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

Optimal. Leaf size=156 \[ -\frac {\sin ^3(c+d x) (5 a B+5 A b+4 b C)}{15 d}+\frac {\sin (c+d x) (5 a B+5 A b+4 b C)}{5 d}+\frac {\sin (c+d x) \cos (c+d x) (4 a A+3 a C+3 b B)}{8 d}+\frac {1}{8} x (4 a A+3 a C+3 b B)+\frac {(a C+b B) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {b C \sin (c+d x) \cos ^4(c+d x)}{5 d} \]

[Out]

1/8*(4*A*a+3*B*b+3*C*a)*x+1/5*(5*A*b+5*B*a+4*C*b)*sin(d*x+c)/d+1/8*(4*A*a+3*B*b+3*C*a)*cos(d*x+c)*sin(d*x+c)/d
+1/4*(B*b+C*a)*cos(d*x+c)^3*sin(d*x+c)/d+1/5*b*C*cos(d*x+c)^4*sin(d*x+c)/d-1/15*(5*A*b+5*B*a+4*C*b)*sin(d*x+c)
^3/d

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Rubi [A]  time = 0.23, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3033, 3023, 2748, 2635, 8, 2633} \[ -\frac {\sin ^3(c+d x) (5 a B+5 A b+4 b C)}{15 d}+\frac {\sin (c+d x) (5 a B+5 A b+4 b C)}{5 d}+\frac {\sin (c+d x) \cos (c+d x) (4 a A+3 a C+3 b B)}{8 d}+\frac {1}{8} x (4 a A+3 a C+3 b B)+\frac {(a C+b B) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {b C \sin (c+d x) \cos ^4(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a*A + 3*b*B + 3*a*C)*x)/8 + ((5*A*b + 5*a*B + 4*b*C)*Sin[c + d*x])/(5*d) + ((4*a*A + 3*b*B + 3*a*C)*Cos[c
+ d*x]*Sin[c + d*x])/(8*d) + ((b*B + a*C)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (b*C*Cos[c + d*x]^4*Sin[c + d*x
])/(5*d) - ((5*A*b + 5*a*B + 4*b*C)*Sin[c + d*x]^3)/(15*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 3033

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

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\frac {b C \cos ^4(c+d x) \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos ^2(c+d x) \left (5 a A+(5 A b+5 a B+4 b C) \cos (c+d x)+5 (b B+a C) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {(b B+a C) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {b C \cos ^4(c+d x) \sin (c+d x)}{5 d}+\frac {1}{20} \int \cos ^2(c+d x) (5 (4 a A+3 b B+3 a C)+4 (5 A b+5 a B+4 b C) \cos (c+d x)) \, dx\\ &=\frac {(b B+a C) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {b C \cos ^4(c+d x) \sin (c+d x)}{5 d}+\frac {1}{4} (4 a A+3 b B+3 a C) \int \cos ^2(c+d x) \, dx+\frac {1}{5} (5 A b+5 a B+4 b C) \int \cos ^3(c+d x) \, dx\\ &=\frac {(4 a A+3 b B+3 a C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {(b B+a C) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {b C \cos ^4(c+d x) \sin (c+d x)}{5 d}+\frac {1}{8} (4 a A+3 b B+3 a C) \int 1 \, dx-\frac {(5 A b+5 a B+4 b C) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {1}{8} (4 a A+3 b B+3 a C) x+\frac {(5 A b+5 a B+4 b C) \sin (c+d x)}{5 d}+\frac {(4 a A+3 b B+3 a C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {(b B+a C) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {b C \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {(5 A b+5 a B+4 b C) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.59, size = 117, normalized size = 0.75 \[ \frac {-160 \sin ^3(c+d x) (a B+A b+2 b C)+480 \sin (c+d x) (a B+A b+b C)+15 (4 (c+d x) (4 a A+3 a C+3 b B)+8 \sin (2 (c+d x)) (a (A+C)+b B)+(a C+b B) \sin (4 (c+d x)))+96 b C \sin ^5(c+d x)}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(480*(A*b + a*B + b*C)*Sin[c + d*x] - 160*(A*b + a*B + 2*b*C)*Sin[c + d*x]^3 + 96*b*C*Sin[c + d*x]^5 + 15*(4*(
4*a*A + 3*b*B + 3*a*C)*(c + d*x) + 8*(b*B + a*(A + C))*Sin[2*(c + d*x)] + (b*B + a*C)*Sin[4*(c + d*x)]))/(480*
d)

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fricas [A]  time = 0.43, size = 121, normalized size = 0.78 \[ \frac {15 \, {\left ({\left (4 \, A + 3 \, C\right )} a + 3 \, B b\right )} d x + {\left (24 \, C b \cos \left (d x + c\right )^{4} + 30 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, B a + {\left (5 \, A + 4 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} + 80 \, B a + 16 \, {\left (5 \, A + 4 \, C\right )} b + 15 \, {\left ({\left (4 \, A + 3 \, C\right )} a + 3 \, B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*((4*A + 3*C)*a + 3*B*b)*d*x + (24*C*b*cos(d*x + c)^4 + 30*(C*a + B*b)*cos(d*x + c)^3 + 8*(5*B*a + (5
*A + 4*C)*b)*cos(d*x + c)^2 + 80*B*a + 16*(5*A + 4*C)*b + 15*((4*A + 3*C)*a + 3*B*b)*cos(d*x + c))*sin(d*x + c
))/d

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giac [A]  time = 0.28, size = 129, normalized size = 0.83 \[ \frac {1}{8} \, {\left (4 \, A a + 3 \, C a + 3 \, B b\right )} x + \frac {C b \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (C a + B b\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (4 \, B a + 4 \, A b + 5 \, C b\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (A a + C a + B b\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (6 \, B a + 6 \, A b + 5 \, C b\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*A*a + 3*C*a + 3*B*b)*x + 1/80*C*b*sin(5*d*x + 5*c)/d + 1/32*(C*a + B*b)*sin(4*d*x + 4*c)/d + 1/48*(4*B*
a + 4*A*b + 5*C*b)*sin(3*d*x + 3*c)/d + 1/4*(A*a + C*a + B*b)*sin(2*d*x + 2*c)/d + 1/8*(6*B*a + 6*A*b + 5*C*b)
*sin(d*x + c)/d

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maple [A]  time = 0.31, size = 173, normalized size = 1.11 \[ \frac {\frac {C b \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+B b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 166, normalized size = 1.06 \[ \frac {120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B b + 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C b}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(120*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a + 15*(12*d*x + 12*
c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b + 15*(12*d*x + 12*c
+ sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*b + 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*
b)/d

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mupad [B]  time = 5.25, size = 259, normalized size = 1.66 \[ \frac {x\,\left (A\,a+\frac {3\,B\,b}{4}+\frac {3\,C\,a}{4}\right )}{2}+\frac {\left (2\,A\,b-A\,a+2\,B\,a-\frac {5\,B\,b}{4}-\frac {5\,C\,a}{4}+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {16\,A\,b}{3}-2\,A\,a+\frac {16\,B\,a}{3}-\frac {B\,b}{2}-\frac {C\,a}{2}+\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,b}{3}+\frac {20\,B\,a}{3}+\frac {116\,C\,b}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (2\,A\,a+\frac {16\,A\,b}{3}+\frac {16\,B\,a}{3}+\frac {B\,b}{2}+\frac {C\,a}{2}+\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A\,a+2\,A\,b+2\,B\,a+\frac {5\,B\,b}{4}+\frac {5\,C\,a}{4}+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*(A*a + (3*B*b)/4 + (3*C*a)/4))/2 + (tan(c/2 + (d*x)/2)^3*(2*A*a + (16*A*b)/3 + (16*B*a)/3 + (B*b)/2 + (C*a)
/2 + (8*C*b)/3) - tan(c/2 + (d*x)/2)^9*(A*a - 2*A*b - 2*B*a + (5*B*b)/4 + (5*C*a)/4 - 2*C*b) - tan(c/2 + (d*x)
/2)^7*(2*A*a - (16*A*b)/3 - (16*B*a)/3 + (B*b)/2 + (C*a)/2 - (8*C*b)/3) + tan(c/2 + (d*x)/2)*(A*a + 2*A*b + 2*
B*a + (5*B*b)/4 + (5*C*a)/4 + 2*C*b) + tan(c/2 + (d*x)/2)^5*((20*A*b)/3 + (20*B*a)/3 + (116*C*b)/15))/(d*(5*ta
n(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d
*x)/2)^10 + 1))

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sympy [A]  time = 2.54, size = 428, normalized size = 2.74 \[ \begin {cases} \frac {A a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 A b \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A b \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {2 B a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 B b x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B b x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 B b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 B b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {3 C a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 C a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {8 C b \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C b \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {C b \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\relax (c )}\right ) \left (A + B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((A*a*x*sin(c + d*x)**2/2 + A*a*x*cos(c + d*x)**2/2 + A*a*sin(c + d*x)*cos(c + d*x)/(2*d) + 2*A*b*sin
(c + d*x)**3/(3*d) + A*b*sin(c + d*x)*cos(c + d*x)**2/d + 2*B*a*sin(c + d*x)**3/(3*d) + B*a*sin(c + d*x)*cos(c
+ d*x)**2/d + 3*B*b*x*sin(c + d*x)**4/8 + 3*B*b*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*B*b*x*cos(c + d*x)**4
/8 + 3*B*b*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 5*B*b*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 3*C*a*x*sin(c + d*x
)**4/8 + 3*C*a*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*C*a*x*cos(c + d*x)**4/8 + 3*C*a*sin(c + d*x)**3*cos(c +
d*x)/(8*d) + 5*C*a*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 8*C*b*sin(c + d*x)**5/(15*d) + 4*C*b*sin(c + d*x)**3*
cos(c + d*x)**2/(3*d) + C*b*sin(c + d*x)*cos(c + d*x)**4/d, Ne(d, 0)), (x*(a + b*cos(c))*(A + B*cos(c) + C*cos
(c)**2)*cos(c)**2, True))

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