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

Optimal. Leaf size=96 \[ -\frac{\left (a^2 C-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b d}+\frac{1}{2} a x (2 A+C)+\frac{C \sin (c+d x) (a+b \cos (c+d x))^2}{3 b d}-\frac{a C \sin (c+d x) \cos (c+d x)}{6 d} \]

[Out]

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

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Rubi [A]  time = 0.0736837, antiderivative size = 96, 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 = {3024, 2734} \[ -\frac{\left (a^2 C-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 b d}+\frac{1}{2} a x (2 A+C)+\frac{C \sin (c+d x) (a+b \cos (c+d x))^2}{3 b d}-\frac{a C \sin (c+d x) \cos (c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.116301, size = 64, normalized size = 0.67 \[ \frac{12 a A d x+3 a C \sin (2 (c+d x))+6 a c C+6 a C d x+3 b (4 A+3 C) \sin (c+d x)+b C \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.016, size = 68, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ({\frac{Cb \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+aC \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +Ab\sin \left ( dx+c \right ) +aA \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))*(A+C*cos(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 0.989892, size = 90, normalized size = 0.94 \begin{align*} \frac{12 \,{\left (d x + c\right )} A a + 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b + 12 \, A b \sin \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.3702, size = 140, normalized size = 1.46 \begin{align*} \frac{3 \,{\left (2 \, A + C\right )} a d x +{\left (2 \, C b \cos \left (d x + c\right )^{2} + 3 \, C a \cos \left (d x + c\right ) + 2 \,{\left (3 \, A + 2 \, C\right )} b\right )} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.664865, size = 121, normalized size = 1.26 \begin{align*} \begin{cases} A a x + \frac{A b \sin{\left (c + d x \right )}}{d} + \frac{C a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{C a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{C a \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{2 C b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{C b \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\left (c \right )}\right ) \left (a + b \cos{\left (c \right )}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.32369, size = 86, normalized size = 0.9 \begin{align*} \frac{1}{2} \,{\left (2 \, A a + C a\right )} x + \frac{C b \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac{C a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (4 \, A b + 3 \, C b\right )} \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*A*a + C*a)*x + 1/12*C*b*sin(3*d*x + 3*c)/d + 1/4*C*a*sin(2*d*x + 2*c)/d + 1/4*(4*A*b + 3*C*b)*sin(d*x +
c)/d