3.32 \(\int (a+a \sin (c+d x))^3 \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0544642, 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} \[ \frac{a^3 \cos ^3(c+d x)}{3 d}-\frac{4 a^3 \cos (c+d x)}{d}-\frac{3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{5 a^3 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^3,x]

[Out]

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

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 (c+d x))^3 \, dx &=\int \left (a^3+3 a^3 \sin (c+d x)+3 a^3 \sin ^2(c+d x)+a^3 \sin ^3(c+d x)\right ) \, dx\\ &=a^3 x+a^3 \int \sin ^3(c+d x) \, dx+\left (3 a^3\right ) \int \sin (c+d x) \, dx+\left (3 a^3\right ) \int \sin ^2(c+d x) \, dx\\ &=a^3 x-\frac{3 a^3 \cos (c+d x)}{d}-\frac{3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\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 (c+d x)\right )}{d}\\ &=\frac{5 a^3 x}{2}-\frac{4 a^3 \cos (c+d x)}{d}+\frac{a^3 \cos ^3(c+d x)}{3 d}-\frac{3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.326201, size = 44, normalized size = 0.7 \[ \frac{a^3 (-9 \sin (2 (c+d x))-45 \cos (c+d x)+\cos (3 (c+d x))+30 c+30 d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^3,x]

[Out]

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

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Maple [A]  time = 0.027, size = 74, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ( -{\frac{{a}^{3} \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{3}}+3\,{a}^{3} \left ( -1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) -3\,{a}^{3}\cos \left ( dx+c \right ) +{a}^{3} \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.07352, size = 97, normalized size = 1.54 \begin{align*} a^{3} x + \frac{{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{3}}{3 \, d} + \frac{3 \,{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3}}{4 \, d} - \frac{3 \, a^{3} \cos \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

a^3*x + 1/3*(cos(d*x + c)^3 - 3*cos(d*x + c))*a^3/d + 3/4*(2*d*x + 2*c - sin(2*d*x + 2*c))*a^3/d - 3*a^3*cos(d
*x + c)/d

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Fricas [A]  time = 1.45802, size = 134, normalized size = 2.13 \begin{align*} \frac{2 \, a^{3} \cos \left (d x + c\right )^{3} + 15 \, a^{3} d x - 9 \, a^{3} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 24 \, a^{3} \cos \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.655117, size = 121, normalized size = 1.92 \begin{align*} \begin{cases} \frac{3 a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{3 a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + a^{3} x - \frac{a^{3} \sin ^{2}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} - \frac{3 a^{3} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} - \frac{2 a^{3} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac{3 a^{3} \cos{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a \sin{\left (c \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(d*x+c))**3,x)

[Out]

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

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Giac [A]  time = 1.64017, size = 74, normalized size = 1.17 \begin{align*} \frac{5}{2} \, a^{3} x + \frac{a^{3} \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac{15 \, a^{3} \cos \left (d x + c\right )}{4 \, d} - \frac{3 \, a^{3} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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