3.969 \(\int \cos ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

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

[Out]

4/3*(A-B)*(a+a*sin(d*x+c))^6/a^4/d-4/7*(3*A-5*B)*(a+a*sin(d*x+c))^7/a^5/d+3/4*(A-3*B)*(a+a*sin(d*x+c))^8/a^6/d
-1/9*(A-7*B)*(a+a*sin(d*x+c))^9/a^7/d-1/10*B*(a+a*sin(d*x+c))^10/a^8/d

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Rubi [A]  time = 0.18, antiderivative size = 134, 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} \[ -\frac {(A-7 B) (a \sin (c+d x)+a)^9}{9 a^7 d}+\frac {3 (A-3 B) (a \sin (c+d x)+a)^8}{4 a^6 d}-\frac {4 (3 A-5 B) (a \sin (c+d x)+a)^7}{7 a^5 d}+\frac {4 (A-B) (a \sin (c+d x)+a)^6}{3 a^4 d}-\frac {B (a \sin (c+d x)+a)^{10}}{10 a^8 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

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

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Mathematica [A]  time = 1.19, size = 86, normalized size = 0.64 \[ -\frac {a^2 (\sin (c+d x)+1)^6 \left (28 (5 A-17 B) \sin ^3(c+d x)+(651 B-525 A) \sin ^2(c+d x)+6 (115 A-61 B) \sin (c+d x)-325 A+126 B \sin ^4(c+d x)+61 B\right )}{1260 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

-1/1260*(a^2*(1 + Sin[c + d*x])^6*(-325*A + 61*B + 6*(115*A - 61*B)*Sin[c + d*x] + (-525*A + 651*B)*Sin[c + d*
x]^2 + 28*(5*A - 17*B)*Sin[c + d*x]^3 + 126*B*Sin[c + d*x]^4))/d

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fricas [A]  time = 0.85, size = 127, normalized size = 0.95 \[ \frac {126 \, B a^{2} \cos \left (d x + c\right )^{10} - 315 \, {\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{8} - 4 \, {\left (35 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{8} - 10 \, {\left (5 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{6} - 12 \, {\left (5 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{4} - 16 \, {\left (5 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} - 32 \, {\left (5 \, A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{1260 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/1260*(126*B*a^2*cos(d*x + c)^10 - 315*(A + B)*a^2*cos(d*x + c)^8 - 4*(35*(A + 2*B)*a^2*cos(d*x + c)^8 - 10*(
5*A + B)*a^2*cos(d*x + c)^6 - 12*(5*A + B)*a^2*cos(d*x + c)^4 - 16*(5*A + B)*a^2*cos(d*x + c)^2 - 32*(5*A + B)
*a^2)*sin(d*x + c))/d

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giac [A]  time = 0.44, size = 239, normalized size = 1.78 \[ \frac {B a^{2} \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} - \frac {A a^{2} \cos \left (8 \, d x + 8 \, c\right )}{512 \, d} + \frac {7 \, A a^{2} \sin \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {{\left (16 \, A a^{2} + 7 \, B a^{2}\right )} \cos \left (6 \, d x + 6 \, c\right )}{1024 \, d} - \frac {{\left (7 \, A a^{2} + 4 \, B a^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )}{128 \, d} - \frac {7 \, {\left (8 \, A a^{2} + 5 \, B a^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} - \frac {{\left (A a^{2} + 2 \, B a^{2}\right )} \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} - \frac {{\left (A a^{2} + 10 \, B a^{2}\right )} \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} + \frac {{\left (5 \, A a^{2} - 4 \, B a^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {7 \, {\left (11 \, A a^{2} + 2 \, B a^{2}\right )} \sin \left (d x + c\right )}{128 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/5120*B*a^2*cos(10*d*x + 10*c)/d - 1/512*A*a^2*cos(8*d*x + 8*c)/d + 7/64*A*a^2*sin(3*d*x + 3*c)/d - 1/1024*(1
6*A*a^2 + 7*B*a^2)*cos(6*d*x + 6*c)/d - 1/128*(7*A*a^2 + 4*B*a^2)*cos(4*d*x + 4*c)/d - 7/512*(8*A*a^2 + 5*B*a^
2)*cos(2*d*x + 2*c)/d - 1/2304*(A*a^2 + 2*B*a^2)*sin(9*d*x + 9*c)/d - 1/1792*(A*a^2 + 10*B*a^2)*sin(7*d*x + 7*
c)/d + 1/320*(5*A*a^2 - 4*B*a^2)*sin(5*d*x + 5*c)/d + 7/128*(11*A*a^2 + 2*B*a^2)*sin(d*x + c)/d

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maple [A]  time = 0.48, size = 231, normalized size = 1.72 \[ \frac {a^{2} A \left (-\frac {\left (\cos ^{8}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{9}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{63}\right )+B \,a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{8}\left (d x +c \right )\right )}{10}-\frac {\left (\cos ^{8}\left (d x +c \right )\right )}{40}\right )-\frac {a^{2} A \left (\cos ^{8}\left (d x +c \right )\right )}{4}+2 B \,a^{2} \left (-\frac {\left (\cos ^{8}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{9}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{63}\right )+\frac {a^{2} A \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}-\frac {B \,a^{2} \left (\cos ^{8}\left (d x +c \right )\right )}{8}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a^2*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)
)+B*a^2*(-1/10*sin(d*x+c)^2*cos(d*x+c)^8-1/40*cos(d*x+c)^8)-1/4*a^2*A*cos(d*x+c)^8+2*B*a^2*(-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))+1/7*a^2*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^2*cos(d*x+c)^8)

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maxima [A]  time = 0.31, size = 168, normalized size = 1.25 \[ -\frac {126 \, B a^{2} \sin \left (d x + c\right )^{10} + 140 \, {\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{9} + 315 \, {\left (A - B\right )} a^{2} \sin \left (d x + c\right )^{8} - 360 \, {\left (A + 3 \, B\right )} a^{2} \sin \left (d x + c\right )^{7} - 1260 \, A a^{2} \sin \left (d x + c\right )^{6} + 1512 \, B a^{2} \sin \left (d x + c\right )^{5} + 630 \, {\left (3 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{4} + 840 \, {\left (A - B\right )} a^{2} \sin \left (d x + c\right )^{3} - 630 \, {\left (2 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{2} - 1260 \, A a^{2} \sin \left (d x + c\right )}{1260 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/1260*(126*B*a^2*sin(d*x + c)^10 + 140*(A + 2*B)*a^2*sin(d*x + c)^9 + 315*(A - B)*a^2*sin(d*x + c)^8 - 360*(
A + 3*B)*a^2*sin(d*x + c)^7 - 1260*A*a^2*sin(d*x + c)^6 + 1512*B*a^2*sin(d*x + c)^5 + 630*(3*A + B)*a^2*sin(d*
x + c)^4 + 840*(A - B)*a^2*sin(d*x + c)^3 - 630*(2*A + B)*a^2*sin(d*x + c)^2 - 1260*A*a^2*sin(d*x + c))/d

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mupad [B]  time = 9.20, size = 168, normalized size = 1.25 \[ -\frac {\frac {2\,a^2\,{\sin \left (c+d\,x\right )}^3\,\left (A-B\right )}{3}-\frac {a^2\,{\sin \left (c+d\,x\right )}^2\,\left (2\,A+B\right )}{2}-A\,a^2\,{\sin \left (c+d\,x\right )}^6+\frac {a^2\,{\sin \left (c+d\,x\right )}^4\,\left (3\,A+B\right )}{2}+\frac {a^2\,{\sin \left (c+d\,x\right )}^8\,\left (A-B\right )}{4}-\frac {2\,a^2\,{\sin \left (c+d\,x\right )}^7\,\left (A+3\,B\right )}{7}+\frac {a^2\,{\sin \left (c+d\,x\right )}^9\,\left (A+2\,B\right )}{9}+\frac {6\,B\,a^2\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {B\,a^2\,{\sin \left (c+d\,x\right )}^{10}}{10}-A\,a^2\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7*(A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2,x)

[Out]

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

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sympy [A]  time = 24.90, size = 440, normalized size = 3.28 \[ \begin {cases} \frac {16 A a^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {8 A a^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {16 A a^{2} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {2 A a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {8 A a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac {2 A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac {A a^{2} \cos ^{8}{\left (c + d x \right )}}{4 d} + \frac {B a^{2} \sin ^{10}{\left (c + d x \right )}}{40 d} + \frac {32 B a^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {B a^{2} \sin ^{8}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8 d} + \frac {16 B a^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {B a^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {4 B a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {B a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{4 d} + \frac {2 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \cos ^{8}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a \sin {\relax (c )} + a\right )^{2} \cos ^{7}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**7*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((16*A*a**2*sin(c + d*x)**9/(315*d) + 8*A*a**2*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 16*A*a**2*sin
(c + d*x)**7/(35*d) + 2*A*a**2*sin(c + d*x)**5*cos(c + d*x)**4/(5*d) + 8*A*a**2*sin(c + d*x)**5*cos(c + d*x)**
2/(5*d) + A*a**2*sin(c + d*x)**3*cos(c + d*x)**6/(3*d) + 2*A*a**2*sin(c + d*x)**3*cos(c + d*x)**4/d + A*a**2*s
in(c + d*x)*cos(c + d*x)**6/d - A*a**2*cos(c + d*x)**8/(4*d) + B*a**2*sin(c + d*x)**10/(40*d) + 32*B*a**2*sin(
c + d*x)**9/(315*d) + B*a**2*sin(c + d*x)**8*cos(c + d*x)**2/(8*d) + 16*B*a**2*sin(c + d*x)**7*cos(c + d*x)**2
/(35*d) + B*a**2*sin(c + d*x)**6*cos(c + d*x)**4/(4*d) + 4*B*a**2*sin(c + d*x)**5*cos(c + d*x)**4/(5*d) + B*a*
*2*sin(c + d*x)**4*cos(c + d*x)**6/(4*d) + 2*B*a**2*sin(c + d*x)**3*cos(c + d*x)**6/(3*d) - B*a**2*cos(c + d*x
)**8/(8*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**2*cos(c)**7, True))

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