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

Optimal. Leaf size=161 \[ \frac{a \left (a^2 (-C)+12 A b^2+8 b^2 C\right ) \sin (c+d x)}{6 b d}-\frac{\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} x \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right )+\frac{C \sin (c+d x) (a+b \cos (c+d x))^3}{4 b d}-\frac{a C \sin (c+d x) (a+b \cos (c+d x))^2}{12 b d} \]

[Out]

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

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Rubi [A]  time = 0.214536, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3024, 2753, 2734} \[ \frac{a \left (a^2 (-C)+12 A b^2+8 b^2 C\right ) \sin (c+d x)}{6 b d}-\frac{\left (2 a^2 C-3 b^2 (4 A+3 C)\right ) \sin (c+d x) \cos (c+d x)}{24 d}+\frac{1}{8} x \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right )+\frac{C \sin (c+d x) (a+b \cos (c+d x))^3}{4 b d}-\frac{a C \sin (c+d x) (a+b \cos (c+d x))^2}{12 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3024

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (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) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !LtQ[
m, -1]

Rule 2753

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

Rule 2734

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

Rubi steps

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

Mathematica [A]  time = 0.367917, size = 106, normalized size = 0.66 \[ \frac{12 (c+d x) \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right )+24 \left (C \left (a^2+b^2\right )+A b^2\right ) \sin (2 (c+d x))+48 a b (4 A+3 C) \sin (c+d x)+16 a b C \sin (3 (c+d x))+3 b^2 C \sin (4 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(4*a^2*(2*A + C) + b^2*(4*A + 3*C))*(c + d*x) + 48*a*b*(4*A + 3*C)*Sin[c + d*x] + 24*(A*b^2 + (a^2 + b^2)*
C)*Sin[2*(c + d*x)] + 16*a*b*C*Sin[3*(c + d*x)] + 3*b^2*C*Sin[4*(c + d*x)])/(96*d)

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Maple [A]  time = 0.019, size = 140, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({b}^{2}C \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{2\,abC \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+A{b}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +{a}^{2}C \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +2\,aAb\sin \left ( dx+c \right ) +A{a}^{2} \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.979583, size = 176, normalized size = 1.09 \begin{align*} \frac{96 \,{\left (d x + c\right )} A a^{2} + 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} - 64 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a b + 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{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 b^{2} + 192 \, A a b \sin \left (d x + c\right )}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.57037, size = 248, normalized size = 1.54 \begin{align*} \frac{3 \,{\left (4 \,{\left (2 \, A + C\right )} a^{2} +{\left (4 \, A + 3 \, C\right )} b^{2}\right )} d x +{\left (6 \, C b^{2} \cos \left (d x + c\right )^{3} + 16 \, C a b \cos \left (d x + c\right )^{2} + 16 \,{\left (3 \, A + 2 \, C\right )} a b + 3 \,{\left (4 \, C a^{2} +{\left (4 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(4*(2*A + C)*a^2 + (4*A + 3*C)*b^2)*d*x + (6*C*b^2*cos(d*x + c)^3 + 16*C*a*b*cos(d*x + c)^2 + 16*(3*A
+ 2*C)*a*b + 3*(4*C*a^2 + (4*A + 3*C)*b^2)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [A]  time = 1.60649, size = 309, normalized size = 1.92 \begin{align*} \begin{cases} A a^{2} x + \frac{2 A a b \sin{\left (c + d x \right )}}{d} + \frac{A b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{A b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{A b^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{C a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{C a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{C a^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{4 C a b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{2 C a b \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 C b^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 C b^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 C b^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 C b^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{5 C b^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text{for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\left (c \right )}\right ) \left (a + b \cos{\left (c \right )}\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28016, size = 157, normalized size = 0.98 \begin{align*} \frac{C b^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{C a b \sin \left (3 \, d x + 3 \, c\right )}{6 \, d} + \frac{1}{8} \,{\left (8 \, A a^{2} + 4 \, C a^{2} + 4 \, A b^{2} + 3 \, C b^{2}\right )} x + \frac{{\left (C a^{2} + A b^{2} + C b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (4 \, A a b + 3 \, C a b\right )} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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