3.246 \(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx\)

Optimal. Leaf size=111 \[ -\frac{a (3 A (c+d)+B (3 c+d)) \cos (e+f x)}{3 f}-\frac{a (3 A d+3 B c-B d) \sin (e+f x) \cos (e+f x)}{6 f}+\frac{1}{2} a x (A (2 c+d)+B (c+d))-\frac{B d \cos (e+f x) (a \sin (e+f x)+a)^2}{3 a f} \]

[Out]

(a*(B*(c + d) + A*(2*c + d))*x)/2 - (a*(3*A*(c + d) + B*(3*c + d))*Cos[e + f*x])/(3*f) - (a*(3*B*c + 3*A*d - B
*d)*Cos[e + f*x]*Sin[e + f*x])/(6*f) - (B*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^2)/(3*a*f)

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Rubi [A]  time = 0.15674, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {2968, 3023, 2734} \[ -\frac{a (3 A (c+d)+B (3 c+d)) \cos (e+f x)}{3 f}-\frac{a (3 A d+3 B c-B d) \sin (e+f x) \cos (e+f x)}{6 f}+\frac{1}{2} a x (A (2 c+d)+B (c+d))-\frac{B d \cos (e+f x) (a \sin (e+f x)+a)^2}{3 a f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c + d*Sin[e + f*x]),x]

[Out]

(a*(B*(c + d) + A*(2*c + d))*x)/2 - (a*(3*A*(c + d) + B*(3*c + d))*Cos[e + f*x])/(3*f) - (a*(3*B*c + 3*A*d - B
*d)*Cos[e + f*x]*Sin[e + f*x])/(6*f) - (B*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^2)/(3*a*f)

Rule 2968

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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+a \sin (e+f x)) (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx &=\int (a+a \sin (e+f x)) \left (A c+(B c+A d) \sin (e+f x)+B d \sin ^2(e+f x)\right ) \, dx\\ &=-\frac{B d \cos (e+f x) (a+a \sin (e+f x))^2}{3 a f}+\frac{\int (a+a \sin (e+f x)) (a (3 A c+2 B d)+a (3 B c+3 A d-B d) \sin (e+f x)) \, dx}{3 a}\\ &=\frac{1}{2} a (B (c+d)+A (2 c+d)) x-\frac{a (3 A (c+d)+B (3 c+d)) \cos (e+f x)}{3 f}-\frac{a (3 B c+3 A d-B d) \cos (e+f x) \sin (e+f x)}{6 f}-\frac{B d \cos (e+f x) (a+a \sin (e+f x))^2}{3 a f}\\ \end{align*}

Mathematica [A]  time = 0.426629, size = 104, normalized size = 0.94 \[ \frac{a (-3 (4 A (c+d)+B (4 c+3 d)) \cos (e+f x)+12 A c f x-3 A d \sin (2 (e+f x))+6 A d f x-3 B c \sin (2 (e+f x))+6 B c f x-3 B d \sin (2 (e+f x))+B d \cos (3 (e+f x))+6 B d f x)}{12 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c + d*Sin[e + f*x]),x]

[Out]

(a*(12*A*c*f*x + 6*B*c*f*x + 6*A*d*f*x + 6*B*d*f*x - 3*(4*A*(c + d) + B*(4*c + 3*d))*Cos[e + f*x] + B*d*Cos[3*
(e + f*x)] - 3*B*c*Sin[2*(e + f*x)] - 3*A*d*Sin[2*(e + f*x)] - 3*B*d*Sin[2*(e + f*x)]))/(12*f)

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Maple [A]  time = 0.047, size = 147, normalized size = 1.3 \begin{align*}{\frac{1}{f} \left ( -{\frac{Bad \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+Aad \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) +Bac \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) +Bad \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) -Aac\cos \left ( fx+e \right ) -Aad\cos \left ( fx+e \right ) -Bac\cos \left ( fx+e \right ) +Aac \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x)

[Out]

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

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Maxima [A]  time = 0.957769, size = 193, normalized size = 1.74 \begin{align*} \frac{12 \,{\left (f x + e\right )} A a c + 3 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a c + 3 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a d + 4 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a d + 3 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a d - 12 \, A a c \cos \left (f x + e\right ) - 12 \, B a c \cos \left (f x + e\right ) - 12 \, A a d \cos \left (f x + e\right )}{12 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.94864, size = 225, normalized size = 2.03 \begin{align*} \frac{2 \, B a d \cos \left (f x + e\right )^{3} + 3 \,{\left ({\left (2 \, A + B\right )} a c +{\left (A + B\right )} a d\right )} f x - 3 \,{\left (B a c +{\left (A + B\right )} a d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 6 \,{\left ({\left (A + B\right )} a c +{\left (A + B\right )} a d\right )} \cos \left (f x + e\right )}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.35109, size = 277, normalized size = 2.5 \begin{align*} \begin{cases} A a c x - \frac{A a c \cos{\left (e + f x \right )}}{f} + \frac{A a d x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{A a d x \cos ^{2}{\left (e + f x \right )}}{2} - \frac{A a d \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{A a d \cos{\left (e + f x \right )}}{f} + \frac{B a c x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{B a c x \cos ^{2}{\left (e + f x \right )}}{2} - \frac{B a c \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{B a c \cos{\left (e + f x \right )}}{f} + \frac{B a d x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{B a d x \cos ^{2}{\left (e + f x \right )}}{2} - \frac{B a d \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{B a d \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{2 B a d \cos ^{3}{\left (e + f x \right )}}{3 f} & \text{for}\: f \neq 0 \\x \left (A + B \sin{\left (e \right )}\right ) \left (c + d \sin{\left (e \right )}\right ) \left (a \sin{\left (e \right )} + a\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x)

[Out]

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

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Giac [A]  time = 1.21923, size = 136, normalized size = 1.23 \begin{align*} \frac{B a d \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} + \frac{1}{2} \,{\left (2 \, A a c + B a c + A a d + B a d\right )} x - \frac{{\left (4 \, A a c + 4 \, B a c + 4 \, A a d + 3 \, B a d\right )} \cos \left (f x + e\right )}{4 \, f} - \frac{{\left (B a c + A a d + B a d\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

1/12*B*a*d*cos(3*f*x + 3*e)/f + 1/2*(2*A*a*c + B*a*c + A*a*d + B*a*d)*x - 1/4*(4*A*a*c + 4*B*a*c + 4*A*a*d + 3
*B*a*d)*cos(f*x + e)/f - 1/4*(B*a*c + A*a*d + B*a*d)*sin(2*f*x + 2*e)/f