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

Optimal. Leaf size=121 \[ \frac{a \left (a^2+4 b^2\right ) \sin (c+d x)}{2 d}+\frac{b \left (2 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3}{8} b x \left (4 a^2+b^2\right )+\frac{\sin (c+d x) (a+b \cos (c+d x))^3}{4 d}+\frac{a \sin (c+d x) (a+b \cos (c+d x))^2}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.116425, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2753, 2734} \[ \frac{a \left (a^2+4 b^2\right ) \sin (c+d x)}{2 d}+\frac{b \left (2 a^2+3 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3}{8} b x \left (4 a^2+b^2\right )+\frac{\sin (c+d x) (a+b \cos (c+d x))^3}{4 d}+\frac{a \sin (c+d x) (a+b \cos (c+d x))^2}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(4*a^2 + b^2)*x)/8 + (a*(a^2 + 4*b^2)*Sin[c + d*x])/(2*d) + (b*(2*a^2 + 3*b^2)*Cos[c + d*x]*Sin[c + d*x])
/(8*d) + (a*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(4*d) + ((a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*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))^3 \, dx &=\frac{(a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} \int (3 b+3 a \cos (c+d x)) (a+b \cos (c+d x))^2 \, dx\\ &=\frac{a (a+b \cos (c+d x))^2 \sin (c+d x)}{4 d}+\frac{(a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{12} \int (a+b \cos (c+d x)) \left (15 a b+3 \left (2 a^2+3 b^2\right ) \cos (c+d x)\right ) \, dx\\ &=\frac{3}{8} b \left (4 a^2+b^2\right ) x+\frac{a \left (a^2+4 b^2\right ) \sin (c+d x)}{2 d}+\frac{b \left (2 a^2+3 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a (a+b \cos (c+d x))^2 \sin (c+d x)}{4 d}+\frac{(a+b \cos (c+d x))^3 \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.263678, size = 100, normalized size = 0.83 \[ \frac{8 a \left (4 a^2+9 b^2\right ) \sin (c+d x)+b \left (8 \left (3 a^2+b^2\right ) \sin (2 (c+d x))+48 a^2 c+48 a^2 d x+8 a b \sin (3 (c+d x))+b^2 \sin (4 (c+d x))+12 b^2 c+12 b^2 d x\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a*(4*a^2 + 9*b^2)*Sin[c + d*x] + b*(48*a^2*c + 12*b^2*c + 48*a^2*d*x + 12*b^2*d*x + 8*(3*a^2 + b^2)*Sin[2*(
c + d*x)] + 8*a*b*Sin[3*(c + d*x)] + b^2*Sin[4*(c + d*x)]))/(32*d)

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Maple [A]  time = 0.033, size = 102, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({b}^{3} \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 ) +a{b}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +3\,{a}^{2}b \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{a}^{3}\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))^3,x)

[Out]

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

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Maxima [A]  time = 0.965021, size = 128, normalized size = 1.06 \begin{align*} \frac{24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} b - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b^{2} +{\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^{3} + 32 \, a^{3} \sin \left (d x + c\right )}{32 \, 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))^3,x, algorithm="maxima")

[Out]

1/32*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2*b - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*a*b^2 + (12*d*x + 12*c
+ sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*b^3 + 32*a^3*sin(d*x + c))/d

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Fricas [A]  time = 1.89638, size = 197, normalized size = 1.63 \begin{align*} \frac{3 \,{\left (4 \, a^{2} b + b^{3}\right )} d x +{\left (2 \, b^{3} \cos \left (d x + c\right )^{3} + 8 \, a b^{2} \cos \left (d x + c\right )^{2} + 8 \, a^{3} + 16 \, a b^{2} + 3 \,{\left (4 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )\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)*(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/8*(3*(4*a^2*b + b^3)*d*x + (2*b^3*cos(d*x + c)^3 + 8*a*b^2*cos(d*x + c)^2 + 8*a^3 + 16*a*b^2 + 3*(4*a^2*b +
b^3)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 1.35326, size = 233, normalized size = 1.93 \begin{align*} \begin{cases} \frac{a^{3} \sin{\left (c + d x \right )}}{d} + \frac{3 a^{2} b x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{3 a^{2} b x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{3 a^{2} b \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{2 a b^{2} \sin ^{3}{\left (c + d x \right )}}{d} + \frac{3 a b^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 b^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 b^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 b^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 b^{3} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 b^{3} \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 )^{3} \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))**3,x)

[Out]

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

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Giac [A]  time = 2.18567, size = 130, normalized size = 1.07 \begin{align*} \frac{b^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{a b^{2} \sin \left (3 \, d x + 3 \, c\right )}{4 \, d} + \frac{3}{8} \,{\left (4 \, a^{2} b + b^{3}\right )} x + \frac{{\left (3 \, a^{2} b + b^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (4 \, a^{3} + 9 \, a 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))^3,x, algorithm="giac")

[Out]

1/32*b^3*sin(4*d*x + 4*c)/d + 1/4*a*b^2*sin(3*d*x + 3*c)/d + 3/8*(4*a^2*b + b^3)*x + 1/4*(3*a^2*b + b^3)*sin(2
*d*x + 2*c)/d + 1/4*(4*a^3 + 9*a*b^2)*sin(d*x + c)/d