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

Optimal. Leaf size=178 \[ -\frac {2 a^3 (b B-a C) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d \sqrt {a-b} \sqrt {a+b}}+\frac {x \left (2 a^2+b^2\right ) (b B-a C)}{2 b^4}-\frac {\left (-3 a^2 C+3 a b B-2 b^2 C\right ) \sin (c+d x)}{3 b^3 d}+\frac {(b B-a C) \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.57, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {3029, 2990, 3049, 3023, 2735, 2659, 205} \[ -\frac {\left (-3 a^2 C+3 a b B-2 b^2 C\right ) \sin (c+d x)}{3 b^3 d}-\frac {2 a^3 (b B-a C) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^4 d \sqrt {a-b} \sqrt {a+b}}+\frac {x \left (2 a^2+b^2\right ) (b B-a C)}{2 b^4}+\frac {(b B-a C) \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac {C \sin (c+d x) \cos ^2(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2 + b^2)*(b*B - a*C)*x)/(2*b^4) - (2*a^3*(b*B - a*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])
/(Sqrt[a - b]*b^4*Sqrt[a + b]*d) - ((3*a*b*B - 3*a^2*C - 2*b^2*C)*Sin[c + d*x])/(3*b^3*d) + ((b*B - a*C)*Cos[c
+ d*x]*Sin[c + d*x])/(2*b^2*d) + (C*Cos[c + d*x]^2*Sin[c + d*x])/(3*b*d)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[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 2990

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

Rule 3023

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

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 3049

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] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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+b \cos (c+d x)} \, dx &=\int \frac {\cos ^3(c+d x) (B+C \cos (c+d x))}{a+b \cos (c+d x)} \, dx\\ &=\frac {C \cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac {\int \frac {\cos (c+d x) \left (2 a C+2 b C \cos (c+d x)+3 (b B-a C) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{3 b}\\ &=\frac {(b B-a C) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {C \cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac {\int \frac {3 a (b B-a C)+b (3 b B+a C) \cos (c+d x)-2 \left (3 a b B-3 a^2 C-2 b^2 C\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^2}\\ &=-\frac {\left (3 a b B-3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^3 d}+\frac {(b B-a C) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {C \cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac {\int \frac {3 a b (b B-a C)+3 \left (2 a^2+b^2\right ) (b B-a C) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^3}\\ &=\frac {\left (2 a^2+b^2\right ) (b B-a C) x}{2 b^4}-\frac {\left (3 a b B-3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^3 d}+\frac {(b B-a C) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {C \cos ^2(c+d x) \sin (c+d x)}{3 b d}-\frac {\left (a^3 (b B-a C)\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{b^4}\\ &=\frac {\left (2 a^2+b^2\right ) (b B-a C) x}{2 b^4}-\frac {\left (3 a b B-3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^3 d}+\frac {(b B-a C) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {C \cos ^2(c+d x) \sin (c+d x)}{3 b d}-\frac {\left (2 a^3 (b B-a C)\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^4 d}\\ &=\frac {\left (2 a^2+b^2\right ) (b B-a C) x}{2 b^4}-\frac {2 a^3 (b B-a C) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^4 \sqrt {a+b} d}-\frac {\left (3 a b B-3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{3 b^3 d}+\frac {(b B-a C) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac {C \cos ^2(c+d x) \sin (c+d x)}{3 b d}\\ \end {align*}

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Mathematica [A]  time = 0.47, size = 152, normalized size = 0.85 \[ \frac {6 \left (2 a^2+b^2\right ) (c+d x) (b B-a C)+3 b \left (4 a^2 C-4 a b B+3 b^2 C\right ) \sin (c+d x)-\frac {24 a^3 (a C-b B) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+3 b^2 (b B-a C) \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(2*a^2 + b^2)*(b*B - a*C)*(c + d*x) - (24*a^3*(-(b*B) + a*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 +
b^2]])/Sqrt[-a^2 + b^2] + 3*b*(-4*a*b*B + 4*a^2*C + 3*b^2*C)*Sin[c + d*x] + 3*b^2*(b*B - a*C)*Sin[2*(c + d*x)
] + b^3*C*Sin[3*(c + d*x)])/(12*b^4*d)

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fricas [A]  time = 0.50, size = 541, normalized size = 3.04 \[ \left [-\frac {3 \, {\left (2 \, C a^{5} - 2 \, B a^{4} b - C a^{3} b^{2} + B a^{2} b^{3} - C a b^{4} + B b^{5}\right )} d x - 3 \, {\left (C a^{4} - B a^{3} b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (6 \, C a^{4} b - 6 \, B a^{3} b^{2} - 2 \, C a^{2} b^{3} + 6 \, B a b^{4} - 4 \, C b^{5} + 2 \, {\left (C a^{2} b^{3} - C b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (C a^{3} b^{2} - B a^{2} b^{3} - C a b^{4} + B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )} d}, -\frac {3 \, {\left (2 \, C a^{5} - 2 \, B a^{4} b - C a^{3} b^{2} + B a^{2} b^{3} - C a b^{4} + B b^{5}\right )} d x - 6 \, {\left (C a^{4} - B a^{3} b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (6 \, C a^{4} b - 6 \, B a^{3} b^{2} - 2 \, C a^{2} b^{3} + 6 \, B a b^{4} - 4 \, C b^{5} + 2 \, {\left (C a^{2} b^{3} - C b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (C a^{3} b^{2} - B a^{2} b^{3} - C a b^{4} + B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} b^{4} - b^{6}\right )} d}\right ] \]

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+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

[-1/6*(3*(2*C*a^5 - 2*B*a^4*b - C*a^3*b^2 + B*a^2*b^3 - C*a*b^4 + B*b^5)*d*x - 3*(C*a^4 - B*a^3*b)*sqrt(-a^2 +
b^2)*log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x
+ c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - (6*C*a^4*b - 6*B*a^3*b^2 - 2*C*a^2*b^3
+ 6*B*a*b^4 - 4*C*b^5 + 2*(C*a^2*b^3 - C*b^5)*cos(d*x + c)^2 - 3*(C*a^3*b^2 - B*a^2*b^3 - C*a*b^4 + B*b^5)*co
s(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d), -1/6*(3*(2*C*a^5 - 2*B*a^4*b - C*a^3*b^2 + B*a^2*b^3 - C*a*b^4
+ B*b^5)*d*x - 6*(C*a^4 - B*a^3*b)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))
) - (6*C*a^4*b - 6*B*a^3*b^2 - 2*C*a^2*b^3 + 6*B*a*b^4 - 4*C*b^5 + 2*(C*a^2*b^3 - C*b^5)*cos(d*x + c)^2 - 3*(C
*a^3*b^2 - B*a^2*b^3 - C*a*b^4 + B*b^5)*cos(d*x + c))*sin(d*x + c))/((a^2*b^4 - b^6)*d)]

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giac [B]  time = 0.27, size = 360, normalized size = 2.02 \[ -\frac {\frac {3 \, {\left (2 \, C a^{3} - 2 \, B a^{2} b + C a b^{2} - B b^{3}\right )} {\left (d x + c\right )}}{b^{4}} + \frac {12 \, {\left (C a^{4} - B a^{3} b\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} - \frac {2 \, {\left (6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \]

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+b*cos(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(3*(2*C*a^3 - 2*B*a^2*b + C*a*b^2 - B*b^3)*(d*x + c)/b^4 + 12*(C*a^4 - B*a^3*b)*(pi*floor(1/2*(d*x + c)/p
i + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(a^2 - b^2)))/(sqrt(a
^2 - b^2)*b^4) - 2*(6*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 6*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 3*C*a*b*tan(1/2*d*x + 1/
2*c)^5 - 3*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*b^2*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 1
2*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 4*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^2*tan(1/2*d*x + 1/2*c) - 6*B*a*b*tan(1
/2*d*x + 1/2*c) - 3*C*a*b*tan(1/2*d*x + 1/2*c) + 3*B*b^2*tan(1/2*d*x + 1/2*c) + 6*C*b^2*tan(1/2*d*x + 1/2*c))/
((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^3))/d

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maple [B]  time = 0.14, size = 641, normalized size = 3.60 \[ -\frac {2 a^{3} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d \,b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 a^{4} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d \,b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C \,a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C \,a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{3 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C \,a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} B}{d \,b^{3}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d b}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C \,a^{3}}{d \,b^{4}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C a}{d \,b^{2}} \]

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+b*cos(d*x+c)),x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 6.43, size = 4568, normalized size = 25.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((tan(c/2 + (d*x)/2)*(B*b^2 + 2*C*a^2 + 2*C*b^2 - 2*B*a*b - C*a*b))/b^3 + (tan(c/2 + (d*x)/2)^5*(2*C*a^2 - B*b
^2 + 2*C*b^2 - 2*B*a*b + C*a*b))/b^3 + (4*tan(c/2 + (d*x)/2)^3*(3*C*a^2 + C*b^2 - 3*B*a*b))/(3*b^3))/(d*(3*tan
(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) + (atan((((2*a^2 + b^2)*(B*b - C*a)*((
8*tan(c/2 + (d*x)/2)*(B^2*b^9 - 8*C^2*a^9 - 3*B^2*a*b^8 + 16*C^2*a^8*b + 7*B^2*a^2*b^7 - 13*B^2*a^3*b^6 + 16*B
^2*a^4*b^5 - 16*B^2*a^5*b^4 + 16*B^2*a^6*b^3 - 8*B^2*a^7*b^2 + C^2*a^2*b^7 - 3*C^2*a^3*b^6 + 7*C^2*a^4*b^5 - 1
3*C^2*a^5*b^4 + 16*C^2*a^6*b^3 - 16*C^2*a^7*b^2 - 2*B*C*a*b^8 + 16*B*C*a^8*b + 6*B*C*a^2*b^7 - 14*B*C*a^3*b^6
+ 26*B*C*a^4*b^5 - 32*B*C*a^5*b^4 + 32*B*C*a^6*b^3 - 32*B*C*a^7*b^2))/b^6 + (((8*(2*B*b^13 + 2*B*a^2*b^11 - 6*
B*a^3*b^10 + 4*B*a^4*b^9 + 2*C*a^2*b^11 - 2*C*a^3*b^10 + 6*C*a^4*b^9 - 4*C*a^5*b^8 - 2*B*a*b^12 - 2*C*a*b^12))
/b^9 - (tan(c/2 + (d*x)/2)*(2*a^2 + b^2)*(B*b - C*a)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8)*4i)/b^10)*(2*a^2 + b^
2)*(B*b - C*a)*1i)/(2*b^4)))/(2*b^4) + ((2*a^2 + b^2)*(B*b - C*a)*((8*tan(c/2 + (d*x)/2)*(B^2*b^9 - 8*C^2*a^9
- 3*B^2*a*b^8 + 16*C^2*a^8*b + 7*B^2*a^2*b^7 - 13*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 16*B^2*a^6*b
^3 - 8*B^2*a^7*b^2 + C^2*a^2*b^7 - 3*C^2*a^3*b^6 + 7*C^2*a^4*b^5 - 13*C^2*a^5*b^4 + 16*C^2*a^6*b^3 - 16*C^2*a^
7*b^2 - 2*B*C*a*b^8 + 16*B*C*a^8*b + 6*B*C*a^2*b^7 - 14*B*C*a^3*b^6 + 26*B*C*a^4*b^5 - 32*B*C*a^5*b^4 + 32*B*C
*a^6*b^3 - 32*B*C*a^7*b^2))/b^6 - (((8*(2*B*b^13 + 2*B*a^2*b^11 - 6*B*a^3*b^10 + 4*B*a^4*b^9 + 2*C*a^2*b^11 -
2*C*a^3*b^10 + 6*C*a^4*b^9 - 4*C*a^5*b^8 - 2*B*a*b^12 - 2*C*a*b^12))/b^9 + (tan(c/2 + (d*x)/2)*(2*a^2 + b^2)*(
B*b - C*a)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8)*4i)/b^10)*(2*a^2 + b^2)*(B*b - C*a)*1i)/(2*b^4)))/(2*b^4))/((16
*(4*C^3*a^11 - 6*C^3*a^10*b + B^3*a^3*b^8 - 2*B^3*a^4*b^7 + 5*B^3*a^5*b^6 - 6*B^3*a^6*b^5 + 6*B^3*a^7*b^4 - 4*
B^3*a^8*b^3 - C^3*a^6*b^5 + 2*C^3*a^7*b^4 - 5*C^3*a^8*b^3 + 6*C^3*a^9*b^2 - 12*B*C^2*a^10*b + 3*B*C^2*a^5*b^6
- 6*B*C^2*a^6*b^5 + 15*B*C^2*a^7*b^4 - 18*B*C^2*a^8*b^3 + 18*B*C^2*a^9*b^2 - 3*B^2*C*a^4*b^7 + 6*B^2*C*a^5*b^6
- 15*B^2*C*a^6*b^5 + 18*B^2*C*a^7*b^4 - 18*B^2*C*a^8*b^3 + 12*B^2*C*a^9*b^2))/b^9 - ((2*a^2 + b^2)*(B*b - C*a
)*((8*tan(c/2 + (d*x)/2)*(B^2*b^9 - 8*C^2*a^9 - 3*B^2*a*b^8 + 16*C^2*a^8*b + 7*B^2*a^2*b^7 - 13*B^2*a^3*b^6 +
16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 16*B^2*a^6*b^3 - 8*B^2*a^7*b^2 + C^2*a^2*b^7 - 3*C^2*a^3*b^6 + 7*C^2*a^4*b^5
- 13*C^2*a^5*b^4 + 16*C^2*a^6*b^3 - 16*C^2*a^7*b^2 - 2*B*C*a*b^8 + 16*B*C*a^8*b + 6*B*C*a^2*b^7 - 14*B*C*a^3*
b^6 + 26*B*C*a^4*b^5 - 32*B*C*a^5*b^4 + 32*B*C*a^6*b^3 - 32*B*C*a^7*b^2))/b^6 + (((8*(2*B*b^13 + 2*B*a^2*b^11
- 6*B*a^3*b^10 + 4*B*a^4*b^9 + 2*C*a^2*b^11 - 2*C*a^3*b^10 + 6*C*a^4*b^9 - 4*C*a^5*b^8 - 2*B*a*b^12 - 2*C*a*b^
12))/b^9 - (tan(c/2 + (d*x)/2)*(2*a^2 + b^2)*(B*b - C*a)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8)*4i)/b^10)*(2*a^2
+ b^2)*(B*b - C*a)*1i)/(2*b^4))*1i)/(2*b^4) + ((2*a^2 + b^2)*(B*b - C*a)*((8*tan(c/2 + (d*x)/2)*(B^2*b^9 - 8*C
^2*a^9 - 3*B^2*a*b^8 + 16*C^2*a^8*b + 7*B^2*a^2*b^7 - 13*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 16*B^
2*a^6*b^3 - 8*B^2*a^7*b^2 + C^2*a^2*b^7 - 3*C^2*a^3*b^6 + 7*C^2*a^4*b^5 - 13*C^2*a^5*b^4 + 16*C^2*a^6*b^3 - 16
*C^2*a^7*b^2 - 2*B*C*a*b^8 + 16*B*C*a^8*b + 6*B*C*a^2*b^7 - 14*B*C*a^3*b^6 + 26*B*C*a^4*b^5 - 32*B*C*a^5*b^4 +
32*B*C*a^6*b^3 - 32*B*C*a^7*b^2))/b^6 - (((8*(2*B*b^13 + 2*B*a^2*b^11 - 6*B*a^3*b^10 + 4*B*a^4*b^9 + 2*C*a^2*
b^11 - 2*C*a^3*b^10 + 6*C*a^4*b^9 - 4*C*a^5*b^8 - 2*B*a*b^12 - 2*C*a*b^12))/b^9 + (tan(c/2 + (d*x)/2)*(2*a^2 +
b^2)*(B*b - C*a)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8)*4i)/b^10)*(2*a^2 + b^2)*(B*b - C*a)*1i)/(2*b^4))*1i)/(2*
b^4)))*(2*a^2 + b^2)*(B*b - C*a))/(b^4*d) + (a^3*atan(((a^3*(-(a + b)*(a - b))^(1/2)*(B*b - C*a)*((8*tan(c/2 +
(d*x)/2)*(B^2*b^9 - 8*C^2*a^9 - 3*B^2*a*b^8 + 16*C^2*a^8*b + 7*B^2*a^2*b^7 - 13*B^2*a^3*b^6 + 16*B^2*a^4*b^5
- 16*B^2*a^5*b^4 + 16*B^2*a^6*b^3 - 8*B^2*a^7*b^2 + C^2*a^2*b^7 - 3*C^2*a^3*b^6 + 7*C^2*a^4*b^5 - 13*C^2*a^5*b
^4 + 16*C^2*a^6*b^3 - 16*C^2*a^7*b^2 - 2*B*C*a*b^8 + 16*B*C*a^8*b + 6*B*C*a^2*b^7 - 14*B*C*a^3*b^6 + 26*B*C*a^
4*b^5 - 32*B*C*a^5*b^4 + 32*B*C*a^6*b^3 - 32*B*C*a^7*b^2))/b^6 + (a^3*(-(a + b)*(a - b))^(1/2)*((8*(2*B*b^13 +
2*B*a^2*b^11 - 6*B*a^3*b^10 + 4*B*a^4*b^9 + 2*C*a^2*b^11 - 2*C*a^3*b^10 + 6*C*a^4*b^9 - 4*C*a^5*b^8 - 2*B*a*b
^12 - 2*C*a*b^12))/b^9 - (8*a^3*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2)*(B*b - C*a)*(8*a*b^10 - 16*a^2*b^9
+ 8*a^3*b^8))/(b^6*(b^6 - a^2*b^4)))*(B*b - C*a))/(b^6 - a^2*b^4))*1i)/(b^6 - a^2*b^4) + (a^3*(-(a + b)*(a -
b))^(1/2)*(B*b - C*a)*((8*tan(c/2 + (d*x)/2)*(B^2*b^9 - 8*C^2*a^9 - 3*B^2*a*b^8 + 16*C^2*a^8*b + 7*B^2*a^2*b^7
- 13*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 16*B^2*a^6*b^3 - 8*B^2*a^7*b^2 + C^2*a^2*b^7 - 3*C^2*a^3
*b^6 + 7*C^2*a^4*b^5 - 13*C^2*a^5*b^4 + 16*C^2*a^6*b^3 - 16*C^2*a^7*b^2 - 2*B*C*a*b^8 + 16*B*C*a^8*b + 6*B*C*a
^2*b^7 - 14*B*C*a^3*b^6 + 26*B*C*a^4*b^5 - 32*B*C*a^5*b^4 + 32*B*C*a^6*b^3 - 32*B*C*a^7*b^2))/b^6 - (a^3*(-(a
+ b)*(a - b))^(1/2)*((8*(2*B*b^13 + 2*B*a^2*b^11 - 6*B*a^3*b^10 + 4*B*a^4*b^9 + 2*C*a^2*b^11 - 2*C*a^3*b^10 +
6*C*a^4*b^9 - 4*C*a^5*b^8 - 2*B*a*b^12 - 2*C*a*b^12))/b^9 + (8*a^3*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2)
*(B*b - C*a)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8))/(b^6*(b^6 - a^2*b^4)))*(B*b - C*a))/(b^6 - a^2*b^4))*1i)/(b^
6 - a^2*b^4))/((16*(4*C^3*a^11 - 6*C^3*a^10*b + B^3*a^3*b^8 - 2*B^3*a^4*b^7 + 5*B^3*a^5*b^6 - 6*B^3*a^6*b^5 +
6*B^3*a^7*b^4 - 4*B^3*a^8*b^3 - C^3*a^6*b^5 + 2*C^3*a^7*b^4 - 5*C^3*a^8*b^3 + 6*C^3*a^9*b^2 - 12*B*C^2*a^10*b
+ 3*B*C^2*a^5*b^6 - 6*B*C^2*a^6*b^5 + 15*B*C^2*a^7*b^4 - 18*B*C^2*a^8*b^3 + 18*B*C^2*a^9*b^2 - 3*B^2*C*a^4*b^7
+ 6*B^2*C*a^5*b^6 - 15*B^2*C*a^6*b^5 + 18*B^2*C*a^7*b^4 - 18*B^2*C*a^8*b^3 + 12*B^2*C*a^9*b^2))/b^9 - (a^3*(-
(a + b)*(a - b))^(1/2)*(B*b - C*a)*((8*tan(c/2 + (d*x)/2)*(B^2*b^9 - 8*C^2*a^9 - 3*B^2*a*b^8 + 16*C^2*a^8*b +
7*B^2*a^2*b^7 - 13*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 16*B^2*a^6*b^3 - 8*B^2*a^7*b^2 + C^2*a^2*b^
7 - 3*C^2*a^3*b^6 + 7*C^2*a^4*b^5 - 13*C^2*a^5*b^4 + 16*C^2*a^6*b^3 - 16*C^2*a^7*b^2 - 2*B*C*a*b^8 + 16*B*C*a^
8*b + 6*B*C*a^2*b^7 - 14*B*C*a^3*b^6 + 26*B*C*a^4*b^5 - 32*B*C*a^5*b^4 + 32*B*C*a^6*b^3 - 32*B*C*a^7*b^2))/b^6
+ (a^3*(-(a + b)*(a - b))^(1/2)*((8*(2*B*b^13 + 2*B*a^2*b^11 - 6*B*a^3*b^10 + 4*B*a^4*b^9 + 2*C*a^2*b^11 - 2*
C*a^3*b^10 + 6*C*a^4*b^9 - 4*C*a^5*b^8 - 2*B*a*b^12 - 2*C*a*b^12))/b^9 - (8*a^3*tan(c/2 + (d*x)/2)*(-(a + b)*(
a - b))^(1/2)*(B*b - C*a)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8))/(b^6*(b^6 - a^2*b^4)))*(B*b - C*a))/(b^6 - a^2*
b^4)))/(b^6 - a^2*b^4) + (a^3*(-(a + b)*(a - b))^(1/2)*(B*b - C*a)*((8*tan(c/2 + (d*x)/2)*(B^2*b^9 - 8*C^2*a^9
- 3*B^2*a*b^8 + 16*C^2*a^8*b + 7*B^2*a^2*b^7 - 13*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 16*B^2*a^6*
b^3 - 8*B^2*a^7*b^2 + C^2*a^2*b^7 - 3*C^2*a^3*b^6 + 7*C^2*a^4*b^5 - 13*C^2*a^5*b^4 + 16*C^2*a^6*b^3 - 16*C^2*a
^7*b^2 - 2*B*C*a*b^8 + 16*B*C*a^8*b + 6*B*C*a^2*b^7 - 14*B*C*a^3*b^6 + 26*B*C*a^4*b^5 - 32*B*C*a^5*b^4 + 32*B*
C*a^6*b^3 - 32*B*C*a^7*b^2))/b^6 - (a^3*(-(a + b)*(a - b))^(1/2)*((8*(2*B*b^13 + 2*B*a^2*b^11 - 6*B*a^3*b^10 +
4*B*a^4*b^9 + 2*C*a^2*b^11 - 2*C*a^3*b^10 + 6*C*a^4*b^9 - 4*C*a^5*b^8 - 2*B*a*b^12 - 2*C*a*b^12))/b^9 + (8*a^
3*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2)*(B*b - C*a)*(8*a*b^10 - 16*a^2*b^9 + 8*a^3*b^8))/(b^6*(b^6 - a^2
*b^4)))*(B*b - C*a))/(b^6 - a^2*b^4)))/(b^6 - a^2*b^4)))*(-(a + b)*(a - b))^(1/2)*(B*b - C*a)*2i)/(d*(b^6 - a^
2*b^4))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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+b*cos(d*x+c)),x)

[Out]

Timed out

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