3.559 \(\int (a+b \cos (c+d x))^3 (a^2-b^2 \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=183 \[ \frac{b \left (-32 a^2 b^2+83 a^4-16 b^4\right ) \sin (c+d x)}{30 d}+\frac{b \left (23 a^2-16 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{60 d}+\frac{a b^2 \left (106 a^2-71 b^2\right ) \sin (c+d x) \cos (c+d x)}{120 d}+\frac{1}{8} a x \left (8 a^2 b^2+8 a^4-9 b^4\right )-\frac{b \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}+\frac{a b \sin (c+d x) (a+b \cos (c+d x))^3}{20 d} \]

[Out]

(a*(8*a^4 + 8*a^2*b^2 - 9*b^4)*x)/8 + (b*(83*a^4 - 32*a^2*b^2 - 16*b^4)*Sin[c + d*x])/(30*d) + (a*b^2*(106*a^2
- 71*b^2)*Cos[c + d*x]*Sin[c + d*x])/(120*d) + (b*(23*a^2 - 16*b^2)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(60*
d) + (a*b*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(20*d) - (b*(a + b*Cos[c + d*x])^4*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.323211, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3016, 2753, 2734} \[ \frac{b \left (-32 a^2 b^2+83 a^4-16 b^4\right ) \sin (c+d x)}{30 d}+\frac{b \left (23 a^2-16 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{60 d}+\frac{a b^2 \left (106 a^2-71 b^2\right ) \sin (c+d x) \cos (c+d x)}{120 d}+\frac{1}{8} a x \left (8 a^2 b^2+8 a^4-9 b^4\right )-\frac{b \sin (c+d x) (a+b \cos (c+d x))^4}{5 d}+\frac{a b \sin (c+d x) (a+b \cos (c+d x))^3}{20 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(8*a^4 + 8*a^2*b^2 - 9*b^4)*x)/8 + (b*(83*a^4 - 32*a^2*b^2 - 16*b^4)*Sin[c + d*x])/(30*d) + (a*b^2*(106*a^2
- 71*b^2)*Cos[c + d*x]*Sin[c + d*x])/(120*d) + (b*(23*a^2 - 16*b^2)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(60*
d) + (a*b*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(20*d) - (b*(a + b*Cos[c + d*x])^4*Sin[c + d*x])/(5*d)

Rule 3016

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
C/b^2, Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[-a + b*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x
] && EqQ[A*b^2 + a^2*C, 0]

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 (a+b \cos (c+d x))^3 \left (a^2-b^2 \cos ^2(c+d x)\right ) \, dx &=-\int (-a+b \cos (c+d x)) (a+b \cos (c+d x))^4 \, dx\\ &=-\frac{b (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}-\frac{1}{5} \int (a+b \cos (c+d x))^3 \left (-5 a^2+4 b^2-a b \cos (c+d x)\right ) \, dx\\ &=\frac{a b (a+b \cos (c+d x))^3 \sin (c+d x)}{20 d}-\frac{b (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}-\frac{1}{20} \int (a+b \cos (c+d x))^2 \left (-a \left (20 a^2-13 b^2\right )-b \left (23 a^2-16 b^2\right ) \cos (c+d x)\right ) \, dx\\ &=\frac{b \left (23 a^2-16 b^2\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{60 d}+\frac{a b (a+b \cos (c+d x))^3 \sin (c+d x)}{20 d}-\frac{b (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}-\frac{1}{60} \int (a+b \cos (c+d x)) \left (-60 a^4-7 a^2 b^2+32 b^4-a b \left (106 a^2-71 b^2\right ) \cos (c+d x)\right ) \, dx\\ &=\frac{1}{8} a \left (8 a^4+8 a^2 b^2-9 b^4\right ) x+\frac{b \left (83 a^4-32 a^2 b^2-16 b^4\right ) \sin (c+d x)}{30 d}+\frac{a b^2 \left (106 a^2-71 b^2\right ) \cos (c+d x) \sin (c+d x)}{120 d}+\frac{b \left (23 a^2-16 b^2\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{60 d}+\frac{a b (a+b \cos (c+d x))^3 \sin (c+d x)}{20 d}-\frac{b (a+b \cos (c+d x))^4 \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.585719, size = 139, normalized size = 0.76 \[ -\frac{-60 a \left (8 a^2 b^2+8 a^4-9 b^4\right ) (c+d x)+10 b^3 \left (8 a^2+5 b^2\right ) \sin (3 (c+d x))-120 a b^2 \left (2 a^2-3 b^2\right ) \sin (2 (c+d x))+60 b \left (12 a^2 b^2-24 a^4+5 b^4\right ) \sin (c+d x)+45 a b^4 \sin (4 (c+d x))+6 b^5 \sin (5 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-60*a*(8*a^4 + 8*a^2*b^2 - 9*b^4)*(c + d*x) + 60*b*(-24*a^4 + 12*a^2*b^2 + 5*b^4)*Sin[c + d*x] - 120*a*b^2*(
2*a^2 - 3*b^2)*Sin[2*(c + d*x)] + 10*b^3*(8*a^2 + 5*b^2)*Sin[3*(c + d*x)] + 45*a*b^4*Sin[4*(c + d*x)] + 6*b^5*
Sin[5*(c + d*x)])/(480*d)

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Maple [A]  time = 0.026, size = 151, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( -{\frac{{b}^{5}\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 ) }-3\,a{b}^{4} \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{2\,{a}^{2}{b}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+2\,{a}^{3}{b}^{2} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,{a}^{4}b\sin \left ( dx+c \right ) +{a}^{5} \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(a^2-b^2*cos(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 1.00753, size = 197, normalized size = 1.08 \begin{align*} \frac{480 \,{\left (d x + c\right )} a^{5} + 240 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} b^{2} + 320 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} b^{3} - 45 \,{\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 b^{4} - 32 \,{\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 )} b^{5} + 1440 \, a^{4} b \sin \left (d x + c\right )}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(a^2-b^2*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

1/480*(480*(d*x + c)*a^5 + 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3*b^2 + 320*(sin(d*x + c)^3 - 3*sin(d*x + c)
)*a^2*b^3 - 45*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a*b^4 - 32*(3*sin(d*x + c)^5 - 10*sin(d
*x + c)^3 + 15*sin(d*x + c))*b^5 + 1440*a^4*b*sin(d*x + c))/d

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Fricas [A]  time = 1.49426, size = 308, normalized size = 1.68 \begin{align*} \frac{15 \,{\left (8 \, a^{5} + 8 \, a^{3} b^{2} - 9 \, a b^{4}\right )} d x -{\left (24 \, b^{5} \cos \left (d x + c\right )^{4} + 90 \, a b^{4} \cos \left (d x + c\right )^{3} - 360 \, a^{4} b + 160 \, a^{2} b^{3} + 64 \, b^{5} + 16 \,{\left (5 \, a^{2} b^{3} + 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2} - 15 \,{\left (8 \, a^{3} b^{2} - 9 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(a^2-b^2*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(15*(8*a^5 + 8*a^3*b^2 - 9*a*b^4)*d*x - (24*b^5*cos(d*x + c)^4 + 90*a*b^4*cos(d*x + c)^3 - 360*a^4*b + 1
60*a^2*b^3 + 64*b^5 + 16*(5*a^2*b^3 + 2*b^5)*cos(d*x + c)^2 - 15*(8*a^3*b^2 - 9*a*b^4)*cos(d*x + c))*sin(d*x +
c))/d

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Sympy [A]  time = 2.75186, size = 321, normalized size = 1.75 \begin{align*} \begin{cases} a^{5} x + \frac{3 a^{4} b \sin{\left (c + d x \right )}}{d} + a^{3} b^{2} x \sin ^{2}{\left (c + d x \right )} + a^{3} b^{2} x \cos ^{2}{\left (c + d x \right )} + \frac{a^{3} b^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} - \frac{4 a^{2} b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} - \frac{2 a^{2} b^{3} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac{9 a b^{4} x \sin ^{4}{\left (c + d x \right )}}{8} - \frac{9 a b^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} - \frac{9 a b^{4} x \cos ^{4}{\left (c + d x \right )}}{8} - \frac{9 a b^{4} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{15 a b^{4} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac{8 b^{5} \sin ^{5}{\left (c + d x \right )}}{15 d} - \frac{4 b^{5} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac{b^{5} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \cos{\left (c \right )}\right )^{3} \left (a^{2} - b^{2} \cos ^{2}{\left (c \right )}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**3*(a**2-b**2*cos(d*x+c)**2),x)

[Out]

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

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Giac [A]  time = 1.42564, size = 198, normalized size = 1.08 \begin{align*} -\frac{b^{5} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac{3 \, a b^{4} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{1}{8} \,{\left (8 \, a^{5} + 8 \, a^{3} b^{2} - 9 \, a b^{4}\right )} x - \frac{{\left (8 \, a^{2} b^{3} + 5 \, b^{5}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{{\left (2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (24 \, a^{4} b - 12 \, a^{2} b^{3} - 5 \, b^{5}\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(a^2-b^2*cos(d*x+c)^2),x, algorithm="giac")

[Out]

-1/80*b^5*sin(5*d*x + 5*c)/d - 3/32*a*b^4*sin(4*d*x + 4*c)/d + 1/8*(8*a^5 + 8*a^3*b^2 - 9*a*b^4)*x - 1/48*(8*a
^2*b^3 + 5*b^5)*sin(3*d*x + 3*c)/d + 1/4*(2*a^3*b^2 - 3*a*b^4)*sin(2*d*x + 2*c)/d + 1/8*(24*a^4*b - 12*a^2*b^3
- 5*b^5)*sin(d*x + c)/d