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

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 105, 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} \[ \frac {(A-5 B) (a \sin (c+d x)+a)^7}{7 a^5 d}-\frac {2 (A-2 B) (a \sin (c+d x)+a)^6}{3 a^4 d}+\frac {4 (A-B) (a \sin (c+d x)+a)^5}{5 a^3 d}+\frac {B (a \sin (c+d x)+a)^8}{8 a^6 d} \]

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 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 ^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*}

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Mathematica [A]  time = 0.36, 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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fricas [A]  time = 0.77, size = 109, normalized size = 1.04 \[ \frac {105 \, B a^{2} \cos \left (d x + c\right )^{8} - 280 \, {\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{6} - 8 \, {\left (15 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{6} - 6 \, {\left (4 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{4} - 8 \, {\left (4 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} - 16 \, {\left (4 \, A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{840 \, d} \]

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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giac [B]  time = 0.33, size = 202, normalized size = 1.92 \[ \frac {B a^{2} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {{\left (4 \, A a^{2} + B a^{2}\right )} \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac {{\left (16 \, A a^{2} + 9 \, B a^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {{\left (20 \, A a^{2} + 13 \, B a^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} - \frac {{\left (A a^{2} + 2 \, B a^{2}\right )} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {{\left (A a^{2} - 6 \, B a^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (19 \, A a^{2} - 2 \, B a^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {5 \, {\left (9 \, A a^{2} + 2 \, B a^{2}\right )} \sin \left (d x + c\right )}{64 \, d} \]

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

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maple [B]  time = 0.47, size = 201, normalized size = 1.91 \[ \frac {a^{2} A \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )+B \,a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{8}-\frac {\left (\cos ^{6}\left (d x +c \right )\right )}{24}\right )-\frac {a^{2} A \left (\cos ^{6}\left (d x +c \right )\right )}{3}+2 B \,a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )+\frac {a^{2} A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}-\frac {B \,a^{2} \left (\cos ^{6}\left (d x +c \right )\right )}{6}}{d} \]

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 = 0.31, size = 142, normalized size = 1.35 \[ \frac {105 \, B a^{2} \sin \left (d x + c\right )^{8} + 120 \, {\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{7} + 140 \, {\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right )^{6} - 168 \, {\left (A + 4 \, B\right )} a^{2} \sin \left (d x + c\right )^{5} - 210 \, {\left (4 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{4} - 280 \, {\left (A - 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{3} + 420 \, {\left (2 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{2} + 840 \, A a^{2} \sin \left (d x + c\right )}{840 \, d} \]

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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mupad [B]  time = 0.12, size = 140, normalized size = 1.33 \[ \frac {\frac {a^2\,{\sin \left (c+d\,x\right )}^2\,\left (2\,A+B\right )}{2}-\frac {a^2\,{\sin \left (c+d\,x\right )}^3\,\left (A-2\,B\right )}{3}-\frac {a^2\,{\sin \left (c+d\,x\right )}^4\,\left (4\,A+B\right )}{4}-\frac {a^2\,{\sin \left (c+d\,x\right )}^5\,\left (A+4\,B\right )}{5}+\frac {a^2\,{\sin \left (c+d\,x\right )}^7\,\left (A+2\,B\right )}{7}+\frac {B\,a^2\,{\sin \left (c+d\,x\right )}^8}{8}+\frac {a^2\,{\sin \left (c+d\,x\right )}^6\,\left (2\,A-B\right )}{6}+A\,a^2\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 10.17, size = 335, normalized size = 3.19 \[ \begin {cases} \frac {8 A a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 A a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {8 A a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} + \frac {4 A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {A a^{2} \cos ^{6}{\left (c + d x \right )}}{3 d} + \frac {B a^{2} \sin ^{8}{\left (c + d x \right )}}{24 d} + \frac {16 B a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {B a^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{6 d} + \frac {8 B a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {B a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {2 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a \sin {\relax (c )} + a\right )^{2} \cos ^{5}{\relax (c )} & \text {otherwise} \end {cases} \]

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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