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

Optimal. Leaf size=81 \[ \frac{a \sin ^7(c+d x)}{7 d}-\frac{2 a \sin ^5(c+d x)}{5 d}+\frac{a \sin ^3(c+d x)}{3 d}+\frac{a \cos ^8(c+d x)}{8 d}-\frac{a \cos ^6(c+d x)}{6 d} \]

[Out]

-(a*Cos[c + d*x]^6)/(6*d) + (a*Cos[c + d*x]^8)/(8*d) + (a*Sin[c + d*x]^3)/(3*d) - (2*a*Sin[c + d*x]^5)/(5*d) +
(a*Sin[c + d*x]^7)/(7*d)

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Rubi [A]  time = 0.125666, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2834, 2564, 270, 2565, 14} \[ \frac{a \sin ^7(c+d x)}{7 d}-\frac{2 a \sin ^5(c+d x)}{5 d}+\frac{a \sin ^3(c+d x)}{3 d}+\frac{a \cos ^8(c+d x)}{8 d}-\frac{a \cos ^6(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Cos[c + d*x]^6)/(6*d) + (a*Cos[c + d*x]^8)/(8*d) + (a*Sin[c + d*x]^3)/(3*d) - (2*a*Sin[c + d*x]^5)/(5*d) +
(a*Sin[c + d*x]^7)/(7*d)

Rule 2834

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]),
x_Symbol] :> Dist[a, Int[Cos[e + f*x]^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[Cos[e + f*x]^p*(d*Sin[e +
f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && IntegerQ[n] && ((LtQ[p, 0]
&& NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] || LtQ[p + 1, -n, 2*p + 1])

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx &=a \int \cos ^5(c+d x) \sin ^2(c+d x) \, dx+a \int \cos ^5(c+d x) \sin ^3(c+d x) \, dx\\ &=-\frac{a \operatorname{Subst}\left (\int x^5 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac{a \operatorname{Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{a \operatorname{Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac{a \operatorname{Subst}\left (\int \left (x^5-x^7\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a \cos ^6(c+d x)}{6 d}+\frac{a \cos ^8(c+d x)}{8 d}+\frac{a \sin ^3(c+d x)}{3 d}-\frac{2 a \sin ^5(c+d x)}{5 d}+\frac{a \sin ^7(c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.276744, size = 87, normalized size = 1.07 \[ -\frac{a (-8400 \sin (c+d x)+560 \sin (3 (c+d x))+1008 \sin (5 (c+d x))+240 \sin (7 (c+d x))+2520 \cos (2 (c+d x))+420 \cos (4 (c+d x))-280 \cos (6 (c+d x))-105 \cos (8 (c+d x)))}{107520 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(2520*Cos[2*(c + d*x)] + 420*Cos[4*(c + d*x)] - 280*Cos[6*(c + d*x)] - 105*Cos[8*(c + d*x)] - 8400*Sin[c +
d*x] + 560*Sin[3*(c + d*x)] + 1008*Sin[5*(c + d*x)] + 240*Sin[7*(c + d*x)]))/(107520*d)

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Maple [A]  time = 0.029, size = 84, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( a \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 ) +a \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{7}}+{\frac{\sin \left ( dx+c \right ) }{35} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \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)^2*(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.05539, size = 97, normalized size = 1.2 \begin{align*} \frac{105 \, a \sin \left (d x + c\right )^{8} + 120 \, a \sin \left (d x + c\right )^{7} - 280 \, a \sin \left (d x + c\right )^{6} - 336 \, a \sin \left (d x + c\right )^{5} + 210 \, a \sin \left (d x + c\right )^{4} + 280 \, a \sin \left (d x + c\right )^{3}}{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)^2*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/840*(105*a*sin(d*x + c)^8 + 120*a*sin(d*x + c)^7 - 280*a*sin(d*x + c)^6 - 336*a*sin(d*x + c)^5 + 210*a*sin(d
*x + c)^4 + 280*a*sin(d*x + c)^3)/d

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Fricas [A]  time = 1.14169, size = 192, normalized size = 2.37 \begin{align*} \frac{105 \, a \cos \left (d x + c\right )^{8} - 140 \, a \cos \left (d x + c\right )^{6} - 8 \,{\left (15 \, a \cos \left (d x + c\right )^{6} - 3 \, a \cos \left (d x + c\right )^{4} - 4 \, a \cos \left (d x + c\right )^{2} - 8 \, a\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)^2*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/840*(105*a*cos(d*x + c)^8 - 140*a*cos(d*x + c)^6 - 8*(15*a*cos(d*x + c)^6 - 3*a*cos(d*x + c)^4 - 4*a*cos(d*x
+ c)^2 - 8*a)*sin(d*x + c))/d

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Sympy [A]  time = 12.2315, size = 136, normalized size = 1.68 \begin{align*} \begin{cases} \frac{a \sin ^{8}{\left (c + d x \right )}}{24 d} + \frac{8 a \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac{a \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{6 d} + \frac{4 a \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac{a \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac{a \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right ) \sin ^{2}{\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)**2*(a+a*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.24026, size = 159, normalized size = 1.96 \begin{align*} \frac{a \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} + \frac{a \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac{a \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac{3 \, a \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} - \frac{a \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac{3 \, a \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac{a \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac{5 \, a \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)^2*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/1024*a*cos(8*d*x + 8*c)/d + 1/384*a*cos(6*d*x + 6*c)/d - 1/256*a*cos(4*d*x + 4*c)/d - 3/128*a*cos(2*d*x + 2*
c)/d - 1/448*a*sin(7*d*x + 7*c)/d - 3/320*a*sin(5*d*x + 5*c)/d - 1/192*a*sin(3*d*x + 3*c)/d + 5/64*a*sin(d*x +
c)/d