3.1539 \(\int \cos (c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=54 \[ \frac{(A b-a B) (a+b \sin (c+d x))^3}{3 b^2 d}+\frac{B (a+b \sin (c+d x))^4}{4 b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.07638, antiderivative size = 54, 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 = {2833, 43} \[ \frac{(A b-a B) (a+b \sin (c+d x))^3}{3 b^2 d}+\frac{B (a+b \sin (c+d x))^4}{4 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*(a + b*Sin[c + d*x])^3)/(3*b^2*d) + (B*(a + b*Sin[c + d*x])^4)/(4*b^2*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))^2 (A+B \sin (c+d x)) \, dx &=\frac{\operatorname{Subst}\left (\int (a+x)^2 \left (A+\frac{B x}{b}\right ) \, dx,x,b \sin (c+d x)\right )}{b d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{(A b-a B) (a+x)^2}{b}+\frac{B (a+x)^3}{b}\right ) \, dx,x,b \sin (c+d x)\right )}{b d}\\ &=\frac{(A b-a B) (a+b \sin (c+d x))^3}{3 b^2 d}+\frac{B (a+b \sin (c+d x))^4}{4 b^2 d}\\ \end{align*}

Mathematica [A]  time = 0.0670486, size = 41, normalized size = 0.76 \[ \frac{(a+b \sin (c+d x))^3 (-a B+4 A b+3 b B \sin (c+d x))}{12 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*Sin[c + d*x])^3*(4*A*b - a*B + 3*b*B*Sin[c + d*x]))/(12*b^2*d)

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Maple [A]  time = 0.033, size = 73, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({\frac{B{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4}}+{\frac{ \left ( A{b}^{2}+2\,Bab \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3}}+{\frac{ \left ( 2\,Aab+B{a}^{2} \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2}}+{a}^{2}A\sin \left ( dx+c \right ) \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))^2*(A+B*sin(d*x+c)),x)

[Out]

1/d*(1/4*B*b^2*sin(d*x+c)^4+1/3*(A*b^2+2*B*a*b)*sin(d*x+c)^3+1/2*(2*A*a*b+B*a^2)*sin(d*x+c)^2+a^2*A*sin(d*x+c)
)

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Maxima [A]  time = 0.976322, size = 100, normalized size = 1.85 \begin{align*} \frac{3 \, B b^{2} \sin \left (d x + c\right )^{4} + 12 \, A a^{2} \sin \left (d x + c\right ) + 4 \,{\left (2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )^{3} + 6 \,{\left (B a^{2} + 2 \, A a b\right )} \sin \left (d x + c\right )^{2}}{12 \, 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))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(3*B*b^2*sin(d*x + c)^4 + 12*A*a^2*sin(d*x + c) + 4*(2*B*a*b + A*b^2)*sin(d*x + c)^3 + 6*(B*a^2 + 2*A*a*b
)*sin(d*x + c)^2)/d

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Fricas [A]  time = 1.38124, size = 213, normalized size = 3.94 \begin{align*} \frac{3 \, B b^{2} \cos \left (d x + c\right )^{4} - 6 \,{\left (B a^{2} + 2 \, A a b + B b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \,{\left (3 \, A a^{2} + 2 \, B a b + A b^{2} -{\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, 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))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(3*B*b^2*cos(d*x + c)^4 - 6*(B*a^2 + 2*A*a*b + B*b^2)*cos(d*x + c)^2 + 4*(3*A*a^2 + 2*B*a*b + A*b^2 - (2*
B*a*b + A*b^2)*cos(d*x + c)^2)*sin(d*x + c))/d

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Sympy [A]  time = 1.30888, size = 143, normalized size = 2.65 \begin{align*} \begin{cases} \frac{A a^{2} \sin{\left (c + d x \right )}}{d} - \frac{A a b \cos ^{2}{\left (c + d x \right )}}{d} + \frac{A b^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} - \frac{B a^{2} \cos ^{2}{\left (c + d x \right )}}{2 d} + \frac{2 B a b \sin ^{3}{\left (c + d x \right )}}{3 d} - \frac{B b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} - \frac{B b^{2} \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 + b \sin{\left (c \right )}\right )^{2} \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))**2*(A+B*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.23297, size = 116, normalized size = 2.15 \begin{align*} \frac{3 \, B b^{2} \sin \left (d x + c\right )^{4} + 8 \, B a b \sin \left (d x + c\right )^{3} + 4 \, A b^{2} \sin \left (d x + c\right )^{3} + 6 \, B a^{2} \sin \left (d x + c\right )^{2} + 12 \, A a b \sin \left (d x + c\right )^{2} + 12 \, A a^{2} \sin \left (d x + c\right )}{12 \, 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))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/12*(3*B*b^2*sin(d*x + c)^4 + 8*B*a*b*sin(d*x + c)^3 + 4*A*b^2*sin(d*x + c)^3 + 6*B*a^2*sin(d*x + c)^2 + 12*A
*a*b*sin(d*x + c)^2 + 12*A*a^2*sin(d*x + c))/d