3.169 \(\int (a+b \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=90 \[ -\frac{2 b \left (4 a^2+b^2\right ) \cos (e+f x)}{3 f}+\frac{1}{2} a x \left (2 a^2+3 b^2\right )-\frac{5 a b^2 \sin (e+f x) \cos (e+f x)}{6 f}-\frac{b \cos (e+f x) (a+b \sin (e+f x))^2}{3 f} \]

[Out]

(a*(2*a^2 + 3*b^2)*x)/2 - (2*b*(4*a^2 + b^2)*Cos[e + f*x])/(3*f) - (5*a*b^2*Cos[e + f*x]*Sin[e + f*x])/(6*f) -
(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^2)/(3*f)

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Rubi [A]  time = 0.0656265, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2656, 2734} \[ -\frac{2 b \left (4 a^2+b^2\right ) \cos (e+f x)}{3 f}+\frac{1}{2} a x \left (2 a^2+3 b^2\right )-\frac{5 a b^2 \sin (e+f x) \cos (e+f x)}{6 f}-\frac{b \cos (e+f x) (a+b \sin (e+f x))^2}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])^3,x]

[Out]

(a*(2*a^2 + 3*b^2)*x)/2 - (2*b*(4*a^2 + b^2)*Cos[e + f*x])/(3*f) - (5*a*b^2*Cos[e + f*x]*Sin[e + f*x])/(6*f) -
(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^2)/(3*f)

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 \sin (e+f x))^3 \, dx &=-\frac{b \cos (e+f x) (a+b \sin (e+f x))^2}{3 f}+\frac{1}{3} \int (a+b \sin (e+f x)) \left (3 a^2+2 b^2+5 a b \sin (e+f x)\right ) \, dx\\ &=\frac{1}{2} a \left (2 a^2+3 b^2\right ) x-\frac{2 b \left (4 a^2+b^2\right ) \cos (e+f x)}{3 f}-\frac{5 a b^2 \cos (e+f x) \sin (e+f x)}{6 f}-\frac{b \cos (e+f x) (a+b \sin (e+f x))^2}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.157422, size = 71, normalized size = 0.79 \[ \frac{6 a \left (2 a^2+3 b^2\right ) (e+f x)-9 b \left (4 a^2+b^2\right ) \cos (e+f x)-9 a b^2 \sin (2 (e+f x))+b^3 \cos (3 (e+f x))}{12 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x])^3,x]

[Out]

(6*a*(2*a^2 + 3*b^2)*(e + f*x) - 9*b*(4*a^2 + b^2)*Cos[e + f*x] + b^3*Cos[3*(e + f*x)] - 9*a*b^2*Sin[2*(e + f*
x)])/(12*f)

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Maple [A]  time = 0.022, size = 76, normalized size = 0.8 \begin{align*}{\frac{1}{f} \left ( -{\frac{{b}^{3} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+3\,a{b}^{2} \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -3\,{a}^{2}b\cos \left ( fx+e \right ) + \left ( fx+e \right ){a}^{3} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(f*x+e))^3,x)

[Out]

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

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Maxima [A]  time = 1.96226, size = 100, normalized size = 1.11 \begin{align*} a^{3} x + \frac{3 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a b^{2}}{4 \, f} + \frac{{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} b^{3}}{3 \, f} - \frac{3 \, a^{2} b \cos \left (f x + e\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.74884, size = 169, normalized size = 1.88 \begin{align*} \frac{2 \, b^{3} \cos \left (f x + e\right )^{3} - 9 \, a b^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 3 \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} f x - 6 \,{\left (3 \, a^{2} b + b^{3}\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))^3,x, algorithm="fricas")

[Out]

1/6*(2*b^3*cos(f*x + e)^3 - 9*a*b^2*cos(f*x + e)*sin(f*x + e) + 3*(2*a^3 + 3*a*b^2)*f*x - 6*(3*a^2*b + b^3)*co
s(f*x + e))/f

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Sympy [A]  time = 1.32388, size = 128, normalized size = 1.42 \begin{align*} \begin{cases} a^{3} x - \frac{3 a^{2} b \cos{\left (e + f x \right )}}{f} + \frac{3 a b^{2} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{3 a b^{2} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac{3 a b^{2} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{b^{3} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{2 b^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} & \text{for}\: f \neq 0 \\x \left (a + b \sin{\left (e \right )}\right )^{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))**3,x)

[Out]

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

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Giac [A]  time = 1.71127, size = 101, normalized size = 1.12 \begin{align*} \frac{b^{3} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac{3 \, a b^{2} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} + \frac{1}{2} \,{\left (2 \, a^{3} + 3 \, a b^{2}\right )} x - \frac{3 \,{\left (4 \, a^{2} b + b^{3}\right )} \cos \left (f x + e\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^3,x, algorithm="giac")

[Out]

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