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

Optimal. Leaf size=71 \[ \frac{2 \left (a^2+b^2\right ) \sin (c+d x)}{3 d}+\frac{\sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac{a b \sin (c+d x) \cos (c+d x)}{3 d}+a b x \]

[Out]

a*b*x + (2*(a^2 + b^2)*Sin[c + d*x])/(3*d) + (a*b*Cos[c + d*x]*Sin[c + d*x])/(3*d) + ((a + b*Cos[c + d*x])^2*S
in[c + d*x])/(3*d)

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Rubi [A]  time = 0.0498053, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2753, 2734} \[ \frac{2 \left (a^2+b^2\right ) \sin (c+d x)}{3 d}+\frac{\sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac{a b \sin (c+d x) \cos (c+d x)}{3 d}+a b x \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*b*x + (2*(a^2 + b^2)*Sin[c + d*x])/(3*d) + (a*b*Cos[c + d*x]*Sin[c + d*x])/(3*d) + ((a + b*Cos[c + d*x])^2*S
in[c + d*x])/(3*d)

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

Mathematica [A]  time = 0.153382, size = 59, normalized size = 0.83 \[ \frac{3 \left (4 a^2+3 b^2\right ) \sin (c+d x)+b (12 a (c+d x)+6 a \sin (2 (c+d x))+b \sin (3 (c+d x)))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(4*a^2 + 3*b^2)*Sin[c + d*x] + b*(12*a*(c + d*x) + 6*a*Sin[2*(c + d*x)] + b*Sin[3*(c + d*x)]))/(12*d)

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Maple [A]  time = 0.033, size = 63, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+2\,ab \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{a}^{2}\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.966371, size = 81, normalized size = 1.14 \begin{align*} \frac{3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b - 2 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} b^{2} + 6 \, a^{2} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.8673, size = 124, normalized size = 1.75 \begin{align*} \frac{3 \, a b d x +{\left (b^{2} \cos \left (d x + c\right )^{2} + 3 \, a b \cos \left (d x + c\right ) + 3 \, a^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.611332, size = 107, normalized size = 1.51 \begin{align*} \begin{cases} \frac{a^{2} \sin{\left (c + d x \right )}}{d} + a b x \sin ^{2}{\left (c + d x \right )} + a b x \cos ^{2}{\left (c + d x \right )} + \frac{a b \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{2 b^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{b^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \cos{\left (c \right )}\right )^{2} \cos{\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)*(a+b*cos(d*x+c))**2,x)

[Out]

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

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Giac [A]  time = 1.39079, size = 81, normalized size = 1.14 \begin{align*} a b x + \frac{b^{2} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{a b \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac{{\left (4 \, a^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*b*x + 1/12*b^2*sin(3*d*x + 3*c)/d + 1/2*a*b*sin(2*d*x + 2*c)/d + 1/4*(4*a^2 + 3*b^2)*sin(d*x + c)/d