3.987 \(\int \cos ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=105 \[ \frac {B (a \sin (c+d x)+a)^9}{9 a^6 d}+\frac {(A-5 B) (a \sin (c+d x)+a)^8}{8 a^5 d}-\frac {4 (A-2 B) (a \sin (c+d x)+a)^7}{7 a^4 d}+\frac {2 (A-B) (a \sin (c+d x)+a)^6}{3 a^3 d} \]

[Out]

2/3*(A-B)*(a+a*sin(d*x+c))^6/a^3/d-4/7*(A-2*B)*(a+a*sin(d*x+c))^7/a^4/d+1/8*(A-5*B)*(a+a*sin(d*x+c))^8/a^5/d+1
/9*B*(a+a*sin(d*x+c))^9/a^6/d

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Rubi [A]  time = 0.15, antiderivative size = 105, 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} \[ \frac {(A-5 B) (a \sin (c+d x)+a)^8}{8 a^5 d}-\frac {4 (A-2 B) (a \sin (c+d x)+a)^7}{7 a^4 d}+\frac {2 (A-B) (a \sin (c+d x)+a)^6}{3 a^3 d}+\frac {B (a \sin (c+d x)+a)^9}{9 a^6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(2*(A - B)*(a + a*Sin[c + d*x])^6)/(3*a^3*d) - (4*(A - 2*B)*(a + a*Sin[c + d*x])^7)/(7*a^4*d) + ((A - 5*B)*(a
+ a*Sin[c + d*x])^8)/(8*a^5*d) + (B*(a + a*Sin[c + d*x])^9)/(9*a^6*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 ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=\frac {\operatorname {Subst}\left (\int (a-x)^2 (a+x)^5 \left (A+\frac {B x}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (4 a^2 (A-B) (a+x)^5-4 a (A-2 B) (a+x)^6+(A-5 B) (a+x)^7+\frac {B (a+x)^8}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d}\\ &=\frac {2 (A-B) (a+a \sin (c+d x))^6}{3 a^3 d}-\frac {4 (A-2 B) (a+a \sin (c+d x))^7}{7 a^4 d}+\frac {(A-5 B) (a+a \sin (c+d x))^8}{8 a^5 d}+\frac {B (a+a \sin (c+d x))^9}{9 a^6 d}\\ \end {align*}

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Mathematica [A]  time = 0.44, size = 70, normalized size = 0.67 \[ \frac {a^3 (\sin (c+d x)+1)^6 \left (21 (3 A-7 B) \sin ^2(c+d x)-6 (27 A-19 B) \sin (c+d x)+111 A+56 B \sin ^3(c+d x)-19 B\right )}{504 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(a^3*(1 + Sin[c + d*x])^6*(111*A - 19*B - 6*(27*A - 19*B)*Sin[c + d*x] + 21*(3*A - 7*B)*Sin[c + d*x]^2 + 56*B*
Sin[c + d*x]^3))/(504*d)

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fricas [A]  time = 0.73, size = 129, normalized size = 1.23 \[ \frac {63 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{8} - 336 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{6} + 8 \, {\left (7 \, B a^{3} \cos \left (d x + c\right )^{8} - {\left (27 \, A + 37 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} + 6 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{4} + 8 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 16 \, {\left (3 \, A + B\right )} a^{3}\right )} \sin \left (d x + c\right )}{504 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/504*(63*(A + 3*B)*a^3*cos(d*x + c)^8 - 336*(A + B)*a^3*cos(d*x + c)^6 + 8*(7*B*a^3*cos(d*x + c)^8 - (27*A +
37*B)*a^3*cos(d*x + c)^6 + 6*(3*A + B)*a^3*cos(d*x + c)^4 + 8*(3*A + B)*a^3*cos(d*x + c)^2 + 16*(3*A + B)*a^3)
*sin(d*x + c))/d

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giac [B]  time = 0.60, size = 230, normalized size = 2.19 \[ \frac {B a^{3} \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac {{\left (A a^{3} + 3 \, B a^{3}\right )} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {{\left (5 \, A a^{3} - B a^{3}\right )} \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac {{\left (25 \, A a^{3} + 11 \, B a^{3}\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {{\left (33 \, A a^{3} + 19 \, B a^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} - \frac {{\left (12 \, A a^{3} + 11 \, B a^{3}\right )} \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac {{\left (A a^{3} + 2 \, B a^{3}\right )} \sin \left (5 \, d x + 5 \, c\right )}{64 \, d} + \frac {{\left (17 \, A a^{3} - 4 \, B a^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {11 \, {\left (10 \, A a^{3} + 3 \, B a^{3}\right )} \sin \left (d x + c\right )}{128 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/2304*B*a^3*sin(9*d*x + 9*c)/d + 1/1024*(A*a^3 + 3*B*a^3)*cos(8*d*x + 8*c)/d - 1/384*(5*A*a^3 - B*a^3)*cos(6*
d*x + 6*c)/d - 1/256*(25*A*a^3 + 11*B*a^3)*cos(4*d*x + 4*c)/d - 1/128*(33*A*a^3 + 19*B*a^3)*cos(2*d*x + 2*c)/d
- 1/1792*(12*A*a^3 + 11*B*a^3)*sin(7*d*x + 7*c)/d - 1/64*(A*a^3 + 2*B*a^3)*sin(5*d*x + 5*c)/d + 1/192*(17*A*a
^3 - 4*B*a^3)*sin(3*d*x + 3*c)/d + 11/128*(10*A*a^3 + 3*B*a^3)*sin(d*x + c)/d

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maple [B]  time = 0.49, size = 305, normalized size = 2.90 \[ \frac {a^{3} A \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{8}-\frac {\left (\cos ^{6}\left (d x +c \right )\right )}{24}\right )+B \,a^{3} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{9}-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{21}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{105}\right )+3 a^{3} A \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )+3 B \,a^{3} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{8}-\frac {\left (\cos ^{6}\left (d x +c \right )\right )}{24}\right )-\frac {a^{3} A \left (\cos ^{6}\left (d x +c \right )\right )}{2}+3 B \,a^{3} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )+\frac {a^{3} A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}-\frac {B \,a^{3} \left (\cos ^{6}\left (d x +c \right )\right )}{6}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a^3*A*(-1/8*sin(d*x+c)^2*cos(d*x+c)^6-1/24*cos(d*x+c)^6)+B*a^3*(-1/9*sin(d*x+c)^3*cos(d*x+c)^6-1/21*sin(d
*x+c)*cos(d*x+c)^6+1/105*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))+3*a^3*A*(-1/7*sin(d*x+c)*cos(d*x+c)^6
+1/35*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))+3*B*a^3*(-1/8*sin(d*x+c)^2*cos(d*x+c)^6-1/24*cos(d*x+c)^
6)-1/2*a^3*A*cos(d*x+c)^6+3*B*a^3*(-1/7*sin(d*x+c)*cos(d*x+c)^6+1/35*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d
*x+c))+1/5*a^3*A*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)-1/6*B*a^3*cos(d*x+c)^6)

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maxima [A]  time = 0.37, size = 158, normalized size = 1.50 \[ \frac {56 \, B a^{3} \sin \left (d x + c\right )^{9} + 63 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{8} + 72 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{7} + 84 \, {\left (A - 5 \, B\right )} a^{3} \sin \left (d x + c\right )^{6} - 504 \, {\left (A + B\right )} a^{3} \sin \left (d x + c\right )^{5} - 126 \, {\left (5 \, A - B\right )} a^{3} \sin \left (d x + c\right )^{4} + 168 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{3} + 252 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{2} + 504 \, A a^{3} \sin \left (d x + c\right )}{504 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/504*(56*B*a^3*sin(d*x + c)^9 + 63*(A + 3*B)*a^3*sin(d*x + c)^8 + 72*(3*A + B)*a^3*sin(d*x + c)^7 + 84*(A - 5
*B)*a^3*sin(d*x + c)^6 - 504*(A + B)*a^3*sin(d*x + c)^5 - 126*(5*A - B)*a^3*sin(d*x + c)^4 + 168*(A + 3*B)*a^3
*sin(d*x + c)^3 + 252*(3*A + B)*a^3*sin(d*x + c)^2 + 504*A*a^3*sin(d*x + c))/d

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mupad [B]  time = 0.14, size = 156, normalized size = 1.49 \[ \frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^2\,\left (3\,A+B\right )}{2}+\frac {a^3\,{\sin \left (c+d\,x\right )}^3\,\left (A+3\,B\right )}{3}+\frac {a^3\,{\sin \left (c+d\,x\right )}^7\,\left (3\,A+B\right )}{7}+\frac {a^3\,{\sin \left (c+d\,x\right )}^6\,\left (A-5\,B\right )}{6}+\frac {a^3\,{\sin \left (c+d\,x\right )}^8\,\left (A+3\,B\right )}{8}+\frac {B\,a^3\,{\sin \left (c+d\,x\right )}^9}{9}-\frac {a^3\,{\sin \left (c+d\,x\right )}^4\,\left (5\,A-B\right )}{4}+A\,a^3\,\sin \left (c+d\,x\right )-a^3\,{\sin \left (c+d\,x\right )}^5\,\left (A+B\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3,x)

[Out]

((a^3*sin(c + d*x)^2*(3*A + B))/2 + (a^3*sin(c + d*x)^3*(A + 3*B))/3 + (a^3*sin(c + d*x)^7*(3*A + B))/7 + (a^3
*sin(c + d*x)^6*(A - 5*B))/6 + (a^3*sin(c + d*x)^8*(A + 3*B))/8 + (B*a^3*sin(c + d*x)^9)/9 - (a^3*sin(c + d*x)
^4*(5*A - B))/4 + A*a^3*sin(c + d*x) - a^3*sin(c + d*x)^5*(A + B))/d

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sympy [A]  time = 16.66, size = 471, normalized size = 4.49 \[ \begin {cases} \frac {A a^{3} \sin ^{8}{\left (c + d x \right )}}{24 d} + \frac {8 A a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {A a^{3} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{6 d} + \frac {4 A a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {8 A a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {A a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {4 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {A a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {A a^{3} \cos ^{6}{\left (c + d x \right )}}{2 d} + \frac {8 B a^{3} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {B a^{3} \sin ^{8}{\left (c + d x \right )}}{8 d} + \frac {4 B a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {8 B a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {B a^{3} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac {B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {4 B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {3 B a^{3} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {B a^{3} \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a \sin {\relax (c )} + a\right )^{3} \cos ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((A*a**3*sin(c + d*x)**8/(24*d) + 8*A*a**3*sin(c + d*x)**7/(35*d) + A*a**3*sin(c + d*x)**6*cos(c + d*
x)**2/(6*d) + 4*A*a**3*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + 8*A*a**3*sin(c + d*x)**5/(15*d) + A*a**3*sin(c
+ d*x)**4*cos(c + d*x)**4/(4*d) + A*a**3*sin(c + d*x)**3*cos(c + d*x)**4/d + 4*A*a**3*sin(c + d*x)**3*cos(c +
d*x)**2/(3*d) + A*a**3*sin(c + d*x)*cos(c + d*x)**4/d - A*a**3*cos(c + d*x)**6/(2*d) + 8*B*a**3*sin(c + d*x)**
9/(315*d) + B*a**3*sin(c + d*x)**8/(8*d) + 4*B*a**3*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 8*B*a**3*sin(c +
d*x)**7/(35*d) + B*a**3*sin(c + d*x)**6*cos(c + d*x)**2/(2*d) + B*a**3*sin(c + d*x)**5*cos(c + d*x)**4/(5*d) +
4*B*a**3*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + 3*B*a**3*sin(c + d*x)**4*cos(c + d*x)**4/(4*d) + B*a**3*sin(
c + d*x)**3*cos(c + d*x)**4/d - B*a**3*cos(c + d*x)**6/(6*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**3*c
os(c)**5, True))

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