3.220 \(\int (a \cos (c+d x)+b \sin (c+d x))^6 \, dx\)

Optimal. Leaf size=161 \[ -\frac{5 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^3}{24 d}-\frac{5 \left (a^2+b^2\right )^2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))}{16 d}+\frac{5}{16} x \left (a^2+b^2\right )^3-\frac{(b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^5}{6 d} \]

[Out]

(5*(a^2 + b^2)^3*x)/16 - (5*(a^2 + b^2)^2*(b*Cos[c + d*x] - a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x]))
/(16*d) - (5*(a^2 + b^2)*(b*Cos[c + d*x] - a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(24*d) - ((b*C
os[c + d*x] - a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x])^5)/(6*d)

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Rubi [A]  time = 0.0789943, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3073, 8} \[ -\frac{5 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^3}{24 d}-\frac{5 \left (a^2+b^2\right )^2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))}{16 d}+\frac{5}{16} x \left (a^2+b^2\right )^3-\frac{(b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^5}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^6,x]

[Out]

(5*(a^2 + b^2)^3*x)/16 - (5*(a^2 + b^2)^2*(b*Cos[c + d*x] - a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x]))
/(16*d) - (5*(a^2 + b^2)*(b*Cos[c + d*x] - a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(24*d) - ((b*C
os[c + d*x] - a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x])^5)/(6*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))^6 \, dx &=-\frac{(b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^5}{6 d}+\frac{1}{6} \left (5 \left (a^2+b^2\right )\right ) \int (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\\ &=-\frac{5 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^3}{24 d}-\frac{(b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^5}{6 d}+\frac{1}{8} \left (5 \left (a^2+b^2\right )^2\right ) \int (a \cos (c+d x)+b \sin (c+d x))^2 \, dx\\ &=-\frac{5 \left (a^2+b^2\right )^2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))}{16 d}-\frac{5 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^3}{24 d}-\frac{(b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^5}{6 d}+\frac{1}{16} \left (5 \left (a^2+b^2\right )^3\right ) \int 1 \, dx\\ &=\frac{5}{16} \left (a^2+b^2\right )^3 x-\frac{5 \left (a^2+b^2\right )^2 (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))}{16 d}-\frac{5 \left (a^2+b^2\right ) (b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^3}{24 d}-\frac{(b \cos (c+d x)-a \sin (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^5}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.707346, size = 192, normalized size = 1.19 \[ \frac{60 \left (a^2+b^2\right )^3 (c+d x)+45 \left (a^2-b^2\right ) \left (a^2+b^2\right )^2 \sin (2 (c+d x))+9 \left (-5 a^4 b^2-5 a^2 b^4+a^6+b^6\right ) \sin (4 (c+d x))+\left (-15 a^4 b^2+15 a^2 b^4+a^6-b^6\right ) \sin (6 (c+d x))-90 a b \left (a^2+b^2\right )^2 \cos (2 (c+d x))-36 a b \left (a^4-b^4\right ) \cos (4 (c+d x))-2 a b \left (-10 a^2 b^2+3 a^4+3 b^4\right ) \cos (6 (c+d x))}{192 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cos[c + d*x] + b*Sin[c + d*x])^6,x]

[Out]

(60*(a^2 + b^2)^3*(c + d*x) - 90*a*b*(a^2 + b^2)^2*Cos[2*(c + d*x)] - 36*a*b*(a^4 - b^4)*Cos[4*(c + d*x)] - 2*
a*b*(3*a^4 - 10*a^2*b^2 + 3*b^4)*Cos[6*(c + d*x)] + 45*(a^2 - b^2)*(a^2 + b^2)^2*Sin[2*(c + d*x)] + 9*(a^6 - 5
*a^4*b^2 - 5*a^2*b^4 + b^6)*Sin[4*(c + d*x)] + (a^6 - 15*a^4*b^2 + 15*a^2*b^4 - b^6)*Sin[6*(c + d*x)])/(192*d)

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Maple [A]  time = 0.141, size = 285, normalized size = 1.8 \begin{align*}{\frac{1}{d} \left ({b}^{6} \left ( -{\frac{\cos \left ( dx+c \right ) }{6} \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +a{b}^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{6}+15\,{a}^{2}{b}^{4} \left ( -1/6\, \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}-1/8\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}+1/16\,\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) +1/16\,dx+c/16 \right ) +20\,{a}^{3}{b}^{3} \left ( -1/6\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}-1/12\, \left ( \cos \left ( dx+c \right ) \right ) ^{4} \right ) +15\,{a}^{4}{b}^{2} \left ( -1/6\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{5}+1/24\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +1/16\,dx+c/16 \right ) -{a}^{5}b \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{a}^{6} \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \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))^6,x)

[Out]

1/d*(b^6*(-1/6*(sin(d*x+c)^5+5/4*sin(d*x+c)^3+15/8*sin(d*x+c))*cos(d*x+c)+5/16*d*x+5/16*c)+a*b^5*sin(d*x+c)^6+
15*a^2*b^4*(-1/6*sin(d*x+c)^3*cos(d*x+c)^3-1/8*sin(d*x+c)*cos(d*x+c)^3+1/16*sin(d*x+c)*cos(d*x+c)+1/16*d*x+1/1
6*c)+20*a^3*b^3*(-1/6*sin(d*x+c)^2*cos(d*x+c)^4-1/12*cos(d*x+c)^4)+15*a^4*b^2*(-1/6*sin(d*x+c)*cos(d*x+c)^5+1/
24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)-a^5*b*cos(d*x+c)^6+a^6*(1/6*(cos(d*x+c)^5+5/4*cos
(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c))

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Maxima [A]  time = 1.02943, size = 321, normalized size = 1.99 \begin{align*} -\frac{192 \, a^{5} b \cos \left (d x + c\right )^{6} - 192 \, a b^{5} \sin \left (d x + c\right )^{6} +{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{6} - 15 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{4} b^{2} + 320 \,{\left (2 \, \sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4}\right )} a^{3} b^{3} + 15 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 12 \, d x - 12 \, c + 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2} b^{4} -{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b^{6}}{192 \, 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))^6,x, algorithm="maxima")

[Out]

-1/192*(192*a^5*b*cos(d*x + c)^6 - 192*a*b^5*sin(d*x + c)^6 + (4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*
d*x + 4*c) - 48*sin(2*d*x + 2*c))*a^6 - 15*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*a^4*b^2
+ 320*(2*sin(d*x + c)^6 - 3*sin(d*x + c)^4)*a^3*b^3 + 15*(4*sin(2*d*x + 2*c)^3 - 12*d*x - 12*c + 3*sin(4*d*x
+ 4*c))*a^2*b^4 - (4*sin(2*d*x + 2*c)^3 + 60*d*x + 60*c + 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*b^6)/d

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Fricas [A]  time = 2.91479, size = 501, normalized size = 3.11 \begin{align*} -\frac{144 \, a b^{5} \cos \left (d x + c\right )^{2} + 16 \,{\left (3 \, a^{5} b - 10 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{6} + 48 \,{\left (5 \, a^{3} b^{3} - 3 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} - 15 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} d x -{\left (8 \,{\left (a^{6} - 15 \, a^{4} b^{2} + 15 \, a^{2} b^{4} - b^{6}\right )} \cos \left (d x + c\right )^{5} + 2 \,{\left (5 \, a^{6} + 15 \, a^{4} b^{2} - 105 \, a^{2} b^{4} + 13 \, b^{6}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (5 \, a^{6} + 15 \, a^{4} b^{2} + 15 \, a^{2} b^{4} - 11 \, b^{6}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, 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))^6,x, algorithm="fricas")

[Out]

-1/48*(144*a*b^5*cos(d*x + c)^2 + 16*(3*a^5*b - 10*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^6 + 48*(5*a^3*b^3 - 3*a*b^5
)*cos(d*x + c)^4 - 15*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d*x - (8*(a^6 - 15*a^4*b^2 + 15*a^2*b^4 - b^6)*cos(d
*x + c)^5 + 2*(5*a^6 + 15*a^4*b^2 - 105*a^2*b^4 + 13*b^6)*cos(d*x + c)^3 + 3*(5*a^6 + 15*a^4*b^2 + 15*a^2*b^4
- 11*b^6)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 6.39096, size = 821, normalized size = 5.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cos(d*x+c)+b*sin(d*x+c))**6,x)

[Out]

Piecewise((5*a**6*x*sin(c + d*x)**6/16 + 15*a**6*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*a**6*x*sin(c + d*x)
**2*cos(c + d*x)**4/16 + 5*a**6*x*cos(c + d*x)**6/16 + 5*a**6*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*a**6*sin
(c + d*x)**3*cos(c + d*x)**3/(6*d) + 11*a**6*sin(c + d*x)*cos(c + d*x)**5/(16*d) - a**5*b*cos(c + d*x)**6/d +
15*a**4*b**2*x*sin(c + d*x)**6/16 + 45*a**4*b**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 45*a**4*b**2*x*sin(c +
d*x)**2*cos(c + d*x)**4/16 + 15*a**4*b**2*x*cos(c + d*x)**6/16 + 15*a**4*b**2*sin(c + d*x)**5*cos(c + d*x)/(1
6*d) + 5*a**4*b**2*sin(c + d*x)**3*cos(c + d*x)**3/(2*d) - 15*a**4*b**2*sin(c + d*x)*cos(c + d*x)**5/(16*d) -
5*a**3*b**3*sin(c + d*x)**2*cos(c + d*x)**4/d - 5*a**3*b**3*cos(c + d*x)**6/(3*d) + 15*a**2*b**4*x*sin(c + d*x
)**6/16 + 45*a**2*b**4*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 45*a**2*b**4*x*sin(c + d*x)**2*cos(c + d*x)**4/1
6 + 15*a**2*b**4*x*cos(c + d*x)**6/16 + 15*a**2*b**4*sin(c + d*x)**5*cos(c + d*x)/(16*d) - 5*a**2*b**4*sin(c +
d*x)**3*cos(c + d*x)**3/(2*d) - 15*a**2*b**4*sin(c + d*x)*cos(c + d*x)**5/(16*d) - 3*a*b**5*sin(c + d*x)**4*c
os(c + d*x)**2/d - 3*a*b**5*sin(c + d*x)**2*cos(c + d*x)**4/d - a*b**5*cos(c + d*x)**6/d + 5*b**6*x*sin(c + d*
x)**6/16 + 15*b**6*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*b**6*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 5*b**
6*x*cos(c + d*x)**6/16 - 11*b**6*sin(c + d*x)**5*cos(c + d*x)/(16*d) - 5*b**6*sin(c + d*x)**3*cos(c + d*x)**3/
(6*d) - 5*b**6*sin(c + d*x)*cos(c + d*x)**5/(16*d), Ne(d, 0)), (x*(a*cos(c) + b*sin(c))**6, True))

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Giac [A]  time = 1.1269, size = 317, normalized size = 1.97 \begin{align*} \frac{5}{16} \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} x - \frac{{\left (3 \, a^{5} b - 10 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac{3 \,{\left (a^{5} b - a b^{5}\right )} \cos \left (4 \, d x + 4 \, c\right )}{16 \, d} - \frac{15 \,{\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \cos \left (2 \, d x + 2 \, c\right )}{32 \, d} + \frac{{\left (a^{6} - 15 \, a^{4} b^{2} + 15 \, a^{2} b^{4} - b^{6}\right )} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{3 \,{\left (a^{6} - 5 \, a^{4} b^{2} - 5 \, a^{2} b^{4} + b^{6}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{15 \,{\left (a^{6} + a^{4} b^{2} - a^{2} b^{4} - b^{6}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, 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))^6,x, algorithm="giac")

[Out]

5/16*(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*x - 1/96*(3*a^5*b - 10*a^3*b^3 + 3*a*b^5)*cos(6*d*x + 6*c)/d - 3/16*(
a^5*b - a*b^5)*cos(4*d*x + 4*c)/d - 15/32*(a^5*b + 2*a^3*b^3 + a*b^5)*cos(2*d*x + 2*c)/d + 1/192*(a^6 - 15*a^4
*b^2 + 15*a^2*b^4 - b^6)*sin(6*d*x + 6*c)/d + 3/64*(a^6 - 5*a^4*b^2 - 5*a^2*b^4 + b^6)*sin(4*d*x + 4*c)/d + 15
/64*(a^6 + a^4*b^2 - a^2*b^4 - b^6)*sin(2*d*x + 2*c)/d