3.1216 \(\int \cos ^5(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=138 \[ \frac{\left (a^2-2 b^2\right ) \sin ^6(c+d x)}{6 d}-\frac{\left (2 a^2-b^2\right ) \sin ^4(c+d x)}{4 d}+\frac{a^2 \sin ^2(c+d x)}{2 d}+\frac{2 a b \sin ^7(c+d x)}{7 d}-\frac{4 a b \sin ^5(c+d x)}{5 d}+\frac{2 a b \sin ^3(c+d x)}{3 d}+\frac{b^2 \sin ^8(c+d x)}{8 d} \]

[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)

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Rubi [A]  time = 0.12784, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2837, 12, 772} \[ \frac{\left (a^2-2 b^2\right ) \sin ^6(c+d x)}{6 d}-\frac{\left (2 a^2-b^2\right ) \sin ^4(c+d x)}{4 d}+\frac{a^2 \sin ^2(c+d x)}{2 d}+\frac{2 a b \sin ^7(c+d x)}{7 d}-\frac{4 a b \sin ^5(c+d x)}{5 d}+\frac{2 a b \sin ^3(c+d x)}{3 d}+\frac{b^2 \sin ^8(c+d x)}{8 d} \]

Antiderivative was successfully verified.

[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 2837

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*x)/b)^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]

Rule 12

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

Rule 772

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]

Rubi steps

\begin{align*} \int \cos ^5(c+d x) \sin (c+d x) (a+b \sin (c+d x))^2 \, dx &=\frac{\operatorname{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{\operatorname{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{\operatorname{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]  time = 0.654, size = 138, normalized size = 1. \[ -\frac{840 \left (10 a^2+3 b^2\right ) \cos (2 (c+d x))+420 \left (8 a^2+b^2\right ) \cos (4 (c+d x))+560 a^2 \cos (6 (c+d x))-16800 a b \sin (c+d x)+1120 a b \sin (3 (c+d x))+2016 a b \sin (5 (c+d x))+480 a b \sin (7 (c+d x))-280 b^2 \cos (6 (c+d x))-105 b^2 \cos (8 (c+d x))-2590 b^2}{107520 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-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*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)

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Maple [A]  time = 0.04, size = 101, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ( -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{6}}+2\,ab \left ( -1/7\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}+1/35\, \left ( 8/3+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) +{b}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{8}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}{24}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/6*a^2*cos(d*x+c)^6+2*a*b*(-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^2*(-1/8*sin(d*x+c)^2*cos(d*x+c)^6-1/24*cos(d*x+c)^6))

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Maxima [A]  time = 0.993132, size = 146, normalized size = 1.06 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.74325, size = 220, normalized size = 1.59 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 12.8605, size = 163, normalized size = 1.18 \begin{align*} \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 ^{8}{\left (c + d x \right )}}{24 d} + \frac{b^{2} \sin ^{6}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{6 d} + \frac{b^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)**8/(24*d) + b**2*sin(c + d*x)**6*co
s(c + d*x)**2/(6*d) + b**2*sin(c + d*x)**4*cos(c + d*x)**4/(4*d), Ne(d, 0)), (x*(a + b*sin(c))**2*sin(c)*cos(c
)**5, True))

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Giac [A]  time = 1.21535, size = 205, normalized size = 1.49 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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