3.438 \(\int (a+a \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=45 \[ -\frac{2 a^2 \cos (e+f x)}{f}-\frac{a^2 \sin (e+f x) \cos (e+f x)}{2 f}+\frac{3 a^2 x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0148743, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {2644} \[ -\frac{2 a^2 \cos (e+f x)}{f}-\frac{a^2 \sin (e+f x) \cos (e+f x)}{2 f}+\frac{3 a^2 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2,x]

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.188047, size = 34, normalized size = 0.76 \[ -\frac{a^2 (-6 (e+f x)+\sin (2 (e+f x))+8 \cos (e+f x))}{4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2,x]

[Out]

-(a^2*(-6*(e + f*x) + 8*Cos[e + f*x] + Sin[2*(e + f*x)]))/(4*f)

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Maple [A]  time = 0.023, size = 52, normalized size = 1.2 \begin{align*}{\frac{1}{f} \left ({a}^{2} \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) -2\,\cos \left ( fx+e \right ){a}^{2}+{a}^{2} \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,x)

[Out]

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

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Maxima [A]  time = 1.10774, size = 63, normalized size = 1.4 \begin{align*} a^{2} x + \frac{{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2}}{4 \, f} - \frac{2 \, a^{2} \cos \left (f x + e\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.58226, size = 97, normalized size = 2.16 \begin{align*} \frac{3 \, a^{2} f x - a^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 4 \, a^{2} \cos \left (f x + e\right )}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.296199, size = 78, normalized size = 1.73 \begin{align*} \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} + a^{2} x - \frac{a^{2} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{2 a^{2} \cos{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \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,x)

[Out]

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

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Giac [A]  time = 1.32693, size = 54, normalized size = 1.2 \begin{align*} \frac{3}{2} \, a^{2} x - \frac{2 \, a^{2} \cos \left (f x + e\right )}{f} - \frac{a^{2} \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,x, algorithm="giac")

[Out]

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