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

Optimal. Leaf size=128 \[ -\frac {2 a \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d \sqrt {a-b} \sqrt {a+b}}+\frac {x \left (2 a^2 C+b^2 (2 A+C)\right )}{2 b^3}-\frac {a C \sin (c+d x)}{b^2 d}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d} \]

[Out]

1/2*(2*a^2*C+b^2*(2*A+C))*x/b^3-a*C*sin(d*x+c)/b^2/d+1/2*C*cos(d*x+c)*sin(d*x+c)/b/d-2*a*(A*b^2+C*a^2)*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 = 0.26, antiderivative size = 126, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3050, 3023, 2735, 2659, 205} \[ -\frac {2 a \left (a^2 C+A b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^3 d \sqrt {a-b} \sqrt {a+b}}+\frac {x \left (\frac {2 a^2 C}{b^2}+2 A+C\right )}{2 b}-\frac {a C \sin (c+d x)}{b^2 d}+\frac {C \sin (c+d x) \cos (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A + C + (2*a^2*C)/b^2)*x)/(2*b) - (2*a*(A*b^2 + a^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])
/(Sqrt[a - b]*b^3*Sqrt[a + b]*d) - (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 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 3050

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*
x] + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 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 (c+d x) \left (A+C \cos ^2(c+d x)\right )}{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 (2 A+C) \cos (c+d x)-2 a C \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{2 b}\\ &=-\frac {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^2 C+b^2 (2 A+C)\right ) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{2 b^2}\\ &=\frac {\left (2 a^2 C+b^2 (2 A+C)\right ) x}{2 b^3}-\frac {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 \left (A b^2+a^2 C\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{b^3}\\ &=\frac {\left (2 a^2 C+b^2 (2 A+C)\right ) x}{2 b^3}-\frac {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 \left (A b^2+a^2 C\right )\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^2 C+b^2 (2 A+C)\right ) x}{2 b^3}-\frac {2 a \left (A b^2+a^2 C\right ) \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 {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 = 0.42, size = 117, normalized size = 0.91 \[ \frac {2 (c+d x) \left (C \left (2 a^2+b^2\right )+2 A b^2\right )+\frac {8 a \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}-4 a b C \sin (c+d x)+b^2 C \sin (2 (c+d x))}{4 b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.86, size = 385, normalized size = 3.01 \[ \left [\frac {{\left (2 \, C a^{4} + {\left (2 \, A - C\right )} a^{2} b^{2} - {\left (2 \, A + C\right )} b^{4}\right )} d x - {\left (C a^{3} + A a b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (2 \, C a^{3} b - 2 \, C a b^{3} - {\left (C a^{2} b^{2} - C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}, \frac {{\left (2 \, C a^{4} + {\left (2 \, A - C\right )} a^{2} b^{2} - {\left (2 \, A + C\right )} b^{4}\right )} d x - 2 \, {\left (C a^{3} + A a b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (2 \, C a^{3} b - 2 \, C a b^{3} - {\left (C a^{2} b^{2} - C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{3} - b^{5}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((2*C*a^4 + (2*A - C)*a^2*b^2 - (2*A + C)*b^4)*d*x - (C*a^3 + A*a*b^2)*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*C*a*b^3 - (C*a^2*b^2 - C*b^4)*cos(d*x + c))*si
n(d*x + c))/((a^2*b^3 - b^5)*d), 1/2*((2*C*a^4 + (2*A - C)*a^2*b^2 - (2*A + C)*b^4)*d*x - 2*(C*a^3 + A*a*b^2)*
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*C*a*b^3 - (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 = 0.50, size = 199, normalized size = 1.55 \[ \frac {\frac {{\left (2 \, C a^{2} + 2 \, A b^{2} + C b^{2}\right )} {\left (d x + c\right )}}{b^{3}} + \frac {4 \, {\left (C a^{3} + A a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{3}} - \frac {2 \, {\left (2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((2*C*a^2 + 2*A*b^2 + C*b^2)*(d*x + c)/b^3 + 4*(C*a^3 + A*a*b^2)*(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 + C*b*tan(1/2*d*x + 1/2*c)^3 + 2*C*a*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 = 0.12, size = 296, normalized size = 2.31 \[ -\frac {2 a \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d b \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 a^{3} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d \,b^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C a}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d b}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C \,a^{2}}{d \,b^{3}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d*a/b/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A-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^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^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*arctan(tan(1/2*d*x+1/2*c))*A+2/d/b^
3*arctan(tan(1/2*d*x+1/2*c))*C*a^2+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)*(A+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 = 4.65, size = 2398, normalized size = 18.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((tan(c/2 + (d*x)/2)*(2*C*a - C*b))/b^2 + (tan(c/2 + (d*x)/2)^3*(2*C*a + C*b))/b^2)/(d*(2*tan(c/2 + (d*x)/2)
^2 + tan(c/2 + (d*x)/2)^4 + 1)) - (atan(((((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b
^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 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*A*C*b^7 - 12*A*C*a*b^6 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a
^5*b^2))/b^4 + ((C*a^2*1i + b^2*(A*1i + (C*1i)/2))*((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*b^8 + 2*C*a^2*b^8 - 6*C*
a^3*b^7 + 4*C*a^4*b^6 - 8*A*a*b^9 - 2*C*a*b^9))/b^6 - (8*tan(c/2 + (d*x)/2)*(C*a^2*1i + b^2*(A*1i + (C*1i)/2))
*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7))/b^3)*(C*a^2*1i + b^2*(A*1i + (C*1i)/2))*1i)/b^3 + (((8*tan(c/2 + (d
*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a
^3*b^4 + 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*A*C*b^7 - 12*A*C*a*b^6 + 20*A*C*
a^2*b^5 - 28*A*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a^5*b^2))/b^4 - ((C*a^2*1i + b^2*(A*1i + (C*1i)/2))*((8*(4*
A*b^10 + 2*C*b^10 + 4*A*a^2*b^8 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 8*A*a*b^9 - 2*C*a*b^9))/b^6 + (8*t
an(c/2 + (d*x)/2)*(C*a^2*1i + b^2*(A*1i + (C*1i)/2))*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7))/b^3)*(C*a^2*1i
+ b^2*(A*1i + (C*1i)/2))*1i)/b^3)/((16*(4*C^3*a^8 - 4*A^3*a*b^7 - 6*C^3*a^7*b + 4*A^3*a^2*b^6 - 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 - A*C^2*a*b^7 - 4*A^2*C*a*b^7 + 2*A*C^2*a^2*b^6 - 9*A*C^2*a^3*b^
5 + 12*A*C^2*a^4*b^4 - 16*A*C^2*a^5*b^3 + 12*A*C^2*a^6*b^2 + 6*A^2*C*a^2*b^6 - 14*A^2*C*a^3*b^5 + 12*A^2*C*a^4
*b^4))/b^6 + (((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 + C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^
6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 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*A*
C*b^7 - 12*A*C*a*b^6 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32*A*C*a^4*b^3 - 16*A*C*a^5*b^2))/b^4 + ((C*a^2*1i +
b^2*(A*1i + (C*1i)/2))*((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*b^8 + 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 8*A*
a*b^9 - 2*C*a*b^9))/b^6 - (8*tan(c/2 + (d*x)/2)*(C*a^2*1i + b^2*(A*1i + (C*1i)/2))*(8*a*b^8 - 16*a^2*b^7 + 8*a
^3*b^6))/b^7))/b^3)*(C*a^2*1i + b^2*(A*1i + (C*1i)/2)))/b^3 - (((8*tan(c/2 + (d*x)/2)*(4*A^2*b^7 - 8*C^2*a^7 +
C^2*b^7 - 12*A^2*a*b^6 - 3*C^2*a*b^6 + 16*C^2*a^6*b + 16*A^2*a^2*b^5 - 8*A^2*a^3*b^4 + 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*A*C*b^7 - 12*A*C*a*b^6 + 20*A*C*a^2*b^5 - 28*A*C*a^3*b^4 + 32*A
*C*a^4*b^3 - 16*A*C*a^5*b^2))/b^4 - ((C*a^2*1i + b^2*(A*1i + (C*1i)/2))*((8*(4*A*b^10 + 2*C*b^10 + 4*A*a^2*b^8
+ 2*C*a^2*b^8 - 6*C*a^3*b^7 + 4*C*a^4*b^6 - 8*A*a*b^9 - 2*C*a*b^9))/b^6 + (8*tan(c/2 + (d*x)/2)*(C*a^2*1i + b
^2*(A*1i + (C*1i)/2))*(8*a*b^8 - 16*a^2*b^7 + 8*a^3*b^6))/b^7))/b^3)*(C*a^2*1i + b^2*(A*1i + (C*1i)/2)))/b^3))
*(C*a^2*1i + b^2*(A*1i + (C*1i)/2))*2i)/(b^3*d) - (log(a*tan(c/2 + (d*x)/2) - b*tan(c/2 + (d*x)/2) + (b^2 - a^
2)^(1/2))*(C*a^3*(b^2 - a^2)^(1/2) + A*a*b^2*(b^2 - a^2)^(1/2)))/(b^3*d*(a^2 - b^2)) - (a*log((8*a*(a - b)*(4*
A^3*b^6 + 4*C^3*a^6 + A*C^2*b^6 + 4*A^2*C*b^6 - 2*C^3*a^5*b + C^3*a^2*b^4 - C^3*a^3*b^3 + 4*C^3*a^4*b^2 - A*C^
2*a*b^5 - 2*A^2*C*a*b^5 + 8*A*C^2*a^2*b^4 - 4*A*C^2*a^3*b^3 + 12*A*C^2*a^4*b^2 + 12*A^2*C*a^2*b^4))/b^6 + (a*(
A*b^2 + C*a^2)*(b^2 - a^2)^(1/2)*((8*tan(c/2 + (d*x)/2)*(a - b)*(4*A^2*b^6 + 8*C^2*a^6 + C^2*b^6 - 8*A^2*a*b^5
- 2*C^2*a*b^5 - 8*C^2*a^5*b + 8*A^2*a^2*b^4 + 5*C^2*a^2*b^4 - 8*C^2*a^3*b^3 + 8*C^2*a^4*b^2 + 4*A*C*b^6 - 8*A
*C*a*b^5 + 12*A*C*a^2*b^4 - 16*A*C*a^3*b^3 + 16*A*C*a^4*b^2))/b^4 + (a*(A*b^2 + C*a^2)*(b^2 - a^2)^(1/2)*(16*(
a - b)^2*(2*A*b^2 + 2*C*a^2 + C*b^2 + C*a*b) + (64*a^2*b^2*tan(c/2 + (d*x)/2)*(A*b^2 + C*a^2)*(b^2 - a^2)^(1/2
)*(a - b)^2)/(b^5 - a^2*b^3)))/(b^5 - a^2*b^3)))/(b^5 - a^2*b^3))*(-(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2))/(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)*(A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c)),x)

[Out]

Timed out

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