3.417 \(\int \cos ^4(c+d x) (a+b \cos (c+d x))^2 \, dx\)

Optimal. Leaf size=150 \[ \frac {\left (6 a^2+5 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {\left (6 a^2+5 b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x \left (6 a^2+5 b^2\right )+\frac {2 a b \sin ^5(c+d x)}{5 d}-\frac {4 a b \sin ^3(c+d x)}{3 d}+\frac {2 a b \sin (c+d x)}{d}+\frac {b^2 \sin (c+d x) \cos ^5(c+d x)}{6 d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2789, 2633, 3014, 2635, 8} \[ \frac {\left (6 a^2+5 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {\left (6 a^2+5 b^2\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x \left (6 a^2+5 b^2\right )+\frac {2 a b \sin ^5(c+d x)}{5 d}-\frac {4 a b \sin ^3(c+d x)}{3 d}+\frac {2 a b \sin (c+d x)}{d}+\frac {b^2 \sin (c+d x) \cos ^5(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + b*Cos[c + d*x])^2,x]

[Out]

((6*a^2 + 5*b^2)*x)/16 + (2*a*b*Sin[c + d*x])/d + ((6*a^2 + 5*b^2)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + ((6*a^2
+ 5*b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) + (b^2*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (4*a*b*Sin[c + d*x]^
3)/(3*d) + (2*a*b*Sin[c + d*x]^5)/(5*d)

Rule 8

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

Rule 2633

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

Rule 2635

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

Rule 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[
e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*
x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+b \cos (c+d x))^2 \, dx &=(2 a b) \int \cos ^5(c+d x) \, dx+\int \cos ^4(c+d x) \left (a^2+b^2 \cos ^2(c+d x)\right ) \, dx\\ &=\frac {b^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{6} \left (6 a^2+5 b^2\right ) \int \cos ^4(c+d x) \, dx-\frac {(2 a b) \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac {2 a b \sin (c+d x)}{d}+\frac {\left (6 a^2+5 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {b^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {4 a b \sin ^3(c+d x)}{3 d}+\frac {2 a b \sin ^5(c+d x)}{5 d}+\frac {1}{8} \left (6 a^2+5 b^2\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {2 a b \sin (c+d x)}{d}+\frac {\left (6 a^2+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (6 a^2+5 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {b^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {4 a b \sin ^3(c+d x)}{3 d}+\frac {2 a b \sin ^5(c+d x)}{5 d}+\frac {1}{16} \left (6 a^2+5 b^2\right ) \int 1 \, dx\\ &=\frac {1}{16} \left (6 a^2+5 b^2\right ) x+\frac {2 a b \sin (c+d x)}{d}+\frac {\left (6 a^2+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {\left (6 a^2+5 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {b^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {4 a b \sin ^3(c+d x)}{3 d}+\frac {2 a b \sin ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 123, normalized size = 0.82 \[ \frac {5 \left (\left (48 a^2+45 b^2\right ) \sin (2 (c+d x))+\left (6 a^2+9 b^2\right ) \sin (4 (c+d x))+72 a^2 c+72 a^2 d x+b^2 \sin (6 (c+d x))+60 b^2 c+60 b^2 d x\right )+384 a b \sin ^5(c+d x)-1280 a b \sin ^3(c+d x)+1920 a b \sin (c+d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Cos[c + d*x])^2,x]

[Out]

(1920*a*b*Sin[c + d*x] - 1280*a*b*Sin[c + d*x]^3 + 384*a*b*Sin[c + d*x]^5 + 5*(72*a^2*c + 60*b^2*c + 72*a^2*d*
x + 60*b^2*d*x + (48*a^2 + 45*b^2)*Sin[2*(c + d*x)] + (6*a^2 + 9*b^2)*Sin[4*(c + d*x)] + b^2*Sin[6*(c + d*x)])
)/(960*d)

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fricas [A]  time = 0.98, size = 110, normalized size = 0.73 \[ \frac {15 \, {\left (6 \, a^{2} + 5 \, b^{2}\right )} d x + {\left (40 \, b^{2} \cos \left (d x + c\right )^{5} + 96 \, a b \cos \left (d x + c\right )^{4} + 128 \, a b \cos \left (d x + c\right )^{2} + 10 \, {\left (6 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 256 \, a b + 15 \, {\left (6 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

1/240*(15*(6*a^2 + 5*b^2)*d*x + (40*b^2*cos(d*x + c)^5 + 96*a*b*cos(d*x + c)^4 + 128*a*b*cos(d*x + c)^2 + 10*(
6*a^2 + 5*b^2)*cos(d*x + c)^3 + 256*a*b + 15*(6*a^2 + 5*b^2)*cos(d*x + c))*sin(d*x + c))/d

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giac [A]  time = 0.55, size = 127, normalized size = 0.85 \[ \frac {1}{16} \, {\left (6 \, a^{2} + 5 \, b^{2}\right )} x + \frac {b^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {a b \sin \left (5 \, d x + 5 \, c\right )}{40 \, d} + \frac {5 \, a b \sin \left (3 \, d x + 3 \, c\right )}{24 \, d} + \frac {5 \, a b \sin \left (d x + c\right )}{4 \, d} + \frac {{\left (2 \, a^{2} + 3 \, b^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (16 \, a^{2} + 15 \, b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

1/16*(6*a^2 + 5*b^2)*x + 1/192*b^2*sin(6*d*x + 6*c)/d + 1/40*a*b*sin(5*d*x + 5*c)/d + 5/24*a*b*sin(3*d*x + 3*c
)/d + 5/4*a*b*sin(d*x + c)/d + 1/64*(2*a^2 + 3*b^2)*sin(4*d*x + 4*c)/d + 1/64*(16*a^2 + 15*b^2)*sin(2*d*x + 2*
c)/d

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maple [A]  time = 0.05, size = 120, normalized size = 0.80 \[ \frac {b^{2} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {2 a b \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+b*cos(d*x+c))^2,x)

[Out]

1/d*(b^2*(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)+2/5*a*b*(8/3+cos(d*x
+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+a^2*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c))

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maxima [A]  time = 0.53, size = 120, normalized size = 0.80 \[ \frac {30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} + 128 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a b - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b^{2}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

1/960*(30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2 + 128*(3*sin(d*x + c)^5 - 10*sin(d*x + c
)^3 + 15*sin(d*x + c))*a*b - 5*(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^2)/d

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mupad [B]  time = 0.67, size = 143, normalized size = 0.95 \[ \frac {3\,a^2\,x}{8}+\frac {5\,b^2\,x}{16}+\frac {a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {15\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {3\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}+\frac {b^2\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {5\,a\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {5\,a\,b\,\sin \left (3\,c+3\,d\,x\right )}{24\,d}+\frac {a\,b\,\sin \left (5\,c+5\,d\,x\right )}{40\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(a + b*cos(c + d*x))^2,x)

[Out]

(3*a^2*x)/8 + (5*b^2*x)/16 + (a^2*sin(2*c + 2*d*x))/(4*d) + (a^2*sin(4*c + 4*d*x))/(32*d) + (15*b^2*sin(2*c +
2*d*x))/(64*d) + (3*b^2*sin(4*c + 4*d*x))/(64*d) + (b^2*sin(6*c + 6*d*x))/(192*d) + (5*a*b*sin(c + d*x))/(4*d)
+ (5*a*b*sin(3*c + 3*d*x))/(24*d) + (a*b*sin(5*c + 5*d*x))/(40*d)

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sympy [A]  time = 4.02, size = 343, normalized size = 2.29 \[ \begin {cases} \frac {3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {16 a b \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {8 a b \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {2 a b \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 b^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 b^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {15 b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {5 b^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {5 b^{2} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {5 b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {11 b^{2} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right )^{2} \cos ^{4}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+b*cos(d*x+c))**2,x)

[Out]

Piecewise((3*a**2*x*sin(c + d*x)**4/8 + 3*a**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*a**2*x*cos(c + d*x)**4/
8 + 3*a**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 5*a**2*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 16*a*b*sin(c + d*x
)**5/(15*d) + 8*a*b*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 2*a*b*sin(c + d*x)*cos(c + d*x)**4/d + 5*b**2*x*si
n(c + d*x)**6/16 + 15*b**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*b**2*x*sin(c + d*x)**2*cos(c + d*x)**4/16
+ 5*b**2*x*cos(c + d*x)**6/16 + 5*b**2*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*b**2*sin(c + d*x)**3*cos(c + d
*x)**3/(6*d) + 11*b**2*sin(c + d*x)*cos(c + d*x)**5/(16*d), Ne(d, 0)), (x*(a + b*cos(c))**2*cos(c)**4, True))

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