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

Optimal. Leaf size=65 \[ \frac{a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 a x}{8}-\frac{b \cos ^4(c+d x)}{4 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0777632, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3090, 2635, 8, 2565, 30} \[ \frac{a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 a x}{8}-\frac{b \cos ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

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 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0909392, size = 62, normalized size = 0.95 \[ \frac{3 a (c+d x)}{8 d}+\frac{a \sin (2 (c+d x))}{4 d}+\frac{a \sin (4 (c+d x))}{32 d}-\frac{b \cos ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*(c + d*x))/(8*d) - (b*Cos[c + d*x]^4)/(4*d) + (a*Sin[2*(c + d*x)])/(4*d) + (a*Sin[4*(c + d*x)])/(32*d)

________________________________________________________________________________________

Maple [A]  time = 0.04, size = 52, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( a \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{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}b}{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.16457, size = 65, normalized size = 1. \begin{align*} -\frac{8 \, b \cos \left (d x + c\right )^{4} -{\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}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 0.480382, size = 127, normalized size = 1.95 \begin{align*} -\frac{2 \, b \cos \left (d x + c\right )^{4} - 3 \, a d x -{\left (2 \, a \cos \left (d x + c\right )^{3} + 3 \, a \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)^3*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 1.17358, size = 150, normalized size = 2.31 \begin{align*} \begin{cases} \frac{3 a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 a \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 a \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{b \sin ^{4}{\left (c + d x \right )}}{4 d} + \frac{b \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + b \sin{\left (c \right )}\right ) \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*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.11555, size = 88, normalized size = 1.35 \begin{align*} \frac{3}{8} \, a x - \frac{b \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac{b \cos \left (2 \, d x + 2 \, c\right )}{8 \, d} + \frac{a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{a \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)^3*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="giac")

[Out]

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