\(\int \sin (e+f x) (a+b \sin (e+f x))^2 \, dx\) [159]

Optimal result

Integrand size = 19, antiderivative size = 71 \[ \int \sin (e+f x) (a+b \sin (e+f x))^2 \, dx=a b x-\frac {2 \left (a^2+b^2\right ) \cos (e+f x)}{3 f}-\frac {a b \cos (e+f x) \sin (e+f x)}{3 f}-\frac {\cos (e+f x) (a+b \sin (e+f x))^2}{3 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2832, 2813} \[ \int \sin (e+f x) (a+b \sin (e+f x))^2 \, dx=-\frac {2 \left (a^2+b^2\right ) \cos (e+f x)}{3 f}-\frac {\cos (e+f x) (a+b \sin (e+f x))^2}{3 f}-\frac {a b \sin (e+f x) \cos (e+f x)}{3 f}+a b x \]

[In]

Int[Sin[e + f*x]*(a + b*Sin[e + f*x])^2,x]

[Out]

a*b*x - (2*(a^2 + b^2)*Cos[e + f*x])/(3*f) - (a*b*Cos[e + f*x]*Sin[e + f*x])/(3*f) - (Cos[e + f*x]*(a + b*Sin[
e + f*x])^2)/(3*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]

Rule 2832

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Sim
p[b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (e+f x) (a+b \sin (e+f x))^2}{3 f}+\frac {1}{3} \int (2 b+2 a \sin (e+f x)) (a+b \sin (e+f x)) \, dx \\ & = a b x-\frac {2 \left (a^2+b^2\right ) \cos (e+f x)}{3 f}-\frac {a b \cos (e+f x) \sin (e+f x)}{3 f}-\frac {\cos (e+f x) (a+b \sin (e+f x))^2}{3 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.83 \[ \int \sin (e+f x) (a+b \sin (e+f x))^2 \, dx=\frac {-3 \left (4 a^2+3 b^2\right ) \cos (e+f x)+b (12 a (e+f x)+b \cos (3 (e+f x))-6 a \sin (2 (e+f x)))}{12 f} \]

[In]

Integrate[Sin[e + f*x]*(a + b*Sin[e + f*x])^2,x]

[Out]

(-3*(4*a^2 + 3*b^2)*Cos[e + f*x] + b*(12*a*(e + f*x) + b*Cos[3*(e + f*x)] - 6*a*Sin[2*(e + f*x)]))/(12*f)

Maple [A] (verified)

Time = 1.28 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {-\cos \left (f x +e \right ) a^{2}+2 a b \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {b^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}}{f}\)	\(64\)
default	\(\frac {-\cos \left (f x +e \right ) a^{2}+2 a b \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {b^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}}{f}\)	\(64\)
risch	\(a b x -\frac {a^{2} \cos \left (f x +e \right )}{f}-\frac {3 b^{2} \cos \left (f x +e \right )}{4 f}+\frac {\cos \left (3 f x +3 e \right ) b^{2}}{12 f}-\frac {a b \sin \left (2 f x +2 e \right )}{2 f}\)	\(67\)
parallelrisch	\(\frac {\cos \left (3 f x +3 e \right ) b^{2}-6 a b \sin \left (2 f x +2 e \right )+\left (-12 a^{2}-9 b^{2}\right ) \cos \left (f x +e \right )+12 a b x f -12 a^{2}-8 b^{2}}{12 f}\)	\(67\)
parts	\(-\frac {a^{2} \cos \left (f x +e \right )}{f}-\frac {b^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3 f}+\frac {2 a b \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\)	\(69\)
norman	\(\frac {a b x +a b x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {6 a^{2}+4 b^{2}}{3 f}-\frac {2 a^{2} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {\left (4 a^{2}+4 b^{2}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {2 a b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {2 a b \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+3 a b x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+3 a b x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}\)	\(165\)

[In]

int(sin(f*x+e)*(a+b*sin(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.77 \[ \int \sin (e+f x) (a+b \sin (e+f x))^2 \, dx=\frac {b^{2} \cos \left (f x + e\right )^{3} + 3 \, a b f x - 3 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, {\left (a^{2} + b^{2}\right )} \cos \left (f x + e\right )}{3 \, f} \]

[In]

integrate(sin(f*x+e)*(a+b*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.51 \[ \int \sin (e+f x) (a+b \sin (e+f x))^2 \, dx=\begin {cases} - \frac {a^{2} \cos {\left (e + f x \right )}}{f} + a b x \sin ^{2}{\left (e + f x \right )} + a b x \cos ^{2}{\left (e + f x \right )} - \frac {a b \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {b^{2} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 b^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin {\left (e \right )}\right )^{2} \sin {\left (e \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(sin(f*x+e)*(a+b*sin(f*x+e))**2,x)

[Out]

Piecewise((-a**2*cos(e + f*x)/f + a*b*x*sin(e + f*x)**2 + a*b*x*cos(e + f*x)**2 - a*b*sin(e + f*x)*cos(e + f*x
)/f - b**2*sin(e + f*x)**2*cos(e + f*x)/f - 2*b**2*cos(e + f*x)**3/(3*f), Ne(f, 0)), (x*(a + b*sin(e))**2*sin(
e), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.87 \[ \int \sin (e+f x) (a+b \sin (e+f x))^2 \, dx=\frac {3 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a b + 2 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} b^{2} - 6 \, a^{2} \cos \left (f x + e\right )}{6 \, f} \]

[In]

integrate(sin(f*x+e)*(a+b*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

1/6*(3*(2*f*x + 2*e - sin(2*f*x + 2*e))*a*b + 2*(cos(f*x + e)^3 - 3*cos(f*x + e))*b^2 - 6*a^2*cos(f*x + e))/f

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.85 \[ \int \sin (e+f x) (a+b \sin (e+f x))^2 \, dx=a b x + \frac {b^{2} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac {a b \sin \left (2 \, f x + 2 \, e\right )}{2 \, f} - \frac {{\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (f x + e\right )}{4 \, f} \]

[In]

integrate(sin(f*x+e)*(a+b*sin(f*x+e))^2,x, algorithm="giac")

[Out]

a*b*x + 1/12*b^2*cos(3*f*x + 3*e)/f - 1/2*a*b*sin(2*f*x + 2*e)/f - 1/4*(4*a^2 + 3*b^2)*cos(f*x + e)/f

Mupad [B] (verification not implemented)

Time = 9.57 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.45 \[ \int \sin (e+f x) (a+b \sin (e+f x))^2 \, dx=a\,b\,x-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (4\,a^2+4\,b^2\right )+2\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+2\,a^2+\frac {4\,b^2}{3}-2\,a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+2\,a\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^3} \]

[In]

int(sin(e + f*x)*(a + b*sin(e + f*x))^2,x)

[Out]

a*b*x - (tan(e/2 + (f*x)/2)^2*(4*a^2 + 4*b^2) + 2*a^2*tan(e/2 + (f*x)/2)^4 + 2*a^2 + (4*b^2)/3 - 2*a*b*tan(e/2
+ (f*x)/2)^5 + 2*a*b*tan(e/2 + (f*x)/2))/(f*(tan(e/2 + (f*x)/2)^2 + 1)^3)