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

Optimal. Leaf size=89 \[ \frac{(a \sin (c+d x)+a)^8}{8 a^6 d}-\frac{5 (a \sin (c+d x)+a)^7}{7 a^5 d}+\frac{4 (a \sin (c+d x)+a)^6}{3 a^4 d}-\frac{4 (a \sin (c+d x)+a)^5}{5 a^3 d} \]

[Out]

(-4*(a + a*Sin[c + d*x])^5)/(5*a^3*d) + (4*(a + a*Sin[c + d*x])^6)/(3*a^4*d) - (5*(a + a*Sin[c + d*x])^7)/(7*a
^5*d) + (a + a*Sin[c + d*x])^8/(8*a^6*d)

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Rubi [A]  time = 0.0843447, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2836, 12, 77} \[ \frac{(a \sin (c+d x)+a)^8}{8 a^6 d}-\frac{5 (a \sin (c+d x)+a)^7}{7 a^5 d}+\frac{4 (a \sin (c+d x)+a)^6}{3 a^4 d}-\frac{4 (a \sin (c+d x)+a)^5}{5 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(a + a*Sin[c + d*x])^5)/(5*a^3*d) + (4*(a + a*Sin[c + d*x])^6)/(3*a^4*d) - (5*(a + a*Sin[c + d*x])^7)/(7*a
^5*d) + (a + a*Sin[c + d*x])^8/(8*a^6*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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.343968, size = 90, normalized size = 1.01 \[ -\frac{a^2 (-16800 \sin (c+d x)+1120 \sin (3 (c+d x))+2016 \sin (5 (c+d x))+480 \sin (7 (c+d x))+10920 \cos (2 (c+d x))+3780 \cos (4 (c+d x))+280 \cos (6 (c+d x))-105 \cos (8 (c+d x))-2590)}{107520 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(-2590 + 10920*Cos[2*(c + d*x)] + 3780*Cos[4*(c + d*x)] + 280*Cos[6*(c + d*x)] - 105*Cos[8*(c + d*x)] -
16800*Sin[c + d*x] + 1120*Sin[3*(c + d*x)] + 2016*Sin[5*(c + d*x)] + 480*Sin[7*(c + d*x)]))/(107520*d)

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Maple [A]  time = 0.034, size = 102, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({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 ) +2\,{a}^{2} \left ( -1/7\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}+1/35\, \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 ) \right ) -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{6}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*(-1/8*sin(d*x+c)^2*cos(d*x+c)^6-1/24*cos(d*x+c)^6)+2*a^2*(-1/7*sin(d*x+c)*cos(d*x+c)^6+1/35*(8/3+cos(
d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))-1/6*a^2*cos(d*x+c)^6)

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Maxima [A]  time = 1.06281, size = 131, normalized size = 1.47 \begin{align*} \frac{105 \, a^{2} \sin \left (d x + c\right )^{8} + 240 \, a^{2} \sin \left (d x + c\right )^{7} - 140 \, a^{2} \sin \left (d x + c\right )^{6} - 672 \, a^{2} \sin \left (d x + c\right )^{5} - 210 \, a^{2} \sin \left (d x + c\right )^{4} + 560 \, a^{2} \sin \left (d x + c\right )^{3} + 420 \, a^{2} \sin \left (d x + c\right )^{2}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(105*a^2*sin(d*x + c)^8 + 240*a^2*sin(d*x + c)^7 - 140*a^2*sin(d*x + c)^6 - 672*a^2*sin(d*x + c)^5 - 210
*a^2*sin(d*x + c)^4 + 560*a^2*sin(d*x + c)^3 + 420*a^2*sin(d*x + c)^2)/d

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Fricas [A]  time = 1.12122, size = 209, normalized size = 2.35 \begin{align*} \frac{105 \, a^{2} \cos \left (d x + c\right )^{8} - 280 \, a^{2} \cos \left (d x + c\right )^{6} - 16 \,{\left (15 \, 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 )}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(105*a^2*cos(d*x + c)^8 - 280*a^2*cos(d*x + c)^6 - 16*(15*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 = 11.8753, size = 163, normalized size = 1.83 \begin{align*} \begin{cases} \frac{a^{2} \sin ^{8}{\left (c + d x \right )}}{24 d} + \frac{16 a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac{a^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{6 d} + \frac{8 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac{a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac{2 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac{a^{2} \cos ^{6}{\left (c + d x \right )}}{6 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right )^{2} \sin{\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)*(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((a**2*sin(c + d*x)**8/(24*d) + 16*a**2*sin(c + d*x)**7/(105*d) + a**2*sin(c + d*x)**6*cos(c + d*x)**
2/(6*d) + 8*a**2*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + a**2*sin(c + d*x)**4*cos(c + d*x)**4/(4*d) + 2*a**2*
sin(c + d*x)**3*cos(c + d*x)**4/(3*d) - a**2*cos(c + d*x)**6/(6*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*sin(c)*cos
(c)**5, True))

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Giac [A]  time = 1.28626, size = 181, normalized size = 2.03 \begin{align*} \frac{a^{2} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac{a^{2} \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac{9 \, a^{2} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac{13 \, a^{2} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} - \frac{a^{2} \sin \left (7 \, d x + 7 \, c\right )}{224 \, d} - \frac{3 \, 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{5 \, a^{2} \sin \left (d x + c\right )}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1024*a^2*cos(8*d*x + 8*c)/d - 1/384*a^2*cos(6*d*x + 6*c)/d - 9/256*a^2*cos(4*d*x + 4*c)/d - 13/128*a^2*cos(2
*d*x + 2*c)/d - 1/224*a^2*sin(7*d*x + 7*c)/d - 3/160*a^2*sin(5*d*x + 5*c)/d - 1/96*a^2*sin(3*d*x + 3*c)/d + 5/
32*a^2*sin(d*x + c)/d