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]

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

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Rubi [A]  time = 0.06, antiderivative size = 94, 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 = {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 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]

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)]

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*}

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Mathematica [A]  time = 0.32, 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]

-1/6*(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)))/(f*Sqrt[Cos[e + f*x]^2])

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fricas [A]  time = 0.46, size = 70, normalized size = 0.74 \[ \frac {2 \, B a^{2} \cos \left (f x + e\right )^{3} + 3 \, {\left (3 \, A + 2 \, B\right )} a^{2} f x - 3 \, {\left (A + 2 \, B\right )} a^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 12 \, {\left (A + B\right )} a^{2} \cos \left (f x + e\right )}{6 \, f} \]

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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giac [A]  time = 0.16, size = 88, normalized size = 0.94 \[ \frac {B a^{2} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} + \frac {1}{2} \, {\left (3 \, A a^{2} + 2 \, B a^{2}\right )} x - \frac {{\left (8 \, A a^{2} + 7 \, B a^{2}\right )} \cos \left (f x + e\right )}{4 \, f} - \frac {{\left (A a^{2} + 2 \, B a^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \]

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

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maple [A]  time = 0.30, size = 117, normalized size = 1.24 \[ \frac {a^{2} A \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\frac {B \,a^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}-2 a^{2} A \cos \left (f x +e \right )+2 B \,a^{2} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+a^{2} A \left (f x +e \right )-B \,a^{2} \cos \left (f x +e \right )}{f} \]

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^2*A*(-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^2*A*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^2*A*(f*x+e)-B*a^2*cos(f*x+e))

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maxima [A]  time = 0.44, size = 114, normalized size = 1.21 \[ \frac {3 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} + 12 \, {\left (f x + e\right )} A a^{2} + 4 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{2} + 6 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} - 24 \, A a^{2} \cos \left (f x + e\right ) - 12 \, B a^{2} \cos \left (f x + e\right )}{12 \, f} \]

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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mupad [B]  time = 13.16, size = 91, normalized size = 0.97 \[ -\frac {\frac {3\,A\,a^2\,\sin \left (2\,e+2\,f\,x\right )}{2}-\frac {B\,a^2\,\cos \left (3\,e+3\,f\,x\right )}{2}+3\,B\,a^2\,\sin \left (2\,e+2\,f\,x\right )+12\,A\,a^2\,\cos \left (e+f\,x\right )+\frac {21\,B\,a^2\,\cos \left (e+f\,x\right )}{2}-9\,A\,a^2\,f\,x-6\,B\,a^2\,f\,x}{6\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((3*A*a^2*sin(2*e + 2*f*x))/2 - (B*a^2*cos(3*e + 3*f*x))/2 + 3*B*a^2*sin(2*e + 2*f*x) + 12*A*a^2*cos(e + f*x)
+ (21*B*a^2*cos(e + f*x))/2 - 9*A*a^2*f*x - 6*B*a^2*f*x)/(6*f)

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sympy [A]  time = 0.95, size = 199, normalized size = 2.12 \[ \begin {cases} \frac {A a^{2} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {A a^{2} x \cos ^{2}{\left (e + f x \right )}}{2} + A a^{2} x - \frac {A a^{2} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {2 A a^{2} \cos {\left (e + f x \right )}}{f} + B a^{2} x \sin ^{2}{\left (e + f x \right )} + B a^{2} x \cos ^{2}{\left (e + f x \right )} - \frac {B a^{2} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {B a^{2} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 B a^{2} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {B a^{2} \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (A + B \sin {\relax (e )}\right ) \left (a \sin {\relax (e )} + a\right )^{2} & \text {otherwise} \end {cases} \]

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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