3.1537 \(\int \cos ^5(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=231 \[ -\frac{\left (-5 a^2 B+2 a A b+b^2 B\right ) (a+b \sin (c+d x))^6}{3 b^6 d}+\frac{2 \left (3 a^2 A b-5 a^3 B+3 a b^2 B-A b^3\right ) (a+b \sin (c+d x))^5}{5 b^6 d}-\frac{\left (a^2-b^2\right ) \left (-5 a^2 B+4 a A b+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^6 d}+\frac{\left (a^2-b^2\right )^2 (A b-a B) (a+b \sin (c+d x))^3}{3 b^6 d}+\frac{(A b-5 a B) (a+b \sin (c+d x))^7}{7 b^6 d}+\frac{B (a+b \sin (c+d x))^8}{8 b^6 d} \]

[Out]

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

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Rubi [A]  time = 0.259003, antiderivative size = 231, 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} \[ -\frac{\left (-5 a^2 B+2 a A b+b^2 B\right ) (a+b \sin (c+d x))^6}{3 b^6 d}+\frac{2 \left (3 a^2 A b-5 a^3 B+3 a b^2 B-A b^3\right ) (a+b \sin (c+d x))^5}{5 b^6 d}-\frac{\left (a^2-b^2\right ) \left (-5 a^2 B+4 a A b+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^6 d}+\frac{\left (a^2-b^2\right )^2 (A b-a B) (a+b \sin (c+d x))^3}{3 b^6 d}+\frac{(A b-5 a B) (a+b \sin (c+d x))^7}{7 b^6 d}+\frac{B (a+b \sin (c+d x))^8}{8 b^6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

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

Mathematica [A]  time = 0.49881, size = 227, normalized size = 0.98 \[ \frac{140 b^6 \left (a^2 B+2 a A b-2 b^2 B\right ) \sin ^6(c+d x)+168 b^6 \left (a^2 A-4 a b B-2 A b^2\right ) \sin ^5(c+d x)+210 b^6 \left (-2 a^2 B-4 a A b+b^2 B\right ) \sin ^4(c+d x)+280 b^6 \left (-2 a^2 A+2 a b B+A b^2\right ) \sin ^3(c+d x)+840 a^2 A b^6 \sin (c+d x)+a^4 B \left (-28 a^2 b^2+3 a^4+210 b^4\right )+120 b^7 (2 a B+A b) \sin ^7(c+d x)+420 a b^6 (a B+2 A b) \sin ^2(c+d x)+105 b^8 B \sin ^8(c+d x)}{840 b^6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^4*(3*a^4 - 28*a^2*b^2 + 210*b^4)*B + 840*a^2*A*b^6*Sin[c + d*x] + 420*a*b^6*(2*A*b + a*B)*Sin[c + d*x]^2 +
280*b^6*(-2*a^2*A + A*b^2 + 2*a*b*B)*Sin[c + d*x]^3 + 210*b^6*(-4*a*A*b - 2*a^2*B + b^2*B)*Sin[c + d*x]^4 + 16
8*b^6*(a^2*A - 2*A*b^2 - 4*a*b*B)*Sin[c + d*x]^5 + 140*b^6*(2*a*A*b + a^2*B - 2*b^2*B)*Sin[c + d*x]^6 + 120*b^
7*(A*b + 2*a*B)*Sin[c + d*x]^7 + 105*b^8*B*Sin[c + d*x]^8)/(840*b^6*d)

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Maple [A]  time = 0.074, size = 199, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{2}A\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }-{\frac{B{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{6}}-{\frac{Aab \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{3}}+2\,Bab \left ( -1/7\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}+1/35\, \left ( 8/3+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) +A{b}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{7}}+{\frac{\sin \left ( dx+c \right ) }{35} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) +B{b}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{8}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{24}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

1/d*(1/5*a^2*A*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)-1/6*B*a^2*cos(d*x+c)^6-1/3*A*a*b*cos(d*x+c)^6+2*
B*a*b*(-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))+A*b^2*(-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))+B*b^2*(-1/8*sin(d*x+c)^2*cos(d*x+c)^6-1/24
*cos(d*x+c)^6))

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Maxima [A]  time = 0.987959, size = 248, normalized size = 1.07 \begin{align*} \frac{105 \, B b^{2} \sin \left (d x + c\right )^{8} + 120 \,{\left (2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )^{7} + 140 \,{\left (B a^{2} + 2 \, A a b - 2 \, B b^{2}\right )} \sin \left (d x + c\right )^{6} + 168 \,{\left (A a^{2} - 4 \, B a b - 2 \, A b^{2}\right )} \sin \left (d x + c\right )^{5} - 210 \,{\left (2 \, B a^{2} + 4 \, A a b - B b^{2}\right )} \sin \left (d x + c\right )^{4} + 840 \, A a^{2} \sin \left (d x + c\right ) - 280 \,{\left (2 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} + 420 \,{\left (B a^{2} + 2 \, A a b\right )} \sin \left (d x + c\right )^{2}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/840*(105*B*b^2*sin(d*x + c)^8 + 120*(2*B*a*b + A*b^2)*sin(d*x + c)^7 + 140*(B*a^2 + 2*A*a*b - 2*B*b^2)*sin(d
*x + c)^6 + 168*(A*a^2 - 4*B*a*b - 2*A*b^2)*sin(d*x + c)^5 - 210*(2*B*a^2 + 4*A*a*b - B*b^2)*sin(d*x + c)^4 +
840*A*a^2*sin(d*x + c) - 280*(2*A*a^2 - 2*B*a*b - A*b^2)*sin(d*x + c)^3 + 420*(B*a^2 + 2*A*a*b)*sin(d*x + c)^2
)/d

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Fricas [A]  time = 1.64761, size = 356, normalized size = 1.54 \begin{align*} \frac{105 \, B b^{2} \cos \left (d x + c\right )^{8} - 140 \,{\left (B a^{2} + 2 \, A a b + B b^{2}\right )} \cos \left (d x + c\right )^{6} - 8 \,{\left (15 \,{\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{6} - 3 \,{\left (7 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{4} - 56 \, A a^{2} - 16 \, B a b - 8 \, A b^{2} - 4 \,{\left (7 \, 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 )}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/840*(105*B*b^2*cos(d*x + c)^8 - 140*(B*a^2 + 2*A*a*b + B*b^2)*cos(d*x + c)^6 - 8*(15*(2*B*a*b + A*b^2)*cos(d
*x + c)^6 - 3*(7*A*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)^4 - 56*A*a^2 - 16*B*a*b - 8*A*b^2 - 4*(7*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 = 13.3417, size = 335, normalized size = 1.45 \begin{align*} \begin{cases} \frac{8 A a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{A a^{2} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac{A a b \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac{8 A b^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac{4 A b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac{A b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac{B a^{2} \cos ^{6}{\left (c + d x \right )}}{6 d} + \frac{16 B a b \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac{8 B a b \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac{2 B a b \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} + \frac{B b^{2} \sin ^{8}{\left (c + d x \right )}}{24 d} + \frac{B b^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{6 d} + \frac{B b^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 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 ^{5}{\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)**5*(a+b*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((8*A*a**2*sin(c + d*x)**5/(15*d) + 4*A*a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + A*a**2*sin(c + d
*x)*cos(c + d*x)**4/d - A*a*b*cos(c + d*x)**6/(3*d) + 8*A*b**2*sin(c + d*x)**7/(105*d) + 4*A*b**2*sin(c + d*x)
**5*cos(c + d*x)**2/(15*d) + A*b**2*sin(c + d*x)**3*cos(c + d*x)**4/(3*d) - B*a**2*cos(c + d*x)**6/(6*d) + 16*
B*a*b*sin(c + d*x)**7/(105*d) + 8*B*a*b*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + 2*B*a*b*sin(c + d*x)**3*cos(c
+ d*x)**4/(3*d) + B*b**2*sin(c + d*x)**8/(24*d) + B*b**2*sin(c + d*x)**6*cos(c + d*x)**2/(6*d) + B*b**2*sin(c
+ d*x)**4*cos(c + d*x)**4/(4*d), Ne(d, 0)), (x*(A + B*sin(c))*(a + b*sin(c))**2*cos(c)**5, True))

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Giac [A]  time = 1.28887, size = 312, normalized size = 1.35 \begin{align*} \frac{B b^{2} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac{{\left (2 \, B a^{2} + 4 \, A a b - B b^{2}\right )} \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac{{\left (8 \, B a^{2} + 16 \, A a b + B b^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac{{\left (10 \, B a^{2} + 20 \, A a b + 3 \, B b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} - \frac{{\left (2 \, B a b + A b^{2}\right )} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac{{\left (4 \, A a^{2} - 6 \, B a b - 3 \, A b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac{{\left (20 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac{5 \,{\left (8 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/1024*B*b^2*cos(8*d*x + 8*c)/d - 1/384*(2*B*a^2 + 4*A*a*b - B*b^2)*cos(6*d*x + 6*c)/d - 1/256*(8*B*a^2 + 16*A
*a*b + B*b^2)*cos(4*d*x + 4*c)/d - 1/128*(10*B*a^2 + 20*A*a*b + 3*B*b^2)*cos(2*d*x + 2*c)/d - 1/448*(2*B*a*b +
A*b^2)*sin(7*d*x + 7*c)/d + 1/320*(4*A*a^2 - 6*B*a*b - 3*A*b^2)*sin(5*d*x + 5*c)/d + 1/192*(20*A*a^2 - 2*B*a*
b - A*b^2)*sin(3*d*x + 3*c)/d + 5/64*(8*A*a^2 + 2*B*a*b + A*b^2)*sin(d*x + c)/d