3.394 \(\int \cos ^4(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=116 \[ \frac{\left (6 a^2+b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{\left (6 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x \left (6 a^2+b^2\right )-\frac{7 a b \cos ^5(c+d x)}{30 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d} \]

[Out]

((6*a^2 + b^2)*x)/16 - (7*a*b*Cos[c + d*x]^5)/(30*d) + ((6*a^2 + b^2)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + ((6*
a^2 + b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) - (b*Cos[c + d*x]^5*(a + b*Sin[c + d*x]))/(6*d)

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Rubi [A]  time = 0.115815, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2692, 2669, 2635, 8} \[ \frac{\left (6 a^2+b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{\left (6 a^2+b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x \left (6 a^2+b^2\right )-\frac{7 a b \cos ^5(c+d x)}{30 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]

[Out]

((6*a^2 + b^2)*x)/16 - (7*a*b*Cos[c + d*x]^5)/(30*d) + ((6*a^2 + b^2)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + ((6*
a^2 + b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) - (b*Cos[c + d*x]^5*(a + b*Sin[c + d*x]))/(6*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 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))^2 \, dx &=-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}+\frac{1}{6} \int \cos ^4(c+d x) \left (6 a^2+b^2+7 a b \sin (c+d x)\right ) \, dx\\ &=-\frac{7 a b \cos ^5(c+d x)}{30 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}+\frac{1}{6} \left (6 a^2+b^2\right ) \int \cos ^4(c+d x) \, dx\\ &=-\frac{7 a b \cos ^5(c+d x)}{30 d}+\frac{\left (6 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}+\frac{1}{8} \left (6 a^2+b^2\right ) \int \cos ^2(c+d x) \, dx\\ &=-\frac{7 a b \cos ^5(c+d x)}{30 d}+\frac{\left (6 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{\left (6 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}+\frac{1}{16} \left (6 a^2+b^2\right ) \int 1 \, dx\\ &=\frac{1}{16} \left (6 a^2+b^2\right ) x-\frac{7 a b \cos ^5(c+d x)}{30 d}+\frac{\left (6 a^2+b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{\left (6 a^2+b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{b \cos ^5(c+d x) (a+b \sin (c+d x))}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.193004, size = 133, normalized size = 1.15 \[ \frac{240 a^2 \sin (2 (c+d x))+30 a^2 \sin (4 (c+d x))+360 a^2 c+360 a^2 d x-240 a b \cos (c+d x)-120 a b \cos (3 (c+d x))-24 a b \cos (5 (c+d x))+15 b^2 \sin (2 (c+d x))-15 b^2 \sin (4 (c+d x))-5 b^2 \sin (6 (c+d x))+60 b^2 c+60 b^2 d x}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]

[Out]

(360*a^2*c + 60*b^2*c + 360*a^2*d*x + 60*b^2*d*x - 240*a*b*Cos[c + d*x] - 120*a*b*Cos[3*(c + d*x)] - 24*a*b*Co
s[5*(c + d*x)] + 240*a^2*Sin[2*(c + d*x)] + 15*b^2*Sin[2*(c + d*x)] + 30*a^2*Sin[4*(c + d*x)] - 15*b^2*Sin[4*(
c + d*x)] - 5*b^2*Sin[6*(c + d*x)])/(960*d)

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Maple [A]  time = 0.043, size = 108, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{6}}+{\frac{\sin \left ( dx+c \right ) }{24} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{dx}{16}}+{\frac{c}{16}} \right ) -{\frac{2\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}+{a}^{2} \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))^2,x)

[Out]

1/d*(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)-2/5*a*b*
cos(d*x+c)^5+a^2*(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 = 0.962841, size = 119, normalized size = 1.03 \begin{align*} -\frac{384 \, a b \cos \left (d x + c\right )^{5} - 30 \,{\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^{2} - 5 \,{\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 )} b^{2}}{960 \, 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))^2,x, algorithm="maxima")

[Out]

-1/960*(384*a*b*cos(d*x + c)^5 - 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2 - 5*(4*sin(2*d
*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*b^2)/d

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Fricas [A]  time = 2.30337, size = 217, normalized size = 1.87 \begin{align*} -\frac{96 \, a b \cos \left (d x + c\right )^{5} - 15 \,{\left (6 \, a^{2} + b^{2}\right )} d x + 5 \,{\left (8 \, b^{2} \cos \left (d x + c\right )^{5} - 2 \,{\left (6 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{3} - 3 \,{\left (6 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, 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))^2,x, algorithm="fricas")

[Out]

-1/240*(96*a*b*cos(d*x + c)^5 - 15*(6*a^2 + b^2)*d*x + 5*(8*b^2*cos(d*x + c)^5 - 2*(6*a^2 + b^2)*cos(d*x + c)^
3 - 3*(6*a^2 + b^2)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 4.7273, size = 287, normalized size = 2.47 \begin{align*} \begin{cases} \frac{3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 a^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{2 a b \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac{b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{3 b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{3 b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{b^{2} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac{b^{2} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )}\right )^{2} \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))**2,x)

[Out]

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

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Giac [A]  time = 1.09804, size = 166, normalized size = 1.43 \begin{align*} \frac{1}{16} \,{\left (6 \, a^{2} + b^{2}\right )} x - \frac{a b \cos \left (5 \, d x + 5 \, c\right )}{40 \, d} - \frac{a b \cos \left (3 \, d x + 3 \, c\right )}{8 \, d} - \frac{a b \cos \left (d x + c\right )}{4 \, d} - \frac{b^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{{\left (2 \, a^{2} - b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{{\left (16 \, a^{2} + 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))^2,x, algorithm="giac")

[Out]

1/16*(6*a^2 + b^2)*x - 1/40*a*b*cos(5*d*x + 5*c)/d - 1/8*a*b*cos(3*d*x + 3*c)/d - 1/4*a*b*cos(d*x + c)/d - 1/1
92*b^2*sin(6*d*x + 6*c)/d + 1/64*(2*a^2 - b^2)*sin(4*d*x + 4*c)/d + 1/64*(16*a^2 + b^2)*sin(2*d*x + 2*c)/d