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

Optimal. Leaf size=91 \[ \frac{1}{2} x \left (2 a^2+b^2+c^2\right )-\frac{(c \cos (d+e x)-b \sin (d+e x)) (a+b \cos (d+e x)+c \sin (d+e x))}{2 e}+\frac{3 a b \sin (d+e x)}{2 e}-\frac{3 a c \cos (d+e x)}{2 e} \]

[Out]

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

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Rubi [A]  time = 0.0463345, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {3120, 2637, 2638} \[ \frac{1}{2} x \left (2 a^2+b^2+c^2\right )-\frac{(c \cos (d+e x)-b \sin (d+e x)) (a+b \cos (d+e x)+c \sin (d+e x))}{2 e}+\frac{3 a b \sin (d+e x)}{2 e}-\frac{3 a c \cos (d+e x)}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.170644, size = 77, normalized size = 0.85 \[ \frac{2 \left (2 a^2+b^2+c^2\right ) (d+e x)+8 a b \sin (d+e x)-8 a c \cos (d+e x)+\left (b^2-c^2\right ) \sin (2 (d+e x))-2 b c \cos (2 (d+e x))}{4 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(2*a^2 + b^2 + c^2)*(d + e*x) - 8*a*c*Cos[d + e*x] - 2*b*c*Cos[2*(d + e*x)] + 8*a*b*Sin[d + e*x] + (b^2 - c
^2)*Sin[2*(d + e*x)])/(4*e)

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Maple [A]  time = 0.051, size = 99, normalized size = 1.1 \begin{align*}{\frac{1}{e} \left ({a}^{2} \left ( ex+d \right ) +2\,ab\sin \left ( ex+d \right ) -2\,ac\cos \left ( ex+d \right ) +{b}^{2} \left ({\frac{\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) }{2}}+{\frac{ex}{2}}+{\frac{d}{2}} \right ) - \left ( \cos \left ( ex+d \right ) \right ) ^{2}bc+{c}^{2} \left ( -{\frac{\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) }{2}}+{\frac{ex}{2}}+{\frac{d}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e*(a^2*(e*x+d)+2*a*b*sin(e*x+d)-2*a*c*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*
b*c+c^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 = 0.981224, size = 135, normalized size = 1.48 \begin{align*} a^{2} x - \frac{b c \cos \left (e x + d\right )^{2}}{e} + \frac{{\left (2 \, e x + 2 \, d + \sin \left (2 \, e x + 2 \, d\right )\right )} b^{2}}{4 \, e} + \frac{{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} c^{2}}{4 \, e} - 2 \, a{\left (\frac{c \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((a+b*cos(e*x+d)+c*sin(e*x+d))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.23692, size = 173, normalized size = 1.9 \begin{align*} -\frac{2 \, b c \cos \left (e x + d\right )^{2} -{\left (2 \, a^{2} + b^{2} + c^{2}\right )} e x + 4 \, a c \cos \left (e x + d\right ) -{\left (4 \, a b +{\left (b^{2} - c^{2}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{2 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.387578, size = 162, normalized size = 1.78 \begin{align*} \begin{cases} a^{2} x + \frac{2 a b \sin{\left (d + e x \right )}}{e} - \frac{2 a c \cos{\left (d + e x \right )}}{e} + \frac{b^{2} x \sin ^{2}{\left (d + e x \right )}}{2} + \frac{b^{2} x \cos ^{2}{\left (d + e x \right )}}{2} + \frac{b^{2} \sin{\left (d + e x \right )} \cos{\left (d + e x \right )}}{2 e} + \frac{b c \sin ^{2}{\left (d + e x \right )}}{e} + \frac{c^{2} x \sin ^{2}{\left (d + e x \right )}}{2} + \frac{c^{2} x \cos ^{2}{\left (d + e x \right )}}{2} - \frac{c^{2} \sin{\left (d + e x \right )} \cos{\left (d + e x \right )}}{2 e} & \text{for}\: e \neq 0 \\x \left (a + b \cos{\left (d \right )} + c \sin{\left (d \right )}\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12857, size = 109, normalized size = 1.2 \begin{align*} -\frac{1}{2} \, b c \cos \left (2 \, x e + 2 \, d\right ) e^{\left (-1\right )} - 2 \, a c \cos \left (x e + d\right ) e^{\left (-1\right )} + 2 \, a b e^{\left (-1\right )} \sin \left (x e + d\right ) + \frac{1}{4} \,{\left (b^{2} - c^{2}\right )} e^{\left (-1\right )} \sin \left (2 \, x e + 2 \, d\right ) + \frac{1}{2} \,{\left (2 \, a^{2} + b^{2} + c^{2}\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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