3.7.73 \(\int (a+b \sin (e+f x)) (c+d \sin (e+f x)) \, dx\) [673]

Optimal. Leaf size=53 \[ \frac {1}{2} (2 a c+b d) x-\frac {(b c+a d) \cos (e+f x)}{f}-\frac {b d \cos (e+f x) \sin (e+f x)}{2 f} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 53, 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 = {2813} \begin {gather*} -\frac {(a d+b c) \cos (e+f x)}{f}+\frac {1}{2} x (2 a c+b d)-\frac {b d \sin (e+f x) \cos (e+f x)}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*c + b*d)*x)/2 - ((b*c + a*d)*Cos[e + f*x])/f - (b*d*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+b \sin (e+f x)) (c+d \sin (e+f x)) \, dx &=\frac {1}{2} (2 a c+b d) x-\frac {(b c+a d) \cos (e+f x)}{f}-\frac {b d \cos (e+f x) \sin (e+f x)}{2 f}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 52, normalized size = 0.98 \begin {gather*} \frac {2 b d e+4 a c f x+2 b d f x-4 (b c+a d) \cos (e+f x)-b d \sin (2 (e+f x))}{4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*d*e + 4*a*c*f*x + 2*b*d*f*x - 4*(b*c + a*d)*Cos[e + f*x] - b*d*Sin[2*(e + f*x)])/(4*f)

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Maple [A]
time = 0.09, size = 59, normalized size = 1.11

method	result	size
risch	\(a c x +\frac {x b d}{2}-\frac {\cos \left (f x +e \right ) a d}{f}-\frac {\cos \left (f x +e \right ) b c}{f}-\frac {b d \sin \left (2 f x +2 e \right )}{4 f}\)	\(53\)
derivativedivides	\(\frac {b d \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-a d \cos \left (f x +e \right )-b c \cos \left (f x +e \right )+a c \left (f x +e \right )}{f}\)	\(59\)
default	\(\frac {b d \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-a d \cos \left (f x +e \right )-b c \cos \left (f x +e \right )+a c \left (f x +e \right )}{f}\)	\(59\)
norman	\(\frac {\left (a c +\frac {b d}{2}\right ) x +\left (a c +\frac {b d}{2}\right ) x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (2 a c +b d \right ) x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {\left (2 a d +2 b c \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {b d \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {2 \left (a d +b c \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {b d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}\)	\(150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.32, size = 62, normalized size = 1.17 \begin {gather*} \frac {4 \, {\left (f x + e\right )} a c + {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} b d - 4 \, b c \cos \left (f x + e\right ) - 4 \, a d \cos \left (f x + e\right )}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 51, normalized size = 0.96 \begin {gather*} -\frac {b d \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (2 \, a c + b d\right )} f x + 2 \, {\left (b c + a d\right )} \cos \left (f x + e\right )}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (44) = 88\).
time = 0.09, size = 94, normalized size = 1.77 \begin {gather*} \begin {cases} a c x - \frac {a d \cos {\left (e + f x \right )}}{f} - \frac {b c \cos {\left (e + f x \right )}}{f} + \frac {b d x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {b d x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {b d \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin {\left (e \right )}\right ) \left (c + d \sin {\left (e \right )}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e)),x)

[Out]

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

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Giac [A]
time = 0.45, size = 48, normalized size = 0.91 \begin {gather*} \frac {1}{2} \, {\left (2 \, a c + b d\right )} x - \frac {b d \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} - \frac {{\left (b c + a d\right )} \cos \left (f x + e\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))*(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 7.72, size = 52, normalized size = 0.98 \begin {gather*} a\,c\,x+\frac {b\,d\,x}{2}-\frac {a\,d\,\cos \left (e+f\,x\right )}{f}-\frac {b\,c\,\cos \left (e+f\,x\right )}{f}-\frac {b\,d\,\sin \left (2\,e+2\,f\,x\right )}{4\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x))*(c + d*sin(e + f*x)),x)

[Out]

a*c*x + (b*d*x)/2 - (a*d*cos(e + f*x))/f - (b*c*cos(e + f*x))/f - (b*d*sin(2*e + 2*f*x))/(4*f)

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