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

Optimal. Leaf size=105 \[ \frac{(A-5 B) (a \sin (c+d x)+a)^7}{7 a^5 d}-\frac{2 (A-2 B) (a \sin (c+d x)+a)^6}{3 a^4 d}+\frac{4 (A-B) (a \sin (c+d x)+a)^5}{5 a^3 d}+\frac{B (a \sin (c+d x)+a)^8}{8 a^6 d} \]

[Out]

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

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Rubi [A]  time = 0.148639, antiderivative size = 105, 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} \[ \frac{(A-5 B) (a \sin (c+d x)+a)^7}{7 a^5 d}-\frac{2 (A-2 B) (a \sin (c+d x)+a)^6}{3 a^4 d}+\frac{4 (A-B) (a \sin (c+d x)+a)^5}{5 a^3 d}+\frac{B (a \sin (c+d x)+a)^8}{8 a^6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.343973, size = 70, normalized size = 0.67 \[ \frac{a^2 (\sin (c+d x)+1)^5 \left (15 (8 A-19 B) \sin ^2(c+d x)-5 (64 A-47 B) \sin (c+d x)+232 A+105 B \sin ^3(c+d x)-47 B\right )}{840 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 + Sin[c + d*x])^5*(232*A - 47*B - 5*(64*A - 47*B)*Sin[c + d*x] + 15*(8*A - 19*B)*Sin[c + d*x]^2 + 105*
B*Sin[c + d*x]^3))/(840*d)

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Maple [B]  time = 0.067, size = 201, normalized size = 1.9 \begin{align*}{\frac{1}{d} \left ({a}^{2}A \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{a}^{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 ) -{\frac{{a}^{2}A \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{3}}+2\,B{a}^{2} \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 ) +{\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}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01849, size = 192, normalized size = 1.83 \begin{align*} \frac{105 \, B a^{2} \sin \left (d x + c\right )^{8} + 120 \,{\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{7} + 140 \,{\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right )^{6} - 168 \,{\left (A + 4 \, B\right )} a^{2} \sin \left (d x + c\right )^{5} - 210 \,{\left (4 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{4} - 280 \,{\left (A - 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{3} + 420 \,{\left (2 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{2} + 840 \, A a^{2} \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+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.9253, size = 277, normalized size = 2.64 \begin{align*} \frac{105 \, B a^{2} \cos \left (d x + c\right )^{8} - 280 \,{\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{6} - 8 \,{\left (15 \,{\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} - 6 \,{\left (4 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{4} - 8 \,{\left (4 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} - 16 \,{\left (4 \, A + B\right )} a^{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+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/840*(105*B*a^2*cos(d*x + c)^8 - 280*(A + B)*a^2*cos(d*x + c)^6 - 8*(15*(A + 2*B)*a^2*cos(d*x + c)^6 - 6*(4*A
+ B)*a^2*cos(d*x + c)^4 - 8*(4*A + B)*a^2*cos(d*x + c)^2 - 16*(4*A + B)*a^2)*sin(d*x + c))/d

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Sympy [A]  time = 14.0238, size = 335, normalized size = 3.19 \begin{align*} \begin{cases} \frac{8 A a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac{4 A a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac{8 A a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 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^{2} \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac{B a^{2} \sin ^{8}{\left (c + d x \right )}}{24 d} + \frac{16 B a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac{B a^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{6 d} + \frac{8 B a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac{B a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac{2 B a^{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} & \text{for}\: d \neq 0 \\x \left (A + B \sin{\left (c \right )}\right ) \left (a \sin{\left (c \right )} + a\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+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((8*A*a**2*sin(c + d*x)**7/(105*d) + 4*A*a**2*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + 8*A*a**2*sin(c
+ d*x)**5/(15*d) + A*a**2*sin(c + d*x)**3*cos(c + d*x)**4/(3*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**2*cos(c + d*x)**6/(3*d) + B*a**2*sin(c + d*x)**8/(24*d) + 1
6*B*a**2*sin(c + d*x)**7/(105*d) + B*a**2*sin(c + d*x)**6*cos(c + d*x)**2/(6*d) + 8*B*a**2*sin(c + d*x)**5*cos
(c + d*x)**2/(15*d) + B*a**2*sin(c + d*x)**4*cos(c + d*x)**4/(4*d) + 2*B*a**2*sin(c + d*x)**3*cos(c + d*x)**4/
(3*d) - B*a**2*cos(c + d*x)**6/(6*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**2*cos(c)**5, True))

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Giac [B]  time = 1.36602, size = 273, normalized size = 2.6 \begin{align*} \frac{B a^{2} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac{{\left (4 \, A a^{2} + B a^{2}\right )} \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac{{\left (16 \, A a^{2} + 9 \, B a^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac{{\left (20 \, A a^{2} + 13 \, B a^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} - \frac{{\left (A a^{2} + 2 \, B a^{2}\right )} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac{{\left (A a^{2} - 6 \, B a^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac{{\left (19 \, A a^{2} - 2 \, B a^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac{5 \,{\left (9 \, A a^{2} + 2 \, B a^{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+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/1024*B*a^2*cos(8*d*x + 8*c)/d - 1/384*(4*A*a^2 + B*a^2)*cos(6*d*x + 6*c)/d - 1/256*(16*A*a^2 + 9*B*a^2)*cos(
4*d*x + 4*c)/d - 1/128*(20*A*a^2 + 13*B*a^2)*cos(2*d*x + 2*c)/d - 1/448*(A*a^2 + 2*B*a^2)*sin(7*d*x + 7*c)/d +
1/320*(A*a^2 - 6*B*a^2)*sin(5*d*x + 5*c)/d + 1/192*(19*A*a^2 - 2*B*a^2)*sin(3*d*x + 3*c)/d + 5/64*(9*A*a^2 +
2*B*a^2)*sin(d*x + c)/d