3.567 \(\int (a+b \cos (c+d x) \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=107 \[ -\frac{b \left (16 a^2+b^2\right ) \cos (2 c+2 d x)}{24 d}+\frac{1}{8} a x \left (8 a^2+3 b^2\right )-\frac{5 a b^2 \sin (2 c+2 d x) \cos (2 c+2 d x)}{48 d}-\frac{b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^2}{48 d} \]

[Out]

(a*(8*a^2 + 3*b^2)*x)/8 - (b*(16*a^2 + b^2)*Cos[2*c + 2*d*x])/(24*d) - (5*a*b^2*Cos[2*c + 2*d*x]*Sin[2*c + 2*d
*x])/(48*d) - (b*Cos[2*c + 2*d*x]*(2*a + b*Sin[2*c + 2*d*x])^2)/(48*d)

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Rubi [A]  time = 0.0823962, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2666, 2656, 2734} \[ -\frac{b \left (16 a^2+b^2\right ) \cos (2 c+2 d x)}{24 d}+\frac{1}{8} a x \left (8 a^2+3 b^2\right )-\frac{5 a b^2 \sin (2 c+2 d x) \cos (2 c+2 d x)}{48 d}-\frac{b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^2}{48 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(8*a^2 + 3*b^2)*x)/8 - (b*(16*a^2 + b^2)*Cos[2*c + 2*d*x])/(24*d) - (5*a*b^2*Cos[2*c + 2*d*x]*Sin[2*c + 2*d
*x])/(48*d) - (b*Cos[2*c + 2*d*x]*(2*a + b*Sin[2*c + 2*d*x])^2)/(48*d)

Rule 2666

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + (b*Sin[2*c + 2*
d*x])/2)^n, x] /; FreeQ[{a, b, c, d, n}, x]

Rule 2656

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[1/n, Int[(a + b*Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c + d*
x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

\begin{align*} \int (a+b \cos (c+d x) \sin (c+d x))^3 \, dx &=\int \left (a+\frac{1}{2} b \sin (2 c+2 d x)\right )^3 \, dx\\ &=-\frac{b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^2}{48 d}+\frac{1}{3} \int \left (a+\frac{1}{2} b \sin (2 c+2 d x)\right ) \left (\frac{1}{2} \left (6 a^2+b^2\right )+\frac{5}{2} a b \sin (2 c+2 d x)\right ) \, dx\\ &=\frac{1}{8} a \left (8 a^2+3 b^2\right ) x-\frac{b \left (16 a^2+b^2\right ) \cos (2 c+2 d x)}{24 d}-\frac{5 a b^2 \cos (2 c+2 d x) \sin (2 c+2 d x)}{48 d}-\frac{b \cos (2 c+2 d x) (2 a+b \sin (2 c+2 d x))^2}{48 d}\\ \end{align*}

Mathematica [A]  time = 0.282551, size = 75, normalized size = 0.7 \[ \frac{6 a \left (4 \left (8 a^2+3 b^2\right ) (c+d x)-3 b^2 \sin (4 (c+d x))\right )-9 \left (16 a^2 b+b^3\right ) \cos (2 (c+d x))+b^3 \cos (6 (c+d x))}{192 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*(16*a^2*b + b^3)*Cos[2*(c + d*x)] + b^3*Cos[6*(c + d*x)] + 6*a*(4*(8*a^2 + 3*b^2)*(c + d*x) - 3*b^2*Sin[4*
(c + d*x)]))/(192*d)

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Maple [A]  time = 0.061, size = 106, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({b}^{3} \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 ) +3\,a{b}^{2} \left ( -1/4\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}+1/8\,\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +1/8\,dx+c/8 \right ) -{\frac{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{2}b}{2}}+{a}^{3} \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00551, size = 108, normalized size = 1.01 \begin{align*} a^{3} x - \frac{3 \, a^{2} b \cos \left (d x + c\right )^{2}}{2 \, d} + \frac{3 \,{\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a b^{2}}{32 \, d} - \frac{{\left (2 \, \sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4}\right )} b^{3}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x - 3/2*a^2*b*cos(d*x + c)^2/d + 3/32*(4*d*x + 4*c - sin(4*d*x + 4*c))*a*b^2/d - 1/12*(2*sin(d*x + c)^6 -
3*sin(d*x + c)^4)*b^3/d

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Fricas [A]  time = 2.4859, size = 228, normalized size = 2.13 \begin{align*} \frac{4 \, b^{3} \cos \left (d x + c\right )^{6} - 6 \, b^{3} \cos \left (d x + c\right )^{4} - 36 \, a^{2} b \cos \left (d x + c\right )^{2} + 3 \,{\left (8 \, a^{3} + 3 \, a b^{2}\right )} d x - 9 \,{\left (2 \, a b^{2} \cos \left (d x + c\right )^{3} - a b^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c)*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/24*(4*b^3*cos(d*x + c)^6 - 6*b^3*cos(d*x + c)^4 - 36*a^2*b*cos(d*x + c)^2 + 3*(8*a^3 + 3*a*b^2)*d*x - 9*(2*a
*b^2*cos(d*x + c)^3 - a*b^2*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 4.36334, size = 190, normalized size = 1.78 \begin{align*} \begin{cases} a^{3} x + \frac{3 a^{2} b \sin ^{2}{\left (c + d x \right )}}{2 d} + \frac{3 a b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 a b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 a b^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{3 a b^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{4 d} - \frac{b^{3} \cos ^{6}{\left (c + d x \right )}}{12 d} & \text{for}\: d \neq 0 \\x \left (a + b \sin{\left (c \right )} \cos{\left (c \right )}\right )^{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c)*sin(d*x+c))**3,x)

[Out]

Piecewise((a**3*x + 3*a**2*b*sin(c + d*x)**2/(2*d) + 3*a*b**2*x*sin(c + d*x)**4/8 + 3*a*b**2*x*sin(c + d*x)**2
*cos(c + d*x)**2/4 + 3*a*b**2*x*cos(c + d*x)**4/8 + 3*a*b**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) - 3*a*b**2*sin
(c + d*x)*cos(c + d*x)**3/(8*d) - b**3*sin(c + d*x)**2*cos(c + d*x)**4/(4*d) - b**3*cos(c + d*x)**6/(12*d), Ne
(d, 0)), (x*(a + b*sin(c)*cos(c))**3, True))

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Giac [A]  time = 1.13105, size = 101, normalized size = 0.94 \begin{align*} \frac{b^{3} \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac{3 \, a b^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{1}{8} \,{\left (8 \, a^{3} + 3 \, a b^{2}\right )} x - \frac{3 \,{\left (16 \, a^{2} b + b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c)*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/192*b^3*cos(6*d*x + 6*c)/d - 3/32*a*b^2*sin(4*d*x + 4*c)/d + 1/8*(8*a^3 + 3*a*b^2)*x - 3/64*(16*a^2*b + b^3)
*cos(2*d*x + 2*c)/d