3.12 \(\int \cos (c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=129 \[ \frac{a^2 (8 A+7 B) \sin (c+d x)}{6 d}+\frac{a^2 (8 A+7 B) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} a^2 x (8 A+7 B)+\frac{(4 A-B) \sin (c+d x) (a \cos (c+d x)+a)^2}{12 d}+\frac{B \sin (c+d x) (a \cos (c+d x)+a)^3}{4 a d} \]

[Out]

(a^2*(8*A + 7*B)*x)/8 + (a^2*(8*A + 7*B)*Sin[c + d*x])/(6*d) + (a^2*(8*A + 7*B)*Cos[c + d*x]*Sin[c + d*x])/(24
*d) + ((4*A - B)*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(12*d) + (B*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(4*a*d)

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Rubi [A]  time = 0.176065, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2968, 3023, 2751, 2644} \[ \frac{a^2 (8 A+7 B) \sin (c+d x)}{6 d}+\frac{a^2 (8 A+7 B) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} a^2 x (8 A+7 B)+\frac{(4 A-B) \sin (c+d x) (a \cos (c+d x)+a)^2}{12 d}+\frac{B \sin (c+d x) (a \cos (c+d x)+a)^3}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(8*A + 7*B)*x)/8 + (a^2*(8*A + 7*B)*Sin[c + d*x])/(6*d) + (a^2*(8*A + 7*B)*Cos[c + d*x]*Sin[c + d*x])/(24
*d) + ((4*A - B)*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(12*d) + (B*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(4*a*d)

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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 2644

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[((2*a^2 + b^2)*x)/2, x] + (-Simp[(2*a*b*Cos[c
+ d*x])/d, x] - Simp[(b^2*Cos[c + d*x]*Sin[c + d*x])/(2*d), x]) /; FreeQ[{a, b, c, d}, x]

Rubi steps

\begin{align*} \int \cos (c+d x) (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx &=\int (a+a \cos (c+d x))^2 \left (A \cos (c+d x)+B \cos ^2(c+d x)\right ) \, dx\\ &=\frac{B (a+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac{\int (a+a \cos (c+d x))^2 (3 a B+a (4 A-B) \cos (c+d x)) \, dx}{4 a}\\ &=\frac{(4 A-B) (a+a \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac{B (a+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}+\frac{1}{12} (8 A+7 B) \int (a+a \cos (c+d x))^2 \, dx\\ &=\frac{1}{8} a^2 (8 A+7 B) x+\frac{a^2 (8 A+7 B) \sin (c+d x)}{6 d}+\frac{a^2 (8 A+7 B) \cos (c+d x) \sin (c+d x)}{24 d}+\frac{(4 A-B) (a+a \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac{B (a+a \cos (c+d x))^3 \sin (c+d x)}{4 a d}\\ \end{align*}

Mathematica [A]  time = 0.331178, size = 86, normalized size = 0.67 \[ \frac{a^2 (24 (7 A+6 B) \sin (c+d x)+48 (A+B) \sin (2 (c+d x))+8 A \sin (3 (c+d x))+96 A d x+16 B \sin (3 (c+d x))+3 B \sin (4 (c+d x))+84 B c+84 B d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(84*B*c + 96*A*d*x + 84*B*d*x + 24*(7*A + 6*B)*Sin[c + d*x] + 48*(A + B)*Sin[2*(c + d*x)] + 8*A*Sin[3*(c
+ d*x)] + 16*B*Sin[3*(c + d*x)] + 3*B*Sin[4*(c + d*x)]))/(96*d)

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Maple [A]  time = 0.051, size = 154, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{2}A \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+B{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 ) +2\,{a}^{2}A \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{\frac{2\,B{a}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{a}^{2}A\sin \left ( dx+c \right ) +B{a}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.990614, size = 194, normalized size = 1.5 \begin{align*} -\frac{32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 48 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 64 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} - 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^{2} - 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 96 \, A a^{2} \sin \left (d x + c\right )}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+a*cos(d*x+c))^2*(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^2 - 48*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 64*(sin(d*x +
c)^3 - 3*sin(d*x + c))*B*a^2 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^2 - 24*(2*d*x + 2
*c + sin(2*d*x + 2*c))*B*a^2 - 96*A*a^2*sin(d*x + c))/d

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Fricas [A]  time = 1.43896, size = 213, normalized size = 1.65 \begin{align*} \frac{3 \,{\left (8 \, A + 7 \, B\right )} a^{2} d x +{\left (6 \, B a^{2} \cos \left (d x + c\right )^{3} + 8 \,{\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (8 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \,{\left (5 \, A + 4 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(8*A + 7*B)*a^2*d*x + (6*B*a^2*cos(d*x + c)^3 + 8*(A + 2*B)*a^2*cos(d*x + c)^2 + 3*(8*A + 7*B)*a^2*cos
(d*x + c) + 8*(5*A + 4*B)*a^2)*sin(d*x + c))/d

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Sympy [A]  time = 2.19487, size = 338, normalized size = 2.62 \begin{align*} \begin{cases} A a^{2} x \sin ^{2}{\left (c + d x \right )} + A a^{2} x \cos ^{2}{\left (c + d x \right )} + \frac{2 A a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{A a^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{A a^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{A a^{2} \sin{\left (c + d x \right )}}{d} + \frac{3 B a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 B a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{B a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{3 B a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{B a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{3 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{4 B a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{5 B a^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{2 B a^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{B a^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (A + B \cos{\left (c \right )}\right ) \left (a \cos{\left (c \right )} + a\right )^{2} \cos{\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)*(a+a*cos(d*x+c))**2*(A+B*cos(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.18655, size = 149, normalized size = 1.16 \begin{align*} \frac{B a^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{1}{8} \,{\left (8 \, A a^{2} + 7 \, B a^{2}\right )} x + \frac{{\left (A a^{2} + 2 \, B a^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{{\left (A a^{2} + B a^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac{{\left (7 \, A a^{2} + 6 \, B a^{2}\right )} \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)*(a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

1/32*B*a^2*sin(4*d*x + 4*c)/d + 1/8*(8*A*a^2 + 7*B*a^2)*x + 1/12*(A*a^2 + 2*B*a^2)*sin(3*d*x + 3*c)/d + 1/2*(A
*a^2 + B*a^2)*sin(2*d*x + 2*c)/d + 1/4*(7*A*a^2 + 6*B*a^2)*sin(d*x + c)/d