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

Optimal. Leaf size=63 \[ \frac{a^3 \cos ^3(e+f x)}{3 f}-\frac{4 a^3 \cos (e+f x)}{f}-\frac{3 a^3 \sin (e+f x) \cos (e+f x)}{2 f}+\frac{5 a^3 x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0503981, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {2645, 2638, 2635, 8, 2633} \[ \frac{a^3 \cos ^3(e+f x)}{3 f}-\frac{4 a^3 \cos (e+f x)}{f}-\frac{3 a^3 \sin (e+f x) \cos (e+f x)}{2 f}+\frac{5 a^3 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^3,x]

[Out]

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

Rule 2645

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(a + b*sin[c + d*x])^n, x], x] /;
FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

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

Mathematica [A]  time = 0.343007, size = 44, normalized size = 0.7 \[ \frac{a^3 (-9 \sin (2 (e+f x))-45 \cos (e+f x)+\cos (3 (e+f x))+30 e+30 f x)}{12 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^3,x]

[Out]

(a^3*(30*e + 30*f*x - 45*Cos[e + f*x] + Cos[3*(e + f*x)] - 9*Sin[2*(e + f*x)]))/(12*f)

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Maple [A]  time = 0.029, size = 74, normalized size = 1.2 \begin{align*}{\frac{1}{f} \left ( -{\frac{{a}^{3} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+3\,{a}^{3} \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -3\,{a}^{3}\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+a*sin(f*x+e))^3,x)

[Out]

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

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Maxima [A]  time = 1.1194, size = 97, normalized size = 1.54 \begin{align*} a^{3} x + \frac{{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{3}}{3 \, f} + \frac{3 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3}}{4 \, f} - \frac{3 \, a^{3} \cos \left (f x + e\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.91581, size = 134, normalized size = 2.13 \begin{align*} \frac{2 \, a^{3} \cos \left (f x + e\right )^{3} + 15 \, a^{3} f x - 9 \, a^{3} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 24 \, a^{3} \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))^3,x, algorithm="fricas")

[Out]

1/6*(2*a^3*cos(f*x + e)^3 + 15*a^3*f*x - 9*a^3*cos(f*x + e)*sin(f*x + e) - 24*a^3*cos(f*x + e))/f

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Sympy [A]  time = 0.625299, size = 121, normalized size = 1.92 \begin{align*} \begin{cases} \frac{3 a^{3} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{3 a^{3} x \cos ^{2}{\left (e + f x \right )}}{2} + a^{3} x - \frac{a^{3} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{3 a^{3} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{2 a^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{3 a^{3} \cos{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \left (a \sin{\left (e \right )} + a\right )^{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.30665, size = 78, normalized size = 1.24 \begin{align*} \frac{5}{2} \, a^{3} x + \frac{a^{3} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac{15 \, a^{3} \cos \left (f x + e\right )}{4 \, f} - \frac{3 \, a^{3} \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))^3,x, algorithm="giac")

[Out]

5/2*a^3*x + 1/12*a^3*cos(3*f*x + 3*e)/f - 15/4*a^3*cos(f*x + e)/f - 3/4*a^3*sin(2*f*x + 2*e)/f