\(\int \cos ^5(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx\) [1216]

Optimal result

Integrand size = 27, antiderivative size = 138 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2 \sin ^2(c+d x)}{2 d}+\frac {2 a b \sin ^3(c+d x)}{3 d}-\frac {\left (2 a^2-b^2\right ) \sin ^4(c+d x)}{4 d}-\frac {4 a b \sin ^5(c+d x)}{5 d}+\frac {\left (a^2-2 b^2\right ) \sin ^6(c+d x)}{6 d}+\frac {2 a b \sin ^7(c+d x)}{7 d}+\frac {b^2 \sin ^8(c+d x)}{8 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2916, 12, 786} \[ \int \cos ^5(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (a^2-2 b^2\right ) \sin ^6(c+d x)}{6 d}-\frac {\left (2 a^2-b^2\right ) \sin ^4(c+d x)}{4 d}+\frac {a^2 \sin ^2(c+d x)}{2 d}+\frac {2 a b \sin ^7(c+d x)}{7 d}-\frac {4 a b \sin ^5(c+d x)}{5 d}+\frac {2 a b \sin ^3(c+d x)}{3 d}+\frac {b^2 \sin ^8(c+d x)}{8 d} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 786

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]

Rule 2916

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/b)*x)^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]

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {-2590 b^2+840 \left (10 a^2+3 b^2\right ) \cos (2 (c+d x))+420 \left (8 a^2+b^2\right ) \cos (4 (c+d x))+560 a^2 \cos (6 (c+d x))-280 b^2 \cos (6 (c+d x))-105 b^2 \cos (8 (c+d x))-16800 a b \sin (c+d x)+1120 a b \sin (3 (c+d x))+2016 a b \sin (5 (c+d x))+480 a b \sin (7 (c+d x))}{107520 d} \]

[In]

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

[Out]

-1/107520*(-2590*b^2 + 840*(10*a^2 + 3*b^2)*Cos[2*(c + d*x)] + 420*(8*a^2 + b^2)*Cos[4*(c + d*x)] + 560*a^2*Co
s[6*(c + d*x)] - 280*b^2*Cos[6*(c + d*x)] - 105*b^2*Cos[8*(c + d*x)] - 16800*a*b*Sin[c + d*x] + 1120*a*b*Sin[3
*(c + d*x)] + 2016*a*b*Sin[5*(c + d*x)] + 480*a*b*Sin[7*(c + d*x)])/d

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {\frac {b^{2} \left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {2 a b \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (a^{2}-2 b^{2}\right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {4 a b \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (-2 a^{2}+b^{2}\right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {2 a b \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{2}}{d}\)	\(106\)
default	\(\frac {\frac {b^{2} \left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {2 a b \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (a^{2}-2 b^{2}\right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}-\frac {4 a b \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (-2 a^{2}+b^{2}\right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {2 a b \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{2}}{d}\)	\(106\)
parallelrisch	\(\frac {105 b^{2} \cos \left (8 d x +8 c \right )-480 a b \sin \left (7 d x +7 c \right )-560 a^{2} \cos \left (6 d x +6 c \right )+280 b^{2} \cos \left (6 d x +6 c \right )-2016 a b \sin \left (5 d x +5 c \right )-1120 a b \sin \left (3 d x +3 c \right )-3360 \cos \left (4 d x +4 c \right ) a^{2}-420 b^{2} \cos \left (4 d x +4 c \right )-8400 a^{2} \cos \left (2 d x +2 c \right )-2520 b^{2} \cos \left (2 d x +2 c \right )+16800 a b \sin \left (d x +c \right )+12320 a^{2}+2555 b^{2}}{107520 d}\)	\(164\)
risch	\(\frac {5 a b \sin \left (d x +c \right )}{32 d}+\frac {b^{2} \cos \left (8 d x +8 c \right )}{1024 d}-\frac {a b \sin \left (7 d x +7 c \right )}{224 d}-\frac {\cos \left (6 d x +6 c \right ) a^{2}}{192 d}+\frac {\cos \left (6 d x +6 c \right ) b^{2}}{384 d}-\frac {3 a b \sin \left (5 d x +5 c \right )}{160 d}-\frac {a^{2} \cos \left (4 d x +4 c \right )}{32 d}-\frac {\cos \left (4 d x +4 c \right ) b^{2}}{256 d}-\frac {a b \sin \left (3 d x +3 c \right )}{96 d}-\frac {5 a^{2} \cos \left (2 d x +2 c \right )}{64 d}-\frac {3 \cos \left (2 d x +2 c \right ) b^{2}}{128 d}\)	\(182\)
norman	\(\frac {\frac {2 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (a^{2}+b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (a^{2}+b^{2}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {10 \left (4 a^{2}+4 b^{2}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (13 a^{2}-8 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (13 a^{2}-8 b^{2}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {16 a b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {16 a b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {1376 a b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d}+\frac {1376 a b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d}+\frac {16 a b \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {16 a b \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}\)	\(291\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.62 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {105 \, b^{2} \cos \left (d x + c\right )^{8} - 140 \, {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{6} - 16 \, {\left (15 \, a b \cos \left (d x + c\right )^{6} - 3 \, a b \cos \left (d x + c\right )^{4} - 4 \, a b \cos \left (d x + c\right )^{2} - 8 \, a b\right )} \sin \left (d x + c\right )}{840 \, d} \]

[In]

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

[Out]

1/840*(105*b^2*cos(d*x + c)^8 - 140*(a^2 + b^2)*cos(d*x + c)^6 - 16*(15*a*b*cos(d*x + c)^6 - 3*a*b*cos(d*x + c
)^4 - 4*a*b*cos(d*x + c)^2 - 8*a*b)*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.67 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.01 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\begin {cases} - \frac {a^{2} \cos ^{6}{\left (c + d x \right )}}{6 d} + \frac {16 a b \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {8 a b \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {2 a b \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac {b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac {b^{2} \cos ^{8}{\left (c + d x \right )}}{24 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{2} \sin {\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {105 \, b^{2} \sin \left (d x + c\right )^{8} + 240 \, a b \sin \left (d x + c\right )^{7} - 672 \, a b \sin \left (d x + c\right )^{5} + 140 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{6} + 560 \, a b \sin \left (d x + c\right )^{3} - 210 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{4} + 420 \, a^{2} \sin \left (d x + c\right )^{2}}{840 \, d} \]

[In]

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

[Out]

1/840*(105*b^2*sin(d*x + c)^8 + 240*a*b*sin(d*x + c)^7 - 672*a*b*sin(d*x + c)^5 + 140*(a^2 - 2*b^2)*sin(d*x +
c)^6 + 560*a*b*sin(d*x + c)^3 - 210*(2*a^2 - b^2)*sin(d*x + c)^4 + 420*a^2*sin(d*x + c)^2)/d

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.10 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {b^{2} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {a b \sin \left (7 \, d x + 7 \, c\right )}{224 \, d} - \frac {3 \, a b \sin \left (5 \, d x + 5 \, c\right )}{160 \, d} - \frac {a b \sin \left (3 \, d x + 3 \, c\right )}{96 \, d} + \frac {5 \, a b \sin \left (d x + c\right )}{32 \, d} - \frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac {{\left (8 \, a^{2} + b^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {{\left (10 \, a^{2} + 3 \, b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} \]

[In]

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

[Out]

1/1024*b^2*cos(8*d*x + 8*c)/d - 1/224*a*b*sin(7*d*x + 7*c)/d - 3/160*a*b*sin(5*d*x + 5*c)/d - 1/96*a*b*sin(3*d
*x + 3*c)/d + 5/32*a*b*sin(d*x + c)/d - 1/384*(2*a^2 - b^2)*cos(6*d*x + 6*c)/d - 1/256*(8*a^2 + b^2)*cos(4*d*x
+ 4*c)/d - 1/128*(10*a^2 + 3*b^2)*cos(2*d*x + 2*c)/d

Mupad [B] (verification not implemented)

Time = 11.33 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78 \[ \int \cos ^5(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {{\sin \left (c+d\,x\right )}^6\,\left (\frac {a^2}{6}-\frac {b^2}{3}\right )-{\sin \left (c+d\,x\right )}^4\,\left (\frac {a^2}{2}-\frac {b^2}{4}\right )+\frac {a^2\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {b^2\,{\sin \left (c+d\,x\right )}^8}{8}+\frac {2\,a\,b\,{\sin \left (c+d\,x\right )}^3}{3}-\frac {4\,a\,b\,{\sin \left (c+d\,x\right )}^5}{5}+\frac {2\,a\,b\,{\sin \left (c+d\,x\right )}^7}{7}}{d} \]

[In]

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

[Out]

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