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

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

[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)

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Rubi [A]  time = 0.220759, antiderivative size = 180, normalized size of antiderivative = 1., 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 = {2791, 2753, 2734} \[ -\frac{\left (-52 a^2 b^2+3 a^4-16 b^4\right ) \sin (c+d x)}{30 b d}-\frac{\left (3 a^2-16 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))^2}{60 b d}-\frac{a \left (6 a^2-71 b^2\right ) \sin (c+d x) \cos (c+d x)}{120 d}+\frac{1}{8} a x \left (4 a^2+9 b^2\right )+\frac{\sin (c+d x) (a+b \cos (c+d x))^4}{5 b d}-\frac{a \sin (c+d x) (a+b \cos (c+d x))^3}{20 b d} \]

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 2791

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]

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 \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*}

Mathematica [A]  time = 0.305776, size = 130, normalized size = 0.72 \[ \frac{60 b \left (18 a^2+5 b^2\right ) \sin (c+d x)+120 \left (a^3+3 a b^2\right ) \sin (2 (c+d x))+120 a^2 b \sin (3 (c+d x))+240 a^3 c+240 a^3 d x+45 a b^2 \sin (4 (c+d x))+540 a b^2 c+540 a b^2 d x+50 b^3 \sin (3 (c+d x))+6 b^3 \sin (5 (c+d x))}{480 d} \]

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 = 0.034, size = 123, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{3}\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}^{2} \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 ) +{a}^{2}b \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +{a}^{3} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

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)

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

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Maxima [A]  time = 0.952877, size = 161, normalized size = 0.89 \begin{align*} \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{align*}

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 = 1.89415, size = 265, normalized size = 1.47 \begin{align*} \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{align*}

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 = 2.59505, size = 284, normalized size = 1.58 \begin{align*} \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{align*}

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 = 1.86832, size = 167, normalized size = 0.93 \begin{align*} \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{align*}

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