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

Optimal. Leaf size=94 \[ \frac{2 a^2 (3 A+2 B) \sin (c+d x)}{3 d}+\frac{a^2 (3 A+2 B) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{1}{2} a^2 x (3 A+2 B)+\frac{B \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

[Out]

(a^2*(3*A + 2*B)*x)/2 + (2*a^2*(3*A + 2*B)*Sin[c + d*x])/(3*d) + (a^2*(3*A + 2*B)*Cos[c + d*x]*Sin[c + d*x])/(
6*d) + (B*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.0586307, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2751, 2644} \[ \frac{2 a^2 (3 A+2 B) \sin (c+d x)}{3 d}+\frac{a^2 (3 A+2 B) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{1}{2} a^2 x (3 A+2 B)+\frac{B \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(3*A + 2*B)*x)/2 + (2*a^2*(3*A + 2*B)*Sin[c + d*x])/(3*d) + (a^2*(3*A + 2*B)*Cos[c + d*x]*Sin[c + d*x])/(
6*d) + (B*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(3*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 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 (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx &=\frac{B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{3} (3 A+2 B) \int (a+a \cos (c+d x))^2 \, dx\\ &=\frac{1}{2} a^2 (3 A+2 B) x+\frac{2 a^2 (3 A+2 B) \sin (c+d x)}{3 d}+\frac{a^2 (3 A+2 B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.179524, size = 61, normalized size = 0.65 \[ \frac{a^2 (3 (8 A+7 B) \sin (c+d x)+3 (A+2 B) \sin (2 (c+d x))+18 A d x+B \sin (3 (c+d x))+12 B d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(18*A*d*x + 12*B*d*x + 3*(8*A + 7*B)*Sin[c + d*x] + 3*(A + 2*B)*Sin[2*(c + d*x)] + B*Sin[3*(c + d*x)]))/(
12*d)

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Maple [A]  time = 0.046, size = 116, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({\frac{B{a}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{a}^{2}A \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +2\,B{a}^{2} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +2\,{a}^{2}A\sin \left ( dx+c \right ) +B{a}^{2}\sin \left ( dx+c \right ) +{a}^{2}A \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)^2*(A+B*cos(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.00034, size = 149, normalized size = 1.59 \begin{align*} \frac{3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 12 \,{\left (d x + c\right )} A a^{2} - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} + 6 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} + 24 \, A a^{2} \sin \left (d x + c\right ) + 12 \, B a^{2} \sin \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 12*(d*x + c)*A*a^2 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^
2 + 6*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2 + 24*A*a^2*sin(d*x + c) + 12*B*a^2*sin(d*x + c))/d

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Fricas [A]  time = 1.40573, size = 165, normalized size = 1.76 \begin{align*} \frac{3 \,{\left (3 \, A + 2 \, B\right )} a^{2} d x +{\left (2 \, B a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right ) + 2 \,{\left (6 \, A + 5 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.792427, size = 199, normalized size = 2.12 \begin{align*} \begin{cases} \frac{A a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{A a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + A a^{2} x + \frac{A a^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{2 A a^{2} \sin{\left (c + d x \right )}}{d} + B a^{2} x \sin ^{2}{\left (c + d x \right )} + B a^{2} x \cos ^{2}{\left (c + d x \right )} + \frac{2 B a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{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 )}}{d} + \frac{B a^{2} \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 )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19751, size = 115, normalized size = 1.22 \begin{align*} \frac{B a^{2} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{1}{2} \,{\left (3 \, A a^{2} + 2 \, B a^{2}\right )} x + \frac{{\left (A a^{2} + 2 \, B a^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (8 \, A a^{2} + 7 \, 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((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

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