3.62 \(\int (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=58 \[ \frac {(b \cos (c+d x)-a \sin (c+d x))^3}{3 d}-\frac {\left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x))}{d} \]

[Out]

-(a^2+b^2)*(b*cos(d*x+c)-a*sin(d*x+c))/d+1/3*(b*cos(d*x+c)-a*sin(d*x+c))^3/d

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {3072} \[ \frac {(b \cos (c+d x)-a \sin (c+d x))^3}{3 d}-\frac {\left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

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

Rule 3072

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

Rubi steps

\begin {align*} \int (a \cos (c+d x)+b \sin (c+d x))^3 \, dx &=-\frac {\operatorname {Subst}\left (\int \left (a^2+b^2-x^2\right ) \, dx,x,b \cos (c+d x)-a \sin (c+d x)\right )}{d}\\ &=-\frac {\left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x))}{d}+\frac {(b \cos (c+d x)-a \sin (c+d x))^3}{3 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.34, size = 81, normalized size = 1.40 \[ \frac {\left (b^3-3 a^2 b\right ) \cos (3 (c+d x))-9 b \left (a^2+b^2\right ) \cos (c+d x)+2 a \sin (c+d x) \left (\left (a^2-3 b^2\right ) \cos (2 (c+d x))+5 a^2+3 b^2\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(-9*b*(a^2 + b^2)*Cos[c + d*x] + (-3*a^2*b + b^3)*Cos[3*(c + d*x)] + 2*a*(5*a^2 + 3*b^2 + (a^2 - 3*b^2)*Cos[2*
(c + d*x)])*Sin[c + d*x])/(12*d)

________________________________________________________________________________________

fricas [A]  time = 0.87, size = 77, normalized size = 1.33 \[ -\frac {3 \, b^{3} \cos \left (d x + c\right ) + {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (2 \, a^{3} + 3 \, a b^{2} + {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/3*(3*b^3*cos(d*x + c) + (3*a^2*b - b^3)*cos(d*x + c)^3 - (2*a^3 + 3*a*b^2 + (a^3 - 3*a*b^2)*cos(d*x + c)^2)
*sin(d*x + c))/d

________________________________________________________________________________________

giac [A]  time = 0.20, size = 91, normalized size = 1.57 \[ -\frac {{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {3 \, {\left (a^{2} b + b^{3}\right )} \cos \left (d x + c\right )}{4 \, d} + \frac {{\left (a^{3} - 3 \, a b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {3 \, {\left (a^{3} + a b^{2}\right )} \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/12*(3*a^2*b - b^3)*cos(3*d*x + 3*c)/d - 3/4*(a^2*b + b^3)*cos(d*x + c)/d + 1/12*(a^3 - 3*a*b^2)*sin(3*d*x +
3*c)/d + 3/4*(a^3 + a*b^2)*sin(d*x + c)/d

________________________________________________________________________________________

maple [A]  time = 9.74, size = 75, normalized size = 1.29 \[ \frac {-\frac {b^{3} \left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}+b^{2} a \left (\sin ^{3}\left (d x +c \right )\right )-\left (\cos ^{3}\left (d x +c \right )\right ) a^{2} b +\frac {a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

1/d*(-1/3*b^3*(2+sin(d*x+c)^2)*cos(d*x+c)+b^2*a*sin(d*x+c)^3-cos(d*x+c)^3*a^2*b+1/3*a^3*(2+cos(d*x+c)^2)*sin(d
*x+c))

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 84, normalized size = 1.45 \[ -\frac {a^{2} b \cos \left (d x + c\right )^{3}}{d} + \frac {a b^{2} \sin \left (d x + c\right )^{3}}{d} - \frac {{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3}}{3 \, d} + \frac {{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} b^{3}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-a^2*b*cos(d*x + c)^3/d + a*b^2*sin(d*x + c)^3/d - 1/3*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^3/d + 1/3*(cos(d*x
+ c)^3 - 3*cos(d*x + c))*b^3/d

________________________________________________________________________________________

mupad [B]  time = 0.57, size = 104, normalized size = 1.79 \[ \frac {\frac {\sin \left (c+d\,x\right )\,a^3\,{\cos \left (c+d\,x\right )}^2}{3}+\frac {2\,\sin \left (c+d\,x\right )\,a^3}{3}-a^2\,b\,{\cos \left (c+d\,x\right )}^3-\sin \left (c+d\,x\right )\,a\,b^2\,{\cos \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )\,a\,b^2+\frac {b^3\,{\cos \left (c+d\,x\right )}^3}{3}-b^3\,\cos \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))^3,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.52, size = 117, normalized size = 2.02 \[ \begin {cases} \frac {2 a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {a^{2} b \cos ^{3}{\left (c + d x \right )}}{d} + \frac {a b^{2} \sin ^{3}{\left (c + d x \right )}}{d} - \frac {b^{3} \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {2 b^{3} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + b \sin {\relax (c )}\right )^{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Piecewise((2*a**3*sin(c + d*x)**3/(3*d) + a**3*sin(c + d*x)*cos(c + d*x)**2/d - a**2*b*cos(c + d*x)**3/d + a*b
**2*sin(c + d*x)**3/d - b**3*sin(c + d*x)**2*cos(c + d*x)/d - 2*b**3*cos(c + d*x)**3/(3*d), Ne(d, 0)), (x*(a*c
os(c) + b*sin(c))**3, True))

________________________________________________________________________________________