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

Optimal. Leaf size=129 \[ \frac {3 a^3 \sin ^5(c+d x)}{5 d}-\frac {7 a^3 \sin ^3(c+d x)}{3 d}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {a^3 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {23 a^3 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {23 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {23 a^3 x}{16} \]

[Out]

23/16*a^3*x+4*a^3*sin(d*x+c)/d+23/16*a^3*cos(d*x+c)*sin(d*x+c)/d+23/24*a^3*cos(d*x+c)^3*sin(d*x+c)/d+1/6*a^3*c
os(d*x+c)^5*sin(d*x+c)/d-7/3*a^3*sin(d*x+c)^3/d+3/5*a^3*sin(d*x+c)^5/d

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Rubi [A]  time = 0.15, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2757, 2633, 2635, 8} \[ \frac {3 a^3 \sin ^5(c+d x)}{5 d}-\frac {7 a^3 \sin ^3(c+d x)}{3 d}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {a^3 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {23 a^3 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {23 a^3 \sin (c+d x) \cos (c+d x)}{16 d}+\frac {23 a^3 x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(23*a^3*x)/16 + (4*a^3*Sin[c + d*x])/d + (23*a^3*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (23*a^3*Cos[c + d*x]^3*Si
n[c + d*x])/(24*d) + (a^3*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (7*a^3*Sin[c + d*x]^3)/(3*d) + (3*a^3*Sin[c + d
*x]^5)/(5*d)

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]

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 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]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (a+a \cos (c+d x))^3 \, dx &=\int \left (a^3 \cos ^3(c+d x)+3 a^3 \cos ^4(c+d x)+3 a^3 \cos ^5(c+d x)+a^3 \cos ^6(c+d x)\right ) \, dx\\ &=a^3 \int \cos ^3(c+d x) \, dx+a^3 \int \cos ^6(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^4(c+d x) \, dx+\left (3 a^3\right ) \int \cos ^5(c+d x) \, dx\\ &=\frac {3 a^3 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {1}{6} \left (5 a^3\right ) \int \cos ^4(c+d x) \, dx+\frac {1}{4} \left (9 a^3\right ) \int \cos ^2(c+d x) \, dx-\frac {a^3 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac {4 a^3 \sin (c+d x)}{d}+\frac {9 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {23 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {7 a^3 \sin ^3(c+d x)}{3 d}+\frac {3 a^3 \sin ^5(c+d x)}{5 d}+\frac {1}{8} \left (5 a^3\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{8} \left (9 a^3\right ) \int 1 \, dx\\ &=\frac {9 a^3 x}{8}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {23 a^3 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {23 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {7 a^3 \sin ^3(c+d x)}{3 d}+\frac {3 a^3 \sin ^5(c+d x)}{5 d}+\frac {1}{16} \left (5 a^3\right ) \int 1 \, dx\\ &=\frac {23 a^3 x}{16}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {23 a^3 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {23 a^3 \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {a^3 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {7 a^3 \sin ^3(c+d x)}{3 d}+\frac {3 a^3 \sin ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 73, normalized size = 0.57 \[ \frac {a^3 (2520 \sin (c+d x)+945 \sin (2 (c+d x))+380 \sin (3 (c+d x))+135 \sin (4 (c+d x))+36 \sin (5 (c+d x))+5 \sin (6 (c+d x))+1380 d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1380*d*x + 2520*Sin[c + d*x] + 945*Sin[2*(c + d*x)] + 380*Sin[3*(c + d*x)] + 135*Sin[4*(c + d*x)] + 36*S
in[5*(c + d*x)] + 5*Sin[6*(c + d*x)]))/(960*d)

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fricas [A]  time = 0.88, size = 89, normalized size = 0.69 \[ \frac {345 \, a^{3} d x + {\left (40 \, a^{3} \cos \left (d x + c\right )^{5} + 144 \, a^{3} \cos \left (d x + c\right )^{4} + 230 \, a^{3} \cos \left (d x + c\right )^{3} + 272 \, a^{3} \cos \left (d x + c\right )^{2} + 345 \, a^{3} \cos \left (d x + c\right ) + 544 \, a^{3}\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(345*a^3*d*x + (40*a^3*cos(d*x + c)^5 + 144*a^3*cos(d*x + c)^4 + 230*a^3*cos(d*x + c)^3 + 272*a^3*cos(d*
x + c)^2 + 345*a^3*cos(d*x + c) + 544*a^3)*sin(d*x + c))/d

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giac [A]  time = 0.62, size = 106, normalized size = 0.82 \[ \frac {23}{16} \, a^{3} x + \frac {a^{3} \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} + \frac {3 \, a^{3} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {9 \, a^{3} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {19 \, a^{3} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {63 \, a^{3} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {21 \, a^{3} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

23/16*a^3*x + 1/192*a^3*sin(6*d*x + 6*c)/d + 3/80*a^3*sin(5*d*x + 5*c)/d + 9/64*a^3*sin(4*d*x + 4*c)/d + 19/48
*a^3*sin(3*d*x + 3*c)/d + 63/64*a^3*sin(2*d*x + 2*c)/d + 21/8*a^3*sin(d*x + c)/d

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maple [A]  time = 0.06, size = 143, normalized size = 1.11 \[ \frac {a^{3} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {3 a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.97, size = 143, normalized size = 1.11 \[ \frac {192 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} + 90 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(192*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^3 - 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^3 - 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^3 + 90*(12*d*x +
12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^3)/d

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mupad [B]  time = 2.87, size = 121, normalized size = 0.94 \[ \frac {23\,a^3\,x}{16}+\frac {\frac {23\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {391\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\frac {759\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{20}+\frac {969\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {211\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8}+\frac {105\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(23*a^3*x)/16 + ((211*a^3*tan(c/2 + (d*x)/2)^3)/8 + (969*a^3*tan(c/2 + (d*x)/2)^5)/20 + (759*a^3*tan(c/2 + (d*
x)/2)^7)/20 + (391*a^3*tan(c/2 + (d*x)/2)^9)/24 + (23*a^3*tan(c/2 + (d*x)/2)^11)/8 + (105*a^3*tan(c/2 + (d*x)/
2))/8)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^6)

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sympy [A]  time = 3.65, size = 379, normalized size = 2.94 \[ \begin {cases} \frac {5 a^{3} x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {15 a^{3} x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {9 a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {15 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {9 a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {5 a^{3} x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {9 a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {5 a^{3} \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {8 a^{3} \sin ^{5}{\left (c + d x \right )}}{5 d} + \frac {5 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} + \frac {4 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {9 a^{3} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {11 a^{3} \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} + \frac {3 a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {15 a^{3} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + a\right )^{3} \cos ^{3}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+a*cos(d*x+c))**3,x)

[Out]

Piecewise((5*a**3*x*sin(c + d*x)**6/16 + 15*a**3*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 9*a**3*x*sin(c + d*x)*
*4/8 + 15*a**3*x*sin(c + d*x)**2*cos(c + d*x)**4/16 + 9*a**3*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 5*a**3*x*co
s(c + d*x)**6/16 + 9*a**3*x*cos(c + d*x)**4/8 + 5*a**3*sin(c + d*x)**5*cos(c + d*x)/(16*d) + 8*a**3*sin(c + d*
x)**5/(5*d) + 5*a**3*sin(c + d*x)**3*cos(c + d*x)**3/(6*d) + 4*a**3*sin(c + d*x)**3*cos(c + d*x)**2/d + 9*a**3
*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 2*a**3*sin(c + d*x)**3/(3*d) + 11*a**3*sin(c + d*x)*cos(c + d*x)**5/(16*
d) + 3*a**3*sin(c + d*x)*cos(c + d*x)**4/d + 15*a**3*sin(c + d*x)*cos(c + d*x)**3/(8*d) + a**3*sin(c + d*x)*co
s(c + d*x)**2/d, Ne(d, 0)), (x*(a*cos(c) + a)**3*cos(c)**3, True))

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