3.389 \(\int (2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2 \, dx\)

Optimal. Leaf size=81 \[ 2 x \left (3 a^2+b^2\right )+\frac{6 a^2 \cos (d+e x)}{e}+\frac{6 a b \sin (d+e x)}{e}+\frac{2 (a (-\sin (d+e x))+a+b \cos (d+e x)) (a \cos (d+e x)+b \sin (d+e x))}{e} \]

[Out]

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

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Rubi [A]  time = 0.0469208, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3120, 2637, 2638} \[ 2 x \left (3 a^2+b^2\right )+\frac{6 a^2 \cos (d+e x)}{e}+\frac{6 a b \sin (d+e x)}{e}+\frac{2 (a (-\sin (d+e x))+a+b \cos (d+e x)) (a \cos (d+e x)+b \sin (d+e x))}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3120

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_), x_Symbol] :> -Simp[((c*Cos[d
+ e*x] - b*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 1))/(e*n), x] + Dist[1/n, Int[Simp[n*a^2 +
(n - 1)*(b^2 + c^2) + a*b*(2*n - 1)*Cos[d + e*x] + a*c*(2*n - 1)*Sin[d + e*x], x]*(a + b*Cos[d + e*x] + c*Sin
[d + e*x])^(n - 2), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && GtQ[n, 1]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int (2 a+2 b \cos (d+e x)-2 a \sin (d+e x))^2 \, dx &=\frac{2 (a+b \cos (d+e x)-a \sin (d+e x)) (a \cos (d+e x)+b \sin (d+e x))}{e}+\frac{1}{2} \int \left (4 \left (3 a^2+b^2\right )+12 a b \cos (d+e x)-12 a^2 \sin (d+e x)\right ) \, dx\\ &=2 \left (3 a^2+b^2\right ) x+\frac{2 (a+b \cos (d+e x)-a \sin (d+e x)) (a \cos (d+e x)+b \sin (d+e x))}{e}-\left (6 a^2\right ) \int \sin (d+e x) \, dx+(6 a b) \int \cos (d+e x) \, dx\\ &=2 \left (3 a^2+b^2\right ) x+\frac{6 a^2 \cos (d+e x)}{e}+\frac{6 a b \sin (d+e x)}{e}+\frac{2 (a+b \cos (d+e x)-a \sin (d+e x)) (a \cos (d+e x)+b \sin (d+e x))}{e}\\ \end{align*}

Mathematica [A]  time = 0.1516, size = 92, normalized size = 1.14 \[ 4 \left (\frac{\left (3 a^2+b^2\right ) (d+e x)}{2 e}-\frac{\left (a^2-b^2\right ) \sin (2 (d+e x))}{4 e}+\frac{2 a^2 \cos (d+e x)}{e}+\frac{2 a b \sin (d+e x)}{e}+\frac{a b \cos (2 (d+e x))}{2 e}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

4*(((3*a^2 + b^2)*(d + e*x))/(2*e) + (2*a^2*Cos[d + e*x])/e + (a*b*Cos[2*(d + e*x)])/(2*e) + (2*a*b*Sin[d + e*
x])/e - ((a^2 - b^2)*Sin[2*(d + e*x)])/(4*e))

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Maple [A]  time = 0.055, size = 100, normalized size = 1.2 \begin{align*} 4\,{\frac{{a}^{2} \left ( ex+d \right ) +2\,ab\sin \left ( ex+d \right ) +2\,{a}^{2}\cos \left ( ex+d \right ) +{b}^{2} \left ( 1/2\,\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) +1/2\,ex+d/2 \right ) + \left ( \cos \left ( ex+d \right ) \right ) ^{2}ab+{a}^{2} \left ( -1/2\,\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) +1/2\,ex+d/2 \right ) }{e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*a+2*b*cos(e*x+d)-2*a*sin(e*x+d))^2,x)

[Out]

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

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Maxima [A]  time = 0.980257, size = 132, normalized size = 1.63 \begin{align*} 4 \, a^{2} x + \frac{4 \, a b \cos \left (e x + d\right )^{2}}{e} + \frac{{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} a^{2}}{e} + \frac{{\left (2 \, e x + 2 \, d + \sin \left (2 \, e x + 2 \, d\right )\right )} b^{2}}{e} + 8 \, a{\left (\frac{a \cos \left (e x + d\right )}{e} + \frac{b \sin \left (e x + d\right )}{e}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a^2*x + 4*a*b*cos(e*x + d)^2/e + (2*e*x + 2*d - sin(2*e*x + 2*d))*a^2/e + (2*e*x + 2*d + sin(2*e*x + 2*d))*b
^2/e + 8*a*(a*cos(e*x + d)/e + b*sin(e*x + d)/e)

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Fricas [A]  time = 2.14909, size = 161, normalized size = 1.99 \begin{align*} \frac{2 \,{\left (2 \, a b \cos \left (e x + d\right )^{2} +{\left (3 \, a^{2} + b^{2}\right )} e x + 4 \, a^{2} \cos \left (e x + d\right ) +{\left (4 \, a b -{\left (a^{2} - b^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )}}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(2*a*b*cos(e*x + d)^2 + (3*a^2 + b^2)*e*x + 4*a^2*cos(e*x + d) + (4*a*b - (a^2 - b^2)*cos(e*x + d))*sin(e*x
+ d))/e

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Sympy [A]  time = 0.367994, size = 170, normalized size = 2.1 \begin{align*} \begin{cases} 2 a^{2} x \sin ^{2}{\left (d + e x \right )} + 2 a^{2} x \cos ^{2}{\left (d + e x \right )} + 4 a^{2} x - \frac{2 a^{2} \sin{\left (d + e x \right )} \cos{\left (d + e x \right )}}{e} + \frac{8 a^{2} \cos{\left (d + e x \right )}}{e} - \frac{4 a b \sin ^{2}{\left (d + e x \right )}}{e} + \frac{8 a b \sin{\left (d + e x \right )}}{e} + 2 b^{2} x \sin ^{2}{\left (d + e x \right )} + 2 b^{2} x \cos ^{2}{\left (d + e x \right )} + \frac{2 b^{2} \sin{\left (d + e x \right )} \cos{\left (d + e x \right )}}{e} & \text{for}\: e \neq 0 \\x \left (- 2 a \sin{\left (d \right )} + 2 a + 2 b \cos{\left (d \right )}\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*a+2*b*cos(e*x+d)-2*a*sin(e*x+d))**2,x)

[Out]

Piecewise((2*a**2*x*sin(d + e*x)**2 + 2*a**2*x*cos(d + e*x)**2 + 4*a**2*x - 2*a**2*sin(d + e*x)*cos(d + e*x)/e
+ 8*a**2*cos(d + e*x)/e - 4*a*b*sin(d + e*x)**2/e + 8*a*b*sin(d + e*x)/e + 2*b**2*x*sin(d + e*x)**2 + 2*b**2*
x*cos(d + e*x)**2 + 2*b**2*sin(d + e*x)*cos(d + e*x)/e, Ne(e, 0)), (x*(-2*a*sin(d) + 2*a + 2*b*cos(d))**2, Tru
e))

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Giac [A]  time = 1.12162, size = 107, normalized size = 1.32 \begin{align*} 2 \, a b \cos \left (2 \, x e + 2 \, d\right ) e^{\left (-1\right )} + 8 \, a^{2} \cos \left (x e + d\right ) e^{\left (-1\right )} + 8 \, a b e^{\left (-1\right )} \sin \left (x e + d\right ) -{\left (a^{2} - b^{2}\right )} e^{\left (-1\right )} \sin \left (2 \, x e + 2 \, d\right ) + 2 \,{\left (3 \, a^{2} + b^{2}\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*b*cos(2*x*e + 2*d)*e^(-1) + 8*a^2*cos(x*e + d)*e^(-1) + 8*a*b*e^(-1)*sin(x*e + d) - (a^2 - b^2)*e^(-1)*sin
(2*x*e + 2*d) + 2*(3*a^2 + b^2)*x