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

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

[Out]

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

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Rubi [A]  time = 0.0623795, 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} \[ -\frac{2 a^2 (3 c+2 d) \cos (e+f x)}{3 f}-\frac{a^2 (3 c+2 d) \sin (e+f x) \cos (e+f x)}{6 f}+\frac{1}{2} a^2 x (3 c+2 d)-\frac{d \cos (e+f x) (a \sin (e+f x)+a)^2}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(3*c + 2*d)*x)/2 - (2*a^2*(3*c + 2*d)*Cos[e + f*x])/(3*f) - (a^2*(3*c + 2*d)*Cos[e + f*x]*Sin[e + f*x])/(
6*f) - (d*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 (c+d \sin (e+f x)) \, dx &=-\frac{d \cos (e+f x) (a+a \sin (e+f x))^2}{3 f}+\frac{1}{3} (3 c+2 d) \int (a+a \sin (e+f x))^2 \, dx\\ &=\frac{1}{2} a^2 (3 c+2 d) x-\frac{2 a^2 (3 c+2 d) \cos (e+f x)}{3 f}-\frac{a^2 (3 c+2 d) \cos (e+f x) \sin (e+f x)}{6 f}-\frac{d \cos (e+f x) (a+a \sin (e+f x))^2}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.331744, size = 106, normalized size = 1.13 \[ -\frac{a^2 \cos (e+f x) \left (6 (3 c+2 d) \sin ^{-1}\left (\frac{\sqrt{1-\sin (e+f x)}}{\sqrt{2}}\right )+\sqrt{\cos ^2(e+f x)} \left (3 (c+2 d) \sin (e+f x)+2 (6 c+5 d)+2 d \sin ^2(e+f x)\right )\right )}{6 f \sqrt{\cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 117, normalized size = 1.2 \begin{align*}{\frac{1}{f} \left ({a}^{2}c \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) -{\frac{{a}^{2}d \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}-2\,{a}^{2}c\cos \left ( fx+e \right ) +2\,{a}^{2}d \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) +{a}^{2}c \left ( fx+e \right ) -{a}^{2}d\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*(c+d*sin(f*x+e)),x)

[Out]

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

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Maxima [A]  time = 1.13387, 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^{2} c + 12 \,{\left (f x + e\right )} a^{2} c + 4 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} d + 6 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} d - 24 \, a^{2} c \cos \left (f x + e\right ) - 12 \, a^{2} d \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*(c+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.59518, size = 192, normalized size = 2.04 \begin{align*} \frac{2 \, a^{2} d \cos \left (f x + e\right )^{3} + 3 \,{\left (3 \, a^{2} c + 2 \, a^{2} d\right )} f x - 3 \,{\left (a^{2} c + 2 \, a^{2} d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 12 \,{\left (a^{2} c + a^{2} d\right )} \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*(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.800184, size = 199, normalized size = 2.12 \begin{align*} \begin{cases} \frac{a^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{a^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} + a^{2} c x - \frac{a^{2} c \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{2 a^{2} c \cos{\left (e + f x \right )}}{f} + a^{2} d x \sin ^{2}{\left (e + f x \right )} + a^{2} d x \cos ^{2}{\left (e + f x \right )} - \frac{a^{2} d \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{a^{2} d \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{2 a^{2} d \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{a^{2} d \cos{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \left (c + d \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*(c+d*sin(f*x+e)),x)

[Out]

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

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Giac [A]  time = 1.3898, size = 147, normalized size = 1.56 \begin{align*} a^{2} c x + \frac{a^{2} d \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac{a^{2} d \cos \left (f x + e\right )}{f} + \frac{1}{2} \,{\left (a^{2} c + 2 \, a^{2} d\right )} x - \frac{{\left (8 \, a^{2} c + 3 \, a^{2} d\right )} \cos \left (f x + e\right )}{4 \, f} - \frac{{\left (a^{2} c + 2 \, a^{2} d\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*(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

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