3.407 \(\int \cos ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=131 \[ -\frac{b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{60 d}+\frac{a \left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} a x \left (4 a^2+3 b^2\right )-\frac{b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}-\frac{7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{20 d} \]

[Out]

(a*(4*a^2 + 3*b^2)*x)/8 - (b*(27*a^2 + 8*b^2)*Cos[c + d*x]^3)/(60*d) + (a*(4*a^2 + 3*b^2)*Cos[c + d*x]*Sin[c +
d*x])/(8*d) - (7*a*b*Cos[c + d*x]^3*(a + b*Sin[c + d*x]))/(20*d) - (b*Cos[c + d*x]^3*(a + b*Sin[c + d*x])^2)/
(5*d)

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Rubi [A]  time = 0.193578, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2692, 2862, 2669, 2635, 8} \[ -\frac{b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{60 d}+\frac{a \left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} a x \left (4 a^2+3 b^2\right )-\frac{b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}-\frac{7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{20 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + b*Sin[c + d*x])^3,x]

[Out]

(a*(4*a^2 + 3*b^2)*x)/8 - (b*(27*a^2 + 8*b^2)*Cos[c + d*x]^3)/(60*d) + (a*(4*a^2 + 3*b^2)*Cos[c + d*x]*Sin[c +
d*x])/(8*d) - (7*a*b*Cos[c + d*x]^3*(a + b*Sin[c + d*x]))/(20*d) - (b*Cos[c + d*x]^3*(a + b*Sin[c + d*x])^2)/
(5*d)

Rule 2692

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[1/(m + p), Int[(g*Cos[e + f*x])^
p*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(m + p) + a*b*(2*m + p - 1)*Sin[e + f*x]), x], x] /; FreeQ[{
a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + p, 0] && (IntegersQ[2*m, 2*p] || IntegerQ[m
])

Rule 2862

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[1/(m + p + 1), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(m + p + 1) + b*d*m + (a*d*
m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && Gt
Q[m, 0] &&  !LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x])

Rule 2669

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+b \sin (c+d x))^3 \, dx &=-\frac{b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac{1}{5} \int \cos ^2(c+d x) (a+b \sin (c+d x)) \left (5 a^2+2 b^2+7 a b \sin (c+d x)\right ) \, dx\\ &=-\frac{7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{20 d}-\frac{b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac{1}{20} \int \cos ^2(c+d x) \left (5 a \left (4 a^2+3 b^2\right )+b \left (27 a^2+8 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac{b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{60 d}-\frac{7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{20 d}-\frac{b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac{1}{4} \left (a \left (4 a^2+3 b^2\right )\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{60 d}+\frac{a \left (4 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac{7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{20 d}-\frac{b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}+\frac{1}{8} \left (a \left (4 a^2+3 b^2\right )\right ) \int 1 \, dx\\ &=\frac{1}{8} a \left (4 a^2+3 b^2\right ) x-\frac{b \left (27 a^2+8 b^2\right ) \cos ^3(c+d x)}{60 d}+\frac{a \left (4 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}-\frac{7 a b \cos ^3(c+d x) (a+b \sin (c+d x))}{20 d}-\frac{b \cos ^3(c+d x) (a+b \sin (c+d x))^2}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.539743, size = 107, normalized size = 0.82 \[ \frac{15 a \left (4 \left (4 a^2+3 b^2\right ) (c+d x)+8 a^2 \sin (2 (c+d x))-3 b^2 \sin (4 (c+d x))\right )-60 b \left (6 a^2+b^2\right ) \cos (c+d x)-10 \left (12 a^2 b+b^3\right ) \cos (3 (c+d x))+6 b^3 \cos (5 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*(a + b*Sin[c + d*x])^3,x]

[Out]

(-60*b*(6*a^2 + b^2)*Cos[c + d*x] - 10*(12*a^2*b + b^3)*Cos[3*(c + d*x)] + 6*b^3*Cos[5*(c + d*x)] + 15*a*(4*(4
*a^2 + 3*b^2)*(c + d*x) + 8*a^2*Sin[2*(c + d*x)] - 3*b^2*Sin[4*(c + d*x)]))/(480*d)

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Maple [A]  time = 0.05, size = 123, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({b}^{3} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{5}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15}} \right ) +3\,a{b}^{2} \left ( -1/4\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +1/8\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/8\,dx+c/8 \right ) -{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{a}^{3} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+b*sin(d*x+c))^3,x)

[Out]

1/d*(b^3*(-1/5*sin(d*x+c)^2*cos(d*x+c)^3-2/15*cos(d*x+c)^3)+3*a*b^2*(-1/4*cos(d*x+c)^3*sin(d*x+c)+1/8*cos(d*x+
c)*sin(d*x+c)+1/8*d*x+1/8*c)-a^2*b*cos(d*x+c)^3+a^3*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]  time = 0.961946, size = 126, normalized size = 0.96 \begin{align*} -\frac{480 \, a^{2} b \cos \left (d x + c\right )^{3} - 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 45 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a b^{2} - 32 \,{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} b^{3}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/480*(480*a^2*b*cos(d*x + c)^3 - 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3 - 45*(4*d*x + 4*c - sin(4*d*x + 4*
c))*a*b^2 - 32*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*b^3)/d

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Fricas [A]  time = 2.23542, size = 232, normalized size = 1.77 \begin{align*} \frac{24 \, b^{3} \cos \left (d x + c\right )^{5} - 40 \,{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{3} + 15 \,{\left (4 \, a^{3} + 3 \, a b^{2}\right )} d x - 15 \,{\left (6 \, a b^{2} \cos \left (d x + c\right )^{3} -{\left (4 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/120*(24*b^3*cos(d*x + c)^5 - 40*(3*a^2*b + b^3)*cos(d*x + c)^3 + 15*(4*a^3 + 3*a*b^2)*d*x - 15*(6*a*b^2*cos(
d*x + c)^3 - (4*a^3 + 3*a*b^2)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 2.59479, size = 236, normalized size = 1.8 \begin{align*} \begin{cases} \frac{a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{a^{3} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} - \frac{a^{2} b \cos ^{3}{\left (c + d x \right )}}{d} + \frac{3 a b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 a b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 a b^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{3 a b^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{2 b^{3} \cos ^{5}{\left (c + d x \right )}}{15 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{3} \cos ^{2}{\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)**2*(a+b*sin(d*x+c))**3,x)

[Out]

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

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Giac [A]  time = 1.08993, size = 153, normalized size = 1.17 \begin{align*} \frac{b^{3} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{3 \, a b^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{a^{3} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{1}{8} \,{\left (4 \, a^{3} + 3 \, a b^{2}\right )} x - \frac{{\left (12 \, a^{2} b + b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac{{\left (6 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/80*b^3*cos(5*d*x + 5*c)/d - 3/32*a*b^2*sin(4*d*x + 4*c)/d + 1/4*a^3*sin(2*d*x + 2*c)/d + 1/8*(4*a^3 + 3*a*b^
2)*x - 1/48*(12*a^2*b + b^3)*cos(3*d*x + 3*c)/d - 1/8*(6*a^2*b + b^3)*cos(d*x + c)/d