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

Optimal. Leaf size=111 \[ -\frac{\left (a^2+2 b^2\right ) \sin ^3(c+d x)}{3 d}+\frac{\left (a^2+b^2\right ) \sin (c+d x)}{d}+\frac{a b \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac{3 a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac{3 a b x}{4}+\frac{b^2 \sin ^5(c+d x)}{5 d} \]

[Out]

(3*a*b*x)/4 + ((a^2 + b^2)*Sin[c + d*x])/d + (3*a*b*Cos[c + d*x]*Sin[c + d*x])/(4*d) + (a*b*Cos[c + d*x]^3*Sin
[c + d*x])/(2*d) - ((a^2 + 2*b^2)*Sin[c + d*x]^3)/(3*d) + (b^2*Sin[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.108267, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2789, 2635, 8, 3013, 373} \[ -\frac{\left (a^2+2 b^2\right ) \sin ^3(c+d x)}{3 d}+\frac{\left (a^2+b^2\right ) \sin (c+d x)}{d}+\frac{a b \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac{3 a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac{3 a b x}{4}+\frac{b^2 \sin ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*b*x)/4 + ((a^2 + b^2)*Sin[c + d*x])/d + (3*a*b*Cos[c + d*x]*Sin[c + d*x])/(4*d) + (a*b*Cos[c + d*x]^3*Sin
[c + d*x])/(2*d) - ((a^2 + 2*b^2)*Sin[c + d*x]^3)/(3*d) + (b^2*Sin[c + d*x]^5)/(5*d)

Rule 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+b \cos (c+d x))^2 \, dx &=(2 a b) \int \cos ^4(c+d x) \, dx+\int \cos ^3(c+d x) \left (a^2+b^2 \cos ^2(c+d x)\right ) \, dx\\ &=\frac{a b \cos ^3(c+d x) \sin (c+d x)}{2 d}+\frac{1}{2} (3 a b) \int \cos ^2(c+d x) \, dx-\frac{\operatorname{Subst}\left (\int \left (1-x^2\right ) \left (a^2+b^2-b^2 x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac{3 a b \cos (c+d x) \sin (c+d x)}{4 d}+\frac{a b \cos ^3(c+d x) \sin (c+d x)}{2 d}+\frac{1}{4} (3 a b) \int 1 \, dx-\frac{\operatorname{Subst}\left (\int \left (a^2 \left (1+\frac{b^2}{a^2}\right )-\left (a^2+2 b^2\right ) x^2+b^2 x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac{3 a b x}{4}+\frac{\left (a^2+b^2\right ) \sin (c+d x)}{d}+\frac{3 a b \cos (c+d x) \sin (c+d x)}{4 d}+\frac{a b \cos ^3(c+d x) \sin (c+d x)}{2 d}-\frac{\left (a^2+2 b^2\right ) \sin ^3(c+d x)}{3 d}+\frac{b^2 \sin ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.134224, size = 85, normalized size = 0.77 \[ \frac{-80 \left (a^2+2 b^2\right ) \sin ^3(c+d x)+240 \left (a^2+b^2\right ) \sin (c+d x)+15 a b (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))+48 b^2 \sin ^5(c+d x)}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(240*(a^2 + b^2)*Sin[c + d*x] - 80*(a^2 + 2*b^2)*Sin[c + d*x]^3 + 48*b^2*Sin[c + d*x]^5 + 15*a*b*(12*(c + d*x)
+ 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(240*d)

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Maple [A]  time = 0.034, size = 95, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{2}\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 ) }+2\,ab \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 ) +{\frac{{a}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.995854, size = 127, normalized size = 1.14 \begin{align*} -\frac{80 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} - 15 \,{\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 - 16 \,{\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^{2}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(80*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*
a*b - 16*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*b^2)/d

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Fricas [A]  time = 1.90697, size = 215, normalized size = 1.94 \begin{align*} \frac{45 \, a b d x +{\left (12 \, b^{2} \cos \left (d x + c\right )^{4} + 30 \, a b \cos \left (d x + c\right )^{3} + 45 \, a b \cos \left (d x + c\right ) + 4 \,{\left (5 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 40 \, a^{2} + 32 \, b^{2}\right )} \sin \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(45*a*b*d*x + (12*b^2*cos(d*x + c)^4 + 30*a*b*cos(d*x + c)^3 + 45*a*b*cos(d*x + c) + 4*(5*a^2 + 4*b^2)*co
s(d*x + c)^2 + 40*a^2 + 32*b^2)*sin(d*x + c))/d

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Sympy [A]  time = 2.35245, size = 221, normalized size = 1.99 \begin{align*} \begin{cases} \frac{2 a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 a b x \sin ^{4}{\left (c + d x \right )}}{4} + \frac{3 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac{3 a b x \cos ^{4}{\left (c + d x \right )}}{4} + \frac{3 a b \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{4 d} + \frac{5 a b \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac{8 b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{b^{2} \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 )^{2} \cos ^{3}{\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)**3*(a+b*cos(d*x+c))**2,x)

[Out]

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

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Giac [A]  time = 1.34549, size = 138, normalized size = 1.24 \begin{align*} \frac{3}{4} \, a b x + \frac{b^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{a b \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} + \frac{a b \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac{{\left (4 \, a^{2} + 5 \, b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{{\left (6 \, a^{2} + 5 \, b^{2}\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)^3*(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

3/4*a*b*x + 1/80*b^2*sin(5*d*x + 5*c)/d + 1/16*a*b*sin(4*d*x + 4*c)/d + 1/2*a*b*sin(2*d*x + 2*c)/d + 1/48*(4*a
^2 + 5*b^2)*sin(3*d*x + 3*c)/d + 1/8*(6*a^2 + 5*b^2)*sin(d*x + c)/d