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

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

[Out]

1/8*(8*a^4-3*b^4)*x+1/6*a*b*(13*a^2-8*b^2)*sin(d*x+c)/d+1/24*b^2*(14*a^2-9*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/12*a
*b*(a+b*cos(d*x+c))^2*sin(d*x+c)/d-1/4*b*(a+b*cos(d*x+c))^3*sin(d*x+c)/d

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Rubi [A]  time = 0.20, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.22, size = 89, normalized size = 0.69 \[ -\frac {-96 a^4 d x-48 a b \left (4 a^2-3 b^2\right ) \sin (c+d x)+16 a b^3 \sin (3 (c+d x))+24 b^4 \sin (2 (c+d x))+3 b^4 \sin (4 (c+d x))+36 b^4 c+36 b^4 d x}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/96*(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)])/d

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fricas [A]  time = 1.18, size = 80, normalized size = 0.62 \[ \frac {3 \, {\left (8 \, a^{4} - 3 \, b^{4}\right )} d x - {\left (6 \, b^{4} \cos \left (d x + c\right )^{3} + 16 \, a b^{3} \cos \left (d x + c\right )^{2} + 9 \, b^{4} \cos \left (d x + c\right ) - 48 \, a^{3} b + 32 \, a b^{3}\right )} \sin \left (d x + c\right )}{24 \, d} \]

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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giac [A]  time = 2.91, size = 91, normalized size = 0.71 \[ -\frac {b^{4} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {a b^{3} \sin \left (3 \, d x + 3 \, c\right )}{6 \, d} - \frac {b^{4} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {1}{8} \, {\left (8 \, a^{4} - 3 \, b^{4}\right )} x + \frac {{\left (4 \, a^{3} b - 3 \, a b^{3}\right )} \sin \left (d x + c\right )}{2 \, d} \]

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

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maple [A]  time = 0.23, size = 87, normalized size = 0.67 \[ \frac {-b^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {2 a \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 a^{3} b \sin \left (d x +c \right )+a^{4} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^2*(-cos(d*x+c)^2*b^2+a^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 = 0.33, size = 84, normalized size = 0.65 \[ \frac {96 \, {\left (d x + c\right )} a^{4} + 64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b^{3} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b^{4} + 192 \, a^{3} b \sin \left (d x + c\right )}{96 \, d} \]

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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mupad [B]  time = 1.35, size = 107, normalized size = 0.83 \[ a^4\,x-\frac {3\,b^4\,x}{8}-\frac {b^4\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )}{4\,d}-\frac {4\,a\,b^3\,\sin \left (c+d\,x\right )}{3\,d}+\frac {2\,a^3\,b\,\sin \left (c+d\,x\right )}{d}-\frac {3\,b^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{8\,d}-\frac {2\,a\,b^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.09, size = 190, normalized size = 1.47 \[ \begin {cases} a^{4} x + \frac {2 a^{3} b \sin {\left (c + d x \right )}}{d} - \frac {4 a b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 a b^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {3 b^{4} x \sin ^{4}{\left (c + d x \right )}}{8} - \frac {3 b^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} - \frac {3 b^{4} x \cos ^{4}{\left (c + d x \right )}}{8} - \frac {3 b^{4} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {5 b^{4} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right )^{2} \left (a^{2} - b^{2} \cos ^{2}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

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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