3.158 \(\int \sin ^2(e+f x) (a+b \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=101 \[ -\frac{\left (4 a^2+3 b^2\right ) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} x \left (4 a^2+3 b^2\right )+\frac{2 a b \cos ^3(e+f x)}{3 f}-\frac{2 a b \cos (e+f x)}{f}-\frac{b^2 \sin ^3(e+f x) \cos (e+f x)}{4 f} \]

[Out]

((4*a^2 + 3*b^2)*x)/8 - (2*a*b*Cos[e + f*x])/f + (2*a*b*Cos[e + f*x]^3)/(3*f) - ((4*a^2 + 3*b^2)*Cos[e + f*x]*
Sin[e + f*x])/(8*f) - (b^2*Cos[e + f*x]*Sin[e + f*x]^3)/(4*f)

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Rubi [A]  time = 0.0890224, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2789, 2633, 3014, 2635, 8} \[ -\frac{\left (4 a^2+3 b^2\right ) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} x \left (4 a^2+3 b^2\right )+\frac{2 a b \cos ^3(e+f x)}{3 f}-\frac{2 a b \cos (e+f x)}{f}-\frac{b^2 \sin ^3(e+f x) \cos (e+f x)}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a^2 + 3*b^2)*x)/8 - (2*a*b*Cos[e + f*x])/f + (2*a*b*Cos[e + f*x]^3)/(3*f) - ((4*a^2 + 3*b^2)*Cos[e + f*x]*
Sin[e + f*x])/(8*f) - (b^2*Cos[e + f*x]*Sin[e + f*x]^3)/(4*f)

Rule 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[
e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*
x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \sin ^2(e+f x) (a+b \sin (e+f x))^2 \, dx &=(2 a b) \int \sin ^3(e+f x) \, dx+\int \sin ^2(e+f x) \left (a^2+b^2 \sin ^2(e+f x)\right ) \, dx\\ &=-\frac{b^2 \cos (e+f x) \sin ^3(e+f x)}{4 f}+\frac{1}{4} \left (4 a^2+3 b^2\right ) \int \sin ^2(e+f x) \, dx-\frac{(2 a b) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{2 a b \cos (e+f x)}{f}+\frac{2 a b \cos ^3(e+f x)}{3 f}-\frac{\left (4 a^2+3 b^2\right ) \cos (e+f x) \sin (e+f x)}{8 f}-\frac{b^2 \cos (e+f x) \sin ^3(e+f x)}{4 f}+\frac{1}{8} \left (4 a^2+3 b^2\right ) \int 1 \, dx\\ &=\frac{1}{8} \left (4 a^2+3 b^2\right ) x-\frac{2 a b \cos (e+f x)}{f}+\frac{2 a b \cos ^3(e+f x)}{3 f}-\frac{\left (4 a^2+3 b^2\right ) \cos (e+f x) \sin (e+f x)}{8 f}-\frac{b^2 \cos (e+f x) \sin ^3(e+f x)}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.152915, size = 117, normalized size = 1.16 \[ \frac{a^2 (e+f x)}{2 f}-\frac{a^2 \sin (2 (e+f x))}{4 f}-\frac{3 a b \cos (e+f x)}{2 f}+\frac{a b \cos (3 (e+f x))}{6 f}+\frac{3 b^2 (e+f x)}{8 f}-\frac{b^2 \sin (2 (e+f x))}{4 f}+\frac{b^2 \sin (4 (e+f x))}{32 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(e + f*x))/(2*f) + (3*b^2*(e + f*x))/(8*f) - (3*a*b*Cos[e + f*x])/(2*f) + (a*b*Cos[3*(e + f*x)])/(6*f) -
(a^2*Sin[2*(e + f*x)])/(4*f) - (b^2*Sin[2*(e + f*x)])/(4*f) + (b^2*Sin[4*(e + f*x)])/(32*f)

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Maple [A]  time = 0.023, size = 89, normalized size = 0.9 \begin{align*}{\frac{1}{f} \left ({b}^{2} \left ( -{\frac{\cos \left ( fx+e \right ) }{4} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\sin \left ( fx+e \right ) }{2}} \right ) }+{\frac{3\,fx}{8}}+{\frac{3\,e}{8}} \right ) -{\frac{2\,ab \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+{a}^{2} \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(b^2*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)-2/3*a*b*(2+sin(f*x+e)^2)*cos(f*x+e)+a^2
*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e))

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Maxima [A]  time = 1.7117, size = 113, normalized size = 1.12 \begin{align*} \frac{24 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} + 64 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a b + 3 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} b^{2}}{96 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(24*(2*f*x + 2*e - sin(2*f*x + 2*e))*a^2 + 64*(cos(f*x + e)^3 - 3*cos(f*x + e))*a*b + 3*(12*f*x + 12*e +
sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*b^2)/f

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Fricas [A]  time = 1.86773, size = 201, normalized size = 1.99 \begin{align*} \frac{16 \, a b \cos \left (f x + e\right )^{3} + 3 \,{\left (4 \, a^{2} + 3 \, b^{2}\right )} f x - 48 \, a b \cos \left (f x + e\right ) + 3 \,{\left (2 \, b^{2} \cos \left (f x + e\right )^{3} -{\left (4 \, a^{2} + 5 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(16*a*b*cos(f*x + e)^3 + 3*(4*a^2 + 3*b^2)*f*x - 48*a*b*cos(f*x + e) + 3*(2*b^2*cos(f*x + e)^3 - (4*a^2 +
5*b^2)*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 2.10618, size = 211, normalized size = 2.09 \begin{align*} \begin{cases} \frac{a^{2} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{a^{2} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac{a^{2} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{2 a b \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{4 a b \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac{3 b^{2} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac{3 b^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{3 b^{2} x \cos ^{4}{\left (e + f x \right )}}{8} - \frac{5 b^{2} \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} - \frac{3 b^{2} \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} & \text{for}\: f \neq 0 \\x \left (a + b \sin{\left (e \right )}\right )^{2} \sin ^{2}{\left (e \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x*sin(e + f*x)**2/2 + a**2*x*cos(e + f*x)**2/2 - a**2*sin(e + f*x)*cos(e + f*x)/(2*f) - 2*a*b*
sin(e + f*x)**2*cos(e + f*x)/f - 4*a*b*cos(e + f*x)**3/(3*f) + 3*b**2*x*sin(e + f*x)**4/8 + 3*b**2*x*sin(e + f
*x)**2*cos(e + f*x)**2/4 + 3*b**2*x*cos(e + f*x)**4/8 - 5*b**2*sin(e + f*x)**3*cos(e + f*x)/(8*f) - 3*b**2*sin
(e + f*x)*cos(e + f*x)**3/(8*f), Ne(f, 0)), (x*(a + b*sin(e))**2*sin(e)**2, True))

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Giac [A]  time = 2.1186, size = 116, normalized size = 1.15 \begin{align*} \frac{1}{8} \,{\left (4 \, a^{2} + 3 \, b^{2}\right )} x + \frac{a b \cos \left (3 \, f x + 3 \, e\right )}{6 \, f} - \frac{3 \, a b \cos \left (f x + e\right )}{2 \, f} + \frac{b^{2} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} - \frac{{\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*a^2 + 3*b^2)*x + 1/6*a*b*cos(3*f*x + 3*e)/f - 3/2*a*b*cos(f*x + e)/f + 1/32*b^2*sin(4*f*x + 4*e)/f - 1/
4*(a^2 + b^2)*sin(2*f*x + 2*e)/f