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

Optimal. Leaf size=70 \[ \frac {(2 B-5 C) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {C x}{a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

C*x/a^2+1/3*(2*B-5*C)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(B-C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2

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Rubi [A]  time = 0.11, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3019, 2735, 2648} \[ \frac {(2 B-5 C) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {C x}{a^2}-\frac {(B-C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2735

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

Rule 3019

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [B]  time = 0.35, size = 153, normalized size = 2.19 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (-6 B \sin \left (c+\frac {d x}{2}\right )+4 B \sin \left (c+\frac {3 d x}{2}\right )+6 B \sin \left (\frac {d x}{2}\right )+12 C \sin \left (c+\frac {d x}{2}\right )-10 C \sin \left (c+\frac {3 d x}{2}\right )+9 C d x \cos \left (c+\frac {d x}{2}\right )+3 C d x \cos \left (c+\frac {3 d x}{2}\right )+3 C d x \cos \left (2 c+\frac {3 d x}{2}\right )-18 C \sin \left (\frac {d x}{2}\right )+9 C d x \cos \left (\frac {d x}{2}\right )\right )}{24 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(9*C*d*x*Cos[(d*x)/2] + 9*C*d*x*Cos[c + (d*x)/2] + 3*C*d*x*Cos[c + (3*d*x)/2] + 3
*C*d*x*Cos[2*c + (3*d*x)/2] + 6*B*Sin[(d*x)/2] - 18*C*Sin[(d*x)/2] - 6*B*Sin[c + (d*x)/2] + 12*C*Sin[c + (d*x)
/2] + 4*B*Sin[c + (3*d*x)/2] - 10*C*Sin[c + (3*d*x)/2]))/(24*a^2*d)

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fricas [A]  time = 0.47, size = 91, normalized size = 1.30 \[ \frac {3 \, C d x \cos \left (d x + c\right )^{2} + 6 \, C d x \cos \left (d x + c\right ) + 3 \, C d x + {\left ({\left (2 \, B - 5 \, C\right )} \cos \left (d x + c\right ) + B - 4 \, C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*C*d*x*cos(d*x + c)^2 + 6*C*d*x*cos(d*x + c) + 3*C*d*x + ((2*B - 5*C)*cos(d*x + c) + B - 4*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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giac [A]  time = 0.44, size = 86, normalized size = 1.23 \[ \frac {\frac {6 \, {\left (d x + c\right )} C}{a^{2}} - \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(d*x + c)*C/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 - 3*B*a^4*tan(1/2*d*x +
1/2*c) + 9*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [A]  time = 0.10, size = 97, normalized size = 1.39 \[ -\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {3 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.94, size = 120, normalized size = 1.71 \[ -\frac {C {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - \frac {B {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2) - B*(3*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/
d

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mupad [B]  time = 1.08, size = 65, normalized size = 0.93 \[ \frac {3\,B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9\,C\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-B\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+C\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+6\,C\,d\,x}{6\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*B*tan(c/2 + (d*x)/2) - 9*C*tan(c/2 + (d*x)/2) - B*tan(c/2 + (d*x)/2)^3 + C*tan(c/2 + (d*x)/2)^3 + 6*C*d*x)/
(6*a^2*d)

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sympy [A]  time = 3.39, size = 107, normalized size = 1.53 \[ \begin {cases} - \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d} + \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d} + \frac {C x}{a^{2}} + \frac {C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d} - \frac {3 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \left (B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right )}{\left (a \cos {\relax (c )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-B*tan(c/2 + d*x/2)**3/(6*a**2*d) + B*tan(c/2 + d*x/2)/(2*a**2*d) + C*x/a**2 + C*tan(c/2 + d*x/2)**
3/(6*a**2*d) - 3*C*tan(c/2 + d*x/2)/(2*a**2*d), Ne(d, 0)), (x*(B*cos(c) + C*cos(c)**2)/(a*cos(c) + a)**2, True
))

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