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

Optimal. Leaf size=178 \[ \frac{\left (5 a^2 (3 A+2 C)+2 b^2 (5 A+4 C)\right ) \sin (c+d x)}{15 d}+\frac{\left (2 a^2 C+b^2 (5 A+4 C)\right ) \sin (c+d x) \cos ^2(c+d x)}{15 d}+\frac{a b (4 A+3 C) \sin (c+d x) \cos (c+d x)}{4 d}+\frac{1}{4} a b x (4 A+3 C)+\frac{a b C \sin (c+d x) \cos ^3(c+d x)}{10 d}+\frac{C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

[Out]

(a*b*(4*A + 3*C)*x)/4 + ((5*a^2*(3*A + 2*C) + 2*b^2*(5*A + 4*C))*Sin[c + d*x])/(15*d) + (a*b*(4*A + 3*C)*Cos[c
+ d*x]*Sin[c + d*x])/(4*d) + ((2*a^2*C + b^2*(5*A + 4*C))*Cos[c + d*x]^2*Sin[c + d*x])/(15*d) + (a*b*C*Cos[c
+ d*x]^3*Sin[c + d*x])/(10*d) + (C*Cos[c + d*x]^2*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(5*d)

________________________________________________________________________________________

Rubi [A]  time = 0.296465, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {3050, 3033, 3023, 2734} \[ \frac{\left (5 a^2 (3 A+2 C)+2 b^2 (5 A+4 C)\right ) \sin (c+d x)}{15 d}+\frac{\left (2 a^2 C+b^2 (5 A+4 C)\right ) \sin (c+d x) \cos ^2(c+d x)}{15 d}+\frac{a b (4 A+3 C) \sin (c+d x) \cos (c+d x)}{4 d}+\frac{1}{4} a b x (4 A+3 C)+\frac{a b C \sin (c+d x) \cos ^3(c+d x)}{10 d}+\frac{C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*(4*A + 3*C)*x)/4 + ((5*a^2*(3*A + 2*C) + 2*b^2*(5*A + 4*C))*Sin[c + d*x])/(15*d) + (a*b*(4*A + 3*C)*Cos[c
+ d*x]*Sin[c + d*x])/(4*d) + ((2*a^2*C + b^2*(5*A + 4*C))*Cos[c + d*x]^2*Sin[c + d*x])/(15*d) + (a*b*C*Cos[c
+ d*x]^3*Sin[c + d*x])/(10*d) + (C*Cos[c + d*x]^2*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(5*d)

Rule 3050

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

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]

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 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]

Rubi steps

\begin{align*} \int \cos (c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac{C \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac{1}{5} \int \cos (c+d x) (a+b \cos (c+d x)) \left (a (5 A+2 C)+b (5 A+4 C) \cos (c+d x)+2 a C \cos ^2(c+d x)\right ) \, dx\\ &=\frac{a b C \cos ^3(c+d x) \sin (c+d x)}{10 d}+\frac{C \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac{1}{20} \int \cos (c+d x) \left (4 a^2 (5 A+2 C)+10 a b (4 A+3 C) \cos (c+d x)+4 \left (2 a^2 C+b^2 (5 A+4 C)\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac{\left (2 a^2 C+b^2 (5 A+4 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{15 d}+\frac{a b C \cos ^3(c+d x) \sin (c+d x)}{10 d}+\frac{C \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac{1}{60} \int \cos (c+d x) \left (4 \left (5 a^2 (3 A+2 C)+2 b^2 (5 A+4 C)\right )+30 a b (4 A+3 C) \cos (c+d x)\right ) \, dx\\ &=\frac{1}{4} a b (4 A+3 C) x+\frac{\left (5 a^2 (3 A+2 C)+2 b^2 (5 A+4 C)\right ) \sin (c+d x)}{15 d}+\frac{a b (4 A+3 C) \cos (c+d x) \sin (c+d x)}{4 d}+\frac{\left (2 a^2 C+b^2 (5 A+4 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{15 d}+\frac{a b C \cos ^3(c+d x) \sin (c+d x)}{10 d}+\frac{C \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.427378, size = 126, normalized size = 0.71 \[ \frac{30 \left (a^2 (8 A+6 C)+b^2 (6 A+5 C)\right ) \sin (c+d x)+5 \left (4 a^2 C+4 A b^2+5 b^2 C\right ) \sin (3 (c+d x))+60 a b (4 A+3 C) (c+d x)+120 a b (A+C) \sin (2 (c+d x))+15 a b C \sin (4 (c+d x))+3 b^2 C \sin (5 (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*a*b*(4*A + 3*C)*(c + d*x) + 30*(b^2*(6*A + 5*C) + a^2*(8*A + 6*C))*Sin[c + d*x] + 120*a*b*(A + C)*Sin[2*(c
+ d*x)] + 5*(4*A*b^2 + 4*a^2*C + 5*b^2*C)*Sin[3*(c + d*x)] + 15*a*b*C*Sin[4*(c + d*x)] + 3*b^2*C*Sin[5*(c + d
*x)])/(240*d)

________________________________________________________________________________________

Maple [A]  time = 0.017, size = 158, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\frac{A{b}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{{b}^{2}C\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+2\,aAb \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +2\,abC \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +A{a}^{2}\sin \left ( dx+c \right ) +{\frac{{a}^{2}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.00523, size = 208, normalized size = 1.17 \begin{align*} -\frac{80 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} - 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A 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 )} C a b + 80 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{2} - 16 \,{\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^{2} - 240 \, A a^{2} \sin \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(80*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2 - 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a*b - 15*(12*d*x +
12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a*b + 80*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b^2 - 16*(3*sin(
d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*b^2 - 240*A*a^2*sin(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 1.57553, size = 300, normalized size = 1.69 \begin{align*} \frac{15 \,{\left (4 \, A + 3 \, C\right )} a b d x +{\left (12 \, C b^{2} \cos \left (d x + c\right )^{4} + 30 \, C a b \cos \left (d x + c\right )^{3} + 15 \,{\left (4 \, A + 3 \, C\right )} a b \cos \left (d x + c\right ) + 20 \,{\left (3 \, A + 2 \, C\right )} a^{2} + 8 \,{\left (5 \, A + 4 \, C\right )} b^{2} + 4 \,{\left (5 \, C a^{2} +{\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(15*(4*A + 3*C)*a*b*d*x + (12*C*b^2*cos(d*x + c)^4 + 30*C*a*b*cos(d*x + c)^3 + 15*(4*A + 3*C)*a*b*cos(d*x
+ c) + 20*(3*A + 2*C)*a^2 + 8*(5*A + 4*C)*b^2 + 4*(5*C*a^2 + (5*A + 4*C)*b^2)*cos(d*x + c)^2)*sin(d*x + c))/d

________________________________________________________________________________________

Sympy [A]  time = 2.98397, size = 350, normalized size = 1.97 \begin{align*} \begin{cases} \frac{A a^{2} \sin{\left (c + d x \right )}}{d} + A a b x \sin ^{2}{\left (c + d x \right )} + A a b x \cos ^{2}{\left (c + d x \right )} + \frac{A a b \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{2 A b^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{A b^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{2 C a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{C a^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 C a b x \sin ^{4}{\left (c + d x \right )}}{4} + \frac{3 C a b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac{3 C a b x \cos ^{4}{\left (c + d x \right )}}{4} + \frac{3 C a b \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{4 d} + \frac{5 C a b \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac{8 C b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 C b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{C b^{2} \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\left (c \right )}\right ) \left (a + b \cos{\left (c \right )}\right )^{2} \cos{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.50016, size = 192, normalized size = 1.08 \begin{align*} \frac{C b^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{C a b \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} + \frac{1}{4} \,{\left (4 \, A a b + 3 \, C a b\right )} x + \frac{{\left (4 \, C a^{2} + 4 \, A b^{2} + 5 \, C b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{{\left (A a b + C a b\right )} \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac{{\left (8 \, A a^{2} + 6 \, C a^{2} + 6 \, A b^{2} + 5 \, C b^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*C*b^2*sin(5*d*x + 5*c)/d + 1/16*C*a*b*sin(4*d*x + 4*c)/d + 1/4*(4*A*a*b + 3*C*a*b)*x + 1/48*(4*C*a^2 + 4*
A*b^2 + 5*C*b^2)*sin(3*d*x + 3*c)/d + 1/2*(A*a*b + C*a*b)*sin(2*d*x + 2*c)/d + 1/8*(8*A*a^2 + 6*C*a^2 + 6*A*b^
2 + 5*C*b^2)*sin(d*x + c)/d