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(e*x+d)/e+6*a*b*sin(e*x+d)/e+2*(a+b*cos(e*x+d)-a*sin(e*x+d))*(a*cos(e*x+d)+b*sin(e*x+
d))/e

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Rubi [A]  time = 0.05, antiderivative size = 81, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.15, 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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fricas [A]  time = 0.90, size = 70, normalized size = 0.86 \[ \frac {2 \, {\left (2 \, a b \cos \left (e x + d\right )^{2} + {\left (3 \, a^{2} + b^{2}\right )} e x + 4 \, a^{2} \cos \left (e x + d\right ) + {\left (4 \, a b - {\left (a^{2} - b^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )\right )}}{e} \]

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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giac [A]  time = 0.14, size = 79, normalized size = 0.98 \[ 2 \, a b \cos \left (2 \, x e + 2 \, d\right ) e^{\left (-1\right )} + 8 \, a^{2} \cos \left (x e + d\right ) e^{\left (-1\right )} + 8 \, a b e^{\left (-1\right )} \sin \left (x e + d\right ) - {\left (a^{2} - b^{2}\right )} e^{\left (-1\right )} \sin \left (2 \, x e + 2 \, d\right ) + 2 \, {\left (3 \, a^{2} + b^{2}\right )} x \]

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

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maple [A]  time = 0.23, size = 100, normalized size = 1.23 \[ \frac {4 a^{2} \left (e x +d \right )+8 a b \sin \left (e x +d \right )+8 a^{2} \cos \left (e x +d \right )+4 b^{2} \left (\frac {\sin \left (e x +d \right ) \cos \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+4 \left (\cos ^{2}\left (e x +d \right )\right ) a b +4 a^{2} \left (-\frac {\sin \left (e x +d \right ) \cos \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )}{e} \]

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 = 0.32, size = 98, normalized size = 1.21 \[ 4 \, a^{2} x + \frac {4 \, a b \cos \left (e x + d\right )^{2}}{e} + \frac {{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} a^{2}}{e} + \frac {{\left (2 \, e x + 2 \, d + \sin \left (2 \, e x + 2 \, d\right )\right )} b^{2}}{e} + 8 \, a {\left (\frac {a \cos \left (e x + d\right )}{e} + \frac {b \sin \left (e x + d\right )}{e}\right )} \]

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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mupad [B]  time = 3.74, size = 128, normalized size = 1.58 \[ \frac {x\,\left (12\,a^2+4\,b^2\right )}{2}+\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (4\,a^2+16\,a\,b-4\,b^2\right )-{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (16\,a\,b-16\,a^2\right )+16\,a^2+\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (-4\,a^2+16\,a\,b+4\,b^2\right )}{e\,\left ({\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*(12*a^2 + 4*b^2))/2 + (tan(d/2 + (e*x)/2)^3*(16*a*b + 4*a^2 - 4*b^2) - tan(d/2 + (e*x)/2)^2*(16*a*b - 16*a^
2) + 16*a^2 + tan(d/2 + (e*x)/2)*(16*a*b - 4*a^2 + 4*b^2))/(e*(2*tan(d/2 + (e*x)/2)^2 + tan(d/2 + (e*x)/2)^4 +
1))

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sympy [A]  time = 0.32, size = 170, normalized size = 2.10 \[ \begin {cases} 2 a^{2} x \sin ^{2}{\left (d + e x \right )} + 2 a^{2} x \cos ^{2}{\left (d + e x \right )} + 4 a^{2} x - \frac {2 a^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} + \frac {8 a^{2} \cos {\left (d + e x \right )}}{e} + \frac {8 a b \sin {\left (d + e x \right )}}{e} + \frac {4 a b \cos ^{2}{\left (d + e x \right )}}{e} + 2 b^{2} x \sin ^{2}{\left (d + e x \right )} + 2 b^{2} x \cos ^{2}{\left (d + e x \right )} + \frac {2 b^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} & \text {for}\: e \neq 0 \\x \left (- 2 a \sin {\relax (d )} + 2 a + 2 b \cos {\relax (d )}\right )^{2} & \text {otherwise} \end {cases} \]

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