3.988 \(\int \cos ^3(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=78 \[ -\frac{(A-3 B) (a \sin (c+d x)+a)^6}{6 a^3 d}+\frac{2 (A-B) (a \sin (c+d x)+a)^5}{5 a^2 d}-\frac{B (a \sin (c+d x)+a)^7}{7 a^4 d} \]

[Out]

(2*(A - B)*(a + a*Sin[c + d*x])^5)/(5*a^2*d) - ((A - 3*B)*(a + a*Sin[c + d*x])^6)/(6*a^3*d) - (B*(a + a*Sin[c
+ d*x])^7)/(7*a^4*d)

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Rubi [A]  time = 0.133522, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {2836, 77} \[ -\frac{(A-3 B) (a \sin (c+d x)+a)^6}{6 a^3 d}+\frac{2 (A-B) (a \sin (c+d x)+a)^5}{5 a^2 d}-\frac{B (a \sin (c+d x)+a)^7}{7 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A - B)*(a + a*Sin[c + d*x])^5)/(5*a^2*d) - ((A - 3*B)*(a + a*Sin[c + d*x])^6)/(6*a^3*d) - (B*(a + a*Sin[c
+ d*x])^7)/(7*a^4*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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.240138, size = 53, normalized size = 0.68 \[ -\frac{a^3 (\sin (c+d x)+1)^5 \left (5 (7 A-9 B) \sin (c+d x)-49 A+30 B \sin ^2(c+d x)+9 B\right )}{210 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(1 + Sin[c + d*x])^5*(-49*A + 9*B + 5*(7*A - 9*B)*Sin[c + d*x] + 30*B*Sin[c + d*x]^2))/(210*d)

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Maple [B]  time = 0.069, size = 265, normalized size = 3.4 \begin{align*}{\frac{1}{d} \left ({a}^{3}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 ) +B{a}^{3} \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 ) +3\,{a}^{3}A \left ( -1/5\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1/15\, \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) +3\,B{a}^{3} \left ( -1/6\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}-1/12\, \left ( \cos \left ( dx+c \right ) \right ) ^{4} \right ) -{\frac{3\,{a}^{3}A \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4}}+3\,B{a}^{3} \left ( -1/5\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1/15\, \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) +{\frac{{a}^{3}A \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}-{\frac{B{a}^{3} \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*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a^3*A*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4)+B*a^3*(-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))+3*a^3*A*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(2+cos(d*x+c)^
2)*sin(d*x+c))+3*B*a^3*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4)-3/4*a^3*A*cos(d*x+c)^4+3*B*a^3*(-1/5
*sin(d*x+c)*cos(d*x+c)^4+1/15*(2+cos(d*x+c)^2)*sin(d*x+c))+1/3*a^3*A*(2+cos(d*x+c)^2)*sin(d*x+c)-1/4*B*a^3*cos
(d*x+c)^4)

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Maxima [A]  time = 1.02209, size = 170, normalized size = 2.18 \begin{align*} -\frac{30 \, B a^{3} \sin \left (d x + c\right )^{7} + 35 \,{\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{6} + 42 \,{\left (3 \, A + 2 \, B\right )} a^{3} \sin \left (d x + c\right )^{5} + 105 \,{\left (A - B\right )} a^{3} \sin \left (d x + c\right )^{4} - 70 \,{\left (2 \, A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{3} - 105 \,{\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{2} - 210 \, A a^{3} \sin \left (d x + c\right )}{210 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/210*(30*B*a^3*sin(d*x + c)^7 + 35*(A + 3*B)*a^3*sin(d*x + c)^6 + 42*(3*A + 2*B)*a^3*sin(d*x + c)^5 + 105*(A
- B)*a^3*sin(d*x + c)^4 - 70*(2*A + 3*B)*a^3*sin(d*x + c)^3 - 105*(3*A + B)*a^3*sin(d*x + c)^2 - 210*A*a^3*si
n(d*x + c))/d

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Fricas [A]  time = 1.79773, size = 286, normalized size = 3.67 \begin{align*} \frac{35 \,{\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} - 210 \,{\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{4} + 2 \,{\left (15 \, B a^{3} \cos \left (d x + c\right )^{6} - 3 \,{\left (21 \, A + 29 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 8 \,{\left (7 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 16 \,{\left (7 \, A + 3 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{210 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(35*(A + 3*B)*a^3*cos(d*x + c)^6 - 210*(A + B)*a^3*cos(d*x + c)^4 + 2*(15*B*a^3*cos(d*x + c)^6 - 3*(21*A
+ 29*B)*a^3*cos(d*x + c)^4 + 8*(7*A + 3*B)*a^3*cos(d*x + c)^2 + 16*(7*A + 3*B)*a^3)*sin(d*x + c))/d

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Sympy [A]  time = 8.32572, size = 313, normalized size = 4.01 \begin{align*} \begin{cases} \frac{A a^{3} \sin ^{6}{\left (c + d x \right )}}{12 d} + \frac{2 A a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac{A a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} + \frac{A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{2 A a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{A a^{3} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac{3 A a^{3} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac{2 B a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac{B a^{3} \sin ^{6}{\left (c + d x \right )}}{4 d} + \frac{B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac{2 B a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac{3 B a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} + \frac{B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac{B a^{3} \cos ^{4}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\x \left (A + B \sin{\left (c \right )}\right ) \left (a \sin{\left (c \right )} + a\right )^{3} \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*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)

[Out]

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

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Giac [B]  time = 1.34238, size = 232, normalized size = 2.97 \begin{align*} -\frac{30 \, B a^{3} \sin \left (d x + c\right )^{7} + 35 \, A a^{3} \sin \left (d x + c\right )^{6} + 105 \, B a^{3} \sin \left (d x + c\right )^{6} + 126 \, A a^{3} \sin \left (d x + c\right )^{5} + 84 \, B a^{3} \sin \left (d x + c\right )^{5} + 105 \, A a^{3} \sin \left (d x + c\right )^{4} - 105 \, B a^{3} \sin \left (d x + c\right )^{4} - 140 \, A a^{3} \sin \left (d x + c\right )^{3} - 210 \, B a^{3} \sin \left (d x + c\right )^{3} - 315 \, A a^{3} \sin \left (d x + c\right )^{2} - 105 \, B a^{3} \sin \left (d x + c\right )^{2} - 210 \, A a^{3} \sin \left (d x + c\right )}{210 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/210*(30*B*a^3*sin(d*x + c)^7 + 35*A*a^3*sin(d*x + c)^6 + 105*B*a^3*sin(d*x + c)^6 + 126*A*a^3*sin(d*x + c)^
5 + 84*B*a^3*sin(d*x + c)^5 + 105*A*a^3*sin(d*x + c)^4 - 105*B*a^3*sin(d*x + c)^4 - 140*A*a^3*sin(d*x + c)^3 -
210*B*a^3*sin(d*x + c)^3 - 315*A*a^3*sin(d*x + c)^2 - 105*B*a^3*sin(d*x + c)^2 - 210*A*a^3*sin(d*x + c))/d