3.1536 \(\int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=349 \[ \frac{3 \left (-7 a^2 B+2 a A b+b^2 B\right ) (a+b \sin (c+d x))^8}{8 b^8 d}-\frac{\left (15 a^2 A b-35 a^3 B+15 a b^2 B-3 A b^3\right ) (a+b \sin (c+d x))^7}{7 b^8 d}+\frac{\left (20 a^3 A b+30 a^2 b^2 B-35 a^4 B-12 a A b^3-3 b^4 B\right ) (a+b \sin (c+d x))^6}{6 b^8 d}-\frac{3 \left (a^2-b^2\right ) \left (5 a^2 A b-7 a^3 B+3 a b^2 B-A b^3\right ) (a+b \sin (c+d x))^5}{5 b^8 d}+\frac{\left (a^2-b^2\right )^2 \left (-7 a^2 B+6 a A b+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^8 d}-\frac{\left (a^2-b^2\right )^3 (A b-a B) (a+b \sin (c+d x))^3}{3 b^8 d}-\frac{(A b-7 a B) (a+b \sin (c+d x))^9}{9 b^8 d}-\frac{B (a+b \sin (c+d x))^{10}}{10 b^8 d} \]

[Out]

-((a^2 - b^2)^3*(A*b - a*B)*(a + b*Sin[c + d*x])^3)/(3*b^8*d) + ((a^2 - b^2)^2*(6*a*A*b - 7*a^2*B + b^2*B)*(a
+ b*Sin[c + d*x])^4)/(4*b^8*d) - (3*(a^2 - b^2)*(5*a^2*A*b - A*b^3 - 7*a^3*B + 3*a*b^2*B)*(a + b*Sin[c + d*x])
^5)/(5*b^8*d) + ((20*a^3*A*b - 12*a*A*b^3 - 35*a^4*B + 30*a^2*b^2*B - 3*b^4*B)*(a + b*Sin[c + d*x])^6)/(6*b^8*
d) - ((15*a^2*A*b - 3*A*b^3 - 35*a^3*B + 15*a*b^2*B)*(a + b*Sin[c + d*x])^7)/(7*b^8*d) + (3*(2*a*A*b - 7*a^2*B
+ b^2*B)*(a + b*Sin[c + d*x])^8)/(8*b^8*d) - ((A*b - 7*a*B)*(a + b*Sin[c + d*x])^9)/(9*b^8*d) - (B*(a + b*Sin
[c + d*x])^10)/(10*b^8*d)

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Rubi [A]  time = 0.392181, antiderivative size = 349, 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 = {2837, 772} \[ \frac{3 \left (-7 a^2 B+2 a A b+b^2 B\right ) (a+b \sin (c+d x))^8}{8 b^8 d}-\frac{\left (15 a^2 A b-35 a^3 B+15 a b^2 B-3 A b^3\right ) (a+b \sin (c+d x))^7}{7 b^8 d}+\frac{\left (20 a^3 A b+30 a^2 b^2 B-35 a^4 B-12 a A b^3-3 b^4 B\right ) (a+b \sin (c+d x))^6}{6 b^8 d}-\frac{3 \left (a^2-b^2\right ) \left (5 a^2 A b-7 a^3 B+3 a b^2 B-A b^3\right ) (a+b \sin (c+d x))^5}{5 b^8 d}+\frac{\left (a^2-b^2\right )^2 \left (-7 a^2 B+6 a A b+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^8 d}-\frac{\left (a^2-b^2\right )^3 (A b-a B) (a+b \sin (c+d x))^3}{3 b^8 d}-\frac{(A b-7 a B) (a+b \sin (c+d x))^9}{9 b^8 d}-\frac{B (a+b \sin (c+d x))^{10}}{10 b^8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2 - b^2)^3*(A*b - a*B)*(a + b*Sin[c + d*x])^3)/(3*b^8*d) + ((a^2 - b^2)^2*(6*a*A*b - 7*a^2*B + b^2*B)*(a
+ b*Sin[c + d*x])^4)/(4*b^8*d) - (3*(a^2 - b^2)*(5*a^2*A*b - A*b^3 - 7*a^3*B + 3*a*b^2*B)*(a + b*Sin[c + d*x])
^5)/(5*b^8*d) + ((20*a^3*A*b - 12*a*A*b^3 - 35*a^4*B + 30*a^2*b^2*B - 3*b^4*B)*(a + b*Sin[c + d*x])^6)/(6*b^8*
d) - ((15*a^2*A*b - 3*A*b^3 - 35*a^3*B + 15*a*b^2*B)*(a + b*Sin[c + d*x])^7)/(7*b^8*d) + (3*(2*a*A*b - 7*a^2*B
+ b^2*B)*(a + b*Sin[c + d*x])^8)/(8*b^8*d) - ((A*b - 7*a*B)*(a + b*Sin[c + d*x])^9)/(9*b^8*d) - (B*(a + b*Sin
[c + d*x])^10)/(10*b^8*d)

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]

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]

Rubi steps

\begin{align*} \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx &=\frac{\operatorname{Subst}\left (\int (a+x)^2 \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 (\frac{\left (-a^2+b^2\right )^3 (A b-a B) (a+x)^2}{b}+\frac{\left (-a^2+b^2\right )^2 \left (6 a A b-7 a^2 B+b^2 B\right ) (a+x)^3}{b}-\frac{3 \left (-a^2+b^2\right ) \left (-5 a^2 A b+A b^3+7 a^3 B-3 a b^2 B\right ) (a+x)^4}{b}+\frac{\left (20 a^3 A b-12 a A b^3-35 a^4 B+30 a^2 b^2 B-3 b^4 B\right ) (a+x)^5}{b}+\frac{\left (-15 a^2 A b+3 A b^3+35 a^3 B-15 a b^2 B\right ) (a+x)^6}{b}-\frac{3 \left (-2 a A b+7 a^2 B-b^2 B\right ) (a+x)^7}{b}+\frac{(-A b+7 a B) (a+x)^8}{b}-\frac{B (a+x)^9}{b}\right ) \, dx,x,b \sin (c+d x)\right )}{b^7 d}\\ &=-\frac{\left (a^2-b^2\right )^3 (A b-a B) (a+b \sin (c+d x))^3}{3 b^8 d}+\frac{\left (a^2-b^2\right )^2 \left (6 a A b-7 a^2 B+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^8 d}-\frac{3 \left (a^2-b^2\right ) \left (5 a^2 A b-A b^3-7 a^3 B+3 a b^2 B\right ) (a+b \sin (c+d x))^5}{5 b^8 d}+\frac{\left (20 a^3 A b-12 a A b^3-35 a^4 B+30 a^2 b^2 B-3 b^4 B\right ) (a+b \sin (c+d x))^6}{6 b^8 d}-\frac{\left (15 a^2 A b-3 A b^3-35 a^3 B+15 a b^2 B\right ) (a+b \sin (c+d x))^7}{7 b^8 d}+\frac{3 \left (2 a A b-7 a^2 B+b^2 B\right ) (a+b \sin (c+d x))^8}{8 b^8 d}-\frac{(A b-7 a B) (a+b \sin (c+d x))^9}{9 b^8 d}-\frac{B (a+b \sin (c+d x))^{10}}{10 b^8 d}\\ \end{align*}

Mathematica [A]  time = 1.46784, size = 295, normalized size = 0.85 \[ \frac{-315 b^8 \left (a^2 B+2 a A b-3 b^2 B\right ) \sin ^8(c+d x)+360 b^8 \left (a^2 (-A)+6 a b B+3 A b^2\right ) \sin ^7(c+d x)+1260 b^8 \left (a^2 B+2 a A b-b^2 B\right ) \sin ^6(c+d x)-1512 b^8 \left (a^2 (-A)+2 a b B+A b^2\right ) \sin ^5(c+d x)+630 b^8 \left (-3 a^2 B-6 a A b+b^2 B\right ) \sin ^4(c+d x)+840 b^8 \left (-3 a^2 A+2 a b B+A b^2\right ) \sin ^3(c+d x)+2520 a^2 A b^8 \sin (c+d x)-3 a^4 B \left (-9 a^4 b^2+42 a^2 b^4+a^6-210 b^6\right )-280 b^9 (2 a B+A b) \sin ^9(c+d x)+1260 a b^8 (a B+2 A b) \sin ^2(c+d x)-252 b^{10} B \sin ^{10}(c+d x)}{2520 b^8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^4*(a^6 - 9*a^4*b^2 + 42*a^2*b^4 - 210*b^6)*B + 2520*a^2*A*b^8*Sin[c + d*x] + 1260*a*b^8*(2*A*b + a*B)*Si
n[c + d*x]^2 + 840*b^8*(-3*a^2*A + A*b^2 + 2*a*b*B)*Sin[c + d*x]^3 + 630*b^8*(-6*a*A*b - 3*a^2*B + b^2*B)*Sin[
c + d*x]^4 - 1512*b^8*(-(a^2*A) + A*b^2 + 2*a*b*B)*Sin[c + d*x]^5 + 1260*b^8*(2*a*A*b + a^2*B - b^2*B)*Sin[c +
d*x]^6 + 360*b^8*(-(a^2*A) + 3*A*b^2 + 6*a*b*B)*Sin[c + d*x]^7 - 315*b^8*(2*a*A*b + a^2*B - 3*b^2*B)*Sin[c +
d*x]^8 - 280*b^9*(A*b + 2*a*B)*Sin[c + d*x]^9 - 252*b^10*B*Sin[c + d*x]^10)/(2520*b^8*d)

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Maple [A]  time = 0.078, size = 229, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{2}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}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{8}}-{\frac{Aab \left ( \cos \left ( dx+c \right ) \right ) ^{8}}{4}}+2\,Bab \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 ) +A{b}^{2} \left ( -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{8}\sin \left ( dx+c \right ) }{9}}+{\frac{\sin \left ( dx+c \right ) }{63} \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 ) +B{b}^{2} \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 ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.999976, size = 321, normalized size = 0.92 \begin{align*} -\frac{252 \, B b^{2} \sin \left (d x + c\right )^{10} + 280 \,{\left (2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )^{9} + 315 \,{\left (B a^{2} + 2 \, A a b - 3 \, B b^{2}\right )} \sin \left (d x + c\right )^{8} + 360 \,{\left (A a^{2} - 6 \, B a b - 3 \, A b^{2}\right )} \sin \left (d x + c\right )^{7} - 1260 \,{\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \sin \left (d x + c\right )^{6} - 1512 \,{\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{5} + 630 \,{\left (3 \, B a^{2} + 6 \, A a b - B b^{2}\right )} \sin \left (d x + c\right )^{4} - 2520 \, A a^{2} \sin \left (d x + c\right ) + 840 \,{\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} - 1260 \,{\left (B a^{2} + 2 \, A a b\right )} \sin \left (d x + c\right )^{2}}{2520 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2520*(252*B*b^2*sin(d*x + c)^10 + 280*(2*B*a*b + A*b^2)*sin(d*x + c)^9 + 315*(B*a^2 + 2*A*a*b - 3*B*b^2)*si
n(d*x + c)^8 + 360*(A*a^2 - 6*B*a*b - 3*A*b^2)*sin(d*x + c)^7 - 1260*(B*a^2 + 2*A*a*b - B*b^2)*sin(d*x + c)^6
- 1512*(A*a^2 - 2*B*a*b - A*b^2)*sin(d*x + c)^5 + 630*(3*B*a^2 + 6*A*a*b - B*b^2)*sin(d*x + c)^4 - 2520*A*a^2*
sin(d*x + c) + 840*(3*A*a^2 - 2*B*a*b - A*b^2)*sin(d*x + c)^3 - 1260*(B*a^2 + 2*A*a*b)*sin(d*x + c)^2)/d

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Fricas [A]  time = 1.73448, size = 425, normalized size = 1.22 \begin{align*} \frac{252 \, B b^{2} \cos \left (d x + c\right )^{10} - 315 \,{\left (B a^{2} + 2 \, A a b + B b^{2}\right )} \cos \left (d x + c\right )^{8} - 8 \,{\left (35 \,{\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{8} - 5 \,{\left (9 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{6} - 6 \,{\left (9 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{4} - 144 \, A a^{2} - 32 \, B a b - 16 \, A b^{2} - 8 \,{\left (9 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2520 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2520*(252*B*b^2*cos(d*x + c)^10 - 315*(B*a^2 + 2*A*a*b + B*b^2)*cos(d*x + c)^8 - 8*(35*(2*B*a*b + A*b^2)*cos
(d*x + c)^8 - 5*(9*A*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)^6 - 6*(9*A*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)^4 - 14
4*A*a^2 - 32*B*a*b - 16*A*b^2 - 8*(9*A*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)^2)*sin(d*x + c))/d

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Sympy [A]  time = 36.3224, size = 389, normalized size = 1.11 \begin{align*} \begin{cases} \frac{16 A a^{2} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac{8 A a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 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 b \cos ^{8}{\left (c + d x \right )}}{4 d} + \frac{16 A b^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac{8 A b^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac{2 A b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac{A b^{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} + \frac{32 B a b \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac{16 B a b \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac{4 B a b \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac{2 B a b \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac{B b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{8 d} - \frac{B b^{2} \cos ^{10}{\left (c + d x \right )}}{40 d} & \text{for}\: d \neq 0 \\x \left (A + B \sin{\left (c \right )}\right ) \left (a + b \sin{\left (c \right )}\right )^{2} \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+b*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.38306, size = 377, normalized size = 1.08 \begin{align*} \frac{B b^{2} \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac{7 \, A a^{2} \sin \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac{{\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac{{\left (8 \, B a^{2} + 16 \, A a b - B b^{2}\right )} \cos \left (6 \, d x + 6 \, c\right )}{1024 \, d} - \frac{{\left (7 \, B a^{2} + 14 \, A a b + B b^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac{7 \,{\left (4 \, B a^{2} + 8 \, A a b + B b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} - \frac{{\left (2 \, B a b + A b^{2}\right )} \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac{{\left (4 \, A a^{2} - 10 \, B a b - 5 \, A b^{2}\right )} \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} + \frac{{\left (7 \, A a^{2} - 4 \, B a b - 2 \, A b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac{7 \,{\left (10 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )}{128 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5120*B*b^2*cos(10*d*x + 10*c)/d + 7/64*A*a^2*sin(3*d*x + 3*c)/d - 1/1024*(B*a^2 + 2*A*a*b - B*b^2)*cos(8*d*x
+ 8*c)/d - 1/1024*(8*B*a^2 + 16*A*a*b - B*b^2)*cos(6*d*x + 6*c)/d - 1/256*(7*B*a^2 + 14*A*a*b + B*b^2)*cos(4*
d*x + 4*c)/d - 7/512*(4*B*a^2 + 8*A*a*b + B*b^2)*cos(2*d*x + 2*c)/d - 1/2304*(2*B*a*b + A*b^2)*sin(9*d*x + 9*c
)/d + 1/1792*(4*A*a^2 - 10*B*a*b - 5*A*b^2)*sin(7*d*x + 7*c)/d + 1/320*(7*A*a^2 - 4*B*a*b - 2*A*b^2)*sin(5*d*x
+ 5*c)/d + 7/128*(10*A*a^2 + 2*B*a*b + A*b^2)*sin(d*x + c)/d