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

Optimal. Leaf size=55 \[ \frac{1}{2} x \left (a^2+b^2\right )-\frac{(b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0192334, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3073, 8} \[ \frac{1}{2} x \left (a^2+b^2\right )-\frac{(b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3073

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.102152, size = 52, normalized size = 0.95 \[ \frac{2 \left (a^2+b^2\right ) (c+d x)+\left (a^2-b^2\right ) \sin (2 (c+d x))-2 a b \cos (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.041, size = 70, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) - \left ( \cos \left ( dx+c \right ) \right ) ^{2}ab+{a}^{2} \left ({\frac{\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.970316, size = 92, normalized size = 1.67 \begin{align*} -\frac{a b \cos \left (d x + c\right )^{2}}{d} + \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{4 \, d} + \frac{{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} b^{2}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.59547, size = 120, normalized size = 2.18 \begin{align*} -\frac{2 \, a b \cos \left (d x + c\right )^{2} -{\left (a^{2} + b^{2}\right )} d x -{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.339938, size = 128, normalized size = 2.33 \begin{align*} \begin{cases} \frac{a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{a^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{a b \sin ^{2}{\left (c + d x \right )}}{d} + \frac{b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} - \frac{b^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + b \sin{\left (c \right )}\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10991, size = 68, normalized size = 1.24 \begin{align*} \frac{1}{2} \,{\left (a^{2} + b^{2}\right )} x - \frac{a b \cos \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac{{\left (a^{2} - b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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