3.253 \(\int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx\)

Optimal. Leaf size=166 \[ -\frac{a^2 (12 A c+8 A d+8 B c+7 B d) \cos (e+f x)}{6 f}-\frac{a^2 (12 A c+8 A d+8 B c+7 B d) \sin (e+f x) \cos (e+f x)}{24 f}+\frac{1}{8} a^2 x (12 A c+8 A d+8 B c+7 B d)-\frac{(4 A d+4 B c-B d) \cos (e+f x) (a \sin (e+f x)+a)^2}{12 f}-\frac{B d \cos (e+f x) (a \sin (e+f x)+a)^3}{4 a f} \]

[Out]

(a^2*(12*A*c + 8*B*c + 8*A*d + 7*B*d)*x)/8 - (a^2*(12*A*c + 8*B*c + 8*A*d + 7*B*d)*Cos[e + f*x])/(6*f) - (a^2*
(12*A*c + 8*B*c + 8*A*d + 7*B*d)*Cos[e + f*x]*Sin[e + f*x])/(24*f) - ((4*B*c + 4*A*d - B*d)*Cos[e + f*x]*(a +
a*Sin[e + f*x])^2)/(12*f) - (B*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^3)/(4*a*f)

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Rubi [A]  time = 0.270738, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {2968, 3023, 2751, 2644} \[ -\frac{a^2 (12 A c+8 A d+8 B c+7 B d) \cos (e+f x)}{6 f}-\frac{a^2 (12 A c+8 A d+8 B c+7 B d) \sin (e+f x) \cos (e+f x)}{24 f}+\frac{1}{8} a^2 x (12 A c+8 A d+8 B c+7 B d)-\frac{(4 A d+4 B c-B d) \cos (e+f x) (a \sin (e+f x)+a)^2}{12 f}-\frac{B d \cos (e+f x) (a \sin (e+f x)+a)^3}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*A*c + 8*B*c + 8*A*d + 7*B*d)*x)/8 - (a^2*(12*A*c + 8*B*c + 8*A*d + 7*B*d)*Cos[e + f*x])/(6*f) - (a^2*
(12*A*c + 8*B*c + 8*A*d + 7*B*d)*Cos[e + f*x]*Sin[e + f*x])/(24*f) - ((4*B*c + 4*A*d - B*d)*Cos[e + f*x]*(a +
a*Sin[e + f*x])^2)/(12*f) - (B*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^3)/(4*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 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2644

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

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c+d \sin (e+f x)) \, dx &=\int (a+a \sin (e+f x))^2 \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))^3}{4 a f}+\frac{\int (a+a \sin (e+f x))^2 (a (4 A c+3 B d)+a (4 B c+4 A d-B d) \sin (e+f x)) \, dx}{4 a}\\ &=-\frac{(4 B c+4 A d-B d) \cos (e+f x) (a+a \sin (e+f x))^2}{12 f}-\frac{B d \cos (e+f x) (a+a \sin (e+f x))^3}{4 a f}+\frac{1}{12} (12 A c+8 B c+8 A d+7 B d) \int (a+a \sin (e+f x))^2 \, dx\\ &=\frac{1}{8} a^2 (12 A c+8 B c+8 A d+7 B d) x-\frac{a^2 (12 A c+8 B c+8 A d+7 B d) \cos (e+f x)}{6 f}-\frac{a^2 (12 A c+8 B c+8 A d+7 B d) \cos (e+f x) \sin (e+f x)}{24 f}-\frac{(4 B c+4 A d-B d) \cos (e+f x) (a+a \sin (e+f x))^2}{12 f}-\frac{B d \cos (e+f x) (a+a \sin (e+f x))^3}{4 a f}\\ \end{align*}

Mathematica [A]  time = 0.747391, size = 160, normalized size = 0.96 \[ -\frac{a^2 \cos (e+f x) \left (6 (12 A c+8 A d+8 B c+7 B d) \sin ^{-1}\left (\frac{\sqrt{1-\sin (e+f x)}}{\sqrt{2}}\right )+\sqrt{\cos ^2(e+f x)} \left (8 (A d+B (c+2 d)) \sin ^2(e+f x)+3 (4 A c+8 A d+8 B c+7 B d) \sin (e+f x)+8 (6 A c+5 A d+5 B c+4 B d)+6 B d \sin ^3(e+f x)\right )\right )}{24 f \sqrt{\cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*Cos[e + f*x]*(6*(12*A*c + 8*B*c + 8*A*d + 7*B*d)*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt[2]] + Sqrt[Cos[e + f
*x]^2]*(8*(6*A*c + 5*B*c + 5*A*d + 4*B*d) + 3*(4*A*c + 8*B*c + 8*A*d + 7*B*d)*Sin[e + f*x] + 8*(A*d + B*(c + 2
*d))*Sin[e + f*x]^2 + 6*B*d*Sin[e + f*x]^3)))/(24*f*Sqrt[Cos[e + f*x]^2])

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Maple [A]  time = 0.051, size = 278, normalized size = 1.7 \begin{align*}{\frac{1}{f} \left ( A{a}^{2}c \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) -{\frac{A{a}^{2}d \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}-{\frac{B{a}^{2}c \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+B{a}^{2}d \left ( -{\frac{\cos \left ( fx+e \right ) }{4} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\sin \left ( fx+e \right ) }{2}} \right ) }+{\frac{3\,fx}{8}}+{\frac{3\,e}{8}} \right ) -2\,A{a}^{2}c\cos \left ( fx+e \right ) +2\,A{a}^{2}d \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) +2\,B{a}^{2}c \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -{\frac{2\,B{a}^{2}d \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+A{a}^{2}c \left ( fx+e \right ) -A{a}^{2}d\cos \left ( fx+e \right ) -B{a}^{2}c\cos \left ( fx+e \right ) +B{a}^{2}d \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x)

[Out]

1/f*(A*a^2*c*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)-1/3*A*a^2*d*(2+sin(f*x+e)^2)*cos(f*x+e)-1/3*B*a^2*c*(2
+sin(f*x+e)^2)*cos(f*x+e)+B*a^2*d*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)-2*A*a^2*c*cos(
f*x+e)+2*A*a^2*d*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)+2*B*a^2*c*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*
e)-2/3*B*a^2*d*(2+sin(f*x+e)^2)*cos(f*x+e)+A*a^2*c*(f*x+e)-A*a^2*d*cos(f*x+e)-B*a^2*c*cos(f*x+e)+B*a^2*d*(-1/2
*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e))

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Maxima [A]  time = 0.966241, size = 362, normalized size = 2.18 \begin{align*} \frac{24 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} c + 96 \,{\left (f x + e\right )} A a^{2} c + 32 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{2} c + 48 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} c + 32 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a^{2} d + 48 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} d + 64 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{2} d + 3 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} d + 24 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} d - 192 \, A a^{2} c \cos \left (f x + e\right ) - 96 \, B a^{2} c \cos \left (f x + e\right ) - 96 \, A a^{2} d \cos \left (f x + e\right )}{96 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

1/96*(24*(2*f*x + 2*e - sin(2*f*x + 2*e))*A*a^2*c + 96*(f*x + e)*A*a^2*c + 32*(cos(f*x + e)^3 - 3*cos(f*x + e)
)*B*a^2*c + 48*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a^2*c + 32*(cos(f*x + e)^3 - 3*cos(f*x + e))*A*a^2*d + 48*(2
*f*x + 2*e - sin(2*f*x + 2*e))*A*a^2*d + 64*(cos(f*x + e)^3 - 3*cos(f*x + e))*B*a^2*d + 3*(12*f*x + 12*e + sin
(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*B*a^2*d + 24*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a^2*d - 192*A*a^2*c*cos(f*
x + e) - 96*B*a^2*c*cos(f*x + e) - 96*A*a^2*d*cos(f*x + e))/f

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Fricas [A]  time = 2.03709, size = 343, normalized size = 2.07 \begin{align*} \frac{8 \,{\left (B a^{2} c +{\left (A + 2 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (4 \,{\left (3 \, A + 2 \, B\right )} a^{2} c +{\left (8 \, A + 7 \, B\right )} a^{2} d\right )} f x - 48 \,{\left ({\left (A + B\right )} a^{2} c +{\left (A + B\right )} a^{2} d\right )} \cos \left (f x + e\right ) + 3 \,{\left (2 \, B a^{2} d \cos \left (f x + e\right )^{3} -{\left (4 \,{\left (A + 2 \, B\right )} a^{2} c +{\left (8 \, A + 9 \, B\right )} a^{2} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

1/24*(8*(B*a^2*c + (A + 2*B)*a^2*d)*cos(f*x + e)^3 + 3*(4*(3*A + 2*B)*a^2*c + (8*A + 7*B)*a^2*d)*f*x - 48*((A
+ B)*a^2*c + (A + B)*a^2*d)*cos(f*x + e) + 3*(2*B*a^2*d*cos(f*x + e)^3 - (4*(A + 2*B)*a^2*c + (8*A + 9*B)*a^2*
d)*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 2.22361, size = 571, normalized size = 3.44 \begin{align*} \begin{cases} \frac{A a^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{A a^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} + A a^{2} c x - \frac{A a^{2} c \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{2 A a^{2} c \cos{\left (e + f x \right )}}{f} + A a^{2} d x \sin ^{2}{\left (e + f x \right )} + A a^{2} d x \cos ^{2}{\left (e + f x \right )} - \frac{A a^{2} d \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{A a^{2} d \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{2 A a^{2} d \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{A a^{2} d \cos{\left (e + f x \right )}}{f} + B a^{2} c x \sin ^{2}{\left (e + f x \right )} + B a^{2} c x \cos ^{2}{\left (e + f x \right )} - \frac{B a^{2} c \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{B a^{2} c \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{2 B a^{2} c \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{B a^{2} c \cos{\left (e + f x \right )}}{f} + \frac{3 B a^{2} d x \sin ^{4}{\left (e + f x \right )}}{8} + \frac{3 B a^{2} d x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{B a^{2} d x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{3 B a^{2} d x \cos ^{4}{\left (e + f x \right )}}{8} + \frac{B a^{2} d x \cos ^{2}{\left (e + f x \right )}}{2} - \frac{5 B a^{2} d \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} - \frac{2 B a^{2} d \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{3 B a^{2} d \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac{B a^{2} d \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{4 B a^{2} 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 )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x)

[Out]

Piecewise((A*a**2*c*x*sin(e + f*x)**2/2 + A*a**2*c*x*cos(e + f*x)**2/2 + A*a**2*c*x - A*a**2*c*sin(e + f*x)*co
s(e + f*x)/(2*f) - 2*A*a**2*c*cos(e + f*x)/f + A*a**2*d*x*sin(e + f*x)**2 + A*a**2*d*x*cos(e + f*x)**2 - A*a**
2*d*sin(e + f*x)**2*cos(e + f*x)/f - A*a**2*d*sin(e + f*x)*cos(e + f*x)/f - 2*A*a**2*d*cos(e + f*x)**3/(3*f) -
A*a**2*d*cos(e + f*x)/f + B*a**2*c*x*sin(e + f*x)**2 + B*a**2*c*x*cos(e + f*x)**2 - B*a**2*c*sin(e + f*x)**2*
cos(e + f*x)/f - B*a**2*c*sin(e + f*x)*cos(e + f*x)/f - 2*B*a**2*c*cos(e + f*x)**3/(3*f) - B*a**2*c*cos(e + f*
x)/f + 3*B*a**2*d*x*sin(e + f*x)**4/8 + 3*B*a**2*d*x*sin(e + f*x)**2*cos(e + f*x)**2/4 + B*a**2*d*x*sin(e + f*
x)**2/2 + 3*B*a**2*d*x*cos(e + f*x)**4/8 + B*a**2*d*x*cos(e + f*x)**2/2 - 5*B*a**2*d*sin(e + f*x)**3*cos(e + f
*x)/(8*f) - 2*B*a**2*d*sin(e + f*x)**2*cos(e + f*x)/f - 3*B*a**2*d*sin(e + f*x)*cos(e + f*x)**3/(8*f) - B*a**2
*d*sin(e + f*x)*cos(e + f*x)/(2*f) - 4*B*a**2*d*cos(e + f*x)**3/(3*f), Ne(f, 0)), (x*(A + B*sin(e))*(c + d*sin
(e))*(a*sin(e) + a)**2, True))

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Giac [A]  time = 1.24558, size = 232, normalized size = 1.4 \begin{align*} \frac{B a^{2} d \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac{1}{8} \,{\left (12 \, A a^{2} c + 8 \, B a^{2} c + 8 \, A a^{2} d + 7 \, B a^{2} d\right )} x + \frac{{\left (B a^{2} c + A a^{2} d + 2 \, B a^{2} d\right )} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac{{\left (8 \, A a^{2} c + 7 \, B a^{2} c + 7 \, A a^{2} d + 6 \, B a^{2} d\right )} \cos \left (f x + e\right )}{4 \, f} - \frac{{\left (A a^{2} c + 2 \, B a^{2} c + 2 \, A a^{2} d + 2 \, B a^{2} 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))^2*(A+B*sin(f*x+e))*(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

1/32*B*a^2*d*sin(4*f*x + 4*e)/f + 1/8*(12*A*a^2*c + 8*B*a^2*c + 8*A*a^2*d + 7*B*a^2*d)*x + 1/12*(B*a^2*c + A*a
^2*d + 2*B*a^2*d)*cos(3*f*x + 3*e)/f - 1/4*(8*A*a^2*c + 7*B*a^2*c + 7*A*a^2*d + 6*B*a^2*d)*cos(f*x + e)/f - 1/
4*(A*a^2*c + 2*B*a^2*c + 2*A*a^2*d + 2*B*a^2*d)*sin(2*f*x + 2*e)/f