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

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

[Out]

1/8*a*(8*a^4+8*a^2*b^2-9*b^4)*x+1/30*b*(83*a^4-32*a^2*b^2-16*b^4)*sin(d*x+c)/d+1/120*a*b^2*(106*a^2-71*b^2)*co
s(d*x+c)*sin(d*x+c)/d+1/60*b*(23*a^2-16*b^2)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d+1/20*a*b*(a+b*cos(d*x+c))^3*sin(d
*x+c)/d-1/5*b*(a+b*cos(d*x+c))^4*sin(d*x+c)/d

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Rubi [A]  time = 0.32, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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 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))^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*}

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Mathematica [A]  time = 0.58, size = 139, normalized size = 0.76 \[ -\frac {-120 a b^2 \left (2 a^2-3 b^2\right ) \sin (2 (c+d x))+10 b^3 \left (8 a^2+5 b^2\right ) \sin (3 (c+d x))-60 a \left (8 a^4+8 a^2 b^2-9 b^4\right ) (c+d x)+60 b \left (-24 a^4+12 a^2 b^2+5 b^4\right ) \sin (c+d x)+45 a b^4 \sin (4 (c+d x))+6 b^5 \sin (5 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/480*(-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)])/d

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fricas [A]  time = 0.80, size = 132, normalized size = 0.72 \[ \frac {15 \, {\left (8 \, a^{5} + 8 \, a^{3} b^{2} - 9 \, a b^{4}\right )} d x - {\left (24 \, b^{5} \cos \left (d x + c\right )^{4} + 90 \, a b^{4} \cos \left (d x + c\right )^{3} - 360 \, a^{4} b + 160 \, a^{2} b^{3} + 64 \, b^{5} + 16 \, {\left (5 \, a^{2} b^{3} + 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2} - 15 \, {\left (8 \, a^{3} b^{2} - 9 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

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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giac [A]  time = 2.91, size = 147, normalized size = 0.80 \[ -\frac {b^{5} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {3 \, a b^{4} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {1}{8} \, {\left (8 \, a^{5} + 8 \, a^{3} b^{2} - 9 \, a b^{4}\right )} x - \frac {{\left (8 \, a^{2} b^{3} + 5 \, b^{5}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (24 \, a^{4} b - 12 \, a^{2} b^{3} - 5 \, b^{5}\right )} \sin \left (d x + c\right )}{8 \, d} \]

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

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maple [A]  time = 0.26, size = 151, normalized size = 0.83 \[ \frac {-\frac {b^{5} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}-3 a \,b^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {2 a^{2} b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 a^{3} b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{4} b \sin \left (d x +c \right )+\left (d x +c \right ) a^{5}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(-cos(d*x+c)^2*b^2+a^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)+(d*x+c)*a^5)

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maxima [A]  time = 0.36, size = 146, normalized size = 0.80 \[ \frac {480 \, {\left (d x + c\right )} a^{5} + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} b^{2} + 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} b^{3} - 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{4} - 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} b^{5} + 1440 \, a^{4} b \sin \left (d x + c\right )}{480 \, d} \]

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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mupad [B]  time = 2.78, size = 406, normalized size = 2.22 \[ \frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (8\,a^4+8\,a^2\,b^2-9\,b^4\right )}{4\,\left (2\,a^5+2\,a^3\,b^2-\frac {9\,a\,b^4}{4}\right )}\right )\,\left (8\,a^4+8\,a^2\,b^2-9\,b^4\right )}{4\,d}-\frac {\left (-6\,a^4\,b+2\,a^3\,b^2+4\,a^2\,b^3-\frac {15\,a\,b^4}{4}+2\,b^5\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-24\,a^4\,b+4\,a^3\,b^2+\frac {32\,a^2\,b^3}{3}-\frac {3\,a\,b^4}{2}+\frac {8\,b^5}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-36\,a^4\,b+\frac {40\,a^2\,b^3}{3}+\frac {116\,b^5}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-24\,a^4\,b-4\,a^3\,b^2+\frac {32\,a^2\,b^3}{3}+\frac {3\,a\,b^4}{2}+\frac {8\,b^5}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-6\,a^4\,b-2\,a^3\,b^2+4\,a^2\,b^3+\frac {15\,a\,b^4}{4}+2\,b^5\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )\,\left (8\,a^4+8\,a^2\,b^2-9\,b^4\right )}{4\,d} \]

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))^3,x)

[Out]

(a*atan((a*tan(c/2 + (d*x)/2)*(8*a^4 - 9*b^4 + 8*a^2*b^2))/(4*(2*a^5 - (9*a*b^4)/4 + 2*a^3*b^2)))*(8*a^4 - 9*b
^4 + 8*a^2*b^2))/(4*d) - (tan(c/2 + (d*x)/2)^5*((116*b^5)/15 - 36*a^4*b + (40*a^2*b^3)/3) + tan(c/2 + (d*x)/2)
*((15*a*b^4)/4 - 6*a^4*b + 2*b^5 + 4*a^2*b^3 - 2*a^3*b^2) + tan(c/2 + (d*x)/2)^9*(2*b^5 - 6*a^4*b - (15*a*b^4)
/4 + 4*a^2*b^3 + 2*a^3*b^2) + tan(c/2 + (d*x)/2)^3*((3*a*b^4)/2 - 24*a^4*b + (8*b^5)/3 + (32*a^2*b^3)/3 - 4*a^
3*b^2) + tan(c/2 + (d*x)/2)^7*((8*b^5)/3 - 24*a^4*b - (3*a*b^4)/2 + (32*a^2*b^3)/3 + 4*a^3*b^2))/(d*(5*tan(c/2
+ (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2
)^10 + 1)) - (a*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2)*(8*a^4 - 9*b^4 + 8*a^2*b^2))/(4*d)

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sympy [A]  time = 2.39, size = 321, normalized size = 1.75 \[ \begin {cases} a^{5} x + \frac {3 a^{4} b \sin {\left (c + d x \right )}}{d} + a^{3} b^{2} x \sin ^{2}{\left (c + d x \right )} + a^{3} b^{2} x \cos ^{2}{\left (c + d x \right )} + \frac {a^{3} b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {4 a^{2} b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} - \frac {2 a^{2} b^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {9 a b^{4} x \sin ^{4}{\left (c + d x \right )}}{8} - \frac {9 a b^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} - \frac {9 a b^{4} x \cos ^{4}{\left (c + d x \right )}}{8} - \frac {9 a b^{4} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {15 a b^{4} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {8 b^{5} \sin ^{5}{\left (c + d x \right )}}{15 d} - \frac {4 b^{5} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {b^{5} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right )^{3} \left (a^{2} - b^{2} \cos ^{2}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**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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