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

Optimal. Leaf size=134 \[ \frac {2 a^2 (b B-a C) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d \sqrt {a-b} \sqrt {a+b}}-\frac {x \left (-2 a^2 C+2 a b B-b^2 C\right )}{2 b^3}+\frac {(b B-a C) \sin (c+d x)}{b^2 d}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3029, 2990, 3023, 2735, 2659, 205} \[ \frac {2 a^2 (b B-a C) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d \sqrt {a-b} \sqrt {a+b}}-\frac {x \left (-2 a^2 C+2 a b B-b^2 C\right )}{2 b^3}+\frac {(b B-a C) \sin (c+d x)}{b^2 d}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.32, size = 121, normalized size = 0.90 \[ \frac {2 (c+d x) \left (2 a^2 C-2 a b B+b^2 C\right )+\frac {8 a^2 (a C-b B) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+4 b (b B-a C) \sin (c+d x)+b^2 C \sin (2 (c+d x))}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-2*a*b*B + 2*a^2*C + b^2*C)*(c + d*x) + (8*a^2*(-(b*B) + a*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2
+ b^2]])/Sqrt[-a^2 + b^2] + 4*b*(b*B - a*C)*Sin[c + d*x] + b^2*C*Sin[2*(c + d*x)])/(4*b^3*d)

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fricas [A]  time = 0.48, size = 426, normalized size = 3.18 \[ \left [\frac {{\left (2 \, C a^{4} - 2 \, B a^{3} b - C a^{2} b^{2} + 2 \, B a b^{3} - C b^{4}\right )} d x + {\left (C a^{3} - B a^{2} b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (2 \, C a^{3} b - 2 \, B a^{2} b^{2} - 2 \, C a b^{3} + 2 \, B b^{4} - {\left (C a^{2} b^{2} - C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}, \frac {{\left (2 \, C a^{4} - 2 \, B a^{3} b - C a^{2} b^{2} + 2 \, B a b^{3} - C b^{4}\right )} d x - 2 \, {\left (C a^{3} - B a^{2} b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (2 \, C a^{3} b - 2 \, B a^{2} b^{2} - 2 \, C a b^{3} + 2 \, B b^{4} - {\left (C a^{2} b^{2} - C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 227, normalized size = 1.69 \[ \frac {\frac {{\left (2 \, C a^{2} - 2 \, B a b + C b^{2}\right )} {\left (d x + c\right )}}{b^{3}} + \frac {4 \, {\left (C a^{3} - B a^{2} b\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{3}} - \frac {2 \, {\left (2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.13, size = 367, normalized size = 2.74 \[ \frac {2 a^{2} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d \,b^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 a^{3} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d \,b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B a}{d \,b^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} C}{d \,b^{3}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*(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 = 5.43, size = 3761, normalized size = 28.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*(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)*(2*B*b - 2*C*a + C*b))/b^2 - (tan(c/2 + (d*x)/2)^3*(2*C*a - 2*B*b + C*b))/b^2)/(d*(2*tan(
c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 + 1)) - (atan(((((((8*(2*C*b^10 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*
b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 - (4*tan(c/2 + (d*x)/2)*(C*a^2*2i + C*b^2*1i - B
*a*b*2i)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7)*(C*a^2*2i + C*b^2*1i - B*a*b*2i))/(2*b^3) - (8*tan(c/2 + (d*
x)/2)*(8*C^2*a^7 - C^2*b^7 + 3*C^2*a*b^6 - 16*C^2*a^6*b - 4*B^2*a^2*b^5 + 12*B^2*a^3*b^4 - 16*B^2*a^4*b^3 + 8*
B^2*a^5*b^2 - 7*C^2*a^2*b^5 + 13*C^2*a^3*b^4 - 16*C^2*a^4*b^3 + 16*C^2*a^5*b^2 + 4*B*C*a*b^6 - 16*B*C*a^6*b -
12*B*C*a^2*b^5 + 20*B*C*a^3*b^4 - 28*B*C*a^4*b^3 + 32*B*C*a^5*b^2))/b^4)*(C*a^2*2i + C*b^2*1i - B*a*b*2i)*1i)/
(2*b^3) - (((((8*(2*C*b^10 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 4*B*a*b^9 -
2*C*a*b^9))/b^6 + (4*tan(c/2 + (d*x)/2)*(C*a^2*2i + C*b^2*1i - B*a*b*2i)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/
b^7)*(C*a^2*2i + C*b^2*1i - B*a*b*2i))/(2*b^3) + (8*tan(c/2 + (d*x)/2)*(8*C^2*a^7 - C^2*b^7 + 3*C^2*a*b^6 - 16
*C^2*a^6*b - 4*B^2*a^2*b^5 + 12*B^2*a^3*b^4 - 16*B^2*a^4*b^3 + 8*B^2*a^5*b^2 - 7*C^2*a^2*b^5 + 13*C^2*a^3*b^4
- 16*C^2*a^4*b^3 + 16*C^2*a^5*b^2 + 4*B*C*a*b^6 - 16*B*C*a^6*b - 12*B*C*a^2*b^5 + 20*B*C*a^3*b^4 - 28*B*C*a^4*
b^3 + 32*B*C*a^5*b^2))/b^4)*(C*a^2*2i + C*b^2*1i - B*a*b*2i)*1i)/(2*b^3))/((16*(4*C^3*a^8 - 6*C^3*a^7*b + 4*B^
3*a^4*b^4 - 4*B^3*a^5*b^3 - C^3*a^3*b^5 + 2*C^3*a^4*b^4 - 5*C^3*a^5*b^3 + 6*C^3*a^6*b^2 - 12*B*C^2*a^7*b + B*C
^2*a^2*b^6 - 2*B*C^2*a^3*b^5 + 9*B*C^2*a^4*b^4 - 12*B*C^2*a^5*b^3 + 16*B*C^2*a^6*b^2 - 4*B^2*C*a^3*b^5 + 6*B^2
*C*a^4*b^4 - 14*B^2*C*a^5*b^3 + 12*B^2*C*a^6*b^2))/b^6 + (((((8*(2*C*b^10 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^
2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 - (4*tan(c/2 + (d*x)/2)*(C*a^2*2i + C*b^2*1i -
B*a*b*2i)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7)*(C*a^2*2i + C*b^2*1i - B*a*b*2i))/(2*b^3) - (8*tan(c/2 + (
d*x)/2)*(8*C^2*a^7 - C^2*b^7 + 3*C^2*a*b^6 - 16*C^2*a^6*b - 4*B^2*a^2*b^5 + 12*B^2*a^3*b^4 - 16*B^2*a^4*b^3 +
8*B^2*a^5*b^2 - 7*C^2*a^2*b^5 + 13*C^2*a^3*b^4 - 16*C^2*a^4*b^3 + 16*C^2*a^5*b^2 + 4*B*C*a*b^6 - 16*B*C*a^6*b
- 12*B*C*a^2*b^5 + 20*B*C*a^3*b^4 - 28*B*C*a^4*b^3 + 32*B*C*a^5*b^2))/b^4)*(C*a^2*2i + C*b^2*1i - B*a*b*2i))/(
2*b^3) + (((((8*(2*C*b^10 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 4*B*a*b^9 -
2*C*a*b^9))/b^6 + (4*tan(c/2 + (d*x)/2)*(C*a^2*2i + C*b^2*1i - B*a*b*2i)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b
^7)*(C*a^2*2i + C*b^2*1i - B*a*b*2i))/(2*b^3) + (8*tan(c/2 + (d*x)/2)*(8*C^2*a^7 - C^2*b^7 + 3*C^2*a*b^6 - 16*
C^2*a^6*b - 4*B^2*a^2*b^5 + 12*B^2*a^3*b^4 - 16*B^2*a^4*b^3 + 8*B^2*a^5*b^2 - 7*C^2*a^2*b^5 + 13*C^2*a^3*b^4 -
16*C^2*a^4*b^3 + 16*C^2*a^5*b^2 + 4*B*C*a*b^6 - 16*B*C*a^6*b - 12*B*C*a^2*b^5 + 20*B*C*a^3*b^4 - 28*B*C*a^4*b
^3 + 32*B*C*a^5*b^2))/b^4)*(C*a^2*2i + C*b^2*1i - B*a*b*2i))/(2*b^3)))*(C*a^2*2i + C*b^2*1i - B*a*b*2i)*1i)/(b
^3*d) + (a^2*atan(((a^2*(-(a + b)*(a - b))^(1/2)*(B*b - C*a)*((8*tan(c/2 + (d*x)/2)*(8*C^2*a^7 - C^2*b^7 + 3*C
^2*a*b^6 - 16*C^2*a^6*b - 4*B^2*a^2*b^5 + 12*B^2*a^3*b^4 - 16*B^2*a^4*b^3 + 8*B^2*a^5*b^2 - 7*C^2*a^2*b^5 + 13
*C^2*a^3*b^4 - 16*C^2*a^4*b^3 + 16*C^2*a^5*b^2 + 4*B*C*a*b^6 - 16*B*C*a^6*b - 12*B*C*a^2*b^5 + 20*B*C*a^3*b^4
- 28*B*C*a^4*b^3 + 32*B*C*a^5*b^2))/b^4 + (a^2*((8*(2*C*b^10 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8 - 6*C*a
^3*b^7 + 4*C*a^4*b^6 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 + (8*a^2*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2)*(B*b -
C*a)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/(b^4*(b^5 - a^2*b^3)))*(-(a + b)*(a - b))^(1/2)*(B*b - C*a))/(b^5 -
a^2*b^3))*1i)/(b^5 - a^2*b^3) + (a^2*(-(a + b)*(a - b))^(1/2)*(B*b - C*a)*((8*tan(c/2 + (d*x)/2)*(8*C^2*a^7 -
C^2*b^7 + 3*C^2*a*b^6 - 16*C^2*a^6*b - 4*B^2*a^2*b^5 + 12*B^2*a^3*b^4 - 16*B^2*a^4*b^3 + 8*B^2*a^5*b^2 - 7*C^2
*a^2*b^5 + 13*C^2*a^3*b^4 - 16*C^2*a^4*b^3 + 16*C^2*a^5*b^2 + 4*B*C*a*b^6 - 16*B*C*a^6*b - 12*B*C*a^2*b^5 + 20
*B*C*a^3*b^4 - 28*B*C*a^4*b^3 + 32*B*C*a^5*b^2))/b^4 - (a^2*((8*(2*C*b^10 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^
2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 - (8*a^2*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))
^(1/2)*(B*b - C*a)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/(b^4*(b^5 - a^2*b^3)))*(-(a + b)*(a - b))^(1/2)*(B*b -
C*a))/(b^5 - a^2*b^3))*1i)/(b^5 - a^2*b^3))/((16*(4*C^3*a^8 - 6*C^3*a^7*b + 4*B^3*a^4*b^4 - 4*B^3*a^5*b^3 - C^
3*a^3*b^5 + 2*C^3*a^4*b^4 - 5*C^3*a^5*b^3 + 6*C^3*a^6*b^2 - 12*B*C^2*a^7*b + B*C^2*a^2*b^6 - 2*B*C^2*a^3*b^5 +
9*B*C^2*a^4*b^4 - 12*B*C^2*a^5*b^3 + 16*B*C^2*a^6*b^2 - 4*B^2*C*a^3*b^5 + 6*B^2*C*a^4*b^4 - 14*B^2*C*a^5*b^3
+ 12*B^2*C*a^6*b^2))/b^6 + (a^2*(-(a + b)*(a - b))^(1/2)*(B*b - C*a)*((8*tan(c/2 + (d*x)/2)*(8*C^2*a^7 - C^2*b
^7 + 3*C^2*a*b^6 - 16*C^2*a^6*b - 4*B^2*a^2*b^5 + 12*B^2*a^3*b^4 - 16*B^2*a^4*b^3 + 8*B^2*a^5*b^2 - 7*C^2*a^2*
b^5 + 13*C^2*a^3*b^4 - 16*C^2*a^4*b^3 + 16*C^2*a^5*b^2 + 4*B*C*a*b^6 - 16*B*C*a^6*b - 12*B*C*a^2*b^5 + 20*B*C*
a^3*b^4 - 28*B*C*a^4*b^3 + 32*B*C*a^5*b^2))/b^4 + (a^2*((8*(2*C*b^10 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2*C*a^2*b^8
- 6*C*a^3*b^7 + 4*C*a^4*b^6 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 + (8*a^2*tan(c/2 + (d*x)/2)*(-(a + b)*(a - b))^(1/2
)*(B*b - C*a)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/(b^4*(b^5 - a^2*b^3)))*(-(a + b)*(a - b))^(1/2)*(B*b - C*a))
/(b^5 - a^2*b^3)))/(b^5 - a^2*b^3) - (a^2*(-(a + b)*(a - b))^(1/2)*(B*b - C*a)*((8*tan(c/2 + (d*x)/2)*(8*C^2*a
^7 - C^2*b^7 + 3*C^2*a*b^6 - 16*C^2*a^6*b - 4*B^2*a^2*b^5 + 12*B^2*a^3*b^4 - 16*B^2*a^4*b^3 + 8*B^2*a^5*b^2 -
7*C^2*a^2*b^5 + 13*C^2*a^3*b^4 - 16*C^2*a^4*b^3 + 16*C^2*a^5*b^2 + 4*B*C*a*b^6 - 16*B*C*a^6*b - 12*B*C*a^2*b^5
+ 20*B*C*a^3*b^4 - 28*B*C*a^4*b^3 + 32*B*C*a^5*b^2))/b^4 - (a^2*((8*(2*C*b^10 + 8*B*a^2*b^8 - 4*B*a^3*b^7 + 2
*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 4*B*a*b^9 - 2*C*a*b^9))/b^6 - (8*a^2*tan(c/2 + (d*x)/2)*(-(a + b)*(a
- b))^(1/2)*(B*b - C*a)*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/(b^4*(b^5 - a^2*b^3)))*(-(a + b)*(a - b))^(1/2)*(B
*b - C*a))/(b^5 - a^2*b^3)))/(b^5 - a^2*b^3)))*(-(a + b)*(a - b))^(1/2)*(B*b - C*a)*2i)/(d*(b^5 - a^2*b^3))

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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)*(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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