3.956 \(\int \cos (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=49 \[ \frac{a (A+B) \sin ^2(c+d x)}{2 d}+\frac{a A \sin (c+d x)}{d}+\frac{a B \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0631775, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {2833, 43} \[ \frac{a (A+B) \sin ^2(c+d x)}{2 d}+\frac{a A \sin (c+d x)}{d}+\frac{a B \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2833

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.394483, size = 46, normalized size = 0.94 \[ -\frac{a (\cos (2 (c+d x)) (3 (A+B)+2 B \sin (c+d x))-2 (6 A+B) \sin (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(-2*(6*A + B)*Sin[c + d*x] + Cos[2*(c + d*x)]*(3*(A + B) + 2*B*Sin[c + d*x])))/(12*d)

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Maple [A]  time = 0.026, size = 44, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{aB \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3}}+{\frac{ \left ( aA+aB \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2}}+A\sin \left ( dx+c \right ) a \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/3*a*B*sin(d*x+c)^3+1/2*(A*a+B*a)*sin(d*x+c)^2+A*sin(d*x+c)*a)

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Maxima [A]  time = 1.00871, size = 57, normalized size = 1.16 \begin{align*} \frac{2 \, B a \sin \left (d x + c\right )^{3} + 3 \,{\left (A + B\right )} a \sin \left (d x + c\right )^{2} + 6 \, A a \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*a*sin(d*x + c)^3 + 3*(A + B)*a*sin(d*x + c)^2 + 6*A*a*sin(d*x + c))/d

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Fricas [A]  time = 1.64018, size = 120, normalized size = 2.45 \begin{align*} -\frac{3 \,{\left (A + B\right )} a \cos \left (d x + c\right )^{2} + 2 \,{\left (B a \cos \left (d x + c\right )^{2} -{\left (3 \, A + B\right )} a\right )} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(A + B)*a*cos(d*x + c)^2 + 2*(B*a*cos(d*x + c)^2 - (3*A + B)*a)*sin(d*x + c))/d

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Sympy [A]  time = 0.630478, size = 75, normalized size = 1.53 \begin{align*} \begin{cases} \frac{A a \sin{\left (c + d x \right )}}{d} - \frac{A a \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac{B a \sin ^{3}{\left (c + d x \right )}}{3 d} - \frac{B a \cos ^{2}{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (A + B \sin{\left (c \right )}\right ) \left (a \sin{\left (c \right )} + a\right ) \cos{\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)*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.26491, size = 70, normalized size = 1.43 \begin{align*} \frac{2 \, B a \sin \left (d x + c\right )^{3} + 3 \, A a \sin \left (d x + c\right )^{2} + 3 \, B a \sin \left (d x + c\right )^{2} + 6 \, A a \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*a*sin(d*x + c)^3 + 3*A*a*sin(d*x + c)^2 + 3*B*a*sin(d*x + c)^2 + 6*A*a*sin(d*x + c))/d