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

Optimal. Leaf size=65 \[ -\frac{a \sin ^7(c+d x)}{7 d}-\frac{a \sin ^6(c+d x)}{6 d}+\frac{a \sin ^5(c+d x)}{5 d}+\frac{a \sin ^4(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.0688219, 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} \[ -\frac{a \sin ^7(c+d x)}{7 d}-\frac{a \sin ^6(c+d x)}{6 d}+\frac{a \sin ^5(c+d x)}{5 d}+\frac{a \sin ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.300185, size = 51, normalized size = 0.78 \[ \frac{a \left (-315 \cos (2 (c+d x))+35 \cos (6 (c+d x))+96 \sin ^5(c+d x) (5 \cos (2 (c+d x))+9)\right )}{6720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-315*Cos[2*(c + d*x)] + 35*Cos[6*(c + d*x)] + 96*(9 + 5*Cos[2*(c + d*x)])*Sin[c + d*x]^5))/(6720*d)

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Maple [A]  time = 0.032, size = 92, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ( a \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{7}}-{\frac{3\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{35}}+{\frac{ \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{35}} \right ) +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 ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(-1/7*sin(d*x+c)^3*cos(d*x+c)^4-3/35*sin(d*x+c)*cos(d*x+c)^4+1/35*(2+cos(d*x+c)^2)*sin(d*x+c))+a*(-1/6*
sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4))

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Maxima [A]  time = 0.985143, size = 68, normalized size = 1.05 \begin{align*} -\frac{60 \, a \sin \left (d x + c\right )^{7} + 70 \, a \sin \left (d x + c\right )^{6} - 84 \, a \sin \left (d x + c\right )^{5} - 105 \, a \sin \left (d x + c\right )^{4}}{420 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/420*(60*a*sin(d*x + c)^7 + 70*a*sin(d*x + c)^6 - 84*a*sin(d*x + c)^5 - 105*a*sin(d*x + c)^4)/d

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Fricas [A]  time = 1.77282, size = 188, normalized size = 2.89 \begin{align*} \frac{70 \, a \cos \left (d x + c\right )^{6} - 105 \, a \cos \left (d x + c\right )^{4} + 12 \,{\left (5 \, a \cos \left (d x + c\right )^{6} - 8 \, 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 )}{420 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(70*a*cos(d*x + c)^6 - 105*a*cos(d*x + c)^4 + 12*(5*a*cos(d*x + c)^6 - 8*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 = 7.57609, size = 90, normalized size = 1.38 \begin{align*} \begin{cases} \frac{2 a \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac{a \sin ^{6}{\left (c + d x \right )}}{12 d} + \frac{a \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac{a \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \right )} + a\right ) \sin ^{3}{\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)**3*(a+a*sin(d*x+c)),x)

[Out]

Piecewise((2*a*sin(c + d*x)**7/(35*d) + a*sin(c + d*x)**6/(12*d) + a*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + a
*sin(c + d*x)**4*cos(c + d*x)**2/(4*d), Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)**3*cos(c)**3, True))

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Giac [A]  time = 1.38701, size = 68, normalized size = 1.05 \begin{align*} -\frac{60 \, a \sin \left (d x + c\right )^{7} + 70 \, a \sin \left (d x + c\right )^{6} - 84 \, a \sin \left (d x + c\right )^{5} - 105 \, a \sin \left (d x + c\right )^{4}}{420 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/420*(60*a*sin(d*x + c)^7 + 70*a*sin(d*x + c)^6 - 84*a*sin(d*x + c)^5 - 105*a*sin(d*x + c)^4)/d