\(\int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx\) [497]

Optimal result

Integrand size = 27, antiderivative size = 81 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cos ^6(c+d x)}{6 d}+\frac {a \cos ^8(c+d x)}{8 d}+\frac {a \sin ^3(c+d x)}{3 d}-\frac {2 a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^7(c+d x)}{7 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 81, 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} \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin ^7(c+d x)}{7 d}-\frac {2 a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^3(c+d x)}{3 d}+\frac {a \cos ^8(c+d x)}{8 d}-\frac {a \cos ^6(c+d x)}{6 d} \]

[In]

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

[Out]

-1/6*(a*Cos[c + d*x]^6)/d + (a*Cos[c + d*x]^8)/(8*d) + (a*Sin[c + d*x]^3)/(3*d) - (2*a*Sin[c + d*x]^5)/(5*d) +
(a*Sin[c + d*x]^7)/(7*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 ^5(c+d x) \sin ^2(c+d x) \, dx+a \int \cos ^5(c+d x) \sin ^3(c+d x) \, dx \\ & = -\frac {a \text {Subst}\left (\int x^5 \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 )^2 \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {a \text {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sin (c+d x)\right )}{d}-\frac {a \text {Subst}\left (\int \left (x^5-x^7\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {a \cos ^6(c+d x)}{6 d}+\frac {a \cos ^8(c+d x)}{8 d}+\frac {a \sin ^3(c+d x)}{3 d}-\frac {2 a \sin ^5(c+d x)}{5 d}+\frac {a \sin ^7(c+d x)}{7 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.07 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a (2520 \cos (2 (c+d x))+420 \cos (4 (c+d x))-280 \cos (6 (c+d x))-105 \cos (8 (c+d x))-8400 \sin (c+d x)+560 \sin (3 (c+d x))+1008 \sin (5 (c+d x))+240 \sin (7 (c+d x)))}{107520 d} \]

[In]

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

[Out]

-1/107520*(a*(2520*Cos[2*(c + d*x)] + 420*Cos[4*(c + d*x)] - 280*Cos[6*(c + d*x)] - 105*Cos[8*(c + d*x)] - 840
0*Sin[c + d*x] + 560*Sin[3*(c + d*x)] + 1008*Sin[5*(c + d*x)] + 240*Sin[7*(c + d*x)]))/d

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {a \left (\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {\left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{3}-\frac {2 \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\)	\(67\)
default	\(\frac {a \left (\frac {\left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {\left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{3}-\frac {2 \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}\right )}{d}\)	\(67\)
parallelrisch	\(\frac {a \left (-\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (1728 \cos \left (2 d x +2 c \right )+105 \sin \left (5 d x +5 c \right )+1050 \sin \left (d x +c \right )+595 \sin \left (3 d x +3 c \right )+240 \cos \left (4 d x +4 c \right )+2512\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{26880 d}\)	\(105\)
risch	\(\frac {5 a \sin \left (d x +c \right )}{64 d}+\frac {a \cos \left (8 d x +8 c \right )}{1024 d}-\frac {a \sin \left (7 d x +7 c \right )}{448 d}+\frac {a \cos \left (6 d x +6 c \right )}{384 d}-\frac {3 a \sin \left (5 d x +5 c \right )}{320 d}-\frac {a \cos \left (4 d x +4 c \right )}{256 d}-\frac {a \sin \left (3 d x +3 c \right )}{192 d}-\frac {3 a \cos \left (2 d x +2 c \right )}{128 d}\)	\(119\)
norman	\(\frac {\frac {8 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {688 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d}+\frac {688 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d}+\frac {8 a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {8 a \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {16 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {16 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {40 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}\)	\(205\)

[In]

int(cos(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

a/d*(1/8*sin(d*x+c)^8+1/7*sin(d*x+c)^7-1/3*sin(d*x+c)^6-2/5*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 = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {105 \, a \cos \left (d x + c\right )^{8} - 140 \, a \cos \left (d x + c\right )^{6} - 8 \, {\left (15 \, a \cos \left (d x + c\right )^{6} - 3 \, a \cos \left (d x + c\right )^{4} - 4 \, a \cos \left (d x + c\right )^{2} - 8 \, a\right )} \sin \left (d x + c\right )}{840 \, d} \]

[In]

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

[Out]

1/840*(105*a*cos(d*x + c)^8 - 140*a*cos(d*x + c)^6 - 8*(15*a*cos(d*x + c)^6 - 3*a*cos(d*x + c)^4 - 4*a*cos(d*x
+ c)^2 - 8*a)*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.64 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.41 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {8 a \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 a \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {a \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac {a \cos ^{8}{\left (c + d x \right )}}{24 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin ^{2}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**5*sin(d*x+c)**2*(a+a*sin(d*x+c)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {105 \, a \sin \left (d x + c\right )^{8} + 120 \, a \sin \left (d x + c\right )^{7} - 280 \, a \sin \left (d x + c\right )^{6} - 336 \, a \sin \left (d x + c\right )^{5} + 210 \, a \sin \left (d x + c\right )^{4} + 280 \, a \sin \left (d x + c\right )^{3}}{840 \, d} \]

[In]

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

[Out]

1/840*(105*a*sin(d*x + c)^8 + 120*a*sin(d*x + c)^7 - 280*a*sin(d*x + c)^6 - 336*a*sin(d*x + c)^5 + 210*a*sin(d
*x + c)^4 + 280*a*sin(d*x + c)^3)/d

Giac [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.46 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} + \frac {a \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac {a \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {3 \, a \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} - \frac {a \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {3 \, a \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {a \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {5 \, a \sin \left (d x + c\right )}{64 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.88 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {\frac {a\,{\sin \left (c+d\,x\right )}^8}{8}+\frac {a\,{\sin \left (c+d\,x\right )}^7}{7}-\frac {a\,{\sin \left (c+d\,x\right )}^6}{3}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {a\,{\sin \left (c+d\,x\right )}^4}{4}+\frac {a\,{\sin \left (c+d\,x\right )}^3}{3}}{d} \]

[In]

int(cos(c + d*x)^5*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 - (2*a*sin(c + d*x)^5)/5 - (a*sin(c + d*x)^6)/3 + (a*sin(c + d*x)
^7)/7 + (a*sin(c + d*x)^8)/8)/d