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

[Out]

2/5*(A-B)*(a+a*sin(d*x+c))^5/a^2/d-1/6*(A-3*B)*(a+a*sin(d*x+c))^6/a^3/d-1/7*B*(a+a*sin(d*x+c))^7/a^4/d

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Rubi [A]  time = 0.13, antiderivative size = 78, normalized size of antiderivative = 1.00, 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 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]))))

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) (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*}

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Mathematica [A]  time = 0.24, 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]

-1/210*(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))/d

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fricas [A]  time = 0.73, size = 115, normalized size = 1.47 \[ \frac {35 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} - 210 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{4} + 2 \, {\left (15 \, B a^{3} \cos \left (d x + c\right )^{6} - 3 \, {\left (21 \, A + 29 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 8 \, {\left (7 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 16 \, {\left (7 \, A + 3 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{210 \, d} \]

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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giac [B]  time = 0.35, size = 172, normalized size = 2.21 \[ -\frac {30 \, B a^{3} \sin \left (d x + c\right )^{7} + 35 \, A a^{3} \sin \left (d x + c\right )^{6} + 105 \, B a^{3} \sin \left (d x + c\right )^{6} + 126 \, A a^{3} \sin \left (d x + c\right )^{5} + 84 \, B a^{3} \sin \left (d x + c\right )^{5} + 105 \, A a^{3} \sin \left (d x + c\right )^{4} - 105 \, B a^{3} \sin \left (d x + c\right )^{4} - 140 \, A a^{3} \sin \left (d x + c\right )^{3} - 210 \, B a^{3} \sin \left (d x + c\right )^{3} - 315 \, A a^{3} \sin \left (d x + c\right )^{2} - 105 \, B a^{3} \sin \left (d x + c\right )^{2} - 210 \, A a^{3} \sin \left (d x + c\right )}{210 \, d} \]

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

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maple [B]  time = 0.48, size = 265, normalized size = 3.40 \[ \frac {a^{3} A \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{12}\right )+B \,a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{7}-\frac {3 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{35}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{35}\right )+3 a^{3} A \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )+3 B \,a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{6}-\frac {\left (\cos ^{4}\left (d x +c \right )\right )}{12}\right )-\frac {3 a^{3} A \left (\cos ^{4}\left (d x +c \right )\right )}{4}+3 B \,a^{3} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )+\frac {a^{3} A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}-\frac {B \,a^{3} \left (\cos ^{4}\left (d x +c \right )\right )}{4}}{d} \]

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 = 0.44, size = 126, normalized size = 1.62 \[ -\frac {30 \, B a^{3} \sin \left (d x + c\right )^{7} + 35 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{6} + 42 \, {\left (3 \, A + 2 \, B\right )} a^{3} \sin \left (d x + c\right )^{5} + 105 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right )^{4} - 70 \, {\left (2 \, A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{3} - 105 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{2} - 210 \, A a^{3} \sin \left (d x + c\right )}{210 \, d} \]

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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mupad [B]  time = 9.14, size = 126, normalized size = 1.62 \[ -\frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^4\,\left (A-B\right )}{2}-\frac {a^3\,{\sin \left (c+d\,x\right )}^2\,\left (3\,A+B\right )}{2}+\frac {a^3\,{\sin \left (c+d\,x\right )}^6\,\left (A+3\,B\right )}{6}+\frac {B\,a^3\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {a^3\,{\sin \left (c+d\,x\right )}^3\,\left (2\,A+3\,B\right )}{3}+\frac {a^3\,{\sin \left (c+d\,x\right )}^5\,\left (3\,A+2\,B\right )}{5}-A\,a^3\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 6.34, size = 313, normalized size = 4.01 \[ \begin {cases} \frac {A a^{3} \sin ^{6}{\left (c + d x \right )}}{12 d} + \frac {2 A a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {A a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} + \frac {A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {2 A a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {3 A a^{3} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {2 B a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {B a^{3} \sin ^{6}{\left (c + d x \right )}}{4 d} + \frac {B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {2 B a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {3 B a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} + \frac {B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {B a^{3} \cos ^{4}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a \sin {\relax (c )} + a\right )^{3} \cos ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(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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