3.247 \(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \, dx\)

Optimal. Leaf size=48 \[ -\frac {a (A+B) \cos (e+f x)}{f}+\frac {1}{2} a x (2 A+B)-\frac {a B \sin (e+f x) \cos (e+f x)}{2 f} \]

[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

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Rubi [A]  time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2734} \[ -\frac {a (A+B) \cos (e+f x)}{f}+\frac {1}{2} a x (2 A+B)-\frac {a B \sin (e+f x) \cos (e+f x)}{2 f} \]

Antiderivative was successfully verified.

[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 2734

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)) (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}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 45, normalized size = 0.94 \[ \frac {a (-4 (A+B) \cos (e+f x)+4 A f x-B \sin (2 (e+f x))+2 B e+2 B f x)}{4 f} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.43, size = 43, normalized size = 0.90 \[ \frac {{\left (2 \, A + B\right )} a f x - B a \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, {\left (A + B\right )} a \cos \left (f x + e\right )}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.13, size = 48, normalized size = 1.00 \[ \frac {1}{2} \, {\left (2 \, A a + B a\right )} x - \frac {B a \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} - \frac {{\left (A a + B a\right )} \cos \left (f x + e\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.16, size = 59, normalized size = 1.23 \[ \frac {a B \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-a A \cos \left (f x +e \right )-a B \cos \left (f x +e \right )+a A \left (f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 57, normalized size = 1.19 \[ \frac {4 \, {\left (f x + e\right )} A a + {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a - 4 \, A a \cos \left (f x + e\right ) - 4 \, B a \cos \left (f x + e\right )}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 13.26, size = 100, normalized size = 2.08 \[ A\,a\,x-\frac {-B\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+\left (2\,A\,a+2\,B\,a\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+B\,a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+2\,A\,a+2\,B\,a}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}+\frac {B\,a\,x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.35, size = 94, normalized size = 1.96 \[ \begin {cases} A a x - \frac {A a \cos {\left (e + f x \right )}}{f} + \frac {B a x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {B a x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {B a \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {B a \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (A + B \sin {\relax (e )}\right ) \left (a \sin {\relax (e )} + a\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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