3.1531 \(\int \cos (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=52 \[ \frac{(a B+A b) \sin ^2(c+d x)}{2 d}+\frac{a A \sin (c+d x)}{d}+\frac{b B \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0537233, antiderivative size = 52, 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} \[ \frac{(a B+A b) \sin ^2(c+d x)}{2 d}+\frac{a A \sin (c+d x)}{d}+\frac{b B \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0802957, size = 45, normalized size = 0.87 \[ \frac{\sin (c+d x) \left (3 (a B+A b) \sin (c+d x)+6 a A+2 b B \sin ^2(c+d x)\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sin[c + d*x]*(6*a*A + 3*(A*b + a*B)*Sin[c + d*x] + 2*b*B*Sin[c + d*x]^2))/(6*d)

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Maple [A]  time = 0.029, size = 44, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{B \left ( \sin \left ( dx+c \right ) \right ) ^{3}b}{3}}+{\frac{ \left ( Ab+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+b*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 0.970237, size = 61, normalized size = 1.17 \begin{align*} \frac{2 \, B b \sin \left (d x + c\right )^{3} + 6 \, A a \sin \left (d x + c\right ) + 3 \,{\left (B a + A b\right )} \sin \left (d x + c\right )^{2}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.38161, size = 123, normalized size = 2.37 \begin{align*} -\frac{3 \,{\left (B a + A b\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (B b \cos \left (d x + c\right )^{2} - 3 \, A a - B b\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+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.600874, size = 75, normalized size = 1.44 \begin{align*} \begin{cases} \frac{A a \sin{\left (c + d x \right )}}{d} - \frac{A b \cos ^{2}{\left (c + d x \right )}}{2 d} - \frac{B a \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac{B b \sin ^{3}{\left (c + d x \right )}}{3 d} & \text{for}\: d \neq 0 \\x \left (A + B \sin{\left (c \right )}\right ) \left (a + b \sin{\left (c \right )}\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+b*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.2267, size = 70, normalized size = 1.35 \begin{align*} \frac{2 \, B b \sin \left (d x + c\right )^{3} + 3 \, B a \sin \left (d x + c\right )^{2} + 3 \, A b \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+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

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