3.680 \(\int (a+b \sin (e+f x))^2 (c+d \sin (e+f x)) \, dx\)

Optimal. Leaf size=107 \[ -\frac{2 \left (a^2 d+3 a b c+b^2 d\right ) \cos (e+f x)}{3 f}+\frac{1}{2} x \left (2 a^2 c+2 a b d+b^2 c\right )-\frac{b (2 a d+3 b c) \sin (e+f x) \cos (e+f x)}{6 f}-\frac{d \cos (e+f x) (a+b \sin (e+f x))^2}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.0936561, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2753, 2734} \[ -\frac{2 \left (a^2 d+3 a b c+b^2 d\right ) \cos (e+f x)}{3 f}+\frac{1}{2} x \left (2 a^2 c+2 a b d+b^2 c\right )-\frac{b (2 a d+3 b c) \sin (e+f x) \cos (e+f x)}{6 f}-\frac{d \cos (e+f x) (a+b \sin (e+f x))^2}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2753

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

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

Mathematica [A]  time = 0.294564, size = 90, normalized size = 0.84 \[ \frac{6 (e+f x) \left (2 a^2 c+2 a b d+b^2 c\right )-3 \left (4 a^2 d+8 a b c+3 b^2 d\right ) \cos (e+f x)-3 b (2 a d+b c) \sin (2 (e+f x))+b^2 d \cos (3 (e+f x))}{12 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.026, size = 115, normalized size = 1.1 \begin{align*}{\frac{1}{f} \left ({a}^{2}c \left ( fx+e \right ) -{a}^{2}d\cos \left ( fx+e \right ) -2\,abc\cos \left ( fx+e \right ) +2\,abd \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) +{b}^{2}c \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) -{\frac{{b}^{2}d \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.12328, size = 151, normalized size = 1.41 \begin{align*} \frac{12 \,{\left (f x + e\right )} a^{2} c + 3 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} b^{2} c + 6 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a b d + 4 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} b^{2} d - 24 \, a b c \cos \left (f x + e\right ) - 12 \, a^{2} d \cos \left (f x + e\right )}{12 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(f*x + e)*a^2*c + 3*(2*f*x + 2*e - sin(2*f*x + 2*e))*b^2*c + 6*(2*f*x + 2*e - sin(2*f*x + 2*e))*a*b*d
+ 4*(cos(f*x + e)^3 - 3*cos(f*x + e))*b^2*d - 24*a*b*c*cos(f*x + e) - 12*a^2*d*cos(f*x + e))/f

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Fricas [A]  time = 1.59181, size = 215, normalized size = 2.01 \begin{align*} \frac{2 \, b^{2} d \cos \left (f x + e\right )^{3} + 3 \,{\left (2 \, a b d +{\left (2 \, a^{2} + b^{2}\right )} c\right )} f x - 3 \,{\left (b^{2} c + 2 \, a b d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 6 \,{\left (2 \, a b c +{\left (a^{2} + b^{2}\right )} 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+b*sin(f*x+e))^2*(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.786323, size = 199, normalized size = 1.86 \begin{align*} \begin{cases} a^{2} c x - \frac{a^{2} d \cos{\left (e + f x \right )}}{f} - \frac{2 a b c \cos{\left (e + f x \right )}}{f} + a b d x \sin ^{2}{\left (e + f x \right )} + a b d x \cos ^{2}{\left (e + f x \right )} - \frac{a b d \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} + \frac{b^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{b^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} - \frac{b^{2} c \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{b^{2} d \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{2 b^{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 )^{2} \left (c + d \sin{\left (e \right )}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.33933, size = 130, normalized size = 1.21 \begin{align*} \frac{b^{2} d \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} + \frac{1}{2} \,{\left (2 \, a^{2} c + b^{2} c + 2 \, a b d\right )} x - \frac{{\left (8 \, a b c + 4 \, a^{2} d + 3 \, b^{2} d\right )} \cos \left (f x + e\right )}{4 \, f} - \frac{{\left (b^{2} c + 2 \, a b 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+b*sin(f*x+e))^2*(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

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