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

Optimal. Leaf size=125 \[ -\frac {a (5 B+4 C) \sin ^3(c+d x)}{15 d}+\frac {a (5 B+4 C) \sin (c+d x)}{5 d}+\frac {a (B+C) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 a (B+C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3}{8} a x (B+C)+\frac {a C \sin (c+d x) \cos ^4(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {3029, 2968, 3023, 2748, 2633, 2635, 8} \[ -\frac {a (5 B+4 C) \sin ^3(c+d x)}{15 d}+\frac {a (5 B+4 C) \sin (c+d x)}{5 d}+\frac {a (B+C) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 a (B+C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3}{8} a x (B+C)+\frac {a C \sin (c+d x) \cos ^4(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*(B + C)*x)/8 + (a*(5*B + 4*C)*Sin[c + d*x])/(5*d) + (3*a*(B + C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*(B
+ C)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (a*C*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) - (a*(5*B + 4*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 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, 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 3029

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

Rubi steps

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

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Mathematica [A]  time = 0.34, size = 102, normalized size = 0.82 \[ \frac {a (60 (6 B+5 C) \sin (c+d x)+120 (B+C) \sin (2 (c+d x))+40 B \sin (3 (c+d x))+15 B \sin (4 (c+d x))+180 B d x+50 C \sin (3 (c+d x))+15 C \sin (4 (c+d x))+6 C \sin (5 (c+d x))+180 C d x)}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(180*B*d*x + 180*C*d*x + 60*(6*B + 5*C)*Sin[c + d*x] + 120*(B + C)*Sin[2*(c + d*x)] + 40*B*Sin[3*(c + d*x)]
+ 50*C*Sin[3*(c + d*x)] + 15*B*Sin[4*(c + d*x)] + 15*C*Sin[4*(c + d*x)] + 6*C*Sin[5*(c + d*x)]))/(480*d)

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fricas [A]  time = 0.48, size = 88, normalized size = 0.70 \[ \frac {45 \, {\left (B + C\right )} a d x + {\left (24 \, C a \cos \left (d x + c\right )^{4} + 30 \, {\left (B + C\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, B + 4 \, C\right )} a \cos \left (d x + c\right )^{2} + 45 \, {\left (B + C\right )} a \cos \left (d x + c\right ) + 16 \, {\left (5 \, B + 4 \, C\right )} a\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(45*(B + C)*a*d*x + (24*C*a*cos(d*x + c)^4 + 30*(B + C)*a*cos(d*x + c)^3 + 8*(5*B + 4*C)*a*cos(d*x + c)^
2 + 45*(B + C)*a*cos(d*x + c) + 16*(5*B + 4*C)*a)*sin(d*x + c))/d

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giac [A]  time = 0.37, size = 112, normalized size = 0.90 \[ \frac {3}{8} \, {\left (B a + C a\right )} x + \frac {C a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (B a + C a\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (4 \, B a + 5 \, C a\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (B a + C a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (6 \, B a + 5 \, C a\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.26, size = 128, normalized size = 1.02 \[ \frac {\frac {a C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.62, size = 124, normalized size = 0.99 \[ -\frac {160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.29, size = 236, normalized size = 1.89 \[ \frac {\left (\frac {3\,B\,a}{4}+\frac {3\,C\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {29\,B\,a}{6}+\frac {13\,C\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,B\,a}{3}+\frac {116\,C\,a}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {35\,B\,a}{6}+\frac {19\,C\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,B\,a}{4}+\frac {13\,C\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,a\,\mathrm {atan}\left (\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B+C\right )}{4\,\left (\frac {3\,B\,a}{4}+\frac {3\,C\,a}{4}\right )}\right )\,\left (B+C\right )}{4\,d}-\frac {3\,a\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )\,\left (B+C\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((13*B*a)/4 + (13*C*a)/4) + tan(c/2 + (d*x)/2)^9*((3*B*a)/4 + (3*C*a)/4) + tan(c/2 + (d*x)
/2)^7*((29*B*a)/6 + (13*C*a)/6) + tan(c/2 + (d*x)/2)^3*((35*B*a)/6 + (19*C*a)/6) + tan(c/2 + (d*x)/2)^5*((20*B
*a)/3 + (116*C*a)/15))/(d*(5*tan(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)) + (3*a*atan((3*a*tan(c/2 + (d*x)/2)*(B + C))/(4*((3*B*a)/4 + (3
*C*a)/4)))*(B + C))/(4*d) - (3*a*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2)*(B + C))/(4*d)

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sympy [A]  time = 2.11, size = 338, normalized size = 2.70 \[ \begin {cases} \frac {3 B a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 B a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 B a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 B a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {B a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 C a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {8 C a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 C a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {C a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 C a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right ) \left (a \cos {\relax (c )} + a\right ) \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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