\(\int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx\) [1215]

Optimal result

Integrand size = 29, antiderivative size = 138 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {a b \sin ^4(c+d x)}{2 d}-\frac {\left (2 a^2-b^2\right ) \sin ^5(c+d x)}{5 d}-\frac {2 a b \sin ^6(c+d x)}{3 d}+\frac {\left (a^2-2 b^2\right ) \sin ^7(c+d x)}{7 d}+\frac {a b \sin ^8(c+d x)}{4 d}+\frac {b^2 \sin ^9(c+d x)}{9 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2916, 12, 962} \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (a^2-2 b^2\right ) \sin ^7(c+d x)}{7 d}-\frac {\left (2 a^2-b^2\right ) \sin ^5(c+d x)}{5 d}+\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {a b \sin ^8(c+d x)}{4 d}-\frac {2 a b \sin ^6(c+d x)}{3 d}+\frac {a b \sin ^4(c+d x)}{2 d}+\frac {b^2 \sin ^9(c+d x)}{9 d} \]

[In]

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

[Out]

(a^2*Sin[c + d*x]^3)/(3*d) + (a*b*Sin[c + d*x]^4)/(2*d) - ((2*a^2 - b^2)*Sin[c + d*x]^5)/(5*d) - (2*a*b*Sin[c
+ d*x]^6)/(3*d) + ((a^2 - 2*b^2)*Sin[c + d*x]^7)/(7*d) + (a*b*Sin[c + d*x]^8)/(4*d) + (b^2*Sin[c + d*x]^9)/(9*
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 962

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 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^2 (a+x)^2 \left (b^2-x^2\right )^2}{b^2} \, dx,x,b \sin (c+d x)\right )}{b^5 d} \\ & = \frac {\text {Subst}\left (\int x^2 (a+x)^2 \left (b^2-x^2\right )^2 \, dx,x,b \sin (c+d x)\right )}{b^7 d} \\ & = \frac {\text {Subst}\left (\int \left (a^2 b^4 x^2+2 a b^4 x^3+b^2 \left (-2 a^2+b^2\right ) x^4-4 a b^2 x^5+\left (a^2-2 b^2\right ) x^6+2 a x^7+x^8\right ) \, dx,x,b \sin (c+d x)\right )}{b^7 d} \\ & = \frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {a b \sin ^4(c+d x)}{2 d}-\frac {\left (2 a^2-b^2\right ) \sin ^5(c+d x)}{5 d}-\frac {2 a b \sin ^6(c+d x)}{3 d}+\frac {\left (a^2-2 b^2\right ) \sin ^7(c+d x)}{7 d}+\frac {a b \sin ^8(c+d x)}{4 d}+\frac {b^2 \sin ^9(c+d x)}{9 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.22 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {-7560 a b \cos (2 (c+d x))-1260 a b \cos (4 (c+d x))+840 a b \cos (6 (c+d x))+315 a b \cos (8 (c+d x))+12600 a^2 \sin (c+d x)+3780 b^2 \sin (c+d x)-840 a^2 \sin (3 (c+d x))-840 b^2 \sin (3 (c+d x))-1512 a^2 \sin (5 (c+d x))-504 b^2 \sin (5 (c+d x))-360 a^2 \sin (7 (c+d x))+90 b^2 \sin (7 (c+d x))+70 b^2 \sin (9 (c+d x))}{161280 d} \]

[In]

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

[Out]

(-7560*a*b*Cos[2*(c + d*x)] - 1260*a*b*Cos[4*(c + d*x)] + 840*a*b*Cos[6*(c + d*x)] + 315*a*b*Cos[8*(c + d*x)]
+ 12600*a^2*Sin[c + d*x] + 3780*b^2*Sin[c + d*x] - 840*a^2*Sin[3*(c + d*x)] - 840*b^2*Sin[3*(c + d*x)] - 1512*
a^2*Sin[5*(c + d*x)] - 504*b^2*Sin[5*(c + d*x)] - 360*a^2*Sin[7*(c + d*x)] + 90*b^2*Sin[7*(c + d*x)] + 70*b^2*
Sin[9*(c + d*x)])/(161280*d)

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {\frac {b^{2} \left (\sin ^{9}\left (d x +c \right )\right )}{9}+\frac {a b \left (\sin ^{8}\left (d x +c \right )\right )}{4}+\frac {\left (a^{2}-2 b^{2}\right ) \left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {2 a b \left (\sin ^{6}\left (d x +c \right )\right )}{3}+\frac {\left (-2 a^{2}+b^{2}\right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {a b \left (\sin ^{4}\left (d x +c \right )\right )}{2}+\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d}\)	\(106\)
default	\(\frac {\frac {b^{2} \left (\sin ^{9}\left (d x +c \right )\right )}{9}+\frac {a b \left (\sin ^{8}\left (d x +c \right )\right )}{4}+\frac {\left (a^{2}-2 b^{2}\right ) \left (\sin ^{7}\left (d x +c \right )\right )}{7}-\frac {2 a b \left (\sin ^{6}\left (d x +c \right )\right )}{3}+\frac {\left (-2 a^{2}+b^{2}\right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {a b \left (\sin ^{4}\left (d x +c \right )\right )}{2}+\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d}\)	\(106\)
parallelrisch	\(-\frac {\left (\sin \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )-3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (360 \cos \left (4 d x +4 c \right ) a^{2}-300 b^{2} \cos \left (4 d x +4 c \right )+2592 a^{2} \cos \left (2 d x +2 c \right )-186 b^{2} \cos \left (2 d x +2 c \right )+3150 a b \sin \left (d x +c \right )-70 b^{2} \cos \left (6 d x +6 c \right )+315 a b \sin \left (5 d x +5 c \right )+1785 a b \sin \left (3 d x +3 c \right )+3768 a^{2}+556 b^{2}\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40320 d}\)	\(165\)
risch	\(\frac {5 a^{2} \sin \left (d x +c \right )}{64 d}+\frac {3 b^{2} \sin \left (d x +c \right )}{128 d}+\frac {b^{2} \sin \left (9 d x +9 c \right )}{2304 d}+\frac {a b \cos \left (8 d x +8 c \right )}{512 d}-\frac {\sin \left (7 d x +7 c \right ) a^{2}}{448 d}+\frac {\sin \left (7 d x +7 c \right ) b^{2}}{1792 d}+\frac {a b \cos \left (6 d x +6 c \right )}{192 d}-\frac {3 \sin \left (5 d x +5 c \right ) a^{2}}{320 d}-\frac {\sin \left (5 d x +5 c \right ) b^{2}}{320 d}-\frac {a b \cos \left (4 d x +4 c \right )}{128 d}-\frac {a^{2} \sin \left (3 d x +3 c \right )}{192 d}-\frac {\sin \left (3 d x +3 c \right ) b^{2}}{192 d}-\frac {3 a b \cos \left (2 d x +2 c \right )}{64 d}\)	\(213\)
norman	\(\frac {\frac {8 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 a^{2} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {16 \left (a^{2}+2 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {16 \left (a^{2}+2 b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {8 \left (31 a^{2}-48 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {8 \left (31 a^{2}-48 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {32 \left (129 a^{2}+218 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{315 d}-\frac {8 a b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a b \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 a b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a b \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {16 a b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {16 a b \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}\)	\(295\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.88 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {315 \, a b \cos \left (d x + c\right )^{8} - 420 \, a b \cos \left (d x + c\right )^{6} + 4 \, {\left (35 \, b^{2} \cos \left (d x + c\right )^{8} - 5 \, {\left (9 \, a^{2} + 10 \, b^{2}\right )} \cos \left (d x + c\right )^{6} + 3 \, {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (3 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} + 24 \, a^{2} + 8 \, b^{2}\right )} \sin \left (d x + c\right )}{1260 \, d} \]

[In]

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

[Out]

1/1260*(315*a*b*cos(d*x + c)^8 - 420*a*b*cos(d*x + c)^6 + 4*(35*b^2*cos(d*x + c)^8 - 5*(9*a^2 + 10*b^2)*cos(d*
x + c)^6 + 3*(3*a^2 + b^2)*cos(d*x + c)^4 + 4*(3*a^2 + b^2)*cos(d*x + c)^2 + 24*a^2 + 8*b^2)*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.95 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.38 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\begin {cases} \frac {8 a^{2} \sin ^{7}{\left (c + d x \right )}}{105 d} + \frac {4 a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{15 d} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{3 d} - \frac {a b \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac {a b \cos ^{8}{\left (c + d x \right )}}{12 d} + \frac {8 b^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {4 b^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{2} \sin ^{2}{\left (c \right )} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((8*a**2*sin(c + d*x)**7/(105*d) + 4*a**2*sin(c + d*x)**5*cos(c + d*x)**2/(15*d) + a**2*sin(c + d*x)*
*3*cos(c + d*x)**4/(3*d) - a*b*sin(c + d*x)**2*cos(c + d*x)**6/(3*d) - a*b*cos(c + d*x)**8/(12*d) + 8*b**2*sin
(c + d*x)**9/(315*d) + 4*b**2*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + b**2*sin(c + d*x)**5*cos(c + d*x)**4/(5
*d), Ne(d, 0)), (x*(a + b*sin(c))**2*sin(c)**2*cos(c)**5, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {140 \, b^{2} \sin \left (d x + c\right )^{9} + 315 \, a b \sin \left (d x + c\right )^{8} - 840 \, a b \sin \left (d x + c\right )^{6} + 180 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )^{7} + 630 \, a b \sin \left (d x + c\right )^{4} - 252 \, {\left (2 \, a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{5} + 420 \, a^{2} \sin \left (d x + c\right )^{3}}{1260 \, d} \]

[In]

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

[Out]

1/1260*(140*b^2*sin(d*x + c)^9 + 315*a*b*sin(d*x + c)^8 - 840*a*b*sin(d*x + c)^6 + 180*(a^2 - 2*b^2)*sin(d*x +
c)^7 + 630*a*b*sin(d*x + c)^4 - 252*(2*a^2 - b^2)*sin(d*x + c)^5 + 420*a^2*sin(d*x + c)^3)/d

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.25 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a b \cos \left (8 \, d x + 8 \, c\right )}{512 \, d} + \frac {a b \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {a b \cos \left (4 \, d x + 4 \, c\right )}{128 \, d} - \frac {3 \, a b \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {b^{2} \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} - \frac {{\left (4 \, a^{2} - b^{2}\right )} \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} - \frac {{\left (3 \, a^{2} + b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac {{\left (a^{2} + b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {{\left (10 \, a^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{128 \, d} \]

[In]

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

[Out]

1/512*a*b*cos(8*d*x + 8*c)/d + 1/192*a*b*cos(6*d*x + 6*c)/d - 1/128*a*b*cos(4*d*x + 4*c)/d - 3/64*a*b*cos(2*d*
x + 2*c)/d + 1/2304*b^2*sin(9*d*x + 9*c)/d - 1/1792*(4*a^2 - b^2)*sin(7*d*x + 7*c)/d - 1/320*(3*a^2 + b^2)*sin
(5*d*x + 5*c)/d - 1/192*(a^2 + b^2)*sin(3*d*x + 3*c)/d + 1/128*(10*a^2 + 3*b^2)*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 11.32 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78 \[ \int \cos ^5(c+d x) \sin ^2(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {{\sin \left (c+d\,x\right )}^7\,\left (\frac {a^2}{7}-\frac {2\,b^2}{7}\right )-{\sin \left (c+d\,x\right )}^5\,\left (\frac {2\,a^2}{5}-\frac {b^2}{5}\right )+\frac {a^2\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {b^2\,{\sin \left (c+d\,x\right )}^9}{9}+\frac {a\,b\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {2\,a\,b\,{\sin \left (c+d\,x\right )}^6}{3}+\frac {a\,b\,{\sin \left (c+d\,x\right )}^8}{4}}{d} \]

[In]

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

[Out]

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