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

Optimal. Leaf size=81 \[ \frac{a \cos ^5(c+d x)}{5 d}-\frac{a \cos ^3(c+d x)}{3 d}-\frac{a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{a \sin (c+d x) \cos (c+d x)}{8 d}+\frac{a x}{8} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.128165, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2838, 2568, 2635, 8, 2565, 14} \[ \frac{a \cos ^5(c+d x)}{5 d}-\frac{a \cos ^3(c+d x)}{3 d}-\frac{a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{a \sin (c+d x) \cos (c+d x)}{8 d}+\frac{a x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2838

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

Rule 2568

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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

Mathematica [A]  time = 0.103072, size = 54, normalized size = 0.67 \[ \frac{a (-15 \sin (4 (c+d x))-60 \cos (c+d x)-10 \cos (3 (c+d x))+6 \cos (5 (c+d x))+60 c+60 d x)}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(60*c + 60*d*x - 60*Cos[c + d*x] - 10*Cos[3*(c + d*x)] + 6*Cos[5*(c + d*x)] - 15*Sin[4*(c + d*x)]))/(480*d)

________________________________________________________________________________________

Maple [A]  time = 0.027, size = 77, 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 ) ^{3}}{5}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15}} \right ) +a \left ( -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) }{4}}+{\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{8}}+{\frac{dx}{8}}+{\frac{c}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)+a*(-1/4*cos(d*x+c)^3*sin(d*x+c)+1/8*cos(d*x+c)*sin(d
*x+c)+1/8*d*x+1/8*c))

________________________________________________________________________________________

Maxima [A]  time = 1.2968, size = 70, normalized size = 0.86 \begin{align*} \frac{32 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a + 15 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(32*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a + 15*(4*d*x + 4*c - sin(4*d*x + 4*c))*a)/d

________________________________________________________________________________________

Fricas [A]  time = 1.65869, size = 162, normalized size = 2. \begin{align*} \frac{24 \, a \cos \left (d x + c\right )^{5} - 40 \, a \cos \left (d x + c\right )^{3} + 15 \, a d x - 15 \,{\left (2 \, a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(24*a*cos(d*x + c)^5 - 40*a*cos(d*x + c)^3 + 15*a*d*x - 15*(2*a*cos(d*x + c)^3 - a*cos(d*x + c))*sin(d*x
+ c))/d

________________________________________________________________________________________

Sympy [A]  time = 2.67902, size = 144, normalized size = 1.78 \begin{align*} \begin{cases} \frac{a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{a \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{a \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{a \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{2 a \cos ^{5}{\left (c + d x \right )}}{15 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right ) \sin ^{2}{\left (c \right )} \cos ^{2}{\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)**2*sin(d*x+c)**2*(a+a*sin(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.36946, size = 84, normalized size = 1.04 \begin{align*} \frac{1}{8} \, a x + \frac{a \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{a \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac{a \cos \left (d x + c\right )}{8 \, d} - \frac{a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a*x + 1/80*a*cos(5*d*x + 5*c)/d - 1/48*a*cos(3*d*x + 3*c)/d - 1/8*a*cos(d*x + c)/d - 1/32*a*sin(4*d*x + 4*
c)/d