3.1538 \(\int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=132 \[ \frac{\left (-3 a^2 B+2 a A b+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^4 d}-\frac{\left (a^2-b^2\right ) (A b-a B) (a+b \sin (c+d x))^3}{3 b^4 d}-\frac{(A b-3 a B) (a+b \sin (c+d x))^5}{5 b^4 d}-\frac{B (a+b \sin (c+d x))^6}{6 b^4 d} \]

[Out]

-((a^2 - b^2)*(A*b - a*B)*(a + b*Sin[c + d*x])^3)/(3*b^4*d) + ((2*a*A*b - 3*a^2*B + b^2*B)*(a + b*Sin[c + d*x]
)^4)/(4*b^4*d) - ((A*b - 3*a*B)*(a + b*Sin[c + d*x])^5)/(5*b^4*d) - (B*(a + b*Sin[c + d*x])^6)/(6*b^4*d)

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Rubi [A]  time = 0.170227, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {2837, 772} \[ \frac{\left (-3 a^2 B+2 a A b+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^4 d}-\frac{\left (a^2-b^2\right ) (A b-a B) (a+b \sin (c+d x))^3}{3 b^4 d}-\frac{(A b-3 a B) (a+b \sin (c+d x))^5}{5 b^4 d}-\frac{B (a+b \sin (c+d x))^6}{6 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2 - b^2)*(A*b - a*B)*(a + b*Sin[c + d*x])^3)/(3*b^4*d) + ((2*a*A*b - 3*a^2*B + b^2*B)*(a + b*Sin[c + d*x]
)^4)/(4*b^4*d) - ((A*b - 3*a*B)*(a + b*Sin[c + d*x])^5)/(5*b^4*d) - (B*(a + b*Sin[c + d*x])^6)/(6*b^4*d)

Rule 2837

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*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \cos ^3(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 ) \left (b^2-x^2\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{\left (-a^2+b^2\right ) (A b-a B) (a+x)^2}{b}+\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) (a+x)^3}{b}+\frac{(-A b+3 a B) (a+x)^4}{b}-\frac{B (a+x)^5}{b}\right ) \, dx,x,b \sin (c+d x)\right )}{b^3 d}\\ &=-\frac{\left (a^2-b^2\right ) (A b-a B) (a+b \sin (c+d x))^3}{3 b^4 d}+\frac{\left (2 a A b-3 a^2 B+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^4 d}-\frac{(A b-3 a B) (a+b \sin (c+d x))^5}{5 b^4 d}-\frac{B (a+b \sin (c+d x))^6}{6 b^4 d}\\ \end{align*}

Mathematica [A]  time = 0.245859, size = 111, normalized size = 0.84 \[ \frac{(a+b \sin (c+d x))^3 \left (3 b \left (a^2 (-B)+2 a A b+5 b^2 B\right ) \sin (c+d x)-2 a^2 A b+a^3 B-6 b^2 (2 A b-a B) \sin ^2(c+d x)-5 a b^2 B+20 A b^3-10 b^3 B \sin ^3(c+d x)\right )}{60 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*Sin[c + d*x])^3*(-2*a^2*A*b + 20*A*b^3 + a^3*B - 5*a*b^2*B + 3*b*(2*a*A*b - a^2*B + 5*b^2*B)*Sin[c + d
*x] - 6*b^2*(2*A*b - a*B)*Sin[c + d*x]^2 - 10*b^3*B*Sin[c + d*x]^3))/(60*b^4*d)

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Maple [A]  time = 0.078, size = 169, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{2}A \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}-{\frac{B{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4}}-{\frac{Aab \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{2}}+2\,Bab \left ( -1/5\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1/15\, \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) +A{b}^{2} \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 ) +B{b}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{6}}-{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{12}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

1/d*(1/3*a^2*A*(2+cos(d*x+c)^2)*sin(d*x+c)-1/4*B*a^2*cos(d*x+c)^4-1/2*A*a*b*cos(d*x+c)^4+2*B*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))+A*b^2*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(2+cos(d*x+c)^2)*s
in(d*x+c))+B*b^2*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4))

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Maxima [A]  time = 0.992413, size = 173, normalized size = 1.31 \begin{align*} -\frac{10 \, B b^{2} \sin \left (d x + c\right )^{6} + 12 \,{\left (2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )^{5} + 15 \,{\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \sin \left (d x + c\right )^{4} - 60 \, A a^{2} \sin \left (d x + c\right ) + 20 \,{\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} - 30 \,{\left (B a^{2} + 2 \, A a b\right )} \sin \left (d x + c\right )^{2}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(10*B*b^2*sin(d*x + c)^6 + 12*(2*B*a*b + A*b^2)*sin(d*x + c)^5 + 15*(B*a^2 + 2*A*a*b - B*b^2)*sin(d*x +
c)^4 - 60*A*a^2*sin(d*x + c) + 20*(A*a^2 - 2*B*a*b - A*b^2)*sin(d*x + c)^3 - 30*(B*a^2 + 2*A*a*b)*sin(d*x + c)
^2)/d

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Fricas [A]  time = 1.47332, size = 284, normalized size = 2.15 \begin{align*} \frac{10 \, B b^{2} \cos \left (d x + c\right )^{6} - 15 \,{\left (B a^{2} + 2 \, A a b + B b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \,{\left (3 \,{\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{4} - 10 \, A a^{2} - 4 \, B a b - 2 \, A b^{2} -{\left (5 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2}\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+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/60*(10*B*b^2*cos(d*x + c)^6 - 15*(B*a^2 + 2*A*a*b + B*b^2)*cos(d*x + c)^4 - 4*(3*(2*B*a*b + A*b^2)*cos(d*x +
c)^4 - 10*A*a^2 - 4*B*a*b - 2*A*b^2 - (5*A*a^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 = 4.73452, size = 228, normalized size = 1.73 \begin{align*} \begin{cases} \frac{2 A a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{A a^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac{A a b \cos ^{4}{\left (c + d x \right )}}{2 d} + \frac{2 A b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{A b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac{B a^{2} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac{4 B a b \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{2 B a b \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{B b^{2} \sin ^{6}{\left (c + d x \right )}}{12 d} + \frac{B b^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\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 ^{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+b*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((2*A*a**2*sin(c + d*x)**3/(3*d) + A*a**2*sin(c + d*x)*cos(c + d*x)**2/d - A*a*b*cos(c + d*x)**4/(2*d
) + 2*A*b**2*sin(c + d*x)**5/(15*d) + A*b**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) - B*a**2*cos(c + d*x)**4/(4
*d) + 4*B*a*b*sin(c + d*x)**5/(15*d) + 2*B*a*b*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + B*b**2*sin(c + d*x)**6/
(12*d) + B*b**2*sin(c + d*x)**4*cos(c + d*x)**2/(4*d), Ne(d, 0)), (x*(A + B*sin(c))*(a + b*sin(c))**2*cos(c)**
3, True))

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Giac [A]  time = 1.28835, size = 227, normalized size = 1.72 \begin{align*} -\frac{10 \, B b^{2} \sin \left (d x + c\right )^{6} + 24 \, B a b \sin \left (d x + c\right )^{5} + 12 \, A b^{2} \sin \left (d x + c\right )^{5} + 15 \, B a^{2} \sin \left (d x + c\right )^{4} + 30 \, A a b \sin \left (d x + c\right )^{4} - 15 \, B b^{2} \sin \left (d x + c\right )^{4} + 20 \, A a^{2} \sin \left (d x + c\right )^{3} - 40 \, B a b \sin \left (d x + c\right )^{3} - 20 \, A b^{2} \sin \left (d x + c\right )^{3} - 30 \, B a^{2} \sin \left (d x + c\right )^{2} - 60 \, A a b \sin \left (d x + c\right )^{2} - 60 \, A a^{2} \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+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/60*(10*B*b^2*sin(d*x + c)^6 + 24*B*a*b*sin(d*x + c)^5 + 12*A*b^2*sin(d*x + c)^5 + 15*B*a^2*sin(d*x + c)^4 +
30*A*a*b*sin(d*x + c)^4 - 15*B*b^2*sin(d*x + c)^4 + 20*A*a^2*sin(d*x + c)^3 - 40*B*a*b*sin(d*x + c)^3 - 20*A*
b^2*sin(d*x + c)^3 - 30*B*a^2*sin(d*x + c)^2 - 60*A*a*b*sin(d*x + c)^2 - 60*A*a^2*sin(d*x + c))/d