\(\int \cos ^7(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx\) [660]

Optimal result

Integrand size = 27, antiderivative size = 97 \[ \int \cos ^7(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cos ^8(c+d x)}{8 d}+\frac {a \cos ^{10}(c+d x)}{10 d}+\frac {a \sin ^3(c+d x)}{3 d}-\frac {3 a \sin ^5(c+d x)}{5 d}+\frac {3 a \sin ^7(c+d x)}{7 d}-\frac {a \sin ^9(c+d x)}{9 d} \]

[Out]

-1/8*a*cos(d*x+c)^8/d+1/10*a*cos(d*x+c)^10/d+1/3*a*sin(d*x+c)^3/d-3/5*a*sin(d*x+c)^5/d+3/7*a*sin(d*x+c)^7/d-1/
9*a*sin(d*x+c)^9/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2913, 2644, 276, 2645, 14} \[ \int \cos ^7(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \sin ^9(c+d x)}{9 d}+\frac {3 a \sin ^7(c+d x)}{7 d}-\frac {3 a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \cos ^{10}(c+d x)}{10 d}-\frac {a \cos ^8(c+d x)}{8 d} \]

[In]

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

[Out]

-1/8*(a*Cos[c + d*x]^8)/d + (a*Cos[c + d*x]^10)/(10*d) + (a*Sin[c + d*x]^3)/(3*d) - (3*a*Sin[c + d*x]^5)/(5*d)
+ (3*a*Sin[c + d*x]^7)/(7*d) - (a*Sin[c + d*x]^9)/(9*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2913

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]),
x_Symbol] :> Dist[a, Int[Cos[e + f*x]^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[Cos[e + f*x]^p*(d*Sin[e +
f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && IntegerQ[n] && ((LtQ[p, 0]
&& NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] || LtQ[p + 1, -n, 2*p + 1])

Rubi steps \begin{align*} \text {integral}& = a \int \cos ^7(c+d x) \sin ^2(c+d x) \, dx+a \int \cos ^7(c+d x) \sin ^3(c+d x) \, dx \\ & = -\frac {a \text {Subst}\left (\int x^7 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {a \text {Subst}\left (\int x^2 \left (1-x^2\right )^3 \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {a \text {Subst}\left (\int \left (x^2-3 x^4+3 x^6-x^8\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac {a \text {Subst}\left (\int \left (x^7-x^9\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {a \cos ^8(c+d x)}{8 d}+\frac {a \cos ^{10}(c+d x)}{10 d}+\frac {a \sin ^3(c+d x)}{3 d}-\frac {3 a \sin ^5(c+d x)}{5 d}+\frac {3 a \sin ^7(c+d x)}{7 d}-\frac {a \sin ^9(c+d x)}{9 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00 \[ \int \cos ^7(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a (4410 \cos (2 (c+d x))+1260 \cos (4 (c+d x))-315 \cos (6 (c+d x))-315 \cos (8 (c+d x))-63 \cos (10 (c+d x))-17640 \sin (c+d x)+2016 \sin (5 (c+d x))+900 \sin (7 (c+d x))+140 \sin (9 (c+d x)))}{322560 d} \]

[In]

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

[Out]

-1/322560*(a*(4410*Cos[2*(c + d*x)] + 1260*Cos[4*(c + d*x)] - 315*Cos[6*(c + d*x)] - 315*Cos[8*(c + d*x)] - 63
*Cos[10*(c + d*x)] - 17640*Sin[c + d*x] + 2016*Sin[5*(c + d*x)] + 900*Sin[7*(c + d*x)] + 140*Sin[9*(c + d*x)])
)/d

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(-\frac {a \left (\frac {\left (\sin ^{10}\left (d x +c \right )\right )}{10}+\frac {\left (\sin ^{9}\left (d x +c \right )\right )}{9}-\frac {3 \left (\sin ^{8}\left (d x +c \right )\right )}{8}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}+\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\)	\(88\)
default	\(-\frac {a \left (\frac {\left (\sin ^{10}\left (d x +c \right )\right )}{10}+\frac {\left (\sin ^{9}\left (d x +c \right )\right )}{9}-\frac {3 \left (\sin ^{8}\left (d x +c \right )\right )}{8}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}+\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\)	\(88\)
parallelrisch	\(-\frac {a \left (\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (5556 \cos \left (2 d x +2 c \right )+63 \sin \left (7 d x +7 c \right )+504 \sin \left (5 d x +5 c \right )+140 \cos \left (6 d x +6 c \right )+2205 \sin \left (d x +c \right )+1638 \sin \left (3 d x +3 c \right )+1320 \cos \left (4 d x +4 c \right )+6424\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{80640 d}\)	\(125\)
risch	\(\frac {7 a \sin \left (d x +c \right )}{128 d}+\frac {a \cos \left (10 d x +10 c \right )}{5120 d}-\frac {a \sin \left (9 d x +9 c \right )}{2304 d}+\frac {a \cos \left (8 d x +8 c \right )}{1024 d}-\frac {5 a \sin \left (7 d x +7 c \right )}{1792 d}+\frac {a \cos \left (6 d x +6 c \right )}{1024 d}-\frac {a \sin \left (5 d x +5 c \right )}{160 d}-\frac {a \cos \left (4 d x +4 c \right )}{256 d}-\frac {7 a \cos \left (2 d x +2 c \right )}{512 d}\)	\(134\)

[In]

int(cos(d*x+c)^7*sin(d*x+c)^2*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-a/d*(1/10*sin(d*x+c)^10+1/9*sin(d*x+c)^9-3/8*sin(d*x+c)^8-3/7*sin(d*x+c)^7+1/2*sin(d*x+c)^6+3/5*sin(d*x+c)^5-
1/4*sin(d*x+c)^4-1/3*sin(d*x+c)^3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.87 \[ \int \cos ^7(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {252 \, a \cos \left (d x + c\right )^{10} - 315 \, a \cos \left (d x + c\right )^{8} - 8 \, {\left (35 \, a \cos \left (d x + c\right )^{8} - 5 \, a \cos \left (d x + c\right )^{6} - 6 \, a \cos \left (d x + c\right )^{4} - 8 \, a \cos \left (d x + c\right )^{2} - 16 \, a\right )} \sin \left (d x + c\right )}{2520 \, d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.29 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.42 \[ \int \cos ^7(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {16 a \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {8 a \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {2 a \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {a \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{8 d} - \frac {a \cos ^{10}{\left (c + d x \right )}}{40 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin ^{2}{\left (c \right )} \cos ^{7}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**7*sin(d*x+c)**2*(a+a*sin(d*x+c)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.97 \[ \int \cos ^7(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {252 \, a \sin \left (d x + c\right )^{10} + 280 \, a \sin \left (d x + c\right )^{9} - 945 \, a \sin \left (d x + c\right )^{8} - 1080 \, a \sin \left (d x + c\right )^{7} + 1260 \, a \sin \left (d x + c\right )^{6} + 1512 \, a \sin \left (d x + c\right )^{5} - 630 \, a \sin \left (d x + c\right )^{4} - 840 \, a \sin \left (d x + c\right )^{3}}{2520 \, d} \]

[In]

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

[Out]

-1/2520*(252*a*sin(d*x + c)^10 + 280*a*sin(d*x + c)^9 - 945*a*sin(d*x + c)^8 - 1080*a*sin(d*x + c)^7 + 1260*a*
sin(d*x + c)^6 + 1512*a*sin(d*x + c)^5 - 630*a*sin(d*x + c)^4 - 840*a*sin(d*x + c)^3)/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.37 \[ \int \cos ^7(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac {a \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} + \frac {a \cos \left (6 \, d x + 6 \, c\right )}{1024 \, d} - \frac {a \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {7 \, a \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} - \frac {a \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} - \frac {5 \, a \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac {a \sin \left (5 \, d x + 5 \, c\right )}{160 \, d} + \frac {7 \, a \sin \left (d x + c\right )}{128 \, d} \]

[In]

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

[Out]

1/5120*a*cos(10*d*x + 10*c)/d + 1/1024*a*cos(8*d*x + 8*c)/d + 1/1024*a*cos(6*d*x + 6*c)/d - 1/256*a*cos(4*d*x
+ 4*c)/d - 7/512*a*cos(2*d*x + 2*c)/d - 1/2304*a*sin(9*d*x + 9*c)/d - 5/1792*a*sin(7*d*x + 7*c)/d - 1/160*a*si
n(5*d*x + 5*c)/d + 7/128*a*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 10.65 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.96 \[ \int \cos ^7(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {-\frac {a\,{\sin \left (c+d\,x\right )}^{10}}{10}-\frac {a\,{\sin \left (c+d\,x\right )}^9}{9}+\frac {3\,a\,{\sin \left (c+d\,x\right )}^8}{8}+\frac {3\,a\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {a\,{\sin \left (c+d\,x\right )}^6}{2}-\frac {3\,a\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \]

[In]

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

[Out]

((a*sin(c + d*x)^3)/3 + (a*sin(c + d*x)^4)/4 - (3*a*sin(c + d*x)^5)/5 - (a*sin(c + d*x)^6)/2 + (3*a*sin(c + d*
x)^7)/7 + (3*a*sin(c + d*x)^8)/8 - (a*sin(c + d*x)^9)/9 - (a*sin(c + d*x)^10)/10)/d