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

Optimal. Leaf size=131 \[ -\frac {a (5 A+4 C) \sin ^3(c+d x)}{15 d}+\frac {a (5 A+4 C) \sin (c+d x)}{5 d}+\frac {a (4 A+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a x (4 A+3 C)+\frac {a C \sin (c+d x) \cos ^4(c+d x)}{5 d}+\frac {a C \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3034, 3023, 2748, 2635, 8, 2633} \[ -\frac {a (5 A+4 C) \sin ^3(c+d x)}{15 d}+\frac {a (5 A+4 C) \sin (c+d x)}{5 d}+\frac {a (4 A+3 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} a x (4 A+3 C)+\frac {a C \sin (c+d x) \cos ^4(c+d x)}{5 d}+\frac {a C \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.29, size = 86, normalized size = 0.66 \[ \frac {a \left (-160 (A+2 C) \sin ^3(c+d x)+480 (A+C) \sin (c+d x)+15 (4 (4 A+3 C) (c+d x)+8 (A+C) \sin (2 (c+d x))+C \sin (4 (c+d x)))+96 C \sin ^5(c+d x)\right )}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 94, normalized size = 0.72 \[ \frac {15 \, {\left (4 \, A + 3 \, C\right )} a d x + {\left (24 \, C a \cos \left (d x + c\right )^{4} + 30 \, C a \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} a \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right ) + 16 \, {\left (5 \, A + 4 \, C\right )} a\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.41, size = 109, normalized size = 0.83 \[ \frac {1}{8} \, {\left (4 \, A a + 3 \, C a\right )} x + \frac {C a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {C a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (4 \, A a + 5 \, C a\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (A a + C a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (6 \, A a + 5 \, C a\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.25, size = 117, normalized size = 0.89 \[ \frac {\frac {a C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*cos(d*x+c))*(A+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*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*A*(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 = 0.31, size = 113, normalized size = 0.86 \[ -\frac {160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480*(160*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a - 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*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 = 2.17, size = 246, normalized size = 1.88 \[ \frac {\left (A\,a+\frac {3\,C\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {10\,A\,a}{3}+\frac {13\,C\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,a}{3}+\frac {116\,C\,a}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {22\,A\,a}{3}+\frac {19\,C\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A\,a+\frac {13\,C\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,A+3\,C\right )}{4\,\left (A\,a+\frac {3\,C\,a}{4}\right )}\right )\,\left (4\,A+3\,C\right )}{4\,d}-\frac {a\,\left (4\,A+3\,C\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(3*A*a + (13*C*a)/4) + tan(c/2 + (d*x)/2)^9*(A*a + (3*C*a)/4) + tan(c/2 + (d*x)/2)^7*((10*
A*a)/3 + (13*C*a)/6) + tan(c/2 + (d*x)/2)^3*((22*A*a)/3 + (19*C*a)/6) + tan(c/2 + (d*x)/2)^5*((20*A*a)/3 + (11
6*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)) + (a*atan((a*tan(c/2 + (d*x)/2)*(4*A + 3*C))/(4*(A*a + (3*C*a)/4)))*(4*A +
3*C))/(4*d) - (a*(4*A + 3*C)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(4*d)

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sympy [A]  time = 2.27, size = 279, normalized size = 2.13 \[ \begin {cases} \frac {A a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {2 A a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {A a \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {3 C a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {8 C a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 C a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {C a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 C a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\relax (c )}\right ) \left (a \cos {\relax (c )} + a\right ) \cos ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+a*cos(d*x+c))*(A+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 + 2*A*a*sin(c + d*x)**3/(3*d) + A*a*sin(c + d*x)*
cos(c + d*x)**2/d + A*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*(A + C*cos(c)**2)*(a*cos(c) + a)*cos(c)**2, True))

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