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

Optimal. Leaf size=49 \[ -\frac{a \sin ^5(c+d x)}{5 d}+\frac{a \sin ^3(c+d x)}{3 d}-\frac{a \cos ^4(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.0811166, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2834, 2565, 30, 2564, 14} \[ -\frac{a \sin ^5(c+d x)}{5 d}+\frac{a \sin ^3(c+d x)}{3 d}-\frac{a \cos ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Cos[c + d*x]^4)/(4*d) + (a*Sin[c + d*x]^3)/(3*d) - (a*Sin[c + d*x]^5)/(5*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 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -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 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 ^3(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx &=a \int \cos ^3(c+d x) \sin (c+d x) \, dx+a \int \cos ^3(c+d x) \sin ^2(c+d x) \, dx\\ &=-\frac{a \operatorname{Subst}\left (\int x^3 \, dx,x,\cos (c+d x)\right )}{d}+\frac{a \operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{a \cos ^4(c+d x)}{4 d}+\frac{a \operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{a \cos ^4(c+d x)}{4 d}+\frac{a \sin ^3(c+d x)}{3 d}-\frac{a \sin ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.099882, size = 58, normalized size = 1.18 \[ -\frac{a (-60 \sin (c+d x)+10 \sin (3 (c+d x))+6 \sin (5 (c+d x))+60 \cos (2 (c+d x))+15 \cos (4 (c+d x))+45)}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(45 + 60*Cos[2*(c + d*x)] + 15*Cos[4*(c + d*x)] - 60*Sin[c + d*x] + 10*Sin[3*(c + d*x)] + 6*Sin[5*(c + d*x
)]))/(480*d)

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Maple [A]  time = 0.023, size = 54, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( 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 ) -{\frac{a \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.968803, size = 68, normalized size = 1.39 \begin{align*} -\frac{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} - 30 \, a \sin \left (d x + c\right )^{2}}{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)*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(12*a*sin(d*x + c)^5 + 15*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3 - 30*a*sin(d*x + c)^2)/d

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Fricas [A]  time = 1.75243, size = 127, normalized size = 2.59 \begin{align*} -\frac{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)*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/60*(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

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Sympy [A]  time = 2.24027, size = 66, normalized size = 1.35 \begin{align*} \begin{cases} \frac{2 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac{a \cos ^{4}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right ) \sin{\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)*(a+a*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.33475, size = 68, normalized size = 1.39 \begin{align*} -\frac{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} - 30 \, a \sin \left (d x + c\right )^{2}}{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)*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/60*(12*a*sin(d*x + c)^5 + 15*a*sin(d*x + c)^4 - 20*a*sin(d*x + c)^3 - 30*a*sin(d*x + c)^2)/d