3.406 \(\int \cos ^4(c+d x) (a+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=158 \[ -\frac{b \left (17 a^2+4 b^2\right ) \cos ^5(c+d x)}{70 d}+\frac{a \left (2 a^2+b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac{3 a \left (2 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{3}{16} a x \left (2 a^2+b^2\right )-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}-\frac{3 a b \cos ^5(c+d x) (a+b \sin (c+d x))}{14 d} \]

[Out]

(3*a*(2*a^2 + b^2)*x)/16 - (b*(17*a^2 + 4*b^2)*Cos[c + d*x]^5)/(70*d) + (3*a*(2*a^2 + b^2)*Cos[c + d*x]*Sin[c
+ d*x])/(16*d) + (a*(2*a^2 + b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(8*d) - (3*a*b*Cos[c + d*x]^5*(a + b*Sin[c + d*
x]))/(14*d) - (b*Cos[c + d*x]^5*(a + b*Sin[c + d*x])^2)/(7*d)

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Rubi [A]  time = 0.216748, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{b \left (17 a^2+4 b^2\right ) \cos ^5(c+d x)}{70 d}+\frac{a \left (2 a^2+b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{8 d}+\frac{3 a \left (2 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{3}{16} a x \left (2 a^2+b^2\right )-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}-\frac{3 a b \cos ^5(c+d x) (a+b \sin (c+d x))}{14 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + b*Sin[c + d*x])^3,x]

[Out]

(3*a*(2*a^2 + b^2)*x)/16 - (b*(17*a^2 + 4*b^2)*Cos[c + d*x]^5)/(70*d) + (3*a*(2*a^2 + b^2)*Cos[c + d*x]*Sin[c
+ d*x])/(16*d) + (a*(2*a^2 + b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(8*d) - (3*a*b*Cos[c + d*x]^5*(a + b*Sin[c + d*
x]))/(14*d) - (b*Cos[c + d*x]^5*(a + b*Sin[c + d*x])^2)/(7*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 ^4(c+d x) (a+b \sin (c+d x))^3 \, dx &=-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{7} \int \cos ^4(c+d x) (a+b \sin (c+d x)) \left (7 a^2+2 b^2+9 a b \sin (c+d x)\right ) \, dx\\ &=-\frac{3 a b \cos ^5(c+d x) (a+b \sin (c+d x))}{14 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{42} \int \cos ^4(c+d x) \left (21 a \left (2 a^2+b^2\right )+3 b \left (17 a^2+4 b^2\right ) \sin (c+d x)\right ) \, dx\\ &=-\frac{b \left (17 a^2+4 b^2\right ) \cos ^5(c+d x)}{70 d}-\frac{3 a b \cos ^5(c+d x) (a+b \sin (c+d x))}{14 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{2} \left (a \left (2 a^2+b^2\right )\right ) \int \cos ^4(c+d x) \, dx\\ &=-\frac{b \left (17 a^2+4 b^2\right ) \cos ^5(c+d x)}{70 d}+\frac{a \left (2 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac{3 a b \cos ^5(c+d x) (a+b \sin (c+d x))}{14 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{8} \left (3 a \left (2 a^2+b^2\right )\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{b \left (17 a^2+4 b^2\right ) \cos ^5(c+d x)}{70 d}+\frac{3 a \left (2 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a \left (2 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac{3 a b \cos ^5(c+d x) (a+b \sin (c+d x))}{14 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}+\frac{1}{16} \left (3 a \left (2 a^2+b^2\right )\right ) \int 1 \, dx\\ &=\frac{3}{16} a \left (2 a^2+b^2\right ) x-\frac{b \left (17 a^2+4 b^2\right ) \cos ^5(c+d x)}{70 d}+\frac{3 a \left (2 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a \left (2 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{8 d}-\frac{3 a b \cos ^5(c+d x) (a+b \sin (c+d x))}{14 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))^2}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.384893, size = 182, normalized size = 1.15 \[ \frac{-105 b \left (8 a^2+b^2\right ) \cos (c+d x)-35 \left (12 a^2 b+b^3\right ) \cos (3 (c+d x))-84 a^2 b \cos (5 (c+d x))+560 a^3 \sin (2 (c+d x))+70 a^3 \sin (4 (c+d x))+840 a^3 c+840 a^3 d x+105 a b^2 \sin (2 (c+d x))-105 a b^2 \sin (4 (c+d x))-35 a b^2 \sin (6 (c+d x))+420 a b^2 c+420 a b^2 d x+7 b^3 \cos (5 (c+d x))+5 b^3 \cos (7 (c+d x))}{2240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sin[c + d*x])^3,x]

[Out]

(840*a^3*c + 420*a*b^2*c + 840*a^3*d*x + 420*a*b^2*d*x - 105*b*(8*a^2 + b^2)*Cos[c + d*x] - 35*(12*a^2*b + b^3
)*Cos[3*(c + d*x)] - 84*a^2*b*Cos[5*(c + d*x)] + 7*b^3*Cos[5*(c + d*x)] + 5*b^3*Cos[7*(c + d*x)] + 560*a^3*Sin
[2*(c + d*x)] + 105*a*b^2*Sin[2*(c + d*x)] + 70*a^3*Sin[4*(c + d*x)] - 105*a*b^2*Sin[4*(c + d*x)] - 35*a*b^2*S
in[6*(c + d*x)])/(2240*d)

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Maple [A]  time = 0.053, size = 145, 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 ) ^{5}}{7}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{35}} \right ) +3\,a{b}^{2} \left ( -1/6\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1/24\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +1/16\,dx+c/16 \right ) -{\frac{3\,{a}^{2}b \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}+{a}^{3} \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+b*sin(d*x+c))^3,x)

[Out]

1/d*(b^3*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)+3*a*b^2*(-1/6*sin(d*x+c)*cos(d*x+c)^5+1/24*(cos(d*
x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)-3/5*a^2*b*cos(d*x+c)^5+a^3*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c
))*sin(d*x+c)+3/8*d*x+3/8*c))

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Maxima [A]  time = 0.959663, size = 158, normalized size = 1. \begin{align*} -\frac{1344 \, a^{2} b \cos \left (d x + c\right )^{5} - 70 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 35 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a b^{2} - 64 \,{\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} b^{3}}{2240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2240*(1344*a^2*b*cos(d*x + c)^5 - 70*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^3 - 35*(4*si
n(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a*b^2 - 64*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*b^3)/d

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Fricas [A]  time = 2.51896, size = 279, normalized size = 1.77 \begin{align*} \frac{80 \, b^{3} \cos \left (d x + c\right )^{7} - 112 \,{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{5} + 105 \,{\left (2 \, a^{3} + a b^{2}\right )} d x - 35 \,{\left (8 \, a b^{2} \cos \left (d x + c\right )^{5} - 2 \,{\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (2 \, a^{3} + a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{560 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/560*(80*b^3*cos(d*x + c)^7 - 112*(3*a^2*b + b^3)*cos(d*x + c)^5 + 105*(2*a^3 + a*b^2)*d*x - 35*(8*a*b^2*cos(
d*x + c)^5 - 2*(2*a^3 + a*b^2)*cos(d*x + c)^3 - 3*(2*a^3 + a*b^2)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 8.27802, size = 348, normalized size = 2.2 \begin{align*} \begin{cases} \frac{3 a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{3} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 a^{3} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{3 a^{2} b \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac{3 a b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{9 a b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{9 a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{3 a b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{3 a b^{2} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} - \frac{3 a b^{2} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac{b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac{2 b^{3} \cos ^{7}{\left (c + d x \right )}}{35 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{3} \cos ^{4}{\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)**4*(a+b*sin(d*x+c))**3,x)

[Out]

Piecewise((3*a**3*x*sin(c + d*x)**4/8 + 3*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*a**3*x*cos(c + d*x)**4/
8 + 3*a**3*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 5*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) - 3*a**2*b*cos(c + d
*x)**5/(5*d) + 3*a*b**2*x*sin(c + d*x)**6/16 + 9*a*b**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 9*a*b**2*x*sin(
c + d*x)**2*cos(c + d*x)**4/16 + 3*a*b**2*x*cos(c + d*x)**6/16 + 3*a*b**2*sin(c + d*x)**5*cos(c + d*x)/(16*d)
+ a*b**2*sin(c + d*x)**3*cos(c + d*x)**3/(2*d) - 3*a*b**2*sin(c + d*x)*cos(c + d*x)**5/(16*d) - b**3*sin(c + d
*x)**2*cos(c + d*x)**5/(5*d) - 2*b**3*cos(c + d*x)**7/(35*d), Ne(d, 0)), (x*(a + b*sin(c))**3*cos(c)**4, True)
)

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Giac [A]  time = 1.10315, size = 234, normalized size = 1.48 \begin{align*} \frac{b^{3} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac{a b^{2} \sin \left (6 \, d x + 6 \, c\right )}{64 \, d} + \frac{3}{16} \,{\left (2 \, a^{3} + a b^{2}\right )} x - \frac{{\left (12 \, a^{2} b - b^{3}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} - \frac{{\left (12 \, a^{2} b + b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac{3 \,{\left (8 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )}{64 \, d} + \frac{{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{{\left (16 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/448*b^3*cos(7*d*x + 7*c)/d - 1/64*a*b^2*sin(6*d*x + 6*c)/d + 3/16*(2*a^3 + a*b^2)*x - 1/320*(12*a^2*b - b^3)
*cos(5*d*x + 5*c)/d - 1/64*(12*a^2*b + b^3)*cos(3*d*x + 3*c)/d - 3/64*(8*a^2*b + b^3)*cos(d*x + c)/d + 1/64*(2
*a^3 - 3*a*b^2)*sin(4*d*x + 4*c)/d + 1/64*(16*a^3 + 3*a*b^2)*sin(2*d*x + 2*c)/d