3.159 \(\int \sin (e+f x) (a+b \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=71 \[ -\frac{2 \left (a^2+b^2\right ) \cos (e+f x)}{3 f}-\frac{\cos (e+f x) (a+b \sin (e+f x))^2}{3 f}-\frac{a b \sin (e+f x) \cos (e+f x)}{3 f}+a b x \]

[Out]

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

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Rubi [A]  time = 0.048996, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2753, 2734} \[ -\frac{2 \left (a^2+b^2\right ) \cos (e+f x)}{3 f}-\frac{\cos (e+f x) (a+b \sin (e+f x))^2}{3 f}-\frac{a b \sin (e+f x) \cos (e+f x)}{3 f}+a b x \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*b*x - (2*(a^2 + b^2)*Cos[e + f*x])/(3*f) - (a*b*Cos[e + f*x]*Sin[e + f*x])/(3*f) - (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 \sin (e+f x) (a+b \sin (e+f x))^2 \, dx &=-\frac{\cos (e+f x) (a+b \sin (e+f x))^2}{3 f}+\frac{1}{3} \int (2 b+2 a \sin (e+f x)) (a+b \sin (e+f x)) \, dx\\ &=a b x-\frac{2 \left (a^2+b^2\right ) \cos (e+f x)}{3 f}-\frac{a b \cos (e+f x) \sin (e+f x)}{3 f}-\frac{\cos (e+f x) (a+b \sin (e+f x))^2}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.210484, size = 59, normalized size = 0.83 \[ \frac{b (12 a (e+f x)-6 a \sin (2 (e+f x))+b \cos (3 (e+f x)))-3 \left (4 a^2+3 b^2\right ) \cos (e+f x)}{12 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.019, size = 64, normalized size = 0.9 \begin{align*}{\frac{1}{f} \left ( -{\frac{{b}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+2\,ab \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -\cos \left ( fx+e \right ){a}^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.60699, size = 84, normalized size = 1.18 \begin{align*} \frac{3 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a b + 2 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} b^{2} - 6 \, a^{2} \cos \left (f x + e\right )}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.97532, size = 139, normalized size = 1.96 \begin{align*} \frac{b^{2} \cos \left (f x + e\right )^{3} + 3 \, a b f x - 3 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \,{\left (a^{2} + b^{2}\right )} \cos \left (f x + e\right )}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.03553, size = 107, normalized size = 1.51 \begin{align*} \begin{cases} - \frac{a^{2} \cos{\left (e + f x \right )}}{f} + a b x \sin ^{2}{\left (e + f x \right )} + a b x \cos ^{2}{\left (e + f x \right )} - \frac{a b \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{b^{2} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{2 b^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} & \text{for}\: f \neq 0 \\x \left (a + b \sin{\left (e \right )}\right )^{2} \sin{\left (e \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.03036, size = 103, normalized size = 1.45 \begin{align*} a b x + \frac{b^{2} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac{b^{2} \cos \left (f x + e\right )}{4 \, f} - \frac{a b \sin \left (2 \, f x + 2 \, e\right )}{2 \, f} - \frac{{\left (2 \, a^{2} + b^{2}\right )} \cos \left (f x + e\right )}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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