3.5.29 \(\int \cos ^2(c+d x) (a+b \cos (c+d x))^3 \, dx\) [429]

Optimal. Leaf size=180 \[ \frac {1}{8} a \left (4 a^2+9 b^2\right ) x-\frac {\left (3 a^4-52 a^2 b^2-16 b^4\right ) \sin (c+d x)}{30 b d}-\frac {a \left (6 a^2-71 b^2\right ) \cos (c+d x) \sin (c+d x)}{120 d}-\frac {\left (3 a^2-16 b^2\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{60 b d}-\frac {a (a+b \cos (c+d x))^3 \sin (c+d x)}{20 b d}+\frac {(a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d} \]

[Out]

1/8*a*(4*a^2+9*b^2)*x-1/30*(3*a^4-52*a^2*b^2-16*b^4)*sin(d*x+c)/b/d-1/120*a*(6*a^2-71*b^2)*cos(d*x+c)*sin(d*x+
c)/d-1/60*(3*a^2-16*b^2)*(a+b*cos(d*x+c))^2*sin(d*x+c)/b/d-1/20*a*(a+b*cos(d*x+c))^3*sin(d*x+c)/b/d+1/5*(a+b*c
os(d*x+c))^4*sin(d*x+c)/b/d

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Rubi [A]
time = 0.15, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2870, 2832, 2813} \begin {gather*} -\frac {\left (3 a^2-16 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{60 b d}-\frac {a \left (6 a^2-71 b^2\right ) \sin (c+d x) \cos (c+d x)}{120 d}+\frac {1}{8} a x \left (4 a^2+9 b^2\right )-\frac {\left (3 a^4-52 a^2 b^2-16 b^4\right ) \sin (c+d x)}{30 b d}+\frac {\sin (c+d x) (a+b \cos (c+d x))^4}{5 b d}-\frac {a \sin (c+d x) (a+b \cos (c+d x))^3}{20 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(4*a^2 + 9*b^2)*x)/8 - ((3*a^4 - 52*a^2*b^2 - 16*b^4)*Sin[c + d*x])/(30*b*d) - (a*(6*a^2 - 71*b^2)*Cos[c +
d*x]*Sin[c + d*x])/(120*d) - ((3*a^2 - 16*b^2)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(60*b*d) - (a*(a + b*Cos[c
+ d*x])^3*Sin[c + d*x])/(20*b*d) + ((a + b*Cos[c + d*x])^4*Sin[c + d*x])/(5*b*d)

Rule 2813

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 2832

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)*Sim
p[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 2870

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]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+b \cos (c+d x))^3 \, dx &=\frac {(a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}+\frac {\int (4 b-a \cos (c+d x)) (a+b \cos (c+d x))^3 \, dx}{5 b}\\ &=-\frac {a (a+b \cos (c+d x))^3 \sin (c+d x)}{20 b d}+\frac {(a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}+\frac {\int (a+b \cos (c+d x))^2 \left (13 a b-\left (3 a^2-16 b^2\right ) \cos (c+d x)\right ) \, dx}{20 b}\\ &=-\frac {\left (3 a^2-16 b^2\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{60 b d}-\frac {a (a+b \cos (c+d x))^3 \sin (c+d x)}{20 b d}+\frac {(a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}+\frac {\int (a+b \cos (c+d x)) \left (b \left (33 a^2+32 b^2\right )-a \left (6 a^2-71 b^2\right ) \cos (c+d x)\right ) \, dx}{60 b}\\ &=\frac {1}{8} a \left (4 a^2+9 b^2\right ) x-\frac {\left (3 a^4-52 a^2 b^2-16 b^4\right ) \sin (c+d x)}{30 b d}-\frac {a \left (6 a^2-71 b^2\right ) \cos (c+d x) \sin (c+d x)}{120 d}-\frac {\left (3 a^2-16 b^2\right ) (a+b \cos (c+d x))^2 \sin (c+d x)}{60 b d}-\frac {a (a+b \cos (c+d x))^3 \sin (c+d x)}{20 b d}+\frac {(a+b \cos (c+d x))^4 \sin (c+d x)}{5 b d}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 130, normalized size = 0.72 \begin {gather*} \frac {240 a^3 c+540 a b^2 c+240 a^3 d x+540 a b^2 d x+60 b \left (18 a^2+5 b^2\right ) \sin (c+d x)+120 \left (a^3+3 a b^2\right ) \sin (2 (c+d x))+120 a^2 b \sin (3 (c+d x))+50 b^3 \sin (3 (c+d x))+45 a b^2 \sin (4 (c+d x))+6 b^3 \sin (5 (c+d x))}{480 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(240*a^3*c + 540*a*b^2*c + 240*a^3*d*x + 540*a*b^2*d*x + 60*b*(18*a^2 + 5*b^2)*Sin[c + d*x] + 120*(a^3 + 3*a*b
^2)*Sin[2*(c + d*x)] + 120*a^2*b*Sin[3*(c + d*x)] + 50*b^3*Sin[3*(c + d*x)] + 45*a*b^2*Sin[4*(c + d*x)] + 6*b^
3*Sin[5*(c + d*x)])/(480*d)

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Maple [A]
time = 0.13, size = 123, normalized size = 0.68

method	result	size
derivativedivides	\(\frac {\frac {b^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 b^{2} a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{2} b \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )+a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(123\)
default	\(\frac {\frac {b^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 b^{2} a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{2} b \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )+a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(123\)
risch	\(\frac {a^{3} x}{2}+\frac {9 a \,b^{2} x}{8}+\frac {9 \sin \left (d x +c \right ) a^{2} b}{4 d}+\frac {5 \sin \left (d x +c \right ) b^{3}}{8 d}+\frac {b^{3} \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 b^{2} a \sin \left (4 d x +4 c \right )}{32 d}+\frac {\sin \left (3 d x +3 c \right ) a^{2} b}{4 d}+\frac {5 \sin \left (3 d x +3 c \right ) b^{3}}{48 d}+\frac {a^{3} \sin \left (2 d x +2 c \right )}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) b^{2} a}{4 d}\)	\(149\)
norman	\(\frac {\left (\frac {1}{2} a^{3}+\frac {9}{8} b^{2} a \right ) x +\left (5 a^{3}+\frac {45}{4} b^{2} a \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (5 a^{3}+\frac {45}{4} b^{2} a \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} a^{3}+\frac {9}{8} b^{2} a \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {5}{2} a^{3}+\frac {45}{8} b^{2} a \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {5}{2} a^{3}+\frac {45}{8} b^{2} a \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\left (4 a^{3}-24 a^{2} b +15 b^{2} a -8 b^{3}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (4 a^{3}+24 a^{2} b +15 b^{2} a +8 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (12 a^{3}-96 a^{2} b +9 b^{2} a -16 b^{3}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {\left (12 a^{3}+96 a^{2} b +9 b^{2} a +16 b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {4 b \left (75 a^{2}+29 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(339\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+b*cos(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 119, normalized size = 0.66 \begin {gather*} \frac {120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} b + 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^{2} + 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^{3}}{480 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(120*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3 - 480*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2*b + 45*(12*d*x + 1
2*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a*b^2 + 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c
))*b^3)/d

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Fricas [A]
time = 0.42, size = 110, normalized size = 0.61 \begin {gather*} \frac {15 \, {\left (4 \, a^{3} + 9 \, a b^{2}\right )} d x + {\left (24 \, b^{3} \cos \left (d x + c\right )^{4} + 90 \, a b^{2} \cos \left (d x + c\right )^{3} + 240 \, a^{2} b + 64 \, b^{3} + 8 \, {\left (15 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(4*a^3 + 9*a*b^2)*d*x + (24*b^3*cos(d*x + c)^4 + 90*a*b^2*cos(d*x + c)^3 + 240*a^2*b + 64*b^3 + 8*(1
5*a^2*b + 4*b^3)*cos(d*x + c)^2 + 15*(4*a^3 + 9*a*b^2)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]
time = 0.31, size = 284, normalized size = 1.58 \begin {gather*} \begin {cases} \frac {a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 a^{2} b \sin ^{3}{\left (c + d x \right )}}{d} + \frac {3 a^{2} b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {9 a b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {9 a b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {9 a b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {9 a b^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {15 a b^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {8 b^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 b^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {b^{3} \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} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.40, size = 124, normalized size = 0.69 \begin {gather*} \frac {b^{3} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {3 \, a b^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {1}{8} \, {\left (4 \, a^{3} + 9 \, a b^{2}\right )} x + \frac {{\left (12 \, a^{2} b + 5 \, b^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (18 \, a^{2} b + 5 \, b^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*b^3*sin(5*d*x + 5*c)/d + 3/32*a*b^2*sin(4*d*x + 4*c)/d + 1/8*(4*a^3 + 9*a*b^2)*x + 1/48*(12*a^2*b + 5*b^3
)*sin(3*d*x + 3*c)/d + 1/4*(a^3 + 3*a*b^2)*sin(2*d*x + 2*c)/d + 1/8*(18*a^2*b + 5*b^3)*sin(d*x + c)/d

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Mupad [B]
time = 2.05, size = 319, normalized size = 1.77 \begin {gather*} \frac {\left (-a^3+6\,a^2\,b-\frac {15\,a\,b^2}{4}+2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-2\,a^3+16\,a^2\,b-\frac {3\,a\,b^2}{2}+\frac {8\,b^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (20\,a^2\,b+\frac {116\,b^3}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (2\,a^3+16\,a^2\,b+\frac {3\,a\,b^2}{2}+\frac {8\,b^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (a^3+6\,a^2\,b+\frac {15\,a\,b^2}{4}+2\,b^3\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\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,a^2+9\,b^2\right )}{4\,\left (a^3+\frac {9\,a\,b^2}{4}\right )}\right )\,\left (4\,a^2+9\,b^2\right )}{4\,d}-\frac {a\,\left (4\,a^2+9\,b^2\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^3*((3*a*b^2)/2 + 16*a^2*b + 2*a^3 + (8*b^3)/3) - tan(c/2 + (d*x)/2)^7*((3*a*b^2)/2 - 16*a^
2*b + 2*a^3 - (8*b^3)/3) + tan(c/2 + (d*x)/2)*((15*a*b^2)/4 + 6*a^2*b + a^3 + 2*b^3) + tan(c/2 + (d*x)/2)^5*(2
0*a^2*b + (116*b^3)/15) - tan(c/2 + (d*x)/2)^9*((15*a*b^2)/4 - 6*a^2*b + a^3 - 2*b^3))/(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((a*tan(c/2 + (d*x)/2)*(4*a^2 + 9*b^2))/(4*((9*a*b^2)/4 + a^3)))*(4*a^2 + 9*b^2))/(4*d) - (a*(4*a^2
+ 9*b^2)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(4*d)

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