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

Optimal. Leaf size=65 \[ -\frac{a \sin ^6(c+d x)}{6 d}-\frac{a \sin ^5(c+d x)}{5 d}+\frac{a \sin ^4(c+d x)}{4 d}+\frac{a \sin ^3(c+d x)}{3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.072234, antiderivative size = 65, 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, 75} \[ -\frac{a \sin ^6(c+d x)}{6 d}-\frac{a \sin ^5(c+d x)}{5 d}+\frac{a \sin ^4(c+d x)}{4 d}+\frac{a \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sin[c + d*x]^3)/(3*d) + (a*Sin[c + d*x]^4)/(4*d) - (a*Sin[c + d*x]^5)/(5*d) - (a*Sin[c + d*x]^6)/(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 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

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

Mathematica [A]  time = 0.214164, size = 51, normalized size = 0.78 \[ \frac{a \left (-45 \cos (2 (c+d x))+5 \cos (6 (c+d x))+32 \sin ^3(c+d x) (3 \cos (2 (c+d x))+7)\right )}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-45*Cos[2*(c + d*x)] + 5*Cos[6*(c + d*x)] + 32*(7 + 3*Cos[2*(c + d*x)])*Sin[c + d*x]^3))/(960*d)

________________________________________________________________________________________

Maple [A]  time = 0.029, size = 74, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( a \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{6}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{12}} \right ) +a \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{ \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{15}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4)+a*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(2+cos(d*x+c)^2
)*sin(d*x+c)))

________________________________________________________________________________________

Maxima [A]  time = 0.974002, size = 68, normalized size = 1.05 \begin{align*} -\frac{10 \, a \sin \left (d x + c\right )^{6} + 12 \, a \sin \left (d x + c\right )^{5} - 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(10*a*sin(d*x + c)^6 + 12*a*sin(d*x + c)^5 - 15*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3)/d

________________________________________________________________________________________

Fricas [A]  time = 1.70878, size = 155, normalized size = 2.38 \begin{align*} \frac{10 \, a \cos \left (d x + c\right )^{6} - 15 \, a \cos \left (d x + c\right )^{4} - 4 \,{\left (3 \, a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(10*a*cos(d*x + c)^6 - 15*a*cos(d*x + c)^4 - 4*(3*a*cos(d*x + c)^4 - a*cos(d*x + c)^2 - 2*a)*sin(d*x + c)
)/d

________________________________________________________________________________________

Sympy [A]  time = 4.29856, size = 90, normalized size = 1.38 \begin{align*} \begin{cases} \frac{a \sin ^{6}{\left (c + d x \right )}}{12 d} + \frac{2 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{a \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} + \frac{a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\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 ^{3}{\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)**3*sin(d*x+c)**2*(a+a*sin(d*x+c)),x)

[Out]

Piecewise((a*sin(c + d*x)**6/(12*d) + 2*a*sin(c + d*x)**5/(15*d) + a*sin(c + d*x)**4*cos(c + d*x)**2/(4*d) + a
*sin(c + d*x)**3*cos(c + d*x)**2/(3*d), Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)**2*cos(c)**3, True))

________________________________________________________________________________________

Giac [A]  time = 1.29402, size = 68, normalized size = 1.05 \begin{align*} -\frac{10 \, a \sin \left (d x + c\right )^{6} + 12 \, a \sin \left (d x + c\right )^{5} - 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(10*a*sin(d*x + c)^6 + 12*a*sin(d*x + c)^5 - 15*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3)/d