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

Optimal. Leaf size=163 \[ \frac{a^2 (4 A+3 C) \sin (c+d x)}{3 d}+\frac{a^2 (4 A+3 C) \sin (c+d x) \cos (c+d x)}{12 d}+\frac{1}{4} a^2 x (4 A+3 C)+\frac{(10 A+3 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{30 d}+\frac{C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^2}{5 d}+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^3}{10 a d} \]

[Out]

(a^2*(4*A + 3*C)*x)/4 + (a^2*(4*A + 3*C)*Sin[c + d*x])/(3*d) + (a^2*(4*A + 3*C)*Cos[c + d*x]*Sin[c + d*x])/(12
*d) + ((10*A + 3*C)*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(30*d) + (C*Cos[c + d*x]^2*(a + a*Cos[c + d*x])^2*Sin
[c + d*x])/(5*d) + (C*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(10*a*d)

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Rubi [A]  time = 0.290271, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {3046, 2968, 3023, 2751, 2644} \[ \frac{a^2 (4 A+3 C) \sin (c+d x)}{3 d}+\frac{a^2 (4 A+3 C) \sin (c+d x) \cos (c+d x)}{12 d}+\frac{1}{4} a^2 x (4 A+3 C)+\frac{(10 A+3 C) \sin (c+d x) (a \cos (c+d x)+a)^2}{30 d}+\frac{C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^2}{5 d}+\frac{C \sin (c+d x) (a \cos (c+d x)+a)^3}{10 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(4*A + 3*C)*x)/4 + (a^2*(4*A + 3*C)*Sin[c + d*x])/(3*d) + (a^2*(4*A + 3*C)*Cos[c + d*x]*Sin[c + d*x])/(12
*d) + ((10*A + 3*C)*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(30*d) + (C*Cos[c + d*x]^2*(a + a*Cos[c + d*x])^2*Sin
[c + d*x])/(5*d) + (C*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(10*a*d)

Rule 3046

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/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Simp
[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b
, c, d, e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-
1)] && NeQ[m + n + 2, 0]

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 2751

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

Rule 2644

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[((2*a^2 + b^2)*x)/2, x] + (-Simp[(2*a*b*Cos[c
+ d*x])/d, x] - Simp[(b^2*Cos[c + d*x]*Sin[c + d*x])/(2*d), x]) /; FreeQ[{a, b, c, d}, x]

Rubi steps

\begin{align*} \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac{C \cos ^2(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac{\int \cos (c+d x) (a+a \cos (c+d x))^2 (a (5 A+2 C)+2 a C \cos (c+d x)) \, dx}{5 a}\\ &=\frac{C \cos ^2(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac{\int (a+a \cos (c+d x))^2 \left (a (5 A+2 C) \cos (c+d x)+2 a C \cos ^2(c+d x)\right ) \, dx}{5 a}\\ &=\frac{C \cos ^2(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{10 a d}+\frac{\int (a+a \cos (c+d x))^2 \left (6 a^2 C+2 a^2 (10 A+3 C) \cos (c+d x)\right ) \, dx}{20 a^2}\\ &=\frac{(10 A+3 C) (a+a \cos (c+d x))^2 \sin (c+d x)}{30 d}+\frac{C \cos ^2(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{10 a d}+\frac{1}{6} (4 A+3 C) \int (a+a \cos (c+d x))^2 \, dx\\ &=\frac{1}{4} a^2 (4 A+3 C) x+\frac{a^2 (4 A+3 C) \sin (c+d x)}{3 d}+\frac{a^2 (4 A+3 C) \cos (c+d x) \sin (c+d x)}{12 d}+\frac{(10 A+3 C) (a+a \cos (c+d x))^2 \sin (c+d x)}{30 d}+\frac{C \cos ^2(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac{C (a+a \cos (c+d x))^3 \sin (c+d x)}{10 a d}\\ \end{align*}

Mathematica [A]  time = 0.374725, size = 97, normalized size = 0.6 \[ \frac{a^2 (30 (14 A+11 C) \sin (c+d x)+120 (A+C) \sin (2 (c+d x))+20 A \sin (3 (c+d x))+240 A d x+45 C \sin (3 (c+d x))+15 C \sin (4 (c+d x))+3 C \sin (5 (c+d x))+120 c C+180 C d x)}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(120*c*C + 240*A*d*x + 180*C*d*x + 30*(14*A + 11*C)*Sin[c + d*x] + 120*(A + C)*Sin[2*(c + d*x)] + 20*A*Si
n[3*(c + d*x)] + 45*C*Sin[3*(c + d*x)] + 15*C*Sin[4*(c + d*x)] + 3*C*Sin[5*(c + d*x)]))/(240*d)

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Maple [A]  time = 0.025, size = 160, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{A{a}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{{a}^{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\,A{a}^{2} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +2\,{a}^{2}C \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+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 1.10604, size = 211, normalized size = 1.29 \begin{align*} -\frac{80 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{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 a^{2} + 80 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} - 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^{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+a*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))*A*a^2 - 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 - 16*(3*sin(d*
x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*a^2 + 80*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2 - 15*(12*d*
x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^2 - 240*A*a^2*sin(d*x + c))/d

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Fricas [A]  time = 1.39308, size = 258, normalized size = 1.58 \begin{align*} \frac{15 \,{\left (4 \, A + 3 \, C\right )} a^{2} d x +{\left (12 \, C a^{2} \cos \left (d x + c\right )^{4} + 30 \, C a^{2} \cos \left (d x + c\right )^{3} + 4 \,{\left (5 \, A + 9 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \,{\left (4 \, A + 3 \, C\right )} a^{2} \cos \left (d x + c\right ) + 4 \,{\left (25 \, A + 18 \, C\right )} a^{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+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

1/60*(15*(4*A + 3*C)*a^2*d*x + (12*C*a^2*cos(d*x + c)^4 + 30*C*a^2*cos(d*x + c)^3 + 4*(5*A + 9*C)*a^2*cos(d*x
+ c)^2 + 15*(4*A + 3*C)*a^2*cos(d*x + c) + 4*(25*A + 18*C)*a^2)*sin(d*x + c))/d

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Sympy [A]  time = 3.09497, size = 350, normalized size = 2.15 \begin{align*} \begin{cases} A a^{2} x \sin ^{2}{\left (c + d x \right )} + A a^{2} x \cos ^{2}{\left (c + d x \right )} + \frac{2 A a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{A a^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{A a^{2} \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{d} + \frac{A a^{2} \sin{\left (c + d x \right )}}{d} + \frac{3 C a^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac{3 C a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac{3 C a^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac{8 C a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 C a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{3 C a^{2} \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{4 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 ^{4}{\left (c + d x \right )}}{d} + \frac{5 C a^{2} \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac{C a^{2} \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\left (c \right )}\right ) \left (a \cos{\left (c \right )} + a\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+a*cos(d*x+c))**2*(A+C*cos(d*x+c)**2),x)

[Out]

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

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Giac [A]  time = 1.17389, size = 174, normalized size = 1.07 \begin{align*} \frac{C a^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{C a^{2} \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} + \frac{1}{4} \,{\left (4 \, A a^{2} + 3 \, C a^{2}\right )} x + \frac{{\left (4 \, A a^{2} + 9 \, C a^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{{\left (A a^{2} + C a^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac{{\left (14 \, A a^{2} + 11 \, C a^{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+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

1/80*C*a^2*sin(5*d*x + 5*c)/d + 1/16*C*a^2*sin(4*d*x + 4*c)/d + 1/4*(4*A*a^2 + 3*C*a^2)*x + 1/48*(4*A*a^2 + 9*
C*a^2)*sin(3*d*x + 3*c)/d + 1/2*(A*a^2 + C*a^2)*sin(2*d*x + 2*c)/d + 1/8*(14*A*a^2 + 11*C*a^2)*sin(d*x + c)/d