3.21 \(\int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=116 \[ -\frac{a^3 (4 A+3 B) \sin ^3(c+d x)}{12 d}+\frac{a^3 (4 A+3 B) \sin (c+d x)}{d}+\frac{3 a^3 (4 A+3 B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{5}{8} a^3 x (4 A+3 B)+\frac{B \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]

[Out]

(5*a^3*(4*A + 3*B)*x)/8 + (a^3*(4*A + 3*B)*Sin[c + d*x])/d + (3*a^3*(4*A + 3*B)*Cos[c + d*x]*Sin[c + d*x])/(8*
d) + (B*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(4*d) - (a^3*(4*A + 3*B)*Sin[c + d*x]^3)/(12*d)

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Rubi [A]  time = 0.0984662, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2751, 2645, 2637, 2635, 8, 2633} \[ -\frac{a^3 (4 A+3 B) \sin ^3(c+d x)}{12 d}+\frac{a^3 (4 A+3 B) \sin (c+d x)}{d}+\frac{3 a^3 (4 A+3 B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{5}{8} a^3 x (4 A+3 B)+\frac{B \sin (c+d x) (a \cos (c+d x)+a)^3}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^3*(4*A + 3*B)*x)/8 + (a^3*(4*A + 3*B)*Sin[c + d*x])/d + (3*a^3*(4*A + 3*B)*Cos[c + d*x]*Sin[c + d*x])/(8*
d) + (B*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(4*d) - (a^3*(4*A + 3*B)*Sin[c + d*x]^3)/(12*d)

Rule 2751

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[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2645

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(a + b*sin[c + d*x])^n, x], x] /;
FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx &=\frac{B (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} (4 A+3 B) \int (a+a \cos (c+d x))^3 \, dx\\ &=\frac{B (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} (4 A+3 B) \int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx\\ &=\frac{1}{4} a^3 (4 A+3 B) x+\frac{B (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} \left (a^3 (4 A+3 B)\right ) \int \cos ^3(c+d x) \, dx+\frac{1}{4} \left (3 a^3 (4 A+3 B)\right ) \int \cos (c+d x) \, dx+\frac{1}{4} \left (3 a^3 (4 A+3 B)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{1}{4} a^3 (4 A+3 B) x+\frac{3 a^3 (4 A+3 B) \sin (c+d x)}{4 d}+\frac{3 a^3 (4 A+3 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{B (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{8} \left (3 a^3 (4 A+3 B)\right ) \int 1 \, dx-\frac{\left (a^3 (4 A+3 B)\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{4 d}\\ &=\frac{5}{8} a^3 (4 A+3 B) x+\frac{a^3 (4 A+3 B) \sin (c+d x)}{d}+\frac{3 a^3 (4 A+3 B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{B (a+a \cos (c+d x))^3 \sin (c+d x)}{4 d}-\frac{a^3 (4 A+3 B) \sin ^3(c+d x)}{12 d}\\ \end{align*}

Mathematica [A]  time = 0.303321, size = 86, normalized size = 0.74 \[ \frac{a^3 (24 (15 A+13 B) \sin (c+d x)+24 (3 A+4 B) \sin (2 (c+d x))+8 A \sin (3 (c+d x))+240 A d x+24 B \sin (3 (c+d x))+3 B \sin (4 (c+d x))+180 B d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(240*A*d*x + 180*B*d*x + 24*(15*A + 13*B)*Sin[c + d*x] + 24*(3*A + 4*B)*Sin[2*(c + d*x)] + 8*A*Sin[3*(c +
d*x)] + 24*B*Sin[3*(c + d*x)] + 3*B*Sin[4*(c + d*x)]))/(96*d)

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Maple [A]  time = 0.047, size = 176, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ({a}^{3}B \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{A{a}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{a}^{3}B \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +3\,A{a}^{3} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,{a}^{3}B \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,A{a}^{3}\sin \left ( dx+c \right ) +{a}^{3}B\sin \left ( dx+c \right ) +A{a}^{3} \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*B*(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*a^3*(2+cos(d*x+c)^2)*sin(d*x+c)+
a^3*B*(2+cos(d*x+c)^2)*sin(d*x+c)+3*A*a^3*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+3*a^3*B*(1/2*cos(d*x+c)*si
n(d*x+c)+1/2*d*x+1/2*c)+3*A*a^3*sin(d*x+c)+a^3*B*sin(d*x+c)+A*a^3*(d*x+c))

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Maxima [A]  time = 1.02594, size = 225, normalized size = 1.94 \begin{align*} -\frac{32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 72 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 96 \,{\left (d x + c\right )} A a^{3} + 96 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 3 \,{\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 a^{3} - 72 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 288 \, A a^{3} \sin \left (d x + c\right ) - 96 \, B a^{3} \sin \left (d x + c\right )}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3 - 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 - 96*(d*x + c)*A
*a^3 + 96*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*
B*a^3 - 72*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 - 288*A*a^3*sin(d*x + c) - 96*B*a^3*sin(d*x + c))/d

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Fricas [A]  time = 1.33383, size = 216, normalized size = 1.86 \begin{align*} \frac{15 \,{\left (4 \, A + 3 \, B\right )} a^{3} d x +{\left (6 \, B a^{3} \cos \left (d x + c\right )^{3} + 8 \,{\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 9 \,{\left (4 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right ) + 8 \,{\left (11 \, A + 9 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(15*(4*A + 3*B)*a^3*d*x + (6*B*a^3*cos(d*x + c)^3 + 8*(A + 3*B)*a^3*cos(d*x + c)^2 + 9*(4*A + 5*B)*a^3*co
s(d*x + c) + 8*(11*A + 9*B)*a^3)*sin(d*x + c))/d

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Sympy [A]  time = 1.6103, size = 371, normalized size = 3.2 \begin{align*} \begin{cases} \frac{3 A a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{3 A a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + A a^{3} x + \frac{2 A a^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{A a^{3} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 A a^{3} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{3 A a^{3} \sin{\left (c + d x \right )}}{d} + \frac{3 B a^{3} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 B a^{3} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 B a^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{3 B a^{3} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 B a^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{3 B a^{3} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{2 B a^{3} \sin ^{3}{\left (c + d x \right )}}{d} + \frac{5 B a^{3} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{3 B a^{3} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 B a^{3} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{B a^{3} \sin{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (A + B \cos{\left (c \right )}\right ) \left (a \cos{\left (c \right )} + a\right )^{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1839, size = 151, normalized size = 1.3 \begin{align*} \frac{B a^{3} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{5}{8} \,{\left (4 \, A a^{3} + 3 \, B a^{3}\right )} x + \frac{{\left (A a^{3} + 3 \, B a^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{{\left (3 \, A a^{3} + 4 \, B a^{3}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (15 \, A a^{3} + 13 \, B a^{3}\right )} \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*B*a^3*sin(4*d*x + 4*c)/d + 5/8*(4*A*a^3 + 3*B*a^3)*x + 1/12*(A*a^3 + 3*B*a^3)*sin(3*d*x + 3*c)/d + 1/4*(3
*A*a^3 + 4*B*a^3)*sin(2*d*x + 2*c)/d + 1/4*(15*A*a^3 + 13*B*a^3)*sin(d*x + c)/d