3.986 \(\int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=134 \[ -\frac{(A-7 B) (a \sin (c+d x)+a)^{10}}{10 a^7 d}+\frac{2 (A-3 B) (a \sin (c+d x)+a)^9}{3 a^6 d}-\frac{(3 A-5 B) (a \sin (c+d x)+a)^8}{2 a^5 d}+\frac{8 (A-B) (a \sin (c+d x)+a)^7}{7 a^4 d}-\frac{B (a \sin (c+d x)+a)^{11}}{11 a^8 d} \]

[Out]

(8*(A - B)*(a + a*Sin[c + d*x])^7)/(7*a^4*d) - ((3*A - 5*B)*(a + a*Sin[c + d*x])^8)/(2*a^5*d) + (2*(A - 3*B)*(
a + a*Sin[c + d*x])^9)/(3*a^6*d) - ((A - 7*B)*(a + a*Sin[c + d*x])^10)/(10*a^7*d) - (B*(a + a*Sin[c + d*x])^11
)/(11*a^8*d)

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Rubi [A]  time = 0.182323, antiderivative size = 134, 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} \[ -\frac{(A-7 B) (a \sin (c+d x)+a)^{10}}{10 a^7 d}+\frac{2 (A-3 B) (a \sin (c+d x)+a)^9}{3 a^6 d}-\frac{(3 A-5 B) (a \sin (c+d x)+a)^8}{2 a^5 d}+\frac{8 (A-B) (a \sin (c+d x)+a)^7}{7 a^4 d}-\frac{B (a \sin (c+d x)+a)^{11}}{11 a^8 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(8*(A - B)*(a + a*Sin[c + d*x])^7)/(7*a^4*d) - ((3*A - 5*B)*(a + a*Sin[c + d*x])^8)/(2*a^5*d) + (2*(A - 3*B)*(
a + a*Sin[c + d*x])^9)/(3*a^6*d) - ((A - 7*B)*(a + a*Sin[c + d*x])^10)/(10*a^7*d) - (B*(a + a*Sin[c + d*x])^11
)/(11*a^8*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 ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=\frac{\operatorname{Subst}\left (\int (a-x)^3 (a+x)^6 \left (A+\frac{B x}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (8 a^3 (A-B) (a+x)^6-4 a^2 (3 A-5 B) (a+x)^7+6 a (A-3 B) (a+x)^8+(-A+7 B) (a+x)^9-\frac{B (a+x)^{10}}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^7 d}\\ &=\frac{8 (A-B) (a+a \sin (c+d x))^7}{7 a^4 d}-\frac{(3 A-5 B) (a+a \sin (c+d x))^8}{2 a^5 d}+\frac{2 (A-3 B) (a+a \sin (c+d x))^9}{3 a^6 d}-\frac{(A-7 B) (a+a \sin (c+d x))^{10}}{10 a^7 d}-\frac{B (a+a \sin (c+d x))^{11}}{11 a^8 d}\\ \end{align*}

Mathematica [A]  time = 1.54663, size = 86, normalized size = 0.64 \[ -\frac{a^3 (\sin (c+d x)+1)^7 \left (21 (11 A-37 B) \sin ^3(c+d x)+(1029 B-847 A) \sin ^2(c+d x)+14 (77 A-39 B) \sin (c+d x)-484 A+210 B \sin ^4(c+d x)+78 B\right )}{2310 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

-(a^3*(1 + Sin[c + d*x])^7*(-484*A + 78*B + 14*(77*A - 39*B)*Sin[c + d*x] + (-847*A + 1029*B)*Sin[c + d*x]^2 +
21*(11*A - 37*B)*Sin[c + d*x]^3 + 210*B*Sin[c + d*x]^4))/(2310*d)

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Maple [B]  time = 0.07, size = 345, normalized size = 2.6 \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 ) ^{8}}{10}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{40}} \right ) +B{a}^{3} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{11}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{8}\sin \left ( dx+c \right ) }{33}}+{\frac{\sin \left ( dx+c \right ) }{231} \left ({\frac{16}{5}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{\frac{6\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{5}} \right ) } \right ) +3\,{a}^{3}A \left ( -1/9\, \left ( \cos \left ( dx+c \right ) \right ) ^{8}\sin \left ( dx+c \right ) +{\frac{\sin \left ( dx+c \right ) }{63} \left ({\frac{16}{5}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{6}+6/5\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+8/5\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) } \right ) +3\,B{a}^{3} \left ( -1/10\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{8}-1/40\, \left ( \cos \left ( dx+c \right ) \right ) ^{8} \right ) -{\frac{3\,{a}^{3}A \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{8}}+3\,B{a}^{3} \left ( -1/9\, \left ( \cos \left ( dx+c \right ) \right ) ^{8}\sin \left ( dx+c \right ) +{\frac{\sin \left ( dx+c \right ) }{63} \left ({\frac{16}{5}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{6}+6/5\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+8/5\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) } \right ) +{\frac{{a}^{3}A\sin \left ( dx+c \right ) }{7} \left ({\frac{16}{5}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{\frac{6\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{5}} \right ) }-{\frac{B{a}^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{8}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a^3*A*(-1/10*sin(d*x+c)^2*cos(d*x+c)^8-1/40*cos(d*x+c)^8)+B*a^3*(-1/11*sin(d*x+c)^3*cos(d*x+c)^8-1/33*cos
(d*x+c)^8*sin(d*x+c)+1/231*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))+3*a^3*A*(-1/9*cos
(d*x+c)^8*sin(d*x+c)+1/63*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))+3*B*a^3*(-1/10*sin
(d*x+c)^2*cos(d*x+c)^8-1/40*cos(d*x+c)^8)-3/8*a^3*A*cos(d*x+c)^8+3*B*a^3*(-1/9*cos(d*x+c)^8*sin(d*x+c)+1/63*(1
6/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))+1/7*a^3*A*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+
8/5*cos(d*x+c)^2)*sin(d*x+c)-1/8*B*a^3*cos(d*x+c)^8)

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Maxima [A]  time = 1.01753, size = 246, normalized size = 1.84 \begin{align*} -\frac{210 \, B a^{3} \sin \left (d x + c\right )^{11} + 231 \,{\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{10} + 770 \, A a^{3} \sin \left (d x + c\right )^{9} - 2310 \, B a^{3} \sin \left (d x + c\right )^{8} - 660 \,{\left (4 \, A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{7} - 2310 \,{\left (A - B\right )} a^{3} \sin \left (d x + c\right )^{6} + 924 \,{\left (3 \, A + 4 \, B\right )} a^{3} \sin \left (d x + c\right )^{5} + 4620 \, A a^{3} \sin \left (d x + c\right )^{4} - 2310 \, B a^{3} \sin \left (d x + c\right )^{3} - 1155 \,{\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{2} - 2310 \, A a^{3} \sin \left (d x + c\right )}{2310 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/2310*(210*B*a^3*sin(d*x + c)^11 + 231*(A + 3*B)*a^3*sin(d*x + c)^10 + 770*A*a^3*sin(d*x + c)^9 - 2310*B*a^3
*sin(d*x + c)^8 - 660*(4*A + 3*B)*a^3*sin(d*x + c)^7 - 2310*(A - B)*a^3*sin(d*x + c)^6 + 924*(3*A + 4*B)*a^3*s
in(d*x + c)^5 + 4620*A*a^3*sin(d*x + c)^4 - 2310*B*a^3*sin(d*x + c)^3 - 1155*(3*A + B)*a^3*sin(d*x + c)^2 - 23
10*A*a^3*sin(d*x + c))/d

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Fricas [A]  time = 2.02686, size = 400, normalized size = 2.99 \begin{align*} \frac{231 \,{\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{10} - 1155 \,{\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{8} + 2 \,{\left (105 \, B a^{3} \cos \left (d x + c\right )^{10} - 35 \,{\left (11 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{8} + 20 \,{\left (11 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} + 24 \,{\left (11 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 32 \,{\left (11 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 64 \,{\left (11 \, A + 3 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{2310 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/2310*(231*(A + 3*B)*a^3*cos(d*x + c)^10 - 1155*(A + B)*a^3*cos(d*x + c)^8 + 2*(105*B*a^3*cos(d*x + c)^10 - 3
5*(11*A + 15*B)*a^3*cos(d*x + c)^8 + 20*(11*A + 3*B)*a^3*cos(d*x + c)^6 + 24*(11*A + 3*B)*a^3*cos(d*x + c)^4 +
32*(11*A + 3*B)*a^3*cos(d*x + c)^2 + 64*(11*A + 3*B)*a^3)*sin(d*x + c))/d

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Sympy [A]  time = 65.6018, size = 530, normalized size = 3.96 \begin{align*} \begin{cases} \frac{16 A a^{3} \sin ^{9}{\left (c + d x \right )}}{105 d} + \frac{24 A a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac{16 A a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac{6 A a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac{8 A a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac{A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} + \frac{2 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac{A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{8 d} + \frac{A a^{3} \sin{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac{A a^{3} \cos ^{10}{\left (c + d x \right )}}{40 d} - \frac{3 A a^{3} \cos ^{8}{\left (c + d x \right )}}{8 d} + \frac{16 B a^{3} \sin ^{11}{\left (c + d x \right )}}{1155 d} + \frac{8 B a^{3} \sin ^{9}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{105 d} + \frac{16 B a^{3} \sin ^{9}{\left (c + d x \right )}}{105 d} + \frac{6 B a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{35 d} + \frac{24 B a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac{B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{5 d} + \frac{6 B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac{B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac{3 B a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{8 d} - \frac{3 B a^{3} \cos ^{10}{\left (c + d x \right )}}{40 d} - \frac{B a^{3} \cos ^{8}{\left (c + d x \right )}}{8 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 ^{7}{\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)**7*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((16*A*a**3*sin(c + d*x)**9/(105*d) + 24*A*a**3*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 16*A*a**3*si
n(c + d*x)**7/(35*d) + 6*A*a**3*sin(c + d*x)**5*cos(c + d*x)**4/(5*d) + 8*A*a**3*sin(c + d*x)**5*cos(c + d*x)*
*2/(5*d) + A*a**3*sin(c + d*x)**3*cos(c + d*x)**6/d + 2*A*a**3*sin(c + d*x)**3*cos(c + d*x)**4/d - A*a**3*sin(
c + d*x)**2*cos(c + d*x)**8/(8*d) + A*a**3*sin(c + d*x)*cos(c + d*x)**6/d - A*a**3*cos(c + d*x)**10/(40*d) - 3
*A*a**3*cos(c + d*x)**8/(8*d) + 16*B*a**3*sin(c + d*x)**11/(1155*d) + 8*B*a**3*sin(c + d*x)**9*cos(c + d*x)**2
/(105*d) + 16*B*a**3*sin(c + d*x)**9/(105*d) + 6*B*a**3*sin(c + d*x)**7*cos(c + d*x)**4/(35*d) + 24*B*a**3*sin
(c + d*x)**7*cos(c + d*x)**2/(35*d) + B*a**3*sin(c + d*x)**5*cos(c + d*x)**6/(5*d) + 6*B*a**3*sin(c + d*x)**5*
cos(c + d*x)**4/(5*d) + B*a**3*sin(c + d*x)**3*cos(c + d*x)**6/d - 3*B*a**3*sin(c + d*x)**2*cos(c + d*x)**8/(8
*d) - 3*B*a**3*cos(c + d*x)**10/(40*d) - B*a**3*cos(c + d*x)**8/(8*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c)
+ a)**3*cos(c)**7, True))

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Giac [B]  time = 1.38131, size = 382, normalized size = 2.85 \begin{align*} \frac{B a^{3} \sin \left (11 \, d x + 11 \, c\right )}{11264 \, d} + \frac{{\left (A a^{3} + 3 \, B a^{3}\right )} \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} - \frac{{\left (A a^{3} - B a^{3}\right )} \cos \left (8 \, d x + 8 \, c\right )}{512 \, d} - \frac{{\left (23 \, A a^{3} + 5 \, B a^{3}\right )} \cos \left (6 \, d x + 6 \, c\right )}{1024 \, d} - \frac{{\left (11 \, A a^{3} + 5 \, B a^{3}\right )} \cos \left (4 \, d x + 4 \, c\right )}{128 \, d} - \frac{7 \,{\left (13 \, A a^{3} + 7 \, B a^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} - \frac{{\left (4 \, A a^{3} + 3 \, B a^{3}\right )} \sin \left (9 \, d x + 9 \, c\right )}{3072 \, d} - \frac{{\left (44 \, A a^{3} + 61 \, B a^{3}\right )} \sin \left (7 \, d x + 7 \, c\right )}{7168 \, d} + \frac{{\left (16 \, A a^{3} - 107 \, B a^{3}\right )} \sin \left (5 \, d x + 5 \, c\right )}{5120 \, d} + \frac{{\left (56 \, A a^{3} - B a^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{512 \, d} + \frac{91 \,{\left (4 \, A a^{3} + B a^{3}\right )} \sin \left (d x + c\right )}{512 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/11264*B*a^3*sin(11*d*x + 11*c)/d + 1/5120*(A*a^3 + 3*B*a^3)*cos(10*d*x + 10*c)/d - 1/512*(A*a^3 - B*a^3)*cos
(8*d*x + 8*c)/d - 1/1024*(23*A*a^3 + 5*B*a^3)*cos(6*d*x + 6*c)/d - 1/128*(11*A*a^3 + 5*B*a^3)*cos(4*d*x + 4*c)
/d - 7/512*(13*A*a^3 + 7*B*a^3)*cos(2*d*x + 2*c)/d - 1/3072*(4*A*a^3 + 3*B*a^3)*sin(9*d*x + 9*c)/d - 1/7168*(4
4*A*a^3 + 61*B*a^3)*sin(7*d*x + 7*c)/d + 1/5120*(16*A*a^3 - 107*B*a^3)*sin(5*d*x + 5*c)/d + 1/512*(56*A*a^3 -
B*a^3)*sin(3*d*x + 3*c)/d + 91/512*(4*A*a^3 + B*a^3)*sin(d*x + c)/d