3.1528 \(\int \cos ^7(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=188 \[ -\frac {(a B+A b) \sin ^8(c+d x)}{8 d}-\frac {(a A-3 b B) \sin ^7(c+d x)}{7 d}+\frac {(a B+A b) \sin ^6(c+d x)}{2 d}+\frac {3 (a A-b B) \sin ^5(c+d x)}{5 d}-\frac {3 (a B+A b) \sin ^4(c+d x)}{4 d}-\frac {(3 a A-b B) \sin ^3(c+d x)}{3 d}+\frac {(a B+A b) \sin ^2(c+d x)}{2 d}+\frac {a A \sin (c+d x)}{d}-\frac {b B \sin ^9(c+d x)}{9 d} \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2837, 772} \[ -\frac {(a B+A b) \sin ^8(c+d x)}{8 d}-\frac {(a A-3 b B) \sin ^7(c+d x)}{7 d}+\frac {(a B+A b) \sin ^6(c+d x)}{2 d}+\frac {3 (a A-b B) \sin ^5(c+d x)}{5 d}-\frac {3 (a B+A b) \sin ^4(c+d x)}{4 d}-\frac {(3 a A-b B) \sin ^3(c+d x)}{3 d}+\frac {(a B+A b) \sin ^2(c+d x)}{2 d}+\frac {a A \sin (c+d x)}{d}-\frac {b B \sin ^9(c+d x)}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2837

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*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.80, size = 151, normalized size = 0.80 \[ \frac {\sin (c+d x) \left (-315 (a B+A b) \sin ^7(c+d x)-360 (a A-3 b B) \sin ^6(c+d x)+1260 (a B+A b) \sin ^5(c+d x)+1512 (a A-b B) \sin ^4(c+d x)-1890 (a B+A b) \sin ^3(c+d x)-840 (3 a A-b B) \sin ^2(c+d x)+1260 (a B+A b) \sin (c+d x)+2520 a A-280 b B \sin ^8(c+d x)\right )}{2520 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sin[c + d*x]*(2520*a*A + 1260*(A*b + a*B)*Sin[c + d*x] - 840*(3*a*A - b*B)*Sin[c + d*x]^2 - 1890*(A*b + a*B)*
Sin[c + d*x]^3 + 1512*(a*A - b*B)*Sin[c + d*x]^4 + 1260*(A*b + a*B)*Sin[c + d*x]^5 - 360*(a*A - 3*b*B)*Sin[c +
d*x]^6 - 315*(A*b + a*B)*Sin[c + d*x]^7 - 280*b*B*Sin[c + d*x]^8))/(2520*d)

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fricas [A]  time = 0.48, size = 106, normalized size = 0.56 \[ -\frac {315 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{8} + 8 \, {\left (35 \, B b \cos \left (d x + c\right )^{8} - 5 \, {\left (9 \, A a + B b\right )} \cos \left (d x + c\right )^{6} - 6 \, {\left (9 \, A a + B b\right )} \cos \left (d x + c\right )^{4} - 8 \, {\left (9 \, A a + B b\right )} \cos \left (d x + c\right )^{2} - 144 \, A a - 16 \, B b\right )} \sin \left (d x + c\right )}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2520*(315*(B*a + A*b)*cos(d*x + c)^8 + 8*(35*B*b*cos(d*x + c)^8 - 5*(9*A*a + B*b)*cos(d*x + c)^6 - 6*(9*A*a
+ B*b)*cos(d*x + c)^4 - 8*(9*A*a + B*b)*cos(d*x + c)^2 - 144*A*a - 16*B*b)*sin(d*x + c))/d

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giac [A]  time = 0.31, size = 182, normalized size = 0.97 \[ -\frac {B b \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac {7 \, A a \sin \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {{\left (B a + A b\right )} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {{\left (B a + A b\right )} \cos \left (6 \, d x + 6 \, c\right )}{128 \, d} - \frac {7 \, {\left (B a + A b\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {7 \, {\left (B a + A b\right )} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} + \frac {{\left (4 \, A a - 5 \, B b\right )} \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} + \frac {{\left (7 \, A a - 2 \, B b\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {7 \, {\left (10 \, A a + B b\right )} \sin \left (d x + c\right )}{128 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2304*B*b*sin(9*d*x + 9*c)/d + 7/64*A*a*sin(3*d*x + 3*c)/d - 1/1024*(B*a + A*b)*cos(8*d*x + 8*c)/d - 1/128*(
B*a + A*b)*cos(6*d*x + 6*c)/d - 7/256*(B*a + A*b)*cos(4*d*x + 4*c)/d - 7/128*(B*a + A*b)*cos(2*d*x + 2*c)/d +
1/1792*(4*A*a - 5*B*b)*sin(7*d*x + 7*c)/d + 1/320*(7*A*a - 2*B*b)*sin(5*d*x + 5*c)/d + 7/128*(10*A*a + B*b)*si
n(d*x + c)/d

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maple [A]  time = 0.46, size = 128, normalized size = 0.68 \[ \frac {B b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{8}\left (d x +c \right )\right )}{9}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{63}\right )-\frac {A b \left (\cos ^{8}\left (d x +c \right )\right )}{8}-\frac {a B \left (\cos ^{8}\left (d x +c \right )\right )}{8}+\frac {a A \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(B*b*(-1/9*sin(d*x+c)*cos(d*x+c)^8+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))-
1/8*A*b*cos(d*x+c)^8-1/8*a*B*cos(d*x+c)^8+1/7*a*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))

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maxima [A]  time = 0.35, size = 151, normalized size = 0.80 \[ -\frac {280 \, B b \sin \left (d x + c\right )^{9} + 315 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{8} + 360 \, {\left (A a - 3 \, B b\right )} \sin \left (d x + c\right )^{7} - 1260 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{6} - 1512 \, {\left (A a - B b\right )} \sin \left (d x + c\right )^{5} + 1890 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{4} + 840 \, {\left (3 \, A a - B b\right )} \sin \left (d x + c\right )^{3} - 2520 \, A a \sin \left (d x + c\right ) - 1260 \, {\left (B a + A b\right )} \sin \left (d x + c\right )^{2}}{2520 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2520*(280*B*b*sin(d*x + c)^9 + 315*(B*a + A*b)*sin(d*x + c)^8 + 360*(A*a - 3*B*b)*sin(d*x + c)^7 - 1260*(B*
a + A*b)*sin(d*x + c)^6 - 1512*(A*a - B*b)*sin(d*x + c)^5 + 1890*(B*a + A*b)*sin(d*x + c)^4 + 840*(3*A*a - B*b
)*sin(d*x + c)^3 - 2520*A*a*sin(d*x + c) - 1260*(B*a + A*b)*sin(d*x + c)^2)/d

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mupad [B]  time = 0.13, size = 156, normalized size = 0.83 \[ -\frac {\frac {B\,b\,{\sin \left (c+d\,x\right )}^9}{9}+\left (\frac {A\,b}{8}+\frac {B\,a}{8}\right )\,{\sin \left (c+d\,x\right )}^8+\left (\frac {A\,a}{7}-\frac {3\,B\,b}{7}\right )\,{\sin \left (c+d\,x\right )}^7+\left (-\frac {A\,b}{2}-\frac {B\,a}{2}\right )\,{\sin \left (c+d\,x\right )}^6+\left (\frac {3\,B\,b}{5}-\frac {3\,A\,a}{5}\right )\,{\sin \left (c+d\,x\right )}^5+\left (\frac {3\,A\,b}{4}+\frac {3\,B\,a}{4}\right )\,{\sin \left (c+d\,x\right )}^4+\left (A\,a-\frac {B\,b}{3}\right )\,{\sin \left (c+d\,x\right )}^3+\left (-\frac {A\,b}{2}-\frac {B\,a}{2}\right )\,{\sin \left (c+d\,x\right )}^2-A\,a\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 15.63, size = 228, normalized size = 1.21 \[ \begin {cases} \frac {16 A a \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {8 A a \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {2 A a \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {A a \sin {\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac {A b \cos ^{8}{\left (c + d x \right )}}{8 d} - \frac {B a \cos ^{8}{\left (c + d x \right )}}{8 d} + \frac {16 B b \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {8 B b \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {2 B b \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {B b \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a + b \sin {\relax (c )}\right ) \cos ^{7}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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