\(\int \sin (e+f x) (a+b \sin (e+f x)) \, dx\) [151]

Optimal result

Integrand size = 17, antiderivative size = 39 \[ \int \sin (e+f x) (a+b \sin (e+f x)) \, dx=\frac {b x}{2}-\frac {a \cos (e+f x)}{f}-\frac {b \cos (e+f x) \sin (e+f x)}{2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2813} \[ \int \sin (e+f x) (a+b \sin (e+f x)) \, dx=-\frac {a \cos (e+f x)}{f}-\frac {b \sin (e+f x) \cos (e+f x)}{2 f}+\frac {b x}{2} \]

[In]

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

[Out]

(b*x)/2 - (a*Cos[e + f*x])/f - (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 {b x}{2}-\frac {a \cos (e+f x)}{f}-\frac {b \cos (e+f x) \sin (e+f x)}{2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90 \[ \int \sin (e+f x) (a+b \sin (e+f x)) \, dx=-\frac {4 a \cos (e+f x)+b (-2 (e+f x)+\sin (2 (e+f x)))}{4 f} \]

[In]

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

[Out]

-1/4*(4*a*Cos[e + f*x] + b*(-2*(e + f*x) + Sin[2*(e + f*x)]))/f

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {b x}{2}-\frac {a \cos \left (f x +e \right )}{f}-\frac {b \sin \left (2 f x +2 e \right )}{4 f}\)	\(33\)
parallelrisch	\(\frac {2 b x f -4 a \cos \left (f x +e \right )-\sin \left (2 f x +2 e \right ) b -4 a}{4 f}\)	\(36\)
derivativedivides	\(\frac {b \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-a \cos \left (f x +e \right )}{f}\)	\(39\)
default	\(\frac {b \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-a \cos \left (f x +e \right )}{f}\)	\(39\)
parts	\(-\frac {a \cos \left (f x +e \right )}{f}+\frac {b \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\)	\(41\)
norman	\(\frac {\frac {b \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+b x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {2 a \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {b x}{2}-\frac {b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {b x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}+\frac {2 a \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}\)	\(116\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87 \[ \int \sin (e+f x) (a+b \sin (e+f x)) \, dx=\frac {b f x - b \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a \cos \left (f x + e\right )}{2 \, f} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (32) = 64\).

Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.69 \[ \int \sin (e+f x) (a+b \sin (e+f x)) \, dx=\begin {cases} - \frac {a \cos {\left (e + f x \right )}}{f} + \frac {b x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {b x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {b \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin {\left (e \right )}\right ) \sin {\left (e \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int \sin (e+f x) (a+b \sin (e+f x)) \, dx=\frac {{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} b - 4 \, a \cos \left (f x + e\right )}{4 \, f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.82 \[ \int \sin (e+f x) (a+b \sin (e+f x)) \, dx=\frac {1}{2} \, b x - \frac {a \cos \left (f x + e\right )}{f} - \frac {b \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.49 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.74 \[ \int \sin (e+f x) (a+b \sin (e+f x)) \, dx=\frac {b\,x}{2}-\frac {-b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+2\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+2\,a}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^2} \]

[In]

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

[Out]

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