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

Optimal. Leaf size=138 \[ \frac {a^2 (12 A+8 B+7 C) \sin (c+d x)}{6 d}+\frac {a^2 (12 A+8 B+7 C) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} a^2 x (12 A+8 B+7 C)+\frac {(4 B-C) \sin (c+d x) (a \cos (c+d x)+a)^2}{12 d}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 a d} \]

[Out]

1/8*a^2*(12*A+8*B+7*C)*x+1/6*a^2*(12*A+8*B+7*C)*sin(d*x+c)/d+1/24*a^2*(12*A+8*B+7*C)*cos(d*x+c)*sin(d*x+c)/d+1
/12*(4*B-C)*(a+a*cos(d*x+c))^2*sin(d*x+c)/d+1/4*C*(a+a*cos(d*x+c))^3*sin(d*x+c)/a/d

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Rubi [A]  time = 0.17, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3023, 2751, 2644} \[ \frac {a^2 (12 A+8 B+7 C) \sin (c+d x)}{6 d}+\frac {a^2 (12 A+8 B+7 C) \sin (c+d x) \cos (c+d x)}{24 d}+\frac {1}{8} a^2 x (12 A+8 B+7 C)+\frac {(4 B-C) \sin (c+d x) (a \cos (c+d x)+a)^2}{12 d}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.39, size = 94, normalized size = 0.68 \[ \frac {a^2 (24 (8 A+7 B+6 C) \sin (c+d x)+24 (A+2 (B+C)) \sin (2 (c+d x))+144 A d x+8 B \sin (3 (c+d x))+96 B d x+16 C \sin (3 (c+d x))+3 C \sin (4 (c+d x))+84 C d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(144*A*d*x + 96*B*d*x + 84*C*d*x + 24*(8*A + 7*B + 6*C)*Sin[c + d*x] + 24*(A + 2*(B + C))*Sin[2*(c + d*x)
] + 8*B*Sin[3*(c + d*x)] + 16*C*Sin[3*(c + d*x)] + 3*C*Sin[4*(c + d*x)]))/(96*d)

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fricas [A]  time = 0.60, size = 99, normalized size = 0.72 \[ \frac {3 \, {\left (12 \, A + 8 \, B + 7 \, C\right )} a^{2} d x + {\left (6 \, C a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 8 \, B + 7 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, {\left (6 \, A + 5 \, B + 4 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{24 \, d} \]

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)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/24*(3*(12*A + 8*B + 7*C)*a^2*d*x + (6*C*a^2*cos(d*x + c)^3 + 8*(B + 2*C)*a^2*cos(d*x + c)^2 + 3*(4*A + 8*B +
7*C)*a^2*cos(d*x + c) + 8*(6*A + 5*B + 4*C)*a^2)*sin(d*x + c))/d

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giac [A]  time = 0.60, size = 129, normalized size = 0.93 \[ \frac {C a^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {1}{8} \, {\left (12 \, A a^{2} + 8 \, B a^{2} + 7 \, C a^{2}\right )} x + \frac {{\left (B a^{2} + 2 \, C a^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (A a^{2} + 2 \, B a^{2} + 2 \, C a^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (8 \, A a^{2} + 7 \, B a^{2} + 6 \, C a^{2}\right )} \sin \left (d x + c\right )}{4 \, d} \]

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)+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.21, size = 203, normalized size = 1.47 \[ \frac {a^{2} C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {2 a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a^{2} A \sin \left (d x +c \right )+B \,a^{2} \sin \left (d x +c \right )+a^{2} A \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^2*C*(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*B*a^2*(2+cos(d*x+c)^2)*sin(d*x+c)+
2/3*a^2*C*(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)+a^2*C*(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 = 0.61, size = 190, normalized size = 1.38 \[ \frac {24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 96 \, {\left (d x + c\right )} A a^{2} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} + 48 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 192 \, A a^{2} \sin \left (d x + c\right ) + 96 \, B a^{2} \sin \left (d x + c\right )}{96 \, d} \]

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)+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

1/96*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 96*(d*x + c)*A*a^2 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*
a^2 + 48*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2 - 64*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2 + 3*(12*d*x + 12*
c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^2 + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2 + 192*A*a^2*sin(d
*x + c) + 96*B*a^2*sin(d*x + c))/d

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mupad [B]  time = 1.26, size = 174, normalized size = 1.26 \[ \frac {3\,A\,a^2\,x}{2}+B\,a^2\,x+\frac {7\,C\,a^2\,x}{8}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {7\,B\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,C\,a^2\,\sin \left (c+d\,x\right )}{2\,d}+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {B\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {C\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{6\,d}+\frac {C\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.40, size = 420, normalized size = 3.04 \[ \begin {cases} \frac {A a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + A a^{2} x + \frac {A a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 A a^{2} \sin {\left (c + d x \right )}}{d} + B a^{2} x \sin ^{2}{\left (c + d x \right )} + B a^{2} x \cos ^{2}{\left (c + d x \right )} + \frac {2 B a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {B a^{2} \sin {\left (c + d x \right )}}{d} + \frac {3 C a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {C a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 C a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {C a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 C a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {4 C a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 C a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {2 C a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {C a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\relax (c )} + a\right )^{2} \left (A + B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

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

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