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

Optimal. Leaf size=103 \[ \frac{a^2 \sin ^5(c+d x)}{5 d}-\frac{a^2 \sin ^3(c+d x)}{d}+\frac{2 a^2 \sin (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac{3 a^2 \sin (c+d x) \cos (c+d x)}{4 d}+\frac{3 a^2 x}{4} \]

[Out]

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

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Rubi [A]  time = 0.104287, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2757, 2633, 2635, 8} \[ \frac{a^2 \sin ^5(c+d x)}{5 d}-\frac{a^2 \sin ^3(c+d x)}{d}+\frac{2 a^2 \sin (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{2 d}+\frac{3 a^2 \sin (c+d x) \cos (c+d x)}{4 d}+\frac{3 a^2 x}{4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

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

Mathematica [A]  time = 0.118239, size = 61, normalized size = 0.59 \[ \frac{a^2 (110 \sin (c+d x)+40 \sin (2 (c+d x))+15 \sin (3 (c+d x))+5 \sin (4 (c+d x))+\sin (5 (c+d x))+60 d x)}{80 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(60*d*x + 110*Sin[c + d*x] + 40*Sin[2*(c + d*x)] + 15*Sin[3*(c + d*x)] + 5*Sin[4*(c + d*x)] + Sin[5*(c +
d*x)]))/(80*d)

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Maple [A]  time = 0.043, size = 96, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{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\,{a}^{2} \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+cos(d*x+c)*a)^2,x)

[Out]

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

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Maxima [A]  time = 1.13985, size = 128, normalized size = 1.24 \begin{align*} \frac{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 )} a^{2} - 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^{2}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.72845, size = 186, normalized size = 1.81 \begin{align*} \frac{15 \, a^{2} d x +{\left (4 \, a^{2} \cos \left (d x + c\right )^{4} + 10 \, a^{2} \cos \left (d x + c\right )^{3} + 12 \, a^{2} \cos \left (d x + c\right )^{2} + 15 \, a^{2} \cos \left (d x + c\right ) + 24 \, a^{2}\right )} \sin \left (d x + c\right )}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(15*a^2*d*x + (4*a^2*cos(d*x + c)^4 + 10*a^2*cos(d*x + c)^3 + 12*a^2*cos(d*x + c)^2 + 15*a^2*cos(d*x + c)
+ 24*a^2)*sin(d*x + c))/d

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Sympy [A]  time = 2.35888, size = 221, normalized size = 2.15 \begin{align*} \begin{cases} \frac{3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac{3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac{3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac{8 a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{3 a^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{4 d} + \frac{2 a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac{5 a^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac{a^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + a\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+a*cos(d*x+c))**2,x)

[Out]

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

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Giac [A]  time = 1.32697, size = 120, normalized size = 1.17 \begin{align*} \frac{3}{4} \, a^{2} x + \frac{a^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{a^{2} \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} + \frac{3 \, a^{2} \sin \left (3 \, d x + 3 \, c\right )}{16 \, d} + \frac{a^{2} \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac{11 \, a^{2} \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+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

3/4*a^2*x + 1/80*a^2*sin(5*d*x + 5*c)/d + 1/16*a^2*sin(4*d*x + 4*c)/d + 3/16*a^2*sin(3*d*x + 3*c)/d + 1/2*a^2*
sin(2*d*x + 2*c)/d + 11/8*a^2*sin(d*x + c)/d