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

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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, 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 8

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

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]

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 = 0.10, 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)

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 52, normalized size = 0.95 \[ -\frac {2 \, a b \cos \left (d x + c\right )^{2} - {\left (a^{2} + b^{2}\right )} d x - {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{2 \, d} \]

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

________________________________________________________________________________________

giac [A]  time = 3.73, size = 50, normalized size = 0.91 \[ \frac {1}{2} \, {\left (a^{2} + b^{2}\right )} x - \frac {a b \cos \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {{\left (a^{2} - b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} \]

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

________________________________________________________________________________________

maple [A]  time = 1.19, size = 70, normalized size = 1.27 \[ \frac {b^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\left (\cos ^{2}\left (d x +c \right )\right ) a b +a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

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

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 68, normalized size = 1.24 \[ -\frac {a b \cos \left (d x + c\right )^{2}}{d} + \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{4 \, d} + \frac {{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} b^{2}}{4 \, d} \]

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

________________________________________________________________________________________

mupad [B]  time = 0.48, size = 63, normalized size = 1.15 \[ \frac {a^2\,x}{2}+\frac {b^2\,x}{2}+\frac {a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}-\frac {b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}-\frac {a\,b\,\cos \left (2\,c+2\,d\,x\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.27, size = 128, normalized size = 2.33 \[ \begin {cases} \frac {a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} - \frac {a b \cos ^{2}{\left (c + d x \right )}}{d} + \frac {b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} - \frac {b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right )^{2} & \text {otherwise} \end {cases} \]

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*co
s(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))

________________________________________________________________________________________