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

Optimal. Leaf size=235 \[ -\frac{a \left (-121 a^2 b^2+4 a^4-128 b^4\right ) \sin (c+d x)}{60 b d}-\frac{\left (4 a^2-25 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^3}{120 b d}-\frac{a \left (4 a^2-53 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{120 b d}-\frac{\left (-178 a^2 b^2+8 a^4-75 b^4\right ) \sin (c+d x) \cos (c+d x)}{240 d}+\frac{1}{16} x \left (36 a^2 b^2+8 a^4+5 b^4\right )+\frac{\sin (c+d x) (a+b \cos (c+d x))^5}{6 b d}-\frac{a \sin (c+d x) (a+b \cos (c+d x))^4}{30 b d} \]

[Out]

((8*a^4 + 36*a^2*b^2 + 5*b^4)*x)/16 - (a*(4*a^4 - 121*a^2*b^2 - 128*b^4)*Sin[c + d*x])/(60*b*d) - ((8*a^4 - 17
8*a^2*b^2 - 75*b^4)*Cos[c + d*x]*Sin[c + d*x])/(240*d) - (a*(4*a^2 - 53*b^2)*(a + b*Cos[c + d*x])^2*Sin[c + d*
x])/(120*b*d) - ((4*a^2 - 25*b^2)*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(120*b*d) - (a*(a + b*Cos[c + d*x])^4*S
in[c + d*x])/(30*b*d) + ((a + b*Cos[c + d*x])^5*Sin[c + d*x])/(6*b*d)

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Rubi [A]  time = 0.320216, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2791, 2753, 2734} \[ -\frac{a \left (-121 a^2 b^2+4 a^4-128 b^4\right ) \sin (c+d x)}{60 b d}-\frac{\left (4 a^2-25 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^3}{120 b d}-\frac{a \left (4 a^2-53 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{120 b d}-\frac{\left (-178 a^2 b^2+8 a^4-75 b^4\right ) \sin (c+d x) \cos (c+d x)}{240 d}+\frac{1}{16} x \left (36 a^2 b^2+8 a^4+5 b^4\right )+\frac{\sin (c+d x) (a+b \cos (c+d x))^5}{6 b d}-\frac{a \sin (c+d x) (a+b \cos (c+d x))^4}{30 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^4 + 36*a^2*b^2 + 5*b^4)*x)/16 - (a*(4*a^4 - 121*a^2*b^2 - 128*b^4)*Sin[c + d*x])/(60*b*d) - ((8*a^4 - 17
8*a^2*b^2 - 75*b^4)*Cos[c + d*x]*Sin[c + d*x])/(240*d) - (a*(4*a^2 - 53*b^2)*(a + b*Cos[c + d*x])^2*Sin[c + d*
x])/(120*b*d) - ((4*a^2 - 25*b^2)*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(120*b*d) - (a*(a + b*Cos[c + d*x])^4*S
in[c + d*x])/(30*b*d) + ((a + b*Cos[c + d*x])^5*Sin[c + d*x])/(6*b*d)

Rule 2791

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
(d^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x
])^m*Simp[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 2753

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

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+b \cos (c+d x))^4 \, dx &=\frac{(a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}+\frac{\int (5 b-a \cos (c+d x)) (a+b \cos (c+d x))^4 \, dx}{6 b}\\ &=-\frac{a (a+b \cos (c+d x))^4 \sin (c+d x)}{30 b d}+\frac{(a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}+\frac{\int (a+b \cos (c+d x))^3 \left (21 a b-\left (4 a^2-25 b^2\right ) \cos (c+d x)\right ) \, dx}{30 b}\\ &=-\frac{\left (4 a^2-25 b^2\right ) (a+b \cos (c+d x))^3 \sin (c+d x)}{120 b d}-\frac{a (a+b \cos (c+d x))^4 \sin (c+d x)}{30 b d}+\frac{(a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}+\frac{\int (a+b \cos (c+d x))^2 \left (3 b \left (24 a^2+25 b^2\right )-3 a \left (4 a^2-53 b^2\right ) \cos (c+d x)\right ) \, dx}{120 b}\\ &=-\frac{a \left (4 a^2-53 b^2\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{120 b d}-\frac{\left (4 a^2-25 b^2\right ) (a+b \cos (c+d x))^3 \sin (c+d x)}{120 b d}-\frac{a (a+b \cos (c+d x))^4 \sin (c+d x)}{30 b d}+\frac{(a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}+\frac{\int (a+b \cos (c+d x)) \left (3 a b \left (64 a^2+181 b^2\right )-3 \left (8 a^4-178 a^2 b^2-75 b^4\right ) \cos (c+d x)\right ) \, dx}{360 b}\\ &=\frac{1}{16} \left (8 a^4+36 a^2 b^2+5 b^4\right ) x-\frac{a \left (4 a^4-121 a^2 b^2-128 b^4\right ) \sin (c+d x)}{60 b d}-\frac{\left (8 a^4-178 a^2 b^2-75 b^4\right ) \cos (c+d x) \sin (c+d x)}{240 d}-\frac{a \left (4 a^2-53 b^2\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{120 b d}-\frac{\left (4 a^2-25 b^2\right ) (a+b \cos (c+d x))^3 \sin (c+d x)}{120 b d}-\frac{a (a+b \cos (c+d x))^4 \sin (c+d x)}{30 b d}+\frac{(a+b \cos (c+d x))^5 \sin (c+d x)}{6 b d}\\ \end{align*}

Mathematica [A]  time = 0.446306, size = 156, normalized size = 0.66 \[ \frac{60 \left (36 a^2 b^2+8 a^4+5 b^4\right ) (c+d x)+45 b^2 \left (4 a^2+b^2\right ) \sin (4 (c+d x))+480 a b \left (6 a^2+5 b^2\right ) \sin (c+d x)+80 a b \left (4 a^2+5 b^2\right ) \sin (3 (c+d x))+15 \left (96 a^2 b^2+16 a^4+15 b^4\right ) \sin (2 (c+d x))+48 a b^3 \sin (5 (c+d x))+5 b^4 \sin (6 (c+d x))}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*(8*a^4 + 36*a^2*b^2 + 5*b^4)*(c + d*x) + 480*a*b*(6*a^2 + 5*b^2)*Sin[c + d*x] + 15*(16*a^4 + 96*a^2*b^2 +
15*b^4)*Sin[2*(c + d*x)] + 80*a*b*(4*a^2 + 5*b^2)*Sin[3*(c + d*x)] + 45*b^2*(4*a^2 + b^2)*Sin[4*(c + d*x)] + 4
8*a*b^3*Sin[5*(c + d*x)] + 5*b^4*Sin[6*(c + d*x)])/(960*d)

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Maple [A]  time = 0.036, size = 174, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ({b}^{4} \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 ) +{\frac{4\,a{b}^{3}\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+6\,{a}^{2}{b}^{2} \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +{\frac{4\,{a}^{3}b \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{a}^{4} \left ({\frac{\cos \left ( dx+c \right ) \sin \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(cos(d*x+c)^2*(a+b*cos(d*x+c))^4,x)

[Out]

1/d*(b^4*(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)+4/5*a*b^3*(8/3+cos(d
*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+6*a^2*b^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)+4/
3*a^3*b*(2+cos(d*x+c)^2)*sin(d*x+c)+a^4*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]  time = 0.979925, size = 230, normalized size = 0.98 \begin{align*} \frac{240 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 1280 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} b + 180 \,{\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} b^{2} + 256 \,{\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^{3} - 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^{4}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(240*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^4 - 1280*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^3*b + 180*(12*d*x +
12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2*b^2 + 256*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*
x + c))*a*b^3 - 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^4)/d

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Fricas [A]  time = 1.98184, size = 358, normalized size = 1.52 \begin{align*} \frac{15 \,{\left (8 \, a^{4} + 36 \, a^{2} b^{2} + 5 \, b^{4}\right )} d x +{\left (40 \, b^{4} \cos \left (d x + c\right )^{5} + 192 \, a b^{3} \cos \left (d x + c\right )^{4} + 640 \, a^{3} b + 512 \, a b^{3} + 10 \,{\left (36 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )^{3} + 64 \,{\left (5 \, a^{3} b + 4 \, a b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (8 \, a^{4} + 36 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(8*a^4 + 36*a^2*b^2 + 5*b^4)*d*x + (40*b^4*cos(d*x + c)^5 + 192*a*b^3*cos(d*x + c)^4 + 640*a^3*b + 5
12*a*b^3 + 10*(36*a^2*b^2 + 5*b^4)*cos(d*x + c)^3 + 64*(5*a^3*b + 4*a*b^3)*cos(d*x + c)^2 + 15*(8*a^4 + 36*a^2
*b^2 + 5*b^4)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 5.04267, size = 459, normalized size = 1.95 \begin{align*} \begin{cases} \frac{a^{4} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{a^{4} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{a^{4} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{8 a^{3} b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{4 a^{3} b \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{9 a^{2} b^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac{9 a^{2} b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac{9 a^{2} b^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac{9 a^{2} b^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{4 d} + \frac{15 a^{2} b^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac{32 a b^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{16 a b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{4 a b^{3} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac{5 b^{4} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{15 b^{4} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{15 b^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{5 b^{4} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{5 b^{4} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{5 b^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac{11 b^{4} \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{\left (c \right )}\right )^{4} \cos ^{2}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**4*x*sin(c + d*x)**2/2 + a**4*x*cos(c + d*x)**2/2 + a**4*sin(c + d*x)*cos(c + d*x)/(2*d) + 8*a**3
*b*sin(c + d*x)**3/(3*d) + 4*a**3*b*sin(c + d*x)*cos(c + d*x)**2/d + 9*a**2*b**2*x*sin(c + d*x)**4/4 + 9*a**2*
b**2*x*sin(c + d*x)**2*cos(c + d*x)**2/2 + 9*a**2*b**2*x*cos(c + d*x)**4/4 + 9*a**2*b**2*sin(c + d*x)**3*cos(c
+ d*x)/(4*d) + 15*a**2*b**2*sin(c + d*x)*cos(c + d*x)**3/(4*d) + 32*a*b**3*sin(c + d*x)**5/(15*d) + 16*a*b**3
*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 4*a*b**3*sin(c + d*x)*cos(c + d*x)**4/d + 5*b**4*x*sin(c + d*x)**6/16
+ 15*b**4*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 15*b**4*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 5*b**4*x*cos(
c + d*x)**6/16 + 5*b**4*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 5*b**4*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + 1
1*b**4*sin(c + d*x)*cos(c + d*x)**5/(16*d), Ne(d, 0)), (x*(a + b*cos(c))**4*cos(c)**2, True))

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Giac [A]  time = 1.30023, size = 227, normalized size = 0.97 \begin{align*} \frac{b^{4} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{a b^{3} \sin \left (5 \, d x + 5 \, c\right )}{20 \, d} + \frac{1}{16} \,{\left (8 \, a^{4} + 36 \, a^{2} b^{2} + 5 \, b^{4}\right )} x + \frac{3 \,{\left (4 \, a^{2} b^{2} + b^{4}\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{{\left (4 \, a^{3} b + 5 \, a b^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{{\left (16 \, a^{4} + 96 \, a^{2} b^{2} + 15 \, b^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac{{\left (6 \, a^{3} b + 5 \, a b^{3}\right )} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*b^4*sin(6*d*x + 6*c)/d + 1/20*a*b^3*sin(5*d*x + 5*c)/d + 1/16*(8*a^4 + 36*a^2*b^2 + 5*b^4)*x + 3/64*(4*a
^2*b^2 + b^4)*sin(4*d*x + 4*c)/d + 1/12*(4*a^3*b + 5*a*b^3)*sin(3*d*x + 3*c)/d + 1/64*(16*a^4 + 96*a^2*b^2 + 1
5*b^4)*sin(2*d*x + 2*c)/d + 1/2*(6*a^3*b + 5*a*b^3)*sin(d*x + c)/d