3.1.32 \(\int (a+a \sin (c+d x))^3 \, dx\) [32]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, 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 = {2724, 2718, 2715, 8, 2713} \begin {gather*} \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} \end {gather*}

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 8

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

Rule 2713

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]

Rule 2715

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] && Integ
erQ[2*n]

Rule 2718

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

Rule 2724

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]

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 \text {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*}

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Mathematica [A]
time = 0.21, size = 44, normalized size = 0.70 \begin {gather*} \frac {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} \end {gather*}

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 = 0.15, size = 74, normalized size = 1.17

method	result	size
risch	\(\frac {5 a^{3} x}{2}-\frac {15 a^{3} \cos \left (d x +c \right )}{4 d}+\frac {a^{3} \cos \left (3 d x +3 c \right )}{12 d}-\frac {3 a^{3} \sin \left (2 d x +2 c \right )}{4 d}\)	\(56\)
derivativedivides	\(\frac {-\frac {a^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}+3 a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-3 a^{3} \cos \left (d x +c \right )+a^{3} \left (d x +c \right )}{d}\)	\(74\)
default	\(\frac {-\frac {a^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}+3 a^{3} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-3 a^{3} \cos \left (d x +c \right )+a^{3} \left (d x +c \right )}{d}\)	\(74\)
norman	\(\frac {\frac {5 a^{3} x}{2}-\frac {22 a^{3}}{3 d}-\frac {3 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {3 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {15 a^{3} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 a^{3} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{3} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {6 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {16 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.29, size = 72, normalized size = 1.14 \begin {gather*} 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 {gather*}

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 = 0.36, size = 54, normalized size = 0.86 \begin {gather*} \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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (58) = 116\).
time = 0.13, size = 121, normalized size = 1.92 \begin {gather*} \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 {gather*}

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 = 6.71, size = 55, normalized size = 0.87 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 8.92, size = 156, normalized size = 2.48 \begin {gather*} \frac {5\,a^3\,x}{2}-\frac {\frac {5\,a^3\,\left (c+d\,x\right )}{2}-3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {a^3\,\left (15\,c+15\,d\,x-44\right )}{6}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {15\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (45\,c+45\,d\,x-36\right )}{6}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {15\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (45\,c+45\,d\,x-96\right )}{6}\right )+3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^3,x)

[Out]

(5*a^3*x)/2 - ((5*a^3*(c + d*x))/2 - 3*a^3*tan(c/2 + (d*x)/2)^5 - (a^3*(15*c + 15*d*x - 44))/6 + tan(c/2 + (d*
x)/2)^4*((15*a^3*(c + d*x))/2 - (a^3*(45*c + 45*d*x - 36))/6) + tan(c/2 + (d*x)/2)^2*((15*a^3*(c + d*x))/2 - (
a^3*(45*c + 45*d*x - 96))/6) + 3*a^3*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*x)/2)^2 + 1)^3)

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