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

Optimal. Leaf size=80 \[ \frac {\sin (c+d x) (a B+A b+b C)}{d}+\frac {1}{2} x (a (2 A+C)+b B)+\frac {(a C+b B) \sin (c+d x) \cos (c+d x)}{2 d}-\frac {b C \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 113, normalized size of antiderivative = 1.41, number of steps used = 2, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {3023, 2734} \[ \frac {\sin (c+d x) \left (a (3 b B-a C)+b^2 (3 A+2 C)\right )}{3 b d}+\frac {1}{2} x (a (2 A+C)+b B)+\frac {(3 b B-a C) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {C \sin (c+d x) (a+b \cos (c+d x))^2}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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+b \cos (c+d x)) \left (A+B \cos (c+d x)+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)+(3 b B-a C) \cos (c+d x)) \, dx}{3 b}\\ &=\frac {1}{2} (b B+a (2 A+C)) x+\frac {\left (b^2 (3 A+2 C)+a (3 b B-a C)\right ) \sin (c+d x)}{3 b d}+\frac {(3 b B-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*}

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Mathematica [A]  time = 0.20, size = 85, normalized size = 1.06 \[ \frac {3 \sin (c+d x) (4 a B+4 A b+3 b C)+12 a A d x+3 (a C+b B) \sin (2 (c+d x))+6 a c C+6 a C d x+6 b B c+6 b B d x+b C \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 70, normalized size = 0.88 \[ \frac {3 \, {\left ({\left (2 \, A + C\right )} a + B b\right )} d x + {\left (2 \, C b \cos \left (d x + c\right )^{2} + 6 \, B a + 2 \, {\left (3 \, A + 2 \, C\right )} b + 3 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 76, normalized size = 0.95 \[ \frac {1}{2} \, {\left (2 \, A a + C a + B b\right )} x + \frac {C b \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (C a + B b\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, B a + 4 \, A b + 3 \, C b\right )} \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.19, size = 102, normalized size = 1.28 \[ \frac {\frac {C b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A b \sin \left (d x +c \right )+a B \sin \left (d x +c \right )+a A \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 98, normalized size = 1.22 \[ \frac {12 \, {\left (d x + c\right )} A a + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b + 12 \, B a \sin \left (d x + c\right ) + 12 \, A b \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 1.79, size = 100, normalized size = 1.25 \[ A\,a\,x+\frac {B\,b\,x}{2}+\frac {C\,a\,x}{2}+\frac {A\,b\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a\,\sin \left (c+d\,x\right )}{d}+\frac {3\,C\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,b\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.60, size = 189, normalized size = 2.36 \[ \begin {cases} A a x + \frac {A b \sin {\left (c + d x \right )}}{d} + \frac {B a \sin {\left (c + d x \right )}}{d} + \frac {B b x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {B b x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {B b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {C a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {C a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {C a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 C b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right ) \left (A + B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((A*a*x + A*b*sin(c + d*x)/d + B*a*sin(c + d*x)/d + B*b*x*sin(c + d*x)**2/2 + B*b*x*cos(c + d*x)**2/2
+ B*b*sin(c + d*x)*cos(c + d*x)/(2*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)*
cos(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 + b*co
s(c))*(A + B*cos(c) + C*cos(c)**2), True))

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