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

Optimal. Leaf size=90 \[ \frac{(A-B) \sin (c+d x)}{a d}+\frac{(A-B) \sin (c+d x)}{a d (\cos (c+d x)+1)}-\frac{x (A-B)}{a}+\frac{B \sin (c+d x) \cos (c+d x)}{2 a d}+\frac{B x}{2 a} \]

[Out]

-(((A - B)*x)/a) + (B*x)/(2*a) + ((A - B)*Sin[c + d*x])/(a*d) + (B*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) + ((A -
B)*Sin[c + d*x])/(a*d*(1 + Cos[c + d*x]))

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Rubi [A]  time = 0.12299, antiderivative size = 99, normalized size of antiderivative = 1.1, 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 = {2977, 2734} \[ \frac{2 (A-B) \sin (c+d x)}{a d}+\frac{(A-B) \sin (c+d x) \cos ^2(c+d x)}{d (a \cos (c+d x)+a)}-\frac{(2 A-3 B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac{x (2 A-3 B)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2977

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

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 ^2(c+d x) (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx &=\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac{\int \cos (c+d x) (2 a (A-B)-a (2 A-3 B) \cos (c+d x)) \, dx}{a^2}\\ &=-\frac{(2 A-3 B) x}{2 a}+\frac{2 (A-B) \sin (c+d x)}{a d}-\frac{(2 A-3 B) \cos (c+d x) \sin (c+d x)}{2 a d}+\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}\\ \end{align*}

Mathematica [B]  time = 0.437828, size = 197, normalized size = 2.19 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-4 d x (2 A-3 B) \cos \left (c+\frac{d x}{2}\right )-4 d x (2 A-3 B) \cos \left (\frac{d x}{2}\right )+4 A \sin \left (c+\frac{d x}{2}\right )+4 A \sin \left (c+\frac{3 d x}{2}\right )+4 A \sin \left (2 c+\frac{3 d x}{2}\right )+20 A \sin \left (\frac{d x}{2}\right )-4 B \sin \left (c+\frac{d x}{2}\right )-3 B \sin \left (c+\frac{3 d x}{2}\right )-3 B \sin \left (2 c+\frac{3 d x}{2}\right )+B \sin \left (2 c+\frac{5 d x}{2}\right )+B \sin \left (3 c+\frac{5 d x}{2}\right )-20 B \sin \left (\frac{d x}{2}\right )\right )}{8 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.069, size = 211, normalized size = 2.3 \begin{align*}{\frac{A}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{B}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-3\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}B}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}A}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{B}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+2\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) }{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A}{da}}+3\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.50592, size = 304, normalized size = 3.38 \begin{align*} -\frac{B{\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{2 \, \sin \left (d x + c\right )}{{\left (a + \frac{a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} - \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)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c)),x, algorithm="maxima")

[Out]

-(B*((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 - 2*sin(d*x + c)/((a + a*sin(d*
x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d

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Fricas [A]  time = 1.42564, size = 204, normalized size = 2.27 \begin{align*} -\frac{{\left (2 \, A - 3 \, B\right )} d x \cos \left (d x + c\right ) +{\left (2 \, A - 3 \, B\right )} d x -{\left (B \cos \left (d x + c\right )^{2} +{\left (2 \, A - B\right )} \cos \left (d x + c\right ) + 4 \, A - 4 \, B\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)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.28966, size = 665, normalized size = 7.39 \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)**2*(A+B*cos(d*x+c))/(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) + 8*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) + 6*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*B*d*x*t
an(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*B*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*B*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*B*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*B*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*B*ta
n(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 + B*cos(c))*c
os(c)**2/(a*cos(c) + a), True))

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Giac [A]  time = 1.17866, size = 167, normalized size = 1.86 \begin{align*} -\frac{\frac{{\left (d x + c\right )}{\left (2 \, A - 3 \, B\right )}}{a} - \frac{2 \,{\left (A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a} - \frac{2 \,{\left (2 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B \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)^2*(A+B*cos(d*x+c))/(a+a*cos(d*x+c)),x, algorithm="giac")

[Out]

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