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

Optimal. Leaf size=122 \[ -\frac {2 \left (a^3 C-2 a b^2 C+b^3 B\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {a (b B-a C) \sin (c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {C x}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3021, 2735, 2659, 205} \[ -\frac {2 \left (a^3 C-2 a b^2 C+b^3 B\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {a (b B-a C) \sin (c+d x)}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {C x}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3021

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^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.53, size = 119, normalized size = 0.98 \[ \frac {-\frac {2 \left (a^3 C-2 a b^2 C+b^3 B\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac {a b (b B-a C) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+C (c+d x)}{b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.51, size = 552, normalized size = 4.52 \[ \left [\frac {2 \, {\left (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} d x - {\left (C a^{4} - 2 \, C a^{2} b^{2} + B a b^{3} + {\left (C a^{3} b - 2 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, {\left (C a^{4} b - B a^{3} b^{2} - C a^{2} b^{3} + B a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d\right )}}, \frac {{\left (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5}\right )} d x \cos \left (d x + c\right ) + {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} d x - {\left (C a^{4} - 2 \, C a^{2} b^{2} + B a b^{3} + {\left (C a^{3} b - 2 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (C a^{4} b - B a^{3} b^{2} - C a^{2} b^{3} + B a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(C*a^4*b - 2*C*a^2*b^3 + C*b^5)*d*x*cos(d*x + c) + 2*(C*a^5 - 2*C*a^3*b^2 + C*a*b^4)*d*x - (C*a^4 - 2*
C*a^2*b^2 + B*a*b^3 + (C*a^3*b - 2*C*a*b^3 + B*b^4)*cos(d*x + c))*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^4*b - B*a^3*b^2 - C*a^2*b^3 + B*a*b^4)*sin(d*x + c))/((a^4*b^3 -
2*a^2*b^5 + b^7)*d*cos(d*x + c) + (a^5*b^2 - 2*a^3*b^4 + a*b^6)*d), ((C*a^4*b - 2*C*a^2*b^3 + C*b^5)*d*x*cos(
d*x + c) + (C*a^5 - 2*C*a^3*b^2 + C*a*b^4)*d*x - (C*a^4 - 2*C*a^2*b^2 + B*a*b^3 + (C*a^3*b - 2*C*a*b^3 + B*b^4
)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (C*a^4*b - B*a^
3*b^2 - C*a^2*b^3 + B*a*b^4)*sin(d*x + c))/((a^4*b^3 - 2*a^2*b^5 + b^7)*d*cos(d*x + c) + (a^5*b^2 - 2*a^3*b^4
+ a*b^6)*d)]

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giac [A]  time = 0.22, size = 199, normalized size = 1.63 \[ \frac {\frac {2 \, {\left (C a^{3} - 2 \, C a b^{2} + B b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {{\left (d x + c\right )} C}{b^{2}} - \frac {2 \, {\left (C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b - b^{3}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.12, size = 320, normalized size = 2.62 \[ \frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{d \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{d b \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {2 b \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 a^{3} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d \,b^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {4 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C a}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*a/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)*B-2/d/b*a^2/(a^2-b^2)*t
an(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b+a+b)*C-2/d*b/(a-b)/(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))*B-2/d*a^3/b^2/(a-b)/(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))*C+4/d/(a-b)/(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))*C*a+2/d/b^2*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((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,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 = 9.13, size = 3775, normalized size = 30.94 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*C*atan(((C*((C*((32*(B*a^2*b^7 - C*b^9 - B*b^9 - B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + B*a*b^8
+ 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a
^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4)*32i)/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))*1i)/b^2 + (32*tan(c/2 + (d*x)/2)
*(B^2*b^6 + 2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 -
4*B*C*a*b^5 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))/b^2 - (C*((C*((32*(B*a^2*b^7 - C*b^9 - B*b^9
- B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + B*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) +
(C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4)*32i)/(b^2*(a*b^4
+ b^5 - a^2*b^3 - a^3*b^2)))*1i)/b^2 - (32*tan(c/2 + (d*x)/2)*(B^2*b^6 + 2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2
*C^2*a^5*b + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 4*B*C*a*b^5 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*
b^3 - a^3*b^2)))/b^2)/((64*(C^3*a^5 - B*C^2*b^5 + B^2*C*b^5 + 2*C^3*a*b^4 - C^3*a^4*b + 2*C^3*a^2*b^3 - 3*C^3*
a^3*b^2 - 3*B*C^2*a*b^4 + B*C^2*a^2*b^3 + B*C^2*a^3*b^2))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (C*((C*((32*(B*a
^2*b^7 - C*b^9 - B*b^9 - B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + B*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6
- a^2*b^4 - a^3*b^3) - (C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*
b^4)*32i)/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)))*1i)/b^2 + (32*tan(c/2 + (d*x)/2)*(B^2*b^6 + 2*C^2*a^6 + C^2
*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 4*B*C*a*b^5 + 2*B*C*a^3*b^3
))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))*1i)/b^2 + (C*((C*((32*(B*a^2*b^7 - C*b^9 - B*b^9 - B*a^3*b^6 + C*a^2*b^7
- 3*C*a^3*b^6 + C*a^5*b^4 + B*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (C*tan(c/2 + (d*x)/2)*(
2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4)*32i)/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^
2)))*1i)/b^2 - (32*tan(c/2 + (d*x)/2)*(B^2*b^6 + 2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + 3*C^2*a^2*b
^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 4*B*C*a*b^5 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))*1i)/b^2)
))/(b^2*d) + (atan(((((32*tan(c/2 + (d*x)/2)*(B^2*b^6 + 2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + 3*C^
2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 4*B*C*a*b^5 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) +
(((32*(B*a^2*b^7 - C*b^9 - B*b^9 - B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + B*a*b^8 + 2*C*a*b^8))/(a*
b^5 + b^6 - a^2*b^4 - a^3*b^3) - (32*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*C*a*b^
2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(
b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*C*a*b^2))/(b^8 - 3*a^
2*b^6 + 3*a^4*b^4 - a^6*b^2))*(-(a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*C*a*b^2)*1i)/(b^8 - 3*a^2*b^6 +
3*a^4*b^4 - a^6*b^2) + (((32*tan(c/2 + (d*x)/2)*(B^2*b^6 + 2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + 3
*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 - 4*B*C*a*b^5 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2)
- (((32*(B*a^2*b^7 - C*b^9 - B*b^9 - B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + B*a*b^8 + 2*C*a*b^8))/
(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (32*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*C*a
*b^2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2
)*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*C*a*b^2))/(b^8 - 3
*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))*(-(a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*C*a*b^2)*1i)/(b^8 - 3*a^2*b^6
+ 3*a^4*b^4 - a^6*b^2))/((64*(C^3*a^5 - B*C^2*b^5 + B^2*C*b^5 + 2*C^3*a*b^4 - C^3*a^4*b + 2*C^3*a^2*b^3 - 3*C
^3*a^3*b^2 - 3*B*C^2*a*b^4 + B*C^2*a^2*b^3 + B*C^2*a^3*b^2))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) + (((32*tan(c/2
+ (d*x)/2)*(B^2*b^6 + 2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2
*a^4*b^2 - 4*B*C*a*b^5 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) + (((32*(B*a^2*b^7 - C*b^9 - B*b^9
- B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + B*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) -
(32*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*C*a*b^2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b
^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6
*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*C*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))*(-(
a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*C*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2) - (((32*tan(c/2
+ (d*x)/2)*(B^2*b^6 + 2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2
*a^4*b^2 - 4*B*C*a*b^5 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) - (((32*(B*a^2*b^7 - C*b^9 - B*b^9
- B*a^3*b^6 + C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + B*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) +
(32*tan(c/2 + (d*x)/2)*(-(a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*C*a*b^2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b
^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/((a*b^4 + b^5 - a^2*b^3 - a^3*b^2)*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6
*b^2)))*(-(a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*C*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2))*(-(
a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*C*a*b^2))/(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)))*(-(a + b)^3*(
a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*C*a*b^2)*2i)/(d*(b^8 - 3*a^2*b^6 + 3*a^4*b^4 - a^6*b^2)) - (2*tan(c/2 + (d*
x)/2)*(C*a^2 - B*a*b))/(d*(a + b)*(a*b - b^2)*(a + b + tan(c/2 + (d*x)/2)^2*(a - b)))

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

[Out]

Timed out

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