3.962 \(\int \cos ^4(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=111 \[ -\frac{a (6 A+B) \cos ^5(c+d x)}{30 d}+\frac{a (6 A+B) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{a (6 A+B) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} a x (6 A+B)-\frac{B \cos ^5(c+d x) (a \sin (c+d x)+a)}{6 d} \]

[Out]

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

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Rubi [A]  time = 0.112641, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2860, 2669, 2635, 8} \[ -\frac{a (6 A+B) \cos ^5(c+d x)}{30 d}+\frac{a (6 A+B) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{a (6 A+B) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} a x (6 A+B)-\frac{B \cos ^5(c+d x) (a \sin (c+d x)+a)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2860

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[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]

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+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx &=-\frac{B \cos ^5(c+d x) (a+a \sin (c+d x))}{6 d}+\frac{1}{6} (6 A+B) \int \cos ^4(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac{a (6 A+B) \cos ^5(c+d x)}{30 d}-\frac{B \cos ^5(c+d x) (a+a \sin (c+d x))}{6 d}+\frac{1}{6} (a (6 A+B)) \int \cos ^4(c+d x) \, dx\\ &=-\frac{a (6 A+B) \cos ^5(c+d x)}{30 d}+\frac{a (6 A+B) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{B \cos ^5(c+d x) (a+a \sin (c+d x))}{6 d}+\frac{1}{8} (a (6 A+B)) \int \cos ^2(c+d x) \, dx\\ &=-\frac{a (6 A+B) \cos ^5(c+d x)}{30 d}+\frac{a (6 A+B) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a (6 A+B) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{B \cos ^5(c+d x) (a+a \sin (c+d x))}{6 d}+\frac{1}{16} (a (6 A+B)) \int 1 \, dx\\ &=\frac{1}{16} a (6 A+B) x-\frac{a (6 A+B) \cos ^5(c+d x)}{30 d}+\frac{a (6 A+B) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a (6 A+B) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac{B \cos ^5(c+d x) (a+a \sin (c+d x))}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.619302, size = 120, normalized size = 1.08 \[ -\frac{a (120 (A+B) \cos (c+d x)+60 (A+B) \cos (3 (c+d x))-240 A \sin (2 (c+d x))-30 A \sin (4 (c+d x))+12 A \cos (5 (c+d x))-360 A d x-15 B \sin (2 (c+d x))+15 B \sin (4 (c+d x))+5 B \sin (6 (c+d x))+12 B \cos (5 (c+d x))-60 B d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(-360*A*d*x - 60*B*d*x + 120*(A + B)*Cos[c + d*x] + 60*(A + B)*Cos[3*(c + d*x)] + 12*A*Cos[5*(c + d*x)] +
12*B*Cos[5*(c + d*x)] - 240*A*Sin[2*(c + d*x)] - 15*B*Sin[2*(c + d*x)] - 30*A*Sin[4*(c + d*x)] + 15*B*Sin[4*(c
+ d*x)] + 5*B*Sin[6*(c + d*x)]))/(960*d)

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Maple [A]  time = 0.055, size = 118, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( aB \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{aA \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}-{\frac{aB \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}+aA \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+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a*B*(-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)-1/5*a*A*
cos(d*x+c)^5-1/5*a*B*cos(d*x+c)^5+a*A*(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 = 1.01507, size = 132, normalized size = 1.19 \begin{align*} -\frac{192 \, A a \cos \left (d x + c\right )^{5} + 192 \, B a \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 a - 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 a}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(192*A*a*cos(d*x + c)^5 + 192*B*a*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*a - 5*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*B*a)/d

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Fricas [A]  time = 1.77524, size = 217, normalized size = 1.95 \begin{align*} -\frac{48 \,{\left (A + B\right )} a \cos \left (d x + c\right )^{5} - 15 \,{\left (6 \, A + B\right )} a d x + 5 \,{\left (8 \, B a \cos \left (d x + c\right )^{5} - 2 \,{\left (6 \, A + B\right )} a \cos \left (d x + c\right )^{3} - 3 \,{\left (6 \, A + B\right )} a \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+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/240*(48*(A + B)*a*cos(d*x + c)^5 - 15*(6*A + B)*a*d*x + 5*(8*B*a*cos(d*x + c)^5 - 2*(6*A + B)*a*cos(d*x + c
)^3 - 3*(6*A + B)*a*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 4.8359, size = 306, normalized size = 2.76 \begin{align*} \begin{cases} \frac{3 A a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 A a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 A a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 A a \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 A a \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{A a \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac{B a x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{3 B a x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{3 B a x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{B a x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{B a \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{B a \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac{B a \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac{B a \cos ^{5}{\left (c + d x \right )}}{5 d} & \text{for}\: d \neq 0 \\x \left (A + B \sin{\left (c \right )}\right ) \left (a \sin{\left (c \right )} + a\right ) \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+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.29168, size = 180, normalized size = 1.62 \begin{align*} \frac{1}{16} \,{\left (6 \, A a + B a\right )} x - \frac{B a \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{{\left (A a + B a\right )} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{{\left (A a + B a\right )} \cos \left (3 \, d x + 3 \, c\right )}{16 \, d} - \frac{{\left (A a + B a\right )} \cos \left (d x + c\right )}{8 \, d} + \frac{{\left (2 \, A a - B a\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{{\left (16 \, A a + B a\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+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*(6*A*a + B*a)*x - 1/192*B*a*sin(6*d*x + 6*c)/d - 1/80*(A*a + B*a)*cos(5*d*x + 5*c)/d - 1/16*(A*a + B*a)*c
os(3*d*x + 3*c)/d - 1/8*(A*a + B*a)*cos(d*x + c)/d + 1/64*(2*A*a - B*a)*sin(4*d*x + 4*c)/d + 1/64*(16*A*a + B*
a)*sin(2*d*x + 2*c)/d