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

Optimal. Leaf size=72 \[ \frac {(A+2 B-5 C) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {C x}{a^2}+\frac {(A-B+C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3019, 2735, 2648} \[ \frac {(A+2 B-5 C) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {C x}{a^2}+\frac {(A-B+C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [B]  time = 0.42, size = 175, normalized size = 2.43 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (2 A \sin \left (c+\frac {3 d x}{2}\right )+6 A \sin \left (\frac {d x}{2}\right )-6 B \sin \left (c+\frac {d x}{2}\right )+4 B \sin \left (c+\frac {3 d x}{2}\right )+6 B \sin \left (\frac {d x}{2}\right )+12 C \sin \left (c+\frac {d x}{2}\right )-10 C \sin \left (c+\frac {3 d x}{2}\right )+9 C d x \cos \left (c+\frac {d x}{2}\right )+3 C d x \cos \left (c+\frac {3 d x}{2}\right )+3 C d x \cos \left (2 c+\frac {3 d x}{2}\right )-18 C \sin \left (\frac {d x}{2}\right )+9 C d x \cos \left (\frac {d x}{2}\right )\right )}{24 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + 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*A*Sin[(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] + 2*A*Sin[c + (3*d*x)/2] + 4*B*Sin[c + (3*d*x)/2] - 10*C*Sin[c + (3*d*x)/2]))/(24*a^2*d
)

________________________________________________________________________________________

fricas [A]  time = 0.50, size = 95, normalized size = 1.32 \[ \frac {3 \, C d x \cos \left (d x + c\right )^{2} + 6 \, C d x \cos \left (d x + c\right ) + 3 \, C d x + {\left ({\left (A + 2 \, B - 5 \, C\right )} \cos \left (d x + c\right ) + 2 \, A + B - 4 \, C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+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 + ((A + 2*B - 5*C)*cos(d*x + c) + 2*A + 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)

________________________________________________________________________________________

giac [A]  time = 0.42, size = 116, normalized size = 1.61 \[ \frac {\frac {6 \, {\left (d x + c\right )} C}{a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+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 + (A*a^4*tan(1/2*d*x + 1/2*c)^3 - 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*A*a^4*tan(1/2*d*x + 1/2*c) + 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

________________________________________________________________________________________

maple [A]  time = 0.12, size = 135, normalized size = 1.88 \[ \frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}-\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {3 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+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*A-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*A
*tan(1/2*d*x+1/2*c)+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

________________________________________________________________________________________

maxima [B]  time = 0.44, size = 164, normalized size = 2.28 \[ -\frac {C {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - \frac {A {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}} - \frac {B {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+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) - A*(3*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/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

________________________________________________________________________________________

mupad [B]  time = 1.30, size = 113, normalized size = 1.57 \[ \frac {\frac {3\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}+\frac {A\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{2}+B\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )-\frac {3\,C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}-\frac {5\,C\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{2}+\frac {9\,C\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (c+d\,x\right )}{2}+\frac {3\,C\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (c+d\,x\right )}{2}}{6\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*A*sin(c/2 + (d*x)/2))/2 + (A*sin((3*c)/2 + (3*d*x)/2))/2 + B*sin((3*c)/2 + (3*d*x)/2) - (3*C*sin(c/2 + (d*
x)/2))/2 - (5*C*sin((3*c)/2 + (3*d*x)/2))/2 + (9*C*cos(c/2 + (d*x)/2)*(c + d*x))/2 + (3*C*cos((3*c)/2 + (3*d*x
)/2)*(c + d*x))/2)/(6*a^2*d*cos(c/2 + (d*x)/2)^3)

________________________________________________________________________________________

sympy [A]  time = 4.51, size = 148, normalized size = 2.06 \[ \begin {cases} \frac {A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d} + \frac {A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d} - \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d} + \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d} + \frac {C x}{a^{2}} + \frac {C \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d} - \frac {3 C \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \cos {\relax (c )} + C \cos ^{2}{\relax (c )}\right )}{\left (a \cos {\relax (c )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________