3.41 \(\int \frac{\cos (c+d x) (A+C \cos ^2(c+d x))}{a+a \cos (c+d x)} \, dx\)

Optimal. Leaf size=98 \[ -\frac{(A+2 C) \sin (c+d x)}{a d}-\frac{(A+C) \sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}+\frac{(2 A+3 C) \sin (c+d x) \cos (c+d x)}{2 a d}+\frac{x (2 A+3 C)}{2 a} \]

[Out]

((2*A + 3*C)*x)/(2*a) - ((A + 2*C)*Sin[c + d*x])/(a*d) + ((2*A + 3*C)*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) - ((A
+ C)*Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))

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Rubi [A]  time = 0.0935063, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {3042, 2734} \[ -\frac{(A+2 C) \sin (c+d x)}{a d}-\frac{(A+C) \sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}+\frac{(2 A+3 C) \sin (c+d x) \cos (c+d x)}{2 a d}+\frac{x (2 A+3 C)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A + 3*C)*x)/(2*a) - ((A + 2*C)*Sin[c + d*x])/(a*d) + ((2*A + 3*C)*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) - ((A
+ C)*Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))

Rule 3042

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x
])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-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 \frac{\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx &=-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac{\int \cos (c+d x) (-a (A+2 C)+a (2 A+3 C) \cos (c+d x)) \, dx}{a^2}\\ &=\frac{(2 A+3 C) x}{2 a}-\frac{(A+2 C) \sin (c+d x)}{a d}+\frac{(2 A+3 C) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{(A+C) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.332249, size = 159, normalized size = 1.62 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (4 d x (2 A+3 C) \cos \left (c+\frac{d x}{2}\right )+4 d x (2 A+3 C) \cos \left (\frac{d x}{2}\right )-16 A \sin \left (\frac{d x}{2}\right )-4 C \sin \left (c+\frac{d x}{2}\right )-3 C \sin \left (c+\frac{3 d x}{2}\right )-3 C \sin \left (2 c+\frac{3 d x}{2}\right )+C \sin \left (2 c+\frac{5 d x}{2}\right )+C \sin \left (3 c+\frac{5 d x}{2}\right )-20 C \sin \left (\frac{d x}{2}\right )\right )}{8 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(4*(2*A + 3*C)*d*x*Cos[(d*x)/2] + 4*(2*A + 3*C)*d*x*Cos[c + (d*x)/2] - 16*A*Sin[(d*
x)/2] - 20*C*Sin[(d*x)/2] - 4*C*Sin[c + (d*x)/2] - 3*C*Sin[c + (3*d*x)/2] - 3*C*Sin[2*c + (3*d*x)/2] + C*Sin[2
*c + (5*d*x)/2] + C*Sin[3*c + (5*d*x)/2]))/(8*a*d*(1 + Cos[c + d*x]))

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Maple [A]  time = 0.03, size = 144, normalized size = 1.5 \begin{align*} -{\frac{A}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-3\,{\frac{C \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{ad \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{C}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A}{ad}}+3\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{ad}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/d*A*tan(1/2*d*x+1/2*c)-1/a/d*C*tan(1/2*d*x+1/2*c)-3/a/d/(tan(1/2*d*x+1/2*c)^2+1)^2*C*tan(1/2*d*x+1/2*c)^3
-1/a/d/(tan(1/2*d*x+1/2*c)^2+1)^2*C*tan(1/2*d*x+1/2*c)+2/a/d*arctan(tan(1/2*d*x+1/2*c))*A+3/a/d*arctan(tan(1/2
*d*x+1/2*c))*C

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Maxima [A]  time = 1.52889, size = 248, normalized size = 2.53 \begin{align*} -\frac{C{\left (\frac{\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac{2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac{3 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - A{\left (\frac{2 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.39718, size = 192, normalized size = 1.96 \begin{align*} \frac{{\left (2 \, A + 3 \, C\right )} d x \cos \left (d x + c\right ) +{\left (2 \, A + 3 \, C\right )} d x +{\left (C \cos \left (d x + c\right )^{2} - C \cos \left (d x + c\right ) - 2 \, A - 4 \, C\right )} \sin \left (d x + c\right )}{2 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.78659, size = 665, normalized size = 6.79 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*A*d*x*tan(c/2 + d*x/2)**4/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 4*A*d
*x*tan(c/2 + d*x/2)**2/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 2*A*d*x/(2*a*d*tan(c/
2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 2*A*tan(c/2 + d*x/2)**5/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*
d*tan(c/2 + d*x/2)**2 + 2*a*d) - 4*A*tan(c/2 + d*x/2)**3/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**
2 + 2*a*d) - 2*A*tan(c/2 + d*x/2)/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 3*C*d*x*ta
n(c/2 + d*x/2)**4/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 6*C*d*x*tan(c/2 + d*x/2)**
2/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) + 3*C*d*x/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d
*tan(c/2 + d*x/2)**2 + 2*a*d) - 2*C*tan(c/2 + d*x/2)**5/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2
+ 2*a*d) - 10*C*tan(c/2 + d*x/2)**3/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d) - 4*C*tan
(c/2 + d*x/2)/(2*a*d*tan(c/2 + d*x/2)**4 + 4*a*d*tan(c/2 + d*x/2)**2 + 2*a*d), Ne(d, 0)), (x*(A + C*cos(c)**2)
*cos(c)/(a*cos(c) + a), True))

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Giac [A]  time = 1.18069, size = 130, normalized size = 1.33 \begin{align*} \frac{\frac{{\left (d x + c\right )}{\left (2 \, A + 3 \, C\right )}}{a} - \frac{2 \,{\left (A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a} - \frac{2 \,{\left (3 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2} a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((d*x + c)*(2*A + 3*C)/a - 2*(A*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*c))/a - 2*(3*C*tan(1/2*d*x + 1/
2*c)^3 + C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a))/d