3.3.30 \(\int (a+a \sin (e+f x)) (c-c \sin (e+f x)) \, dx\) [230]

Optimal. Leaf size=29 \[ \frac {a c x}{2}+\frac {a c \cos (e+f x) \sin (e+f x)}{2 f} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {2813} \begin {gather*} \frac {a c \sin (e+f x) \cos (e+f x)}{2 f}+\frac {a c x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2813

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+a \sin (e+f x)) (c-c \sin (e+f x)) \, dx &=\frac {a c x}{2}+\frac {a c \cos (e+f x) \sin (e+f x)}{2 f}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 25, normalized size = 0.86 \begin {gather*} \frac {a c (2 (e+f x)+\sin (2 (e+f x)))}{4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c*(2*(e + f*x) + Sin[2*(e + f*x)]))/(4*f)

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Maple [A]
time = 0.10, size = 40, normalized size = 1.38

method	result	size
risch	\(\frac {a c x}{2}+\frac {c a \sin \left (2 f x +2 e \right )}{4 f}\)	\(23\)
derivativedivides	\(\frac {-c a \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+c a \left (f x +e \right )}{f}\)	\(40\)
default	\(\frac {-c a \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+c a \left (f x +e \right )}{f}\)	\(40\)
norman	\(\frac {a c x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {c a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {a c x}{2}+\frac {a c x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}-\frac {c a \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 40, normalized size = 1.38 \begin {gather*} -\frac {{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a c - 4 \, {\left (f x + e\right )} a c}{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)),x, algorithm="maxima")

[Out]

-1/4*((2*f*x + 2*e - sin(2*f*x + 2*e))*a*c - 4*(f*x + e)*a*c)/f

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Fricas [A]
time = 0.32, size = 28, normalized size = 0.97 \begin {gather*} \frac {a c f x + a c \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, 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)),x, algorithm="fricas")

[Out]

1/2*(a*c*f*x + a*c*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. 70 vs. \(2 (26) = 52\).
time = 0.08, size = 70, normalized size = 2.41 \begin {gather*} \begin {cases} - \frac {a c x \sin ^{2}{\left (e + f x \right )}}{2} - \frac {a c x \cos ^{2}{\left (e + f x \right )}}{2} + a c x + \frac {a c \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \left (a \sin {\left (e \right )} + a\right ) \left (- c \sin {\left (e \right )} + c\right ) & \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)),x)

[Out]

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

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Giac [A]
time = 0.41, size = 23, normalized size = 0.79 \begin {gather*} \frac {1}{2} \, a c x + \frac {a c \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)),x, algorithm="giac")

[Out]

1/2*a*c*x + 1/4*a*c*sin(2*f*x + 2*e)/f

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Mupad [B]
time = 7.15, size = 54, normalized size = 1.86 \begin {gather*} \frac {a\,c\,x}{2}-\frac {a\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3-a\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^2} \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)),x)

[Out]

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

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