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

Optimal. Leaf size=78 \[ -\frac{(A-3 B) (a \sin (c+d x)+a)^5}{5 a^3 d}+\frac{(A-B) (a \sin (c+d x)+a)^4}{2 a^2 d}-\frac{B (a \sin (c+d x)+a)^6}{6 a^4 d} \]

[Out]

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

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Rubi [A]  time = 0.109171, antiderivative size = 78, 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 = {2836, 77} \[ -\frac{(A-3 B) (a \sin (c+d x)+a)^5}{5 a^3 d}+\frac{(A-B) (a \sin (c+d x)+a)^4}{2 a^2 d}-\frac{B (a \sin (c+d x)+a)^6}{6 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.405951, size = 66, normalized size = 0.85 \[ \frac{a^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^8 (-4 (3 A-4 B) \sin (c+d x)+18 A+5 B \cos (2 (c+d x))-9 B)}{60 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^8*(18*A - 9*B + 5*B*Cos[2*(c + d*x)] - 4*(3*A - 4*B)*Sin[c + d*x]))
/(60*d)

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Maple [B]  time = 0.068, size = 171, normalized size = 2.2 \begin{align*}{\frac{1}{d} \left ({a}^{2}A \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{a}^{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 ) -{\frac{{a}^{2}A \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{2}}+2\,B{a}^{2} \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 ) +{\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}} \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))^2*(A+B*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 0.978895, size = 130, normalized size = 1.67 \begin{align*} -\frac{5 \, B a^{2} \sin \left (d x + c\right )^{6} + 6 \,{\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )^{5} + 15 \, A a^{2} \sin \left (d x + c\right )^{4} - 20 \, B a^{2} \sin \left (d x + c\right )^{3} - 15 \,{\left (2 \, A + B\right )} a^{2} \sin \left (d x + c\right )^{2} - 30 \, A a^{2} \sin \left (d x + c\right )}{30 \, 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))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.79339, size = 224, normalized size = 2.87 \begin{align*} \frac{5 \, B a^{2} \cos \left (d x + c\right )^{6} - 15 \,{\left (A + B\right )} a^{2} \cos \left (d x + c\right )^{4} - 2 \,{\left (3 \,{\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} - 2 \,{\left (3 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} - 4 \,{\left (3 \, A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{30 \, 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))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/30*(5*B*a^2*cos(d*x + c)^6 - 15*(A + B)*a^2*cos(d*x + c)^4 - 2*(3*(A + 2*B)*a^2*cos(d*x + c)^4 - 2*(3*A + B)
*a^2*cos(d*x + c)^2 - 4*(3*A + B)*a^2)*sin(d*x + c))/d

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Sympy [A]  time = 4.96456, size = 228, normalized size = 2.92 \begin{align*} \begin{cases} \frac{2 A a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \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^{2} \cos ^{4}{\left (c + d x \right )}}{2 d} + \frac{B a^{2} \sin ^{6}{\left (c + d x \right )}}{12 d} + \frac{4 B a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{B a^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} + \frac{2 B a^{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} & \text{for}\: d \neq 0 \\x \left (A + B \sin{\left (c \right )}\right ) \left (a \sin{\left (c \right )} + a\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+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.28829, size = 157, normalized size = 2.01 \begin{align*} -\frac{5 \, B a^{2} \sin \left (d x + c\right )^{6} + 6 \, A a^{2} \sin \left (d x + c\right )^{5} + 12 \, B a^{2} \sin \left (d x + c\right )^{5} + 15 \, A a^{2} \sin \left (d x + c\right )^{4} - 20 \, B a^{2} \sin \left (d x + c\right )^{3} - 30 \, A a^{2} \sin \left (d x + c\right )^{2} - 15 \, B a^{2} \sin \left (d x + c\right )^{2} - 30 \, A a^{2} \sin \left (d x + c\right )}{30 \, 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))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

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