\(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx\) [247]

Optimal result

Integrand size = 21, antiderivative size = 48 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=\frac {1}{2} a (2 A+B) x-\frac {a (A+B) \cos (e+f x)}{f}-\frac {a B \cos (e+f x) \sin (e+f x)}{2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2813} \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=-\frac {a (A+B) \cos (e+f x)}{f}+\frac {1}{2} a x (2 A+B)-\frac {a B \sin (e+f x) \cos (e+f x)}{2 f} \]

[In]

Int[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]),x]

[Out]

(a*(2*A + B)*x)/2 - (a*(A + B)*Cos[e + f*x])/f - (a*B*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*} \text {integral}& = \frac {1}{2} a (2 A+B) x-\frac {a (A+B) \cos (e+f x)}{f}-\frac {a B \cos (e+f x) \sin (e+f x)}{2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=\frac {a (2 B e+4 A f x+2 B f x-4 (A+B) \cos (e+f x)-B \sin (2 (e+f x)))}{4 f} \]

[In]

Integrate[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]),x]

[Out]

(a*(2*B*e + 4*A*f*x + 2*B*f*x - 4*(A + B)*Cos[e + f*x] - B*Sin[2*(e + f*x)]))/(4*f)

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.92

method	result	size
parallelrisch	\(\frac {\left (-\frac {B \sin \left (2 f x +2 e \right )}{4}+\left (-A -B \right ) \cos \left (f x +e \right )+f x A +\frac {f x B}{2}+A +B \right ) a}{f}\)	\(44\)
parts	\(a x A -\frac {\left (a A +B a \right ) \cos \left (f x +e \right )}{f}+\frac {B a \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\)	\(52\)
risch	\(a x A +\frac {a B x}{2}-\frac {a \cos \left (f x +e \right ) A}{f}-\frac {a \cos \left (f x +e \right ) B}{f}-\frac {B a \sin \left (2 f x +2 e \right )}{4 f}\)	\(53\)
derivativedivides	\(\frac {B a \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-a A \cos \left (f x +e \right )-B a \cos \left (f x +e \right )+a A \left (f x +e \right )}{f}\)	\(59\)
default	\(\frac {B a \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-a A \cos \left (f x +e \right )-B a \cos \left (f x +e \right )+a A \left (f x +e \right )}{f}\)	\(59\)
norman	\(\frac {\left (a A +\frac {1}{2} B a \right ) x +\left (a A +\frac {1}{2} B a \right ) x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (2 a A +B a \right ) x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {\left (2 a A +2 B a \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {B a \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {2 \left (a A +B a \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {B a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}\)	\(150\)

[In]

int((a+a*sin(f*x+e))*(A+B*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

(-1/4*B*sin(2*f*x+2*e)+(-A-B)*cos(f*x+e)+f*x*A+1/2*f*x*B+A+B)*a/f

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.90 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=\frac {{\left (2 \, A + B\right )} a f x - B a \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, {\left (A + B\right )} a \cos \left (f x + e\right )}{2 \, f} \]

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e)),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (42) = 84\).

Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.96 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=\begin {cases} A a x - \frac {A a \cos {\left (e + f x \right )}}{f} + \frac {B a x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {B a x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {B a \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {B a \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (A + B \sin {\left (e \right )}\right ) \left (a \sin {\left (e \right )} + a\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=\frac {4 \, {\left (f x + e\right )} A a + {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a - 4 \, A a \cos \left (f x + e\right ) - 4 \, B a \cos \left (f x + e\right )}{4 \, f} \]

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e)),x, algorithm="maxima")

[Out]

1/4*(4*(f*x + e)*A*a + (2*f*x + 2*e - sin(2*f*x + 2*e))*B*a - 4*A*a*cos(f*x + e) - 4*B*a*cos(f*x + e))/f

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=\frac {1}{2} \, {\left (2 \, A a + B a\right )} x - \frac {B a \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} - \frac {{\left (A a + B a\right )} \cos \left (f x + e\right )}{f} \]

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e)),x, algorithm="giac")

[Out]

1/2*(2*A*a + B*a)*x - 1/4*B*a*sin(2*f*x + 2*e)/f - (A*a + B*a)*cos(f*x + e)/f

Mupad [B] (verification not implemented)

Time = 12.55 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.08 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx=A\,a\,x-\frac {-B\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+\left (2\,A\,a+2\,B\,a\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+B\,a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+2\,A\,a+2\,B\,a}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}+\frac {B\,a\,x}{2} \]

[In]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x)),x)

[Out]

A*a*x - (2*A*a + 2*B*a + tan(e/2 + (f*x)/2)^2*(2*A*a + 2*B*a) - B*a*tan(e/2 + (f*x)/2)^3 + B*a*tan(e/2 + (f*x)
/2))/(f*(2*tan(e/2 + (f*x)/2)^2 + tan(e/2 + (f*x)/2)^4 + 1)) + (B*a*x)/2