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

Optimal. Leaf size=101 \[ \frac{\left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (4 a^2+3 b^2\right )-\frac{2 a b \sin ^3(c+d x)}{3 d}+\frac{2 a b \sin (c+d x)}{d}+\frac{b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[Out]

((4*a^2 + 3*b^2)*x)/8 + (2*a*b*Sin[c + d*x])/d + ((4*a^2 + 3*b^2)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (b^2*Cos[
c + d*x]^3*Sin[c + d*x])/(4*d) - (2*a*b*Sin[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.0920934, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2789, 2633, 3014, 2635, 8} \[ \frac{\left (4 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (4 a^2+3 b^2\right )-\frac{2 a b \sin ^3(c+d x)}{3 d}+\frac{2 a b \sin (c+d x)}{d}+\frac{b^2 \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a^2 + 3*b^2)*x)/8 + (2*a*b*Sin[c + d*x])/d + ((4*a^2 + 3*b^2)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (b^2*Cos[
c + d*x]^3*Sin[c + d*x])/(4*d) - (2*a*b*Sin[c + d*x]^3)/(3*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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[
e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*
x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

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]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+b \cos (c+d x))^2 \, dx &=(2 a b) \int \cos ^3(c+d x) \, dx+\int \cos ^2(c+d x) \left (a^2+b^2 \cos ^2(c+d x)\right ) \, dx\\ &=\frac{b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{1}{4} \left (4 a^2+3 b^2\right ) \int \cos ^2(c+d x) \, dx-\frac{(2 a b) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac{2 a b \sin (c+d x)}{d}+\frac{\left (4 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{2 a b \sin ^3(c+d x)}{3 d}+\frac{1}{8} \left (4 a^2+3 b^2\right ) \int 1 \, dx\\ &=\frac{1}{8} \left (4 a^2+3 b^2\right ) x+\frac{2 a b \sin (c+d x)}{d}+\frac{\left (4 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{b^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{2 a b \sin ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.156146, size = 86, normalized size = 0.85 \[ \frac{24 \left (a^2+b^2\right ) \sin (2 (c+d x))+48 a^2 c+48 a^2 d x-64 a b \sin ^3(c+d x)+192 a b \sin (c+d x)+3 b^2 \sin (4 (c+d x))+36 b^2 c+36 b^2 d x}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(48*a^2*c + 36*b^2*c + 48*a^2*d*x + 36*b^2*d*x + 192*a*b*Sin[c + d*x] - 64*a*b*Sin[c + d*x]^3 + 24*(a^2 + b^2)
*Sin[2*(c + d*x)] + 3*b^2*Sin[4*(c + d*x)])/(96*d)

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Maple [A]  time = 0.034, size = 89, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +{\frac{2\,ab \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{a}^{2} \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))^2,x)

[Out]

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

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Maxima [A]  time = 0.984981, size = 111, normalized size = 1.1 \begin{align*} \frac{24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 64 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b + 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^{2}}{96 \, 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))^2,x, algorithm="maxima")

[Out]

1/96*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2 - 64*(sin(d*x + c)^3 - 3*sin(d*x + c))*a*b + 3*(12*d*x + 12*c +
sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*b^2)/d

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Fricas [A]  time = 1.85724, size = 184, normalized size = 1.82 \begin{align*} \frac{3 \,{\left (4 \, a^{2} + 3 \, b^{2}\right )} d x +{\left (6 \, b^{2} \cos \left (d x + c\right )^{3} + 16 \, a b \cos \left (d x + c\right )^{2} + 32 \, a b + 3 \,{\left (4 \, a^{2} + 3 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, 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))^2,x, algorithm="fricas")

[Out]

1/24*(3*(4*a^2 + 3*b^2)*d*x + (6*b^2*cos(d*x + c)^3 + 16*a*b*cos(d*x + c)^2 + 32*a*b + 3*(4*a^2 + 3*b^2)*cos(d
*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 1.24016, size = 211, normalized size = 2.09 \begin{align*} \begin{cases} \frac{a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{a^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{4 a b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{2 a b \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 b^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 b^{2} \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{\left (c \right )}\right )^{2} \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))**2,x)

[Out]

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

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Giac [A]  time = 1.74242, size = 111, normalized size = 1.1 \begin{align*} \frac{1}{8} \,{\left (4 \, a^{2} + 3 \, b^{2}\right )} x + \frac{b^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{a b \sin \left (3 \, d x + 3 \, c\right )}{6 \, d} + \frac{3 \, a b \sin \left (d x + c\right )}{2 \, d} + \frac{{\left (a^{2} + b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, 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))^2,x, algorithm="giac")

[Out]

1/8*(4*a^2 + 3*b^2)*x + 1/32*b^2*sin(4*d*x + 4*c)/d + 1/6*a*b*sin(3*d*x + 3*c)/d + 3/2*a*b*sin(d*x + c)/d + 1/
4*(a^2 + b^2)*sin(2*d*x + 2*c)/d