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]

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

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Rubi [A]  time = 0.09, antiderivative size = 101, normalized size of antiderivative = 1.00, 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 8

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

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*}

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Mathematica [A]  time = 0.16, 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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fricas [A]  time = 0.53, size = 84, normalized size = 0.83 \[ \frac {16 \, a b \cos \left (f x + e\right )^{3} + 3 \, {\left (4 \, a^{2} + 3 \, b^{2}\right )} f x - 48 \, a b \cos \left (f x + e\right ) + 3 \, {\left (2 \, b^{2} \cos \left (f x + e\right )^{3} - {\left (4 \, a^{2} + 5 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \]

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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giac [A]  time = 0.88, size = 86, normalized size = 0.85 \[ \frac {1}{8} \, {\left (4 \, a^{2} + 3 \, b^{2}\right )} x + \frac {a b \cos \left (3 \, f x + 3 \, e\right )}{6 \, f} - \frac {3 \, a b \cos \left (f x + e\right )}{2 \, f} + \frac {b^{2} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} - \frac {{\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \]

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

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maple [A]  time = 0.21, size = 89, normalized size = 0.88 \[ \frac {b^{2} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {2 a b \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+a^{2} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f} \]

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 = 1.00, size = 84, normalized size = 0.83 \[ \frac {24 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} + 64 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a b + 3 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} b^{2}}{96 \, f} \]

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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mupad [B]  time = 6.93, size = 85, normalized size = 0.84 \[ \frac {\frac {3\,b^2\,\sin \left (4\,e+4\,f\,x\right )}{4}-6\,b^2\,\sin \left (2\,e+2\,f\,x\right )-6\,a^2\,\sin \left (2\,e+2\,f\,x\right )-36\,a\,b\,\cos \left (e+f\,x\right )+4\,a\,b\,\cos \left (3\,e+3\,f\,x\right )+12\,a^2\,f\,x+9\,b^2\,f\,x}{24\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*b^2*sin(4*e + 4*f*x))/4 - 6*b^2*sin(2*e + 2*f*x) - 6*a^2*sin(2*e + 2*f*x) - 36*a*b*cos(e + f*x) + 4*a*b*co
s(3*e + 3*f*x) + 12*a^2*f*x + 9*b^2*f*x)/(24*f)

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sympy [A]  time = 1.73, size = 211, normalized size = 2.09 \[ \begin {cases} \frac {a^{2} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {a^{2} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {a^{2} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {2 a b \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {4 a b \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac {3 b^{2} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {3 b^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {3 b^{2} x \cos ^{4}{\left (e + f x \right )}}{8} - \frac {5 b^{2} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {3 b^{2} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin {\relax (e )}\right )^{2} \sin ^{2}{\relax (e )} & \text {otherwise} \end {cases} \]

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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