3.963 \(\int \cos ^2(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=84 \[ -\frac {a (4 A+B) \cos ^3(c+d x)}{12 d}+\frac {a (4 A+B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a x (4 A+B)-\frac {B \cos ^3(c+d x) (a \sin (c+d x)+a)}{4 d} \]

[Out]

1/8*a*(4*A+B)*x-1/12*a*(4*A+B)*cos(d*x+c)^3/d+1/8*a*(4*A+B)*cos(d*x+c)*sin(d*x+c)/d-1/4*B*cos(d*x+c)^3*(a+a*si
n(d*x+c))/d

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Rubi [A]  time = 0.10, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2860, 2669, 2635, 8} \[ -\frac {a (4 A+B) \cos ^3(c+d x)}{12 d}+\frac {a (4 A+B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a x (4 A+B)-\frac {B \cos ^3(c+d x) (a \sin (c+d x)+a)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(4*A + B)*x)/8 - (a*(4*A + B)*Cos[c + d*x]^3)/(12*d) + (a*(4*A + B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (B*C
os[c + d*x]^3*(a + a*Sin[c + d*x]))/(4*d)

Rule 8

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

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 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 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]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx &=-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))}{4 d}+\frac {1}{4} (4 A+B) \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx\\ &=-\frac {a (4 A+B) \cos ^3(c+d x)}{12 d}-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))}{4 d}+\frac {1}{4} (a (4 A+B)) \int \cos ^2(c+d x) \, dx\\ &=-\frac {a (4 A+B) \cos ^3(c+d x)}{12 d}+\frac {a (4 A+B) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))}{4 d}+\frac {1}{8} (a (4 A+B)) \int 1 \, dx\\ &=\frac {1}{8} a (4 A+B) x-\frac {a (4 A+B) \cos ^3(c+d x)}{12 d}+\frac {a (4 A+B) \cos (c+d x) \sin (c+d x)}{8 d}-\frac {B \cos ^3(c+d x) (a+a \sin (c+d x))}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.67, size = 64, normalized size = 0.76 \[ -\frac {a (24 (A+B) \cos (c+d x)+8 (A+B) \cos (3 (c+d x))-12 d x (4 A+B)-24 A \sin (2 (c+d x))+3 B \sin (4 (c+d x)))}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/96*(a*(-12*(4*A + B)*d*x + 24*(A + B)*Cos[c + d*x] + 8*(A + B)*Cos[3*(c + d*x)] - 24*A*Sin[2*(c + d*x)] + 3
*B*Sin[4*(c + d*x)]))/d

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fricas [A]  time = 0.79, size = 65, normalized size = 0.77 \[ -\frac {8 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, A + B\right )} a d x + 3 \, {\left (2 \, B a \cos \left (d x + c\right )^{3} - {\left (4 \, A + B\right )} a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(8*(A + B)*a*cos(d*x + c)^3 - 3*(4*A + B)*a*d*x + 3*(2*B*a*cos(d*x + c)^3 - (4*A + B)*a*cos(d*x + c))*si
n(d*x + c))/d

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giac [A]  time = 0.18, size = 83, normalized size = 0.99 \[ \frac {1}{8} \, {\left (4 \, A a + B a\right )} x - \frac {B a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {A a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} - \frac {{\left (A a + B a\right )} \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {{\left (A a + B a\right )} \cos \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*A*a + B*a)*x - 1/32*B*a*sin(4*d*x + 4*c)/d + 1/4*A*a*sin(2*d*x + 2*c)/d - 1/12*(A*a + B*a)*cos(3*d*x +
3*c)/d - 1/4*(A*a + B*a)*cos(d*x + c)/d

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maple [A]  time = 0.36, size = 96, normalized size = 1.14 \[ \frac {a B \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {a A \left (\cos ^{3}\left (d x +c \right )\right )}{3}-\frac {a B \left (\cos ^{3}\left (d x +c \right )\right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a*B*(-1/4*sin(d*x+c)*cos(d*x+c)^3+1/8*cos(d*x+c)*sin(d*x+c)+1/8*d*x+1/8*c)-1/3*a*A*cos(d*x+c)^3-1/3*a*B*c
os(d*x+c)^3+a*A*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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maxima [A]  time = 0.38, size = 74, normalized size = 0.88 \[ -\frac {32 \, A a \cos \left (d x + c\right )^{3} + 32 \, B a \cos \left (d x + c\right )^{3} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 3 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} B a}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(32*A*a*cos(d*x + c)^3 + 32*B*a*cos(d*x + c)^3 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a - 3*(4*d*x + 4*
c - sin(4*d*x + 4*c))*B*a)/d

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mupad [B]  time = 10.55, size = 276, normalized size = 3.29 \[ \frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,A+B\right )}{4\,\left (A\,a+\frac {B\,a}{4}\right )}\right )\,\left (4\,A+B\right )}{4\,d}-\frac {a\,\left (4\,A+B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d}-\frac {\left (A\,a-\frac {B\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (2\,A\,a+2\,B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (A\,a+\frac {7\,B\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (2\,A\,a+2\,B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-A\,a-\frac {7\,B\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {2\,A\,a}{3}+\frac {2\,B\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (\frac {B\,a}{4}-A\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {2\,A\,a}{3}+\frac {2\,B\,a}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(A + B*sin(c + d*x))*(a + a*sin(c + d*x)),x)

[Out]

(a*atan((a*tan(c/2 + (d*x)/2)*(4*A + B))/(4*(A*a + (B*a)/4)))*(4*A + B))/(4*d) - (a*(4*A + B)*(atan(tan(c/2 +
(d*x)/2)) - (d*x)/2))/(4*d) - ((2*A*a)/3 + (2*B*a)/3 - tan(c/2 + (d*x)/2)*(A*a - (B*a)/4) + tan(c/2 + (d*x)/2)
^4*(2*A*a + 2*B*a) + tan(c/2 + (d*x)/2)^6*(2*A*a + 2*B*a) + tan(c/2 + (d*x)/2)^2*((2*A*a)/3 + (2*B*a)/3) + tan
(c/2 + (d*x)/2)^7*(A*a - (B*a)/4) - tan(c/2 + (d*x)/2)^3*(A*a + (7*B*a)/4) + tan(c/2 + (d*x)/2)^5*(A*a + (7*B*
a)/4))/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1
))

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sympy [A]  time = 1.01, size = 199, normalized size = 2.37 \[ \begin {cases} \frac {A a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {A a \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {B a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {B a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {B a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {B a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {B a \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\relax (c )}\right ) \left (a \sin {\relax (c )} + a\right ) \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

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

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