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]

1/4*a*sin(d*x+c)^4/d+1/5*a*sin(d*x+c)^5/d-1/6*a*sin(d*x+c)^6/d-1/7*a*sin(d*x+c)^7/d

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Rubi [A]  time = 0.07, antiderivative size = 65, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.30, 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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fricas [A]  time = 0.49, size = 72, normalized size = 1.11 \[ \frac {70 \, a \cos \left (d x + c\right )^{6} - 105 \, a \cos \left (d x + c\right )^{4} + 12 \, {\left (5 \, a \cos \left (d x + c\right )^{6} - 8 \, a \cos \left (d x + c\right )^{4} + a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right )}{420 \, d} \]

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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giac [A]  time = 0.21, size = 50, normalized size = 0.77 \[ -\frac {60 \, a \sin \left (d x + c\right )^{7} + 70 \, a \sin \left (d x + c\right )^{6} - 84 \, a \sin \left (d x + c\right )^{5} - 105 \, a \sin \left (d x + c\right )^{4}}{420 \, d} \]

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

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maple [A]  time = 0.24, size = 92, normalized size = 1.42 \[ \frac {a \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{7}-\frac {3 \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{35}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{35}\right )+a \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{12}\right )}{d} \]

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*cos(d*x+c)^4*sin(d*x+c)+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 = 0.31, size = 50, normalized size = 0.77 \[ -\frac {60 \, a \sin \left (d x + c\right )^{7} + 70 \, a \sin \left (d x + c\right )^{6} - 84 \, a \sin \left (d x + c\right )^{5} - 105 \, a \sin \left (d x + c\right )^{4}}{420 \, d} \]

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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mupad [B]  time = 0.07, size = 49, normalized size = 0.75 \[ \frac {-\frac {a\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {a\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {a\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a\,{\sin \left (c+d\,x\right )}^4}{4}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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 + (a*sin(c + d*x)^5)/5 - (a*sin(c + d*x)^6)/6 - (a*sin(c + d*x)^7)/7)/d

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sympy [A]  time = 7.83, size = 90, normalized size = 1.38 \[ \begin {cases} \frac {2 a \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {a \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} - \frac {a \cos ^{6}{\left (c + d x \right )}}{12 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\relax (c )} + a\right ) \sin ^{3}{\relax (c )} \cos ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

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

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