3.566 \(\int \cos ^5(c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=160 \[ \frac{a^2 \sin ^{n+1}(c+d x)}{d (n+1)}+\frac{2 a^2 \sin ^{n+2}(c+d x)}{d (n+2)}-\frac{a^2 \sin ^{n+3}(c+d x)}{d (n+3)}-\frac{4 a^2 \sin ^{n+4}(c+d x)}{d (n+4)}-\frac{a^2 \sin ^{n+5}(c+d x)}{d (n+5)}+\frac{2 a^2 \sin ^{n+6}(c+d x)}{d (n+6)}+\frac{a^2 \sin ^{n+7}(c+d x)}{d (n+7)} \]

[Out]

(a^2*Sin[c + d*x]^(1 + n))/(d*(1 + n)) + (2*a^2*Sin[c + d*x]^(2 + n))/(d*(2 + n)) - (a^2*Sin[c + d*x]^(3 + n))
/(d*(3 + n)) - (4*a^2*Sin[c + d*x]^(4 + n))/(d*(4 + n)) - (a^2*Sin[c + d*x]^(5 + n))/(d*(5 + n)) + (2*a^2*Sin[
c + d*x]^(6 + n))/(d*(6 + n)) + (a^2*Sin[c + d*x]^(7 + n))/(d*(7 + n))

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Rubi [A]  time = 0.166758, antiderivative size = 160, 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, 88} \[ \frac{a^2 \sin ^{n+1}(c+d x)}{d (n+1)}+\frac{2 a^2 \sin ^{n+2}(c+d x)}{d (n+2)}-\frac{a^2 \sin ^{n+3}(c+d x)}{d (n+3)}-\frac{4 a^2 \sin ^{n+4}(c+d x)}{d (n+4)}-\frac{a^2 \sin ^{n+5}(c+d x)}{d (n+5)}+\frac{2 a^2 \sin ^{n+6}(c+d x)}{d (n+6)}+\frac{a^2 \sin ^{n+7}(c+d x)}{d (n+7)} \]

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]

(a^2*Sin[c + d*x]^(1 + n))/(d*(1 + n)) + (2*a^2*Sin[c + d*x]^(2 + n))/(d*(2 + n)) - (a^2*Sin[c + d*x]^(3 + n))
/(d*(3 + n)) - (4*a^2*Sin[c + d*x]^(4 + n))/(d*(4 + n)) - (a^2*Sin[c + d*x]^(5 + n))/(d*(5 + n)) + (2*a^2*Sin[
c + d*x]^(6 + n))/(d*(6 + n)) + (a^2*Sin[c + d*x]^(7 + n))/(d*(7 + 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 88

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]))

Rubi steps

\begin{align*} \int \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 (a+x)^4 \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^6 \left (\frac{x}{a}\right )^n+2 a^6 \left (\frac{x}{a}\right )^{1+n}-a^6 \left (\frac{x}{a}\right )^{2+n}-4 a^6 \left (\frac{x}{a}\right )^{3+n}-a^6 \left (\frac{x}{a}\right )^{4+n}+2 a^6 \left (\frac{x}{a}\right )^{5+n}+a^6 \left (\frac{x}{a}\right )^{6+n}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{a^2 \sin ^{1+n}(c+d x)}{d (1+n)}+\frac{2 a^2 \sin ^{2+n}(c+d x)}{d (2+n)}-\frac{a^2 \sin ^{3+n}(c+d x)}{d (3+n)}-\frac{4 a^2 \sin ^{4+n}(c+d x)}{d (4+n)}-\frac{a^2 \sin ^{5+n}(c+d x)}{d (5+n)}+\frac{2 a^2 \sin ^{6+n}(c+d x)}{d (6+n)}+\frac{a^2 \sin ^{7+n}(c+d x)}{d (7+n)}\\ \end{align*}

Mathematica [A]  time = 0.389313, size = 110, normalized size = 0.69 \[ \frac{a^2 \sin ^{n+1}(c+d x) \left (\frac{\sin ^6(c+d x)}{n+7}+\frac{2 \sin ^5(c+d x)}{n+6}-\frac{\sin ^4(c+d x)}{n+5}-\frac{4 \sin ^3(c+d x)}{n+4}-\frac{\sin ^2(c+d x)}{n+3}+\frac{2 \sin (c+d x)}{n+2}+\frac{1}{n+1}\right )}{d} \]

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]

(a^2*Sin[c + d*x]^(1 + n)*((1 + n)^(-1) + (2*Sin[c + d*x])/(2 + n) - Sin[c + d*x]^2/(3 + n) - (4*Sin[c + d*x]^
3)/(4 + n) - Sin[c + d*x]^4/(5 + n) + (2*Sin[c + d*x]^5)/(6 + n) + Sin[c + d*x]^6/(7 + n)))/d

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Maple [F]  time = 7.947, size = 0, normalized size = 0. \begin{align*} \int \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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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]

Exception raised: ValueError

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Fricas [B]  time = 1.41732, size = 1143, normalized size = 7.14 \begin{align*} -\frac{{\left (2 \,{\left (a^{2} n^{6} + 22 \, a^{2} n^{5} + 190 \, a^{2} n^{4} + 820 \, a^{2} n^{3} + 1849 \, a^{2} n^{2} + 2038 \, a^{2} n + 840 \, a^{2}\right )} \cos \left (d x + c\right )^{6} - 16 \, a^{2} n^{4} - 256 \, a^{2} n^{3} - 2 \,{\left (a^{2} n^{6} + 18 \, a^{2} n^{5} + 118 \, a^{2} n^{4} + 348 \, a^{2} n^{3} + 457 \, a^{2} n^{2} + 210 \, a^{2} n\right )} \cos \left (d x + c\right )^{4} - 1376 \, a^{2} n^{2} - 2816 \, a^{2} n - 8 \,{\left (a^{2} n^{5} + 16 \, a^{2} n^{4} + 86 \, a^{2} n^{3} + 176 \, a^{2} n^{2} + 105 \, a^{2} n\right )} \cos \left (d x + c\right )^{2} - 1680 \, a^{2} +{\left ({\left (a^{2} n^{6} + 21 \, a^{2} n^{5} + 175 \, a^{2} n^{4} + 735 \, a^{2} n^{3} + 1624 \, a^{2} n^{2} + 1764 \, a^{2} n + 720 \, a^{2}\right )} \cos \left (d x + c\right )^{6} - 16 \, a^{2} n^{4} - 256 \, a^{2} n^{3} - 2 \,{\left (a^{2} n^{6} + 20 \, a^{2} n^{5} + 159 \, a^{2} n^{4} + 640 \, a^{2} n^{3} + 1364 \, a^{2} n^{2} + 1440 \, a^{2} n + 576 \, a^{2}\right )} \cos \left (d x + c\right )^{4} - 1472 \, a^{2} n^{2} - 3584 \, a^{2} n - 8 \,{\left (a^{2} n^{5} + 17 \, a^{2} n^{4} + 108 \, a^{2} n^{3} + 316 \, a^{2} n^{2} + 416 \, a^{2} n + 192 \, a^{2}\right )} \cos \left (d x + c\right )^{2} - 3072 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{d n^{7} + 28 \, d n^{6} + 322 \, d n^{5} + 1960 \, d n^{4} + 6769 \, d n^{3} + 13132 \, d n^{2} + 13068 \, d n + 5040 \, 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*(a^2*n^6 + 22*a^2*n^5 + 190*a^2*n^4 + 820*a^2*n^3 + 1849*a^2*n^2 + 2038*a^2*n + 840*a^2)*cos(d*x + c)^6 -
16*a^2*n^4 - 256*a^2*n^3 - 2*(a^2*n^6 + 18*a^2*n^5 + 118*a^2*n^4 + 348*a^2*n^3 + 457*a^2*n^2 + 210*a^2*n)*cos(
d*x + c)^4 - 1376*a^2*n^2 - 2816*a^2*n - 8*(a^2*n^5 + 16*a^2*n^4 + 86*a^2*n^3 + 176*a^2*n^2 + 105*a^2*n)*cos(d
*x + c)^2 - 1680*a^2 + ((a^2*n^6 + 21*a^2*n^5 + 175*a^2*n^4 + 735*a^2*n^3 + 1624*a^2*n^2 + 1764*a^2*n + 720*a^
2)*cos(d*x + c)^6 - 16*a^2*n^4 - 256*a^2*n^3 - 2*(a^2*n^6 + 20*a^2*n^5 + 159*a^2*n^4 + 640*a^2*n^3 + 1364*a^2*
n^2 + 1440*a^2*n + 576*a^2)*cos(d*x + c)^4 - 1472*a^2*n^2 - 3584*a^2*n - 8*(a^2*n^5 + 17*a^2*n^4 + 108*a^2*n^3
+ 316*a^2*n^2 + 416*a^2*n + 192*a^2)*cos(d*x + c)^2 - 3072*a^2)*sin(d*x + c))*sin(d*x + c)^n/(d*n^7 + 28*d*n^
6 + 322*d*n^5 + 1960*d*n^4 + 6769*d*n^3 + 13132*d*n^2 + 13068*d*n + 5040*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 [B]  time = 1.25401, size = 778, normalized size = 4.86 \begin{align*} \text{result too large to display} \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]

((n^2*sin(d*x + c)^n*sin(d*x + c)^7 + 8*n*sin(d*x + c)^n*sin(d*x + c)^7 - 2*n^2*sin(d*x + c)^n*sin(d*x + c)^5
+ 15*sin(d*x + c)^n*sin(d*x + c)^7 - 20*n*sin(d*x + c)^n*sin(d*x + c)^5 + n^2*sin(d*x + c)^n*sin(d*x + c)^3 -
42*sin(d*x + c)^n*sin(d*x + c)^5 + 12*n*sin(d*x + c)^n*sin(d*x + c)^3 + 35*sin(d*x + c)^n*sin(d*x + c)^3)*a^2/
(n^3 + 15*n^2 + 71*n + 105) + 2*(n^2*sin(d*x + c)^n*sin(d*x + c)^6 + 6*n*sin(d*x + c)^n*sin(d*x + c)^6 - 2*n^2
*sin(d*x + c)^n*sin(d*x + c)^4 + 8*sin(d*x + c)^n*sin(d*x + c)^6 - 16*n*sin(d*x + c)^n*sin(d*x + c)^4 + n^2*si
n(d*x + c)^n*sin(d*x + c)^2 - 24*sin(d*x + c)^n*sin(d*x + c)^4 + 10*n*sin(d*x + c)^n*sin(d*x + c)^2 + 24*sin(d
*x + c)^n*sin(d*x + c)^2)*a^2/(n^3 + 12*n^2 + 44*n + 48) + (n^2*sin(d*x + c)^n*sin(d*x + c)^5 + 4*n*sin(d*x +
c)^n*sin(d*x + c)^5 - 2*n^2*sin(d*x + c)^n*sin(d*x + c)^3 + 3*sin(d*x + c)^n*sin(d*x + c)^5 - 12*n*sin(d*x + c
)^n*sin(d*x + c)^3 + n^2*sin(d*x + c)^n*sin(d*x + c) - 10*sin(d*x + c)^n*sin(d*x + c)^3 + 8*n*sin(d*x + c)^n*s
in(d*x + c) + 15*sin(d*x + c)^n*sin(d*x + c))*a^2/(n^3 + 9*n^2 + 23*n + 15))/d