3.955 \(\int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=78 \[ -\frac{(A-3 B) (a \sin (c+d x)+a)^4}{4 a^3 d}+\frac{2 (A-B) (a \sin (c+d x)+a)^3}{3 a^2 d}-\frac{B (a \sin (c+d x)+a)^5}{5 a^4 d} \]

[Out]

(2*(A - B)*(a + a*Sin[c + d*x])^3)/(3*a^2*d) - ((A - 3*B)*(a + a*Sin[c + d*x])^4)/(4*a^3*d) - (B*(a + a*Sin[c
+ d*x])^5)/(5*a^4*d)

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Rubi [A]  time = 0.0940424, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {2836, 77} \[ -\frac{(A-3 B) (a \sin (c+d x)+a)^4}{4 a^3 d}+\frac{2 (A-B) (a \sin (c+d x)+a)^3}{3 a^2 d}-\frac{B (a \sin (c+d x)+a)^5}{5 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A - B)*(a + a*Sin[c + d*x])^3)/(3*a^2*d) - ((A - 3*B)*(a + a*Sin[c + d*x])^4)/(4*a^3*d) - (B*(a + a*Sin[c
+ d*x])^5)/(5*a^4*d)

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx &=\frac{\operatorname{Subst}\left (\int (a-x) (a+x)^2 \left (A+\frac{B x}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (2 a (A-B) (a+x)^2+(-A+3 B) (a+x)^3-\frac{B (a+x)^4}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac{2 (A-B) (a+a \sin (c+d x))^3}{3 a^2 d}-\frac{(A-3 B) (a+a \sin (c+d x))^4}{4 a^3 d}-\frac{B (a+a \sin (c+d x))^5}{5 a^4 d}\\ \end{align*}

Mathematica [A]  time = 0.804491, size = 78, normalized size = 1. \[ -\frac{a (-4 (100 A+11 B) \sin (c+d x)+3 \cos (4 (c+d x)) (5 (A+B)+4 B \sin (c+d x))+\cos (2 (c+d x)) ((32 B-80 A) \sin (c+d x)+60 (A+B)))}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(-4*(100*A + 11*B)*Sin[c + d*x] + 3*Cos[4*(c + d*x)]*(5*(A + B) + 4*B*Sin[c + d*x]) + Cos[2*(c + d*x)]*(60
*(A + B) + (-80*A + 32*B)*Sin[c + d*x])))/(480*d)

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Maple [A]  time = 0.054, size = 88, 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 ) ^{4}}{5}}+{\frac{ \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{15}} \right ) -{\frac{aA \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4}}-{\frac{aB \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4}}+{\frac{aA \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 0.972714, size = 97, normalized size = 1.24 \begin{align*} -\frac{12 \, B a \sin \left (d x + c\right )^{5} + 15 \,{\left (A + B\right )} a \sin \left (d x + c\right )^{4} + 20 \,{\left (A - B\right )} a \sin \left (d x + c\right )^{3} - 30 \,{\left (A + B\right )} a \sin \left (d x + c\right )^{2} - 60 \, A a \sin \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(12*B*a*sin(d*x + c)^5 + 15*(A + B)*a*sin(d*x + c)^4 + 20*(A - B)*a*sin(d*x + c)^3 - 30*(A + B)*a*sin(d*
x + c)^2 - 60*A*a*sin(d*x + c))/d

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Fricas [A]  time = 1.66021, size = 167, normalized size = 2.14 \begin{align*} -\frac{15 \,{\left (A + B\right )} a \cos \left (d x + c\right )^{4} + 4 \,{\left (3 \, B a \cos \left (d x + c\right )^{4} -{\left (5 \, A + B\right )} a \cos \left (d x + c\right )^{2} - 2 \,{\left (5 \, A + B\right )} a\right )} \sin \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*(A + B)*a*cos(d*x + c)^4 + 4*(3*B*a*cos(d*x + c)^4 - (5*A + B)*a*cos(d*x + c)^2 - 2*(5*A + B)*a)*sin
(d*x + c))/d

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Sympy [A]  time = 2.44826, size = 128, normalized size = 1.64 \begin{align*} \begin{cases} \frac{2 A a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{A a \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac{A a \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac{2 B a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{B a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac{B a \cos ^{4}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\x \left (A + B \sin{\left (c \right )}\right ) \left (a \sin{\left (c \right )} + a\right ) \cos ^{3}{\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)**3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.27429, size = 135, normalized size = 1.73 \begin{align*} -\frac{12 \, B a \sin \left (d x + c\right )^{5} + 15 \, A a \sin \left (d x + c\right )^{4} + 15 \, B a \sin \left (d x + c\right )^{4} + 20 \, A a \sin \left (d x + c\right )^{3} - 20 \, B a \sin \left (d x + c\right )^{3} - 30 \, A a \sin \left (d x + c\right )^{2} - 30 \, B a \sin \left (d x + c\right )^{2} - 60 \, A a \sin \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(12*B*a*sin(d*x + c)^5 + 15*A*a*sin(d*x + c)^4 + 15*B*a*sin(d*x + c)^4 + 20*A*a*sin(d*x + c)^3 - 20*B*a*
sin(d*x + c)^3 - 30*A*a*sin(d*x + c)^2 - 30*B*a*sin(d*x + c)^2 - 60*A*a*sin(d*x + c))/d