n+3}-\frac{(n+4) \sin (c+d x)}{n+2}+\sin ^3(c+d x)+\frac{n+4}{n+1}\right )}{a d (n+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.957, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{n}}{a+a\sin \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.35226, size = 167, normalized size = 1.84 \begin{align*} \frac{{\left ({\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} \sin \left (d x + c\right )^{4} -{\left (n^{3} + 7 \, n^{2} + 14 \, n + 8\right )} \sin \left (d x + c\right )^{3} -{\left (n^{3} + 8 \, n^{2} + 19 \, n + 12\right )} \sin \left (d x + c\right )^{2} +{\left (n^{3} + 9 \, n^{2} + 26 \, n + 24\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{{\left (n^{4} + 10 \, n^{3} + 35 \, n^{2} + 50 \, n + 24\right )} a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((n^3 + 6*n^2 + 11*n + 6)*sin(d*x + c)^4 - (n^3 + 7*n^2 + 14*n + 8)*sin(d*x + c)^3 - (n^3 + 8*n^2 + 19*n + 12)
*sin(d*x + c)^2 + (n^3 + 9*n^2 + 26*n + 24)*sin(d*x + c))*sin(d*x + c)^n/((n^4 + 10*n^3 + 35*n^2 + 50*n + 24)*
a*d)

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Fricas [A]  time = 1.14804, size = 332, normalized size = 3.65 \begin{align*} \frac{{\left ({\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} \cos \left (d x + c\right )^{4} -{\left (n^{3} + 4 \, n^{2} + 3 \, n\right )} \cos \left (d x + c\right )^{2} - 2 \, n^{2} +{\left ({\left (n^{3} + 7 \, n^{2} + 14 \, n + 8\right )} \cos \left (d x + c\right )^{2} + 2 \, n^{2} + 12 \, n + 16\right )} \sin \left (d x + c\right ) - 8 \, n - 6\right )} \sin \left (d x + c\right )^{n}}{a d n^{4} + 10 \, a d n^{3} + 35 \, a d n^{2} + 50 \, a d n + 24 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((n^3 + 6*n^2 + 11*n + 6)*cos(d*x + c)^4 - (n^3 + 4*n^2 + 3*n)*cos(d*x + c)^2 - 2*n^2 + ((n^3 + 7*n^2 + 14*n +
8)*cos(d*x + c)^2 + 2*n^2 + 12*n + 16)*sin(d*x + c) - 8*n - 6)*sin(d*x + c)^n/(a*d*n^4 + 10*a*d*n^3 + 35*a*d*
n^2 + 50*a*d*n + 24*a*d)

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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*sin(d*x+c)**n/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.33454, size = 124, normalized size = 1.36 \begin{align*} \frac{\frac{\sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{4}}{n + 4} - \frac{\sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3}}{n + 3} - \frac{\sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2}}{n + 2} + \frac{\sin \left (d x + c\right )^{n + 1}}{n + 1}}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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