3.569 \(\int \frac{\cos ^5(c+d x) \sin ^n(c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=68 \[ \frac{\sin ^{n+1}(c+d x)}{a^2 d (n+1)}-\frac{2 \sin ^{n+2}(c+d x)}{a^2 d (n+2)}+\frac{\sin ^{n+3}(c+d x)}{a^2 d (n+3)} \]

[Out]

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

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Rubi [A]  time = 0.127235, antiderivative size = 68, 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, 43} \[ \frac{\sin ^{n+1}(c+d x)}{a^2 d (n+1)}-\frac{2 \sin ^{n+2}(c+d x)}{a^2 d (n+2)}+\frac{\sin ^{n+3}(c+d x)}{a^2 d (n+3)} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^2,x]

[Out]

Sin[c + d*x]^(1 + n)/(a^2*d*(1 + n)) - (2*Sin[c + d*x]^(2 + n))/(a^2*d*(2 + n)) + Sin[c + d*x]^(3 + n)/(a^2*d*
(3 + 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.120364, size = 50, normalized size = 0.74 \[ \frac{\sin ^{n+1}(c+d x) \left (\frac{\sin ^2(c+d x)}{n+3}-\frac{2 \sin (c+d x)}{n+2}+\frac{1}{n+1}\right )}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^n)/(a + a*Sin[c + d*x])^2,x]

[Out]

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

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Maple [F]  time = 1.755, 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}}{ \left ( a+a\sin \left ( dx+c \right ) \right ) ^{2}}}\, 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))^2,x)

[Out]

int(cos(d*x+c)^5*sin(d*x+c)^n/(a+a*sin(d*x+c))^2,x)

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Maxima [A]  time = 1.25538, size = 109, normalized size = 1.6 \begin{align*} \frac{{\left ({\left (n^{2} + 3 \, n + 2\right )} \sin \left (d x + c\right )^{3} - 2 \,{\left (n^{2} + 4 \, n + 3\right )} \sin \left (d x + c\right )^{2} +{\left (n^{2} + 5 \, n + 6\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{{\left (n^{3} + 6 \, n^{2} + 11 \, n + 6\right )} a^{2} 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))^2,x, algorithm="maxima")

[Out]

((n^2 + 3*n + 2)*sin(d*x + c)^3 - 2*(n^2 + 4*n + 3)*sin(d*x + c)^2 + (n^2 + 5*n + 6)*sin(d*x + c))*sin(d*x + c
)^n/((n^3 + 6*n^2 + 11*n + 6)*a^2*d)

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Fricas [A]  time = 1.14455, size = 248, normalized size = 3.65 \begin{align*} \frac{{\left (2 \,{\left (n^{2} + 4 \, n + 3\right )} \cos \left (d x + c\right )^{2} - 2 \, n^{2} -{\left ({\left (n^{2} + 3 \, n + 2\right )} \cos \left (d x + c\right )^{2} - 2 \, n^{2} - 8 \, n - 8\right )} \sin \left (d x + c\right ) - 8 \, n - 6\right )} \sin \left (d x + c\right )^{n}}{a^{2} d n^{3} + 6 \, a^{2} d n^{2} + 11 \, a^{2} d n + 6 \, a^{2} 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))^2,x, algorithm="fricas")

[Out]

(2*(n^2 + 4*n + 3)*cos(d*x + c)^2 - 2*n^2 - ((n^2 + 3*n + 2)*cos(d*x + c)^2 - 2*n^2 - 8*n - 8)*sin(d*x + c) -
8*n - 6)*sin(d*x + c)^n/(a^2*d*n^3 + 6*a^2*d*n^2 + 11*a^2*d*n + 6*a^2*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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.35094, size = 93, normalized size = 1.37 \begin{align*} \frac{\frac{\sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3}}{n + 3} - \frac{2 \, \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^{2} 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))^2,x, algorithm="giac")

[Out]

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