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

Optimal. Leaf size=108 \[ \frac {a (3 A+2 C) \sin (c+d x)}{3 d}+\frac {a C \sin (c+d x) \cos ^2(c+d x)}{3 d}+\frac {b (4 A+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} b x (4 A+3 C)+\frac {b C \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[Out]

1/8*b*(4*A+3*C)*x+1/3*a*(3*A+2*C)*sin(d*x+c)/d+1/8*b*(4*A+3*C)*cos(d*x+c)*sin(d*x+c)/d+1/3*a*C*cos(d*x+c)^2*si
n(d*x+c)/d+1/4*b*C*cos(d*x+c)^3*sin(d*x+c)/d

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Rubi [A]  time = 0.11, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3034, 3023, 2734} \[ \frac {a (3 A+2 C) \sin (c+d x)}{3 d}+\frac {a C \sin (c+d x) \cos ^2(c+d x)}{3 d}+\frac {b (4 A+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} b x (4 A+3 C)+\frac {b C \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3034

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m
+ 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*d*(C*(m + 2) + A*(m
+ 3))*Sin[e + f*x] - (2*a*C*d - b*c*C*(m + 3))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 84, normalized size = 0.78 \[ \frac {24 a (4 A+3 C) \sin (c+d x)+8 a C \sin (3 (c+d x))+24 b (A+C) \sin (2 (c+d x))+48 A b c+48 A b d x+3 b C \sin (4 (c+d x))+36 b c C+36 b C d x}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(48*A*b*c + 36*b*c*C + 48*A*b*d*x + 36*b*C*d*x + 24*a*(4*A + 3*C)*Sin[c + d*x] + 24*b*(A + C)*Sin[2*(c + d*x)]
+ 8*a*C*Sin[3*(c + d*x)] + 3*b*C*Sin[4*(c + d*x)])/(96*d)

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fricas [A]  time = 0.80, size = 76, normalized size = 0.70 \[ \frac {3 \, {\left (4 \, A + 3 \, C\right )} b d x + {\left (6 \, C b \cos \left (d x + c\right )^{3} + 8 \, C a \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 3 \, C\right )} b \cos \left (d x + c\right ) + 8 \, {\left (3 \, A + 2 \, C\right )} a\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.37, size = 86, normalized size = 0.80 \[ \frac {1}{8} \, {\left (4 \, A b + 3 \, C b\right )} x + \frac {C b \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {C a \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (A b + C b\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, A a + 3 \, C a\right )} \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*A*b + 3*C*b)*x + 1/32*C*b*sin(4*d*x + 4*c)/d + 1/12*C*a*sin(3*d*x + 3*c)/d + 1/4*(A*b + C*b)*sin(2*d*x
+ 2*c)/d + 1/4*(4*A*a + 3*C*a)*sin(d*x + c)/d

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maple [A]  time = 0.22, size = 96, normalized size = 0.89 \[ \frac {C b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a A \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 90, normalized size = 0.83 \[ -\frac {32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C b - 96 \, A a \sin \left (d x + c\right )}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.44, size = 243, normalized size = 2.25 \[ \frac {\left (2\,A\,a-A\,b+2\,C\,a-\frac {5\,C\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (6\,A\,a-A\,b+\frac {10\,C\,a}{3}+\frac {3\,C\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (6\,A\,a+A\,b+\frac {10\,C\,a}{3}-\frac {3\,C\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+A\,b+2\,C\,a+\frac {5\,C\,b}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,A+3\,C\right )}{4\,\left (A\,b+\frac {3\,C\,b}{4}\right )}\right )\,\left (4\,A+3\,C\right )}{4\,d}-\frac {b\,\left (4\,A+3\,C\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(2*A*a + A*b + 2*C*a + (5*C*b)/4) + tan(c/2 + (d*x)/2)^7*(2*A*a - A*b + 2*C*a - (5*C*b)/4)
+ tan(c/2 + (d*x)/2)^3*(6*A*a + A*b + (10*C*a)/3 - (3*C*b)/4) + tan(c/2 + (d*x)/2)^5*(6*A*a - A*b + (10*C*a)/
3 + (3*C*b)/4))/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)
/2)^8 + 1)) + (b*atan((b*tan(c/2 + (d*x)/2)*(4*A + 3*C))/(4*(A*b + (3*C*b)/4)))*(4*A + 3*C))/(4*d) - (b*(4*A +
3*C)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(4*d)

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sympy [A]  time = 1.08, size = 226, normalized size = 2.09 \[ \begin {cases} \frac {A a \sin {\left (c + d x \right )}}{d} + \frac {A b x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A b x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 C a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 C b x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C b x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 C b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 C b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\relax (c )}\right ) \left (a + b \cos {\relax (c )}\right ) \cos {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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