3.3.29 \(\int (a+a \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx\) [229]

Optimal. Leaf size=52 \[ \frac {1}{2} a c^2 x+\frac {a c^2 \cos ^3(e+f x)}{3 f}+\frac {a c^2 \cos (e+f x) \sin (e+f x)}{2 f} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2815, 2748, 2715, 8} \begin {gather*} \frac {a c^2 \cos ^3(e+f x)}{3 f}+\frac {a c^2 \sin (e+f x) \cos (e+f x)}{2 f}+\frac {1}{2} a c^2 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

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 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2815

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rubi steps

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

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Mathematica [A]
time = 0.20, size = 42, normalized size = 0.81 \begin {gather*} \frac {a c^2 (6 f x+3 \cos (e+f x)+\cos (3 (e+f x))+3 \sin (2 (e+f x)))}{12 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.16, size = 77, normalized size = 1.48

method	result	size
risch	\(\frac {a \,c^{2} x}{2}+\frac {a \,c^{2} \cos \left (f x +e \right )}{4 f}+\frac {a \,c^{2} \cos \left (3 f x +3 e \right )}{12 f}+\frac {a \,c^{2} \sin \left (2 f x +2 e \right )}{4 f}\)	\(60\)
derivativedivides	\(\frac {-\frac {a \,c^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}-a \,c^{2} \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+a \,c^{2} \cos \left (f x +e \right )+a \,c^{2} \left (f x +e \right )}{f}\)	\(77\)
default	\(\frac {-\frac {a \,c^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}-a \,c^{2} \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+a \,c^{2} \cos \left (f x +e \right )+a \,c^{2} \left (f x +e \right )}{f}\)	\(77\)
norman	\(\frac {\frac {2 a \,c^{2} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {a \,c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {2 a \,c^{2}}{3 f}+\frac {a \,c^{2} x}{2}-\frac {a \,c^{2} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {3 a \,c^{2} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {3 a \,c^{2} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {a \,c^{2} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))*(c-c*sin(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 83, normalized size = 1.60 \begin {gather*} \frac {4 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a c^{2} - 3 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a c^{2} + 12 \, {\left (f x + e\right )} a c^{2} + 12 \, a c^{2} \cos \left (f x + e\right )}{12 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 49, normalized size = 0.94 \begin {gather*} \frac {2 \, a c^{2} \cos \left (f x + e\right )^{3} + 3 \, a c^{2} f x + 3 \, a c^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{6 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (46) = 92\).
time = 0.13, size = 133, normalized size = 2.56 \begin {gather*} \begin {cases} - \frac {a c^{2} x \sin ^{2}{\left (e + f x \right )}}{2} - \frac {a c^{2} x \cos ^{2}{\left (e + f x \right )}}{2} + a c^{2} x - \frac {a c^{2} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {a c^{2} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {2 a c^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac {a c^{2} \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (a \sin {\left (e \right )} + a\right ) \left (- c \sin {\left (e \right )} + c\right )^{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.44, size = 62, normalized size = 1.19 \begin {gather*} \frac {1}{2} \, a c^{2} x + \frac {a c^{2} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} + \frac {a c^{2} \cos \left (f x + e\right )}{4 \, f} + \frac {a c^{2} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 8.96, size = 125, normalized size = 2.40 \begin {gather*} \frac {a\,c^2\,x}{2}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {3\,a\,c^2\,\left (e+f\,x\right )}{2}-\frac {a\,c^2\,\left (9\,e+9\,f\,x+12\right )}{6}\right )-a\,c^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+\frac {a\,c^2\,\left (e+f\,x\right )}{2}+a\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5-\frac {a\,c^2\,\left (3\,e+3\,f\,x+4\right )}{6}}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*c^2*x)/2 - (tan(e/2 + (f*x)/2)^4*((3*a*c^2*(e + f*x))/2 - (a*c^2*(9*e + 9*f*x + 12))/6) - a*c^2*tan(e/2 + (
f*x)/2) + (a*c^2*(e + f*x))/2 + a*c^2*tan(e/2 + (f*x)/2)^5 - (a*c^2*(3*e + 3*f*x + 4))/6)/(f*(tan(e/2 + (f*x)/
2)^2 + 1)^3)

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