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

Optimal. Leaf size=133 \[ \frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {B \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac {2 \left (a^3 (-C)+2 a^2 b B-b^3 B\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.39, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3029, 3000, 3001, 3770, 2659, 205} \[ -\frac {2 \left (2 a^2 b B+a^3 (-C)-b^3 B\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {B \tanh ^{-1}(\sin (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*a^2*b*B - b^3*B - a^3*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*(a - b)^(3/2)*(a + b)
^(3/2)*d) + (B*ArcTanh[Sin[c + d*x]])/(a^2*d) + (b*(b*B - a*C)*Sin[c + d*x])/(a*(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 3000

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((A*b^2 - a*b*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*
Sin[e + f*x])^(1 + n))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m
+ n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B)*(m + n + 3)*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{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] && RationalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n,
-1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.60, size = 191, normalized size = 1.44 \[ \frac {\cos (c+d x) (B \sec (c+d x)+C) \left (\frac {2 \left (a^3 C-2 a^2 b B+b^3 B\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}+\frac {a b (b B-a C) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}-B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+B \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{a^2 d (B+C \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 5.06, size = 684, normalized size = 5.14 \[ \left [\frac {{\left (C a^{4} - 2 \, B a^{3} b + B a b^{3} + {\left (C a^{3} b - 2 \, B a^{2} b^{2} + 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 ) + {\left (B a^{5} - 2 \, B a^{3} b^{2} + B a b^{4} + {\left (B a^{4} b - 2 \, B a^{2} b^{3} + B b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{5} - 2 \, B a^{3} b^{2} + B a b^{4} + {\left (B a^{4} b - 2 \, B a^{2} b^{3} + B b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\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^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d\right )}}, \frac {2 \, {\left (C a^{4} - 2 \, B a^{3} b + B a b^{3} + {\left (C a^{3} b - 2 \, B a^{2} b^{2} + 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 (B a^{5} - 2 \, B a^{3} b^{2} + B a b^{4} + {\left (B a^{4} b - 2 \, B a^{2} b^{3} + B b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{5} - 2 \, B a^{3} b^{2} + B a b^{4} + {\left (B a^{4} b - 2 \, B a^{2} b^{3} + B b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\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^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.47, size = 225, normalized size = 1.69 \[ -\frac {\frac {2 \, {\left (C a^{3} - 2 \, B a^{2} b + B b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}}} - \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} + \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, {\left (C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(C*a^3 - 2*B*a^2*b + 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^4 - a^2*b^2)*sqrt(a^2 - b^2)) - B*log(abs(tan(1/2*d*x +
1/2*c) + 1))/a^2 + B*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 + 2*(C*a*b*tan(1/2*d*x + 1/2*c) - B*b^2*tan(1/2*d*
x + 1/2*c))/((a^3 - a*b^2)*(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 = 0.24, size = 342, normalized size = 2.57 \[ \frac {2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{d a \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{d \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {4 b \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) b^{3} B}{d \,a^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C a}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{d \,a^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*sec(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 = 9.19, size = 3763, normalized size = 28.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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