3.511 \(\int \cos ^5(c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=127 \[ \frac{a^2 \sin ^{10}(c+d x)}{10 d}+\frac{2 a^2 \sin ^9(c+d x)}{9 d}-\frac{a^2 \sin ^8(c+d x)}{8 d}-\frac{4 a^2 \sin ^7(c+d x)}{7 d}-\frac{a^2 \sin ^6(c+d x)}{6 d}+\frac{2 a^2 \sin ^5(c+d x)}{5 d}+\frac{a^2 \sin ^4(c+d x)}{4 d} \]

[Out]

(a^2*Sin[c + d*x]^4)/(4*d) + (2*a^2*Sin[c + d*x]^5)/(5*d) - (a^2*Sin[c + d*x]^6)/(6*d) - (4*a^2*Sin[c + d*x]^7
)/(7*d) - (a^2*Sin[c + d*x]^8)/(8*d) + (2*a^2*Sin[c + d*x]^9)/(9*d) + (a^2*Sin[c + d*x]^10)/(10*d)

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Rubi [A]  time = 0.125219, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2836, 12, 88} \[ \frac{a^2 \sin ^{10}(c+d x)}{10 d}+\frac{2 a^2 \sin ^9(c+d x)}{9 d}-\frac{a^2 \sin ^8(c+d x)}{8 d}-\frac{4 a^2 \sin ^7(c+d x)}{7 d}-\frac{a^2 \sin ^6(c+d x)}{6 d}+\frac{2 a^2 \sin ^5(c+d x)}{5 d}+\frac{a^2 \sin ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*Sin[c + d*x]^4)/(4*d) + (2*a^2*Sin[c + d*x]^5)/(5*d) - (a^2*Sin[c + d*x]^6)/(6*d) - (4*a^2*Sin[c + d*x]^7
)/(7*d) - (a^2*Sin[c + d*x]^8)/(8*d) + (2*a^2*Sin[c + d*x]^9)/(9*d) + (a^2*Sin[c + d*x]^10)/(10*d)

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 ^3(c+d x) (a+a \sin (c+d x))^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a-x)^2 x^3 (a+x)^4}{a^3} \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int (a-x)^2 x^3 (a+x)^4 \, dx,x,a \sin (c+d x)\right )}{a^8 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^6 x^3+2 a^5 x^4-a^4 x^5-4 a^3 x^6-a^2 x^7+2 a x^8+x^9\right ) \, dx,x,a \sin (c+d x)\right )}{a^8 d}\\ &=\frac{a^2 \sin ^4(c+d x)}{4 d}+\frac{2 a^2 \sin ^5(c+d x)}{5 d}-\frac{a^2 \sin ^6(c+d x)}{6 d}-\frac{4 a^2 \sin ^7(c+d x)}{7 d}-\frac{a^2 \sin ^8(c+d x)}{8 d}+\frac{2 a^2 \sin ^9(c+d x)}{9 d}+\frac{a^2 \sin ^{10}(c+d x)}{10 d}\\ \end{align*}

Mathematica [A]  time = 0.795818, size = 110, normalized size = 0.87 \[ -\frac{a^2 (-15120 \sin (c+d x)+3360 \sin (3 (c+d x))+2016 \sin (5 (c+d x))-360 \sin (7 (c+d x))-280 \sin (9 (c+d x))+10710 \cos (2 (c+d x))+1260 \cos (4 (c+d x))-1365 \cos (6 (c+d x))-315 \cos (8 (c+d x))+63 \cos (10 (c+d x))-2625)}{322560 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*Sin[c + d*x]^3*(a + a*Sin[c + d*x])^2,x]

[Out]

-(a^2*(-2625 + 10710*Cos[2*(c + d*x)] + 1260*Cos[4*(c + d*x)] - 1365*Cos[6*(c + d*x)] - 315*Cos[8*(c + d*x)] +
63*Cos[10*(c + d*x)] - 15120*Sin[c + d*x] + 3360*Sin[3*(c + d*x)] + 2016*Sin[5*(c + d*x)] - 360*Sin[7*(c + d*
x)] - 280*Sin[9*(c + d*x)]))/(322560*d)

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Maple [A]  time = 0.04, size = 158, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{10}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{20}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{60}} \right ) +2\,{a}^{2} \left ( -1/9\, \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{6}-1/21\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{\frac{ \left ( 8/3+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{105}} \right ) +{a}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{8}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{24}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x)

[Out]

1/d*(a^2*(-1/10*sin(d*x+c)^4*cos(d*x+c)^6-1/20*sin(d*x+c)^2*cos(d*x+c)^6-1/60*cos(d*x+c)^6)+2*a^2*(-1/9*sin(d*
x+c)^3*cos(d*x+c)^6-1/21*sin(d*x+c)*cos(d*x+c)^6+1/105*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))+a^2*(-1
/8*sin(d*x+c)^2*cos(d*x+c)^6-1/24*cos(d*x+c)^6))

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Maxima [A]  time = 1.17302, size = 131, normalized size = 1.03 \begin{align*} \frac{252 \, a^{2} \sin \left (d x + c\right )^{10} + 560 \, a^{2} \sin \left (d x + c\right )^{9} - 315 \, a^{2} \sin \left (d x + c\right )^{8} - 1440 \, a^{2} \sin \left (d x + c\right )^{7} - 420 \, a^{2} \sin \left (d x + c\right )^{6} + 1008 \, a^{2} \sin \left (d x + c\right )^{5} + 630 \, a^{2} \sin \left (d x + c\right )^{4}}{2520 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/2520*(252*a^2*sin(d*x + c)^10 + 560*a^2*sin(d*x + c)^9 - 315*a^2*sin(d*x + c)^8 - 1440*a^2*sin(d*x + c)^7 -
420*a^2*sin(d*x + c)^6 + 1008*a^2*sin(d*x + c)^5 + 630*a^2*sin(d*x + c)^4)/d

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Fricas [A]  time = 1.12418, size = 279, normalized size = 2.2 \begin{align*} -\frac{252 \, a^{2} \cos \left (d x + c\right )^{10} - 945 \, a^{2} \cos \left (d x + c\right )^{8} + 840 \, a^{2} \cos \left (d x + c\right )^{6} - 16 \,{\left (35 \, a^{2} \cos \left (d x + c\right )^{8} - 50 \, a^{2} \cos \left (d x + c\right )^{6} + 3 \, a^{2} \cos \left (d x + c\right )^{4} + 4 \, a^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{2}\right )} \sin \left (d x + c\right )}{2520 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/2520*(252*a^2*cos(d*x + c)^10 - 945*a^2*cos(d*x + c)^8 + 840*a^2*cos(d*x + c)^6 - 16*(35*a^2*cos(d*x + c)^8
- 50*a^2*cos(d*x + c)^6 + 3*a^2*cos(d*x + c)^4 + 4*a^2*cos(d*x + c)^2 + 8*a^2)*sin(d*x + c))/d

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Sympy [A]  time = 32.7615, size = 189, normalized size = 1.49 \begin{align*} \begin{cases} \frac{16 a^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac{8 a^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac{2 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} - \frac{a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac{a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{12 d} - \frac{a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac{a^{2} \cos ^{10}{\left (c + d x \right )}}{60 d} - \frac{a^{2} \cos ^{8}{\left (c + d x \right )}}{24 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{2} \sin ^{3}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*sin(d*x+c)**3*(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((16*a**2*sin(c + d*x)**9/(315*d) + 8*a**2*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 2*a**2*sin(c + d*
x)**5*cos(c + d*x)**4/(5*d) - a**2*sin(c + d*x)**4*cos(c + d*x)**6/(6*d) - a**2*sin(c + d*x)**2*cos(c + d*x)**
8/(12*d) - a**2*sin(c + d*x)**2*cos(c + d*x)**6/(6*d) - a**2*cos(c + d*x)**10/(60*d) - a**2*cos(c + d*x)**8/(2
4*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)**3*cos(c)**5, True))

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Giac [A]  time = 1.30571, size = 227, normalized size = 1.79 \begin{align*} -\frac{a^{2} \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac{a^{2} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} + \frac{13 \, a^{2} \cos \left (6 \, d x + 6 \, c\right )}{3072 \, d} - \frac{a^{2} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac{17 \, a^{2} \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} + \frac{a^{2} \sin \left (9 \, d x + 9 \, c\right )}{1152 \, d} + \frac{a^{2} \sin \left (7 \, d x + 7 \, c\right )}{896 \, d} - \frac{a^{2} \sin \left (5 \, d x + 5 \, c\right )}{160 \, d} - \frac{a^{2} \sin \left (3 \, d x + 3 \, c\right )}{96 \, d} + \frac{3 \, a^{2} \sin \left (d x + c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/5120*a^2*cos(10*d*x + 10*c)/d + 1/1024*a^2*cos(8*d*x + 8*c)/d + 13/3072*a^2*cos(6*d*x + 6*c)/d - 1/256*a^2*
cos(4*d*x + 4*c)/d - 17/512*a^2*cos(2*d*x + 2*c)/d + 1/1152*a^2*sin(9*d*x + 9*c)/d + 1/896*a^2*sin(7*d*x + 7*c
)/d - 1/160*a^2*sin(5*d*x + 5*c)/d - 1/96*a^2*sin(3*d*x + 3*c)/d + 3/64*a^2*sin(d*x + c)/d