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

Optimal. Leaf size=129 \[ \frac{a b \left (13 a^2-8 b^2\right ) \sin (c+d x)}{6 d}+\frac{b^2 \left (14 a^2-9 b^2\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} x \left (8 a^4-3 b^4\right )-\frac{b \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}+\frac{a b \sin (c+d x) (a+b \cos (c+d x))^2}{12 d} \]

[Out]

((8*a^4 - 3*b^4)*x)/8 + (a*b*(13*a^2 - 8*b^2)*Sin[c + d*x])/(6*d) + (b^2*(14*a^2 - 9*b^2)*Cos[c + d*x]*Sin[c +
d*x])/(24*d) + (a*b*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(12*d) - (b*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*
d)

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Rubi [A]  time = 0.201382, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3016, 2753, 2734} \[ \frac{a b \left (13 a^2-8 b^2\right ) \sin (c+d x)}{6 d}+\frac{b^2 \left (14 a^2-9 b^2\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} x \left (8 a^4-3 b^4\right )-\frac{b \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}+\frac{a b \sin (c+d x) (a+b \cos (c+d x))^2}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^4 - 3*b^4)*x)/8 + (a*b*(13*a^2 - 8*b^2)*Sin[c + d*x])/(6*d) + (b^2*(14*a^2 - 9*b^2)*Cos[c + d*x]*Sin[c +
d*x])/(24*d) + (a*b*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(12*d) - (b*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*
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))^2 \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))^3 \, dx\\ &=-\frac{b (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac{1}{4} \int (a+b \cos (c+d x))^2 \left (-4 a^2+3 b^2-a b \cos (c+d x)\right ) \, dx\\ &=\frac{a b (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}-\frac{b (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac{1}{12} \int (a+b \cos (c+d x)) \left (-a \left (12 a^2-7 b^2\right )-b \left (14 a^2-9 b^2\right ) \cos (c+d x)\right ) \, dx\\ &=\frac{1}{8} \left (8 a^4-3 b^4\right ) x+\frac{a b \left (13 a^2-8 b^2\right ) \sin (c+d x)}{6 d}+\frac{b^2 \left (14 a^2-9 b^2\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac{a b (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}-\frac{b (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.219931, size = 89, normalized size = 0.69 \[ -\frac{-48 a b \left (4 a^2-3 b^2\right ) \sin (c+d x)-96 a^4 d x+16 a b^3 \sin (3 (c+d x))+24 b^4 \sin (2 (c+d x))+3 b^4 \sin (4 (c+d x))+36 b^4 c+36 b^4 d x}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(36*b^4*c - 96*a^4*d*x + 36*b^4*d*x - 48*a*b*(4*a^2 - 3*b^2)*Sin[c + d*x] + 24*b^4*Sin[2*(c + d*x)] + 16*a*b^
3*Sin[3*(c + d*x)] + 3*b^4*Sin[4*(c + d*x)])/(96*d)

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Maple [A]  time = 0.021, size = 87, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ( -{b}^{4} \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) -{\frac{2\,a{b}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+2\,{a}^{3}b\sin \left ( dx+c \right ) +{a}^{4} \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))^2*(a^2-b^2*cos(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 0.997582, size = 113, normalized size = 0.88 \begin{align*} \frac{96 \,{\left (d x + c\right )} a^{4} + 64 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b^{3} - 3 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b^{4} + 192 \, a^{3} b \sin \left (d x + c\right )}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(96*(d*x + c)*a^4 + 64*(sin(d*x + c)^3 - 3*sin(d*x + c))*a*b^3 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*
sin(2*d*x + 2*c))*b^4 + 192*a^3*b*sin(d*x + c))/d

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Fricas [A]  time = 1.40704, size = 188, normalized size = 1.46 \begin{align*} \frac{3 \,{\left (8 \, a^{4} - 3 \, b^{4}\right )} d x -{\left (6 \, b^{4} \cos \left (d x + c\right )^{3} + 16 \, a b^{3} \cos \left (d x + c\right )^{2} + 9 \, b^{4} \cos \left (d x + c\right ) - 48 \, a^{3} b + 32 \, a b^{3}\right )} \sin \left (d x + c\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(8*a^4 - 3*b^4)*d*x - (6*b^4*cos(d*x + c)^3 + 16*a*b^3*cos(d*x + c)^2 + 9*b^4*cos(d*x + c) - 48*a^3*b
+ 32*a*b^3)*sin(d*x + c))/d

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Sympy [A]  time = 1.32987, size = 190, normalized size = 1.47 \begin{align*} \begin{cases} a^{4} x + \frac{2 a^{3} b \sin{\left (c + d x \right )}}{d} - \frac{4 a b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} - \frac{2 a b^{3} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac{3 b^{4} x \sin ^{4}{\left (c + d x \right )}}{8} - \frac{3 b^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} - \frac{3 b^{4} x \cos ^{4}{\left (c + d x \right )}}{8} - \frac{3 b^{4} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} - \frac{5 b^{4} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text{for}\: d \neq 0 \\x \left (a + b \cos{\left (c \right )}\right )^{2} \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))**2*(a**2-b**2*cos(d*x+c)**2),x)

[Out]

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

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Giac [A]  time = 1.53148, size = 123, normalized size = 0.95 \begin{align*} -\frac{b^{4} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac{a b^{3} \sin \left (3 \, d x + 3 \, c\right )}{6 \, d} - \frac{b^{4} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{1}{8} \,{\left (8 \, a^{4} - 3 \, b^{4}\right )} x + \frac{{\left (4 \, a^{3} b - 3 \, 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((a+b*cos(d*x+c))^2*(a^2-b^2*cos(d*x+c)^2),x, algorithm="giac")

[Out]

-1/32*b^4*sin(4*d*x + 4*c)/d - 1/6*a*b^3*sin(3*d*x + 3*c)/d - 1/4*b^4*sin(2*d*x + 2*c)/d + 1/8*(8*a^4 - 3*b^4)
*x + 1/2*(4*a^3*b - 3*a*b^3)*sin(d*x + c)/d