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

Optimal. Leaf size=147 \[ \frac{2 (5 B-8 C) \sin (c+d x)}{3 a^2 d}+\frac{(5 B-8 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{(4 B-7 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac{x (4 B-7 C)}{2 a^2}+\frac{(B-C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

-((4*B - 7*C)*x)/(2*a^2) + (2*(5*B - 8*C)*Sin[c + d*x])/(3*a^2*d) - ((4*B - 7*C)*Cos[c + d*x]*Sin[c + d*x])/(2
*a^2*d) + ((5*B - 8*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) + ((B - C)*Cos[c + d*x]^3*Sin
[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

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Rubi [A]  time = 0.36259, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.075, Rules used = {3029, 2977, 2734} \[ \frac{2 (5 B-8 C) \sin (c+d x)}{3 a^2 d}+\frac{(5 B-8 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{(4 B-7 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac{x (4 B-7 C)}{2 a^2}+\frac{(B-C) \sin (c+d x) \cos ^3(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*B - 7*C)*x)/(2*a^2) + (2*(5*B - 8*C)*Sin[c + d*x])/(3*a^2*d) - ((4*B - 7*C)*Cos[c + d*x]*Sin[c + d*x])/(2
*a^2*d) + ((5*B - 8*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) + ((B - C)*Cos[c + d*x]^3*Sin
[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

Rule 3029

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

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) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx &=\int \frac{\cos ^3(c+d x) (B+C \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx\\ &=\frac{(B-C) \cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{\cos ^2(c+d x) (3 a (B-C)-a (2 B-5 C) \cos (c+d x))}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=\frac{(5 B-8 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac{(B-C) \cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \cos (c+d x) \left (2 a^2 (5 B-8 C)-3 a^2 (4 B-7 C) \cos (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{(4 B-7 C) x}{2 a^2}+\frac{2 (5 B-8 C) \sin (c+d x)}{3 a^2 d}-\frac{(4 B-7 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{(5 B-8 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}+\frac{(B-C) \cos ^3(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end{align*}

Mathematica [B]  time = 0.860781, size = 315, normalized size = 2.14 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-36 d x (4 B-7 C) \cos \left (c+\frac{d x}{2}\right )-120 B \sin \left (c+\frac{d x}{2}\right )+164 B \sin \left (c+\frac{3 d x}{2}\right )+36 B \sin \left (2 c+\frac{3 d x}{2}\right )+12 B \sin \left (2 c+\frac{5 d x}{2}\right )+12 B \sin \left (3 c+\frac{5 d x}{2}\right )-48 B d x \cos \left (c+\frac{3 d x}{2}\right )-48 B d x \cos \left (2 c+\frac{3 d x}{2}\right )-36 d x (4 B-7 C) \cos \left (\frac{d x}{2}\right )+264 B \sin \left (\frac{d x}{2}\right )+147 C \sin \left (c+\frac{d x}{2}\right )-239 C \sin \left (c+\frac{3 d x}{2}\right )-63 C \sin \left (2 c+\frac{3 d x}{2}\right )-15 C \sin \left (2 c+\frac{5 d x}{2}\right )-15 C \sin \left (3 c+\frac{5 d x}{2}\right )+3 C \sin \left (3 c+\frac{7 d x}{2}\right )+3 C \sin \left (4 c+\frac{7 d x}{2}\right )+84 C d x \cos \left (c+\frac{3 d x}{2}\right )+84 C d x \cos \left (2 c+\frac{3 d x}{2}\right )-381 C \sin \left (\frac{d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-36*(4*B - 7*C)*d*x*Cos[(d*x)/2] - 36*(4*B - 7*C)*d*x*Cos[c + (d*x)/2] - 48*B*d*x*
Cos[c + (3*d*x)/2] + 84*C*d*x*Cos[c + (3*d*x)/2] - 48*B*d*x*Cos[2*c + (3*d*x)/2] + 84*C*d*x*Cos[2*c + (3*d*x)/
2] + 264*B*Sin[(d*x)/2] - 381*C*Sin[(d*x)/2] - 120*B*Sin[c + (d*x)/2] + 147*C*Sin[c + (d*x)/2] + 164*B*Sin[c +
(3*d*x)/2] - 239*C*Sin[c + (3*d*x)/2] + 36*B*Sin[2*c + (3*d*x)/2] - 63*C*Sin[2*c + (3*d*x)/2] + 12*B*Sin[2*c
+ (5*d*x)/2] - 15*C*Sin[2*c + (5*d*x)/2] + 12*B*Sin[3*c + (5*d*x)/2] - 15*C*Sin[3*c + (5*d*x)/2] + 3*C*Sin[3*c
+ (7*d*x)/2] + 3*C*Sin[4*c + (7*d*x)/2]))/(48*a^2*d*(1 + Cos[c + d*x])^2)

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Maple [A]  time = 0.031, size = 252, normalized size = 1.7 \begin{align*} -{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{5\,B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{7\,C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-5\,{\frac{C \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}B}{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-3\,{\frac{C\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-4\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{d{a}^{2}}}+7\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*B+1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3+5/2/d/a^2*B*tan(1/2*d*x+1/2*c)-7/2/d/a^2*C*
tan(1/2*d*x+1/2*c)-5/d/a^2/(tan(1/2*d*x+1/2*c)^2+1)^2*C*tan(1/2*d*x+1/2*c)^3+2/d/a^2/(tan(1/2*d*x+1/2*c)^2+1)^
2*tan(1/2*d*x+1/2*c)^3*B-3/d/a^2/(tan(1/2*d*x+1/2*c)^2+1)^2*C*tan(1/2*d*x+1/2*c)+2/d/a^2/(tan(1/2*d*x+1/2*c)^2
+1)^2*B*tan(1/2*d*x+1/2*c)-4/d/a^2*arctan(tan(1/2*d*x+1/2*c))*B+7/d/a^2*arctan(tan(1/2*d*x+1/2*c))*C

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Maxima [B]  time = 2.03039, size = 382, normalized size = 2.6 \begin{align*} -\frac{C{\left (\frac{6 \,{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{42 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - B{\left (\frac{\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{24 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac{12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac{a^{2} \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 )}}\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(C*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - s
in(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) - B*((15*sin(d*x + c
)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a
^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d

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Fricas [A]  time = 1.68669, size = 343, normalized size = 2.33 \begin{align*} -\frac{3 \,{\left (4 \, B - 7 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \,{\left (4 \, B - 7 \, C\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (4 \, B - 7 \, C\right )} d x -{\left (3 \, C \cos \left (d x + c\right )^{3} + 6 \,{\left (B - C\right )} \cos \left (d x + c\right )^{2} +{\left (28 \, B - 43 \, C\right )} \cos \left (d x + c\right ) + 20 \, B - 32 \, C\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(4*B - 7*C)*d*x*cos(d*x + c)^2 + 6*(4*B - 7*C)*d*x*cos(d*x + c) + 3*(4*B - 7*C)*d*x - (3*C*cos(d*x + c
)^3 + 6*(B - C)*cos(d*x + c)^2 + (28*B - 43*C)*cos(d*x + c) + 20*B - 32*C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2
+ 2*a^2*d*cos(d*x + c) + a^2*d)

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Sympy [A]  time = 17.4149, size = 848, normalized size = 5.77 \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*(B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))**2,x)

[Out]

Piecewise((-12*B*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**
2*d) - 24*B*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d)
- 12*B*d*x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - B*tan(c/2 + d*x/2)**7/(
6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 13*B*tan(c/2 + d*x/2)**5/(6*a**2*d*
tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 41*B*tan(c/2 + d*x/2)**3/(6*a**2*d*tan(c/2 +
d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 27*B*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**4 +
12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 21*C*d*x*tan(c/2 + d*x/2)**4/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a
**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 42*C*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d
*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 21*C*d*x/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*
a**2*d) + C*tan(c/2 + d*x/2)**7/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 19
*C*tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 71*C*tan(c/
2 + d*x/2)**3/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 39*C*tan(c/2 + d*x/2
)/(6*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d), Ne(d, 0)), (x*(B*cos(c) + C*cos(c
)**2)*cos(c)**2/(a*cos(c) + a)**2, True))

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Giac [A]  time = 1.39133, size = 221, normalized size = 1.5 \begin{align*} -\frac{\frac{3 \,{\left (d x + c\right )}{\left (4 \, B - 7 \, C\right )}}{a^{2}} - \frac{6 \,{\left (2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B \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}} + \frac{B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 21 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(d*x + c)*(4*B - 7*C)/a^2 - 6*(2*B*tan(1/2*d*x + 1/2*c)^3 - 5*C*tan(1/2*d*x + 1/2*c)^3 + 2*B*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^2) + (B*a^4*tan(1/2*d*x + 1/2*c)^3 -
C*a^4*tan(1/2*d*x + 1/2*c)^3 - 15*B*a^4*tan(1/2*d*x + 1/2*c) + 21*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d