3.13 \(\int \cos ^4(c+d x) (a+a \cos (c+d x))^2 \, dx\)

Optimal. Leaf size=129 \[ \frac{2 a^2 \sin ^5(c+d x)}{5 d}-\frac{4 a^2 \sin ^3(c+d x)}{3 d}+\frac{2 a^2 \sin (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{11 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{11 a^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{11 a^2 x}{16} \]

[Out]

(11*a^2*x)/16 + (2*a^2*Sin[c + d*x])/d + (11*a^2*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (11*a^2*Cos[c + d*x]^3*Si
n[c + d*x])/(24*d) + (a^2*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (4*a^2*Sin[c + d*x]^3)/(3*d) + (2*a^2*Sin[c + d
*x]^5)/(5*d)

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Rubi [A]  time = 0.129558, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2757, 2635, 8, 2633} \[ \frac{2 a^2 \sin ^5(c+d x)}{5 d}-\frac{4 a^2 \sin ^3(c+d x)}{3 d}+\frac{2 a^2 \sin (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac{11 a^2 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{11 a^2 \sin (c+d x) \cos (c+d x)}{16 d}+\frac{11 a^2 x}{16} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Cos[c + d*x])^2,x]

[Out]

(11*a^2*x)/16 + (2*a^2*Sin[c + d*x])/d + (11*a^2*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (11*a^2*Cos[c + d*x]^3*Si
n[c + d*x])/(24*d) + (a^2*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (4*a^2*Sin[c + d*x]^3)/(3*d) + (2*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 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 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]

Rubi steps

\begin{align*} \int \cos ^4(c+d x) (a+a \cos (c+d x))^2 \, dx &=\int \left (a^2 \cos ^4(c+d x)+2 a^2 \cos ^5(c+d x)+a^2 \cos ^6(c+d x)\right ) \, dx\\ &=a^2 \int \cos ^4(c+d x) \, dx+a^2 \int \cos ^6(c+d x) \, dx+\left (2 a^2\right ) \int \cos ^5(c+d x) \, dx\\ &=\frac{a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac{1}{4} \left (3 a^2\right ) \int \cos ^2(c+d x) \, dx+\frac{1}{6} \left (5 a^2\right ) \int \cos ^4(c+d x) \, dx-\frac{\left (2 a^2\right ) \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)}{8 d}+\frac{11 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{4 a^2 \sin ^3(c+d x)}{3 d}+\frac{2 a^2 \sin ^5(c+d x)}{5 d}+\frac{1}{8} \left (3 a^2\right ) \int 1 \, dx+\frac{1}{8} \left (5 a^2\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{3 a^2 x}{8}+\frac{2 a^2 \sin (c+d x)}{d}+\frac{11 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac{11 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{4 a^2 \sin ^3(c+d x)}{3 d}+\frac{2 a^2 \sin ^5(c+d x)}{5 d}+\frac{1}{16} \left (5 a^2\right ) \int 1 \, dx\\ &=\frac{11 a^2 x}{16}+\frac{2 a^2 \sin (c+d x)}{d}+\frac{11 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac{11 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac{4 a^2 \sin ^3(c+d x)}{3 d}+\frac{2 a^2 \sin ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.182994, size = 73, normalized size = 0.57 \[ \frac{a^2 (1200 \sin (c+d x)+465 \sin (2 (c+d x))+200 \sin (3 (c+d x))+75 \sin (4 (c+d x))+24 \sin (5 (c+d x))+5 \sin (6 (c+d x))+660 d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Cos[c + d*x])^2,x]

[Out]

(a^2*(660*d*x + 1200*Sin[c + d*x] + 465*Sin[2*(c + d*x)] + 200*Sin[3*(c + d*x)] + 75*Sin[4*(c + d*x)] + 24*Sin
[5*(c + d*x)] + 5*Sin[6*(c + d*x)]))/(960*d)

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Maple [A]  time = 0.046, size = 121, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +{\frac{2\,{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 ) }+{a}^{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 ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+cos(d*x+c)*a)^2,x)

[Out]

1/d*(a^2*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)+2/5*a^2*(8/3+cos(d*x
+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+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))

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Maxima [A]  time = 1.14151, size = 163, normalized size = 1.26 \begin{align*} \frac{128 \,{\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} - 5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} + 30 \,{\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}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

1/960*(128*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^2 - 5*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 6
0*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a^2 + 30*(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 = 1.61933, size = 228, normalized size = 1.77 \begin{align*} \frac{165 \, a^{2} d x +{\left (40 \, a^{2} \cos \left (d x + c\right )^{5} + 96 \, a^{2} \cos \left (d x + c\right )^{4} + 110 \, a^{2} \cos \left (d x + c\right )^{3} + 128 \, a^{2} \cos \left (d x + c\right )^{2} + 165 \, a^{2} \cos \left (d x + c\right ) + 256 \, a^{2}\right )} \sin \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

1/240*(165*a^2*d*x + (40*a^2*cos(d*x + c)^5 + 96*a^2*cos(d*x + c)^4 + 110*a^2*cos(d*x + c)^3 + 128*a^2*cos(d*x
+ c)^2 + 165*a^2*cos(d*x + c) + 256*a^2)*sin(d*x + c))/d

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Sympy [A]  time = 4.57973, size = 343, normalized size = 2.66 \begin{align*} \begin{cases} \frac{5 a^{2} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac{15 a^{2} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac{3 a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{15 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac{3 a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{5 a^{2} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac{3 a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{5 a^{2} \sin ^{5}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{16 d} + \frac{16 a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{5 a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac{8 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 )}}{8 d} + \frac{11 a^{2} \sin{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac{2 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 )}}{8 d} & \text{for}\: d \neq 0 \\x \left (a \cos{\left (c \right )} + a\right )^{2} \cos ^{4}{\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)**4*(a+a*cos(d*x+c))**2,x)

[Out]

Piecewise((5*a**2*x*sin(c + d*x)**6/16 + 15*a**2*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*a**2*x*sin(c + d*x)*
*4/8 + 15*a**2*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 3*a**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 5*a**2*x*co
s(c + d*x)**6/16 + 3*a**2*x*cos(c + d*x)**4/8 + 5*a**2*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 16*a**2*sin(c + d
*x)**5/(15*d) + 5*a**2*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + 8*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)/(8*d) + 11*a**2*sin(c + d*x)*cos(c + d*x)**5/(16*d) + 2*a**2*sin(c + d*x)*
cos(c + d*x)**4/d + 5*a**2*sin(c + d*x)*cos(c + d*x)**3/(8*d), Ne(d, 0)), (x*(a*cos(c) + a)**2*cos(c)**4, True
))

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Giac [A]  time = 1.3619, size = 143, normalized size = 1.11 \begin{align*} \frac{11}{16} \, a^{2} x + \frac{a^{2} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac{a^{2} \sin \left (5 \, d x + 5 \, c\right )}{40 \, d} + \frac{5 \, a^{2} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac{5 \, a^{2} \sin \left (3 \, d x + 3 \, c\right )}{24 \, d} + \frac{31 \, a^{2} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac{5 \, a^{2} \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)^4*(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

11/16*a^2*x + 1/192*a^2*sin(6*d*x + 6*c)/d + 1/40*a^2*sin(5*d*x + 5*c)/d + 5/64*a^2*sin(4*d*x + 4*c)/d + 5/24*
a^2*sin(3*d*x + 3*c)/d + 31/64*a^2*sin(2*d*x + 2*c)/d + 5/4*a^2*sin(d*x + c)/d