3.254 \(\int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) \, dx\)

Optimal. Leaf size=94 \[ -\frac{2 a^2 (3 A+2 B) \cos (e+f x)}{3 f}-\frac{a^2 (3 A+2 B) \sin (e+f x) \cos (e+f x)}{6 f}+\frac{1}{2} a^2 x (3 A+2 B)-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.0607967, antiderivative size = 94, normalized size of antiderivative = 1., 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 = {2751, 2644} \[ -\frac{2 a^2 (3 A+2 B) \cos (e+f x)}{3 f}-\frac{a^2 (3 A+2 B) \sin (e+f x) \cos (e+f x)}{6 f}+\frac{1}{2} a^2 x (3 A+2 B)-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^2}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2751

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[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2644

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[((2*a^2 + b^2)*x)/2, x] + (-Simp[(2*a*b*Cos[c
+ d*x])/d, x] - Simp[(b^2*Cos[c + d*x]*Sin[c + d*x])/(2*d), x]) /; FreeQ[{a, b, c, d}, x]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) \, dx &=-\frac{B \cos (e+f x) (a+a \sin (e+f x))^2}{3 f}+\frac{1}{3} (3 A+2 B) \int (a+a \sin (e+f x))^2 \, dx\\ &=\frac{1}{2} a^2 (3 A+2 B) x-\frac{2 a^2 (3 A+2 B) \cos (e+f x)}{3 f}-\frac{a^2 (3 A+2 B) \cos (e+f x) \sin (e+f x)}{6 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^2}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.319729, size = 106, normalized size = 1.13 \[ -\frac{a^2 \cos (e+f x) \left (6 (3 A+2 B) \sin ^{-1}\left (\frac{\sqrt{1-\sin (e+f x)}}{\sqrt{2}}\right )+\sqrt{\cos ^2(e+f x)} \left (3 (A+2 B) \sin (e+f x)+2 (6 A+5 B)+2 B \sin ^2(e+f x)\right )\right )}{6 f \sqrt{\cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*Cos[e + f*x]*(6*(3*A + 2*B)*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt[2]] + Sqrt[Cos[e + f*x]^2]*(2*(6*A + 5*B)
+ 3*(A + 2*B)*Sin[e + f*x] + 2*B*Sin[e + f*x]^2)))/(6*f*Sqrt[Cos[e + f*x]^2])

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Maple [A]  time = 0.04, size = 117, normalized size = 1.2 \begin{align*}{\frac{1}{f} \left ( A{a}^{2} \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) -{\frac{B{a}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}-2\,A{a}^{2}\cos \left ( fx+e \right ) +2\,B{a}^{2} \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) +A{a}^{2} \left ( fx+e \right ) -B{a}^{2}\cos \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.976525, size = 154, normalized size = 1.64 \begin{align*} \frac{3 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} + 12 \,{\left (f x + e\right )} A a^{2} + 4 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{2} + 6 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} - 24 \, A a^{2} \cos \left (f x + e\right ) - 12 \, B a^{2} \cos \left (f x + e\right )}{12 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.9591, size = 176, normalized size = 1.87 \begin{align*} \frac{2 \, B a^{2} \cos \left (f x + e\right )^{3} + 3 \,{\left (3 \, A + 2 \, B\right )} a^{2} f x - 3 \,{\left (A + 2 \, B\right )} a^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 12 \,{\left (A + B\right )} a^{2} \cos \left (f x + e\right )}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*a^2*cos(f*x + e)^3 + 3*(3*A + 2*B)*a^2*f*x - 3*(A + 2*B)*a^2*cos(f*x + e)*sin(f*x + e) - 12*(A + B)*a
^2*cos(f*x + e))/f

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Sympy [A]  time = 0.918068, size = 199, normalized size = 2.12 \begin{align*} \begin{cases} \frac{A a^{2} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{A a^{2} x \cos ^{2}{\left (e + f x \right )}}{2} + A a^{2} x - \frac{A a^{2} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{2 A a^{2} \cos{\left (e + f x \right )}}{f} + B a^{2} x \sin ^{2}{\left (e + f x \right )} + B a^{2} x \cos ^{2}{\left (e + f x \right )} - \frac{B a^{2} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{B a^{2} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{2 B a^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{B a^{2} \cos{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \left (A + B \sin{\left (e \right )}\right ) \left (a \sin{\left (e \right )} + a\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.2568, size = 119, normalized size = 1.27 \begin{align*} \frac{B a^{2} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} + \frac{1}{2} \,{\left (3 \, A a^{2} + 2 \, B a^{2}\right )} x - \frac{{\left (8 \, A a^{2} + 7 \, B a^{2}\right )} \cos \left (f x + e\right )}{4 \, f} - \frac{{\left (A a^{2} + 2 \, B a^{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((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e)),x, algorithm="giac")

[Out]

1/12*B*a^2*cos(3*f*x + 3*e)/f + 1/2*(3*A*a^2 + 2*B*a^2)*x - 1/4*(8*A*a^2 + 7*B*a^2)*cos(f*x + e)/f - 1/4*(A*a^
2 + 2*B*a^2)*sin(2*f*x + 2*e)/f