3.680 \(\int (a+b \sin (e+f x))^2 (c+d \sin (e+f x)) \, dx\)

Optimal. Leaf size=107 \[ -\frac {2 \left (a^2 d+3 a b c+b^2 d\right ) \cos (e+f x)}{3 f}+\frac {1}{2} x \left (2 a^2 c+2 a b d+b^2 c\right )-\frac {b (2 a d+3 b c) \sin (e+f x) \cos (e+f x)}{6 f}-\frac {d \cos (e+f x) (a+b \sin (e+f x))^2}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2753, 2734} \[ -\frac {2 \left (a^2 d+3 a b c+b^2 d\right ) \cos (e+f x)}{3 f}+\frac {1}{2} x \left (2 a^2 c+2 a b d+b^2 c\right )-\frac {b (2 a d+3 b c) \sin (e+f x) \cos (e+f x)}{6 f}-\frac {d \cos (e+f x) (a+b \sin (e+f x))^2}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2734

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 2753

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)*Simp[
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*} \int (a+b \sin (e+f x))^2 (c+d \sin (e+f x)) \, dx &=-\frac {d \cos (e+f x) (a+b \sin (e+f x))^2}{3 f}+\frac {1}{3} \int (a+b \sin (e+f x)) (3 a c+2 b d+(3 b c+2 a d) \sin (e+f x)) \, dx\\ &=\frac {1}{2} \left (2 a^2 c+b^2 c+2 a b d\right ) x-\frac {2 \left (3 a b c+a^2 d+b^2 d\right ) \cos (e+f x)}{3 f}-\frac {b (3 b c+2 a d) \cos (e+f x) \sin (e+f x)}{6 f}-\frac {d \cos (e+f x) (a+b \sin (e+f x))^2}{3 f}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 90, normalized size = 0.84 \[ \frac {6 (e+f x) \left (2 a^2 c+2 a b d+b^2 c\right )-3 \left (4 a^2 d+8 a b c+3 b^2 d\right ) \cos (e+f x)-3 b (2 a d+b c) \sin (2 (e+f x))+b^2 d \cos (3 (e+f x))}{12 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(2*a^2*c + b^2*c + 2*a*b*d)*(e + f*x) - 3*(8*a*b*c + 4*a^2*d + 3*b^2*d)*Cos[e + f*x] + b^2*d*Cos[3*(e + f*x
)] - 3*b*(b*c + 2*a*d)*Sin[2*(e + f*x)])/(12*f)

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fricas [A]  time = 0.45, size = 89, normalized size = 0.83 \[ \frac {2 \, b^{2} d \cos \left (f x + e\right )^{3} + 3 \, {\left (2 \, a b d + {\left (2 \, a^{2} + b^{2}\right )} c\right )} f x - 3 \, {\left (b^{2} c + 2 \, a b d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 6 \, {\left (2 \, a b c + {\left (a^{2} + b^{2}\right )} d\right )} \cos \left (f x + e\right )}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.37, size = 96, normalized size = 0.90 \[ \frac {b^{2} d \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} + \frac {1}{2} \, {\left (2 \, a^{2} c + b^{2} c + 2 \, a b d\right )} x - \frac {{\left (8 \, a b c + 4 \, a^{2} d + 3 \, b^{2} d\right )} \cos \left (f x + e\right )}{4 \, f} - \frac {{\left (b^{2} c + 2 \, a b d\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.20, size = 115, normalized size = 1.07 \[ \frac {a^{2} c \left (f x +e \right )-a^{2} d \cos \left (f x +e \right )-2 a b c \cos \left (f x +e \right )+2 a b d \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+b^{2} c \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {b^{2} d \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 112, normalized size = 1.05 \[ \frac {12 \, {\left (f x + e\right )} a^{2} c + 3 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} b^{2} c + 6 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a b d + 4 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} b^{2} d - 24 \, a b c \cos \left (f x + e\right ) - 12 \, a^{2} d \cos \left (f x + e\right )}{12 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(f*x + e)*a^2*c + 3*(2*f*x + 2*e - sin(2*f*x + 2*e))*b^2*c + 6*(2*f*x + 2*e - sin(2*f*x + 2*e))*a*b*d
+ 4*(cos(f*x + e)^3 - 3*cos(f*x + e))*b^2*d - 24*a*b*c*cos(f*x + e) - 12*a^2*d*cos(f*x + e))/f

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mupad [B]  time = 7.77, size = 108, normalized size = 1.01 \[ -\frac {\frac {3\,b^2\,c\,\sin \left (2\,e+2\,f\,x\right )}{2}-\frac {b^2\,d\,\cos \left (3\,e+3\,f\,x\right )}{2}+6\,a^2\,d\,\cos \left (e+f\,x\right )+\frac {9\,b^2\,d\,\cos \left (e+f\,x\right )}{2}+3\,a\,b\,d\,\sin \left (2\,e+2\,f\,x\right )-6\,a^2\,c\,f\,x-3\,b^2\,c\,f\,x+12\,a\,b\,c\,\cos \left (e+f\,x\right )-6\,a\,b\,d\,f\,x}{6\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((3*b^2*c*sin(2*e + 2*f*x))/2 - (b^2*d*cos(3*e + 3*f*x))/2 + 6*a^2*d*cos(e + f*x) + (9*b^2*d*cos(e + f*x))/2
+ 3*a*b*d*sin(2*e + 2*f*x) - 6*a^2*c*f*x - 3*b^2*c*f*x + 12*a*b*c*cos(e + f*x) - 6*a*b*d*f*x)/(6*f)

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sympy [A]  time = 0.83, size = 199, normalized size = 1.86 \[ \begin {cases} a^{2} c x - \frac {a^{2} d \cos {\left (e + f x \right )}}{f} - \frac {2 a b c \cos {\left (e + f x \right )}}{f} + a b d x \sin ^{2}{\left (e + f x \right )} + a b d x \cos ^{2}{\left (e + f x \right )} - \frac {a b d \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {b^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {b^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {b^{2} c \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {b^{2} d \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 b^{2} d \cos ^{3}{\left (e + f x \right )}}{3 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin {\relax (e )}\right )^{2} \left (c + d \sin {\relax (e )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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