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

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

[Out]

-2*b*(2*A*a^2-A*b^2+C*a^2)*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+A*
arctanh(sin(d*x+c))/a^2/d+(A*b^2+C*a^2)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.33, antiderivative size = 134, normalized size of antiderivative = 1.00, 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 = {3056, 3001, 3770, 2659, 205} \[ -\frac {2 b \left (2 a^2 A+a^2 C-A b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {A \tanh ^{-1}(\sin (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*(2*a^2*A - A*b^2 + a^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*(a - b)^(3/2)*(a + b)
^(3/2)*d) + (A*ArcTanh[Sin[c + d*x]])/(a^2*d) + ((A*b^2 + a^2*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 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 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx &=\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\int \frac {\left (A \left (a^2-b^2\right )-a 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 {\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {A \int \sec (c+d x) \, dx}{a^2}+\frac {\left (b \left (A b^2-a^2 (2 A+C)\right )\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac {A \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {\left (2 b \left (A b^2-a^2 (2 A+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 b \left (2 a^2 A-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 )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {A \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 1.85, size = 306, normalized size = 2.28 \[ \frac {2 \cos (c+d x) (A \sec (c+d x)+C \cos (c+d x)) \left (\frac {a \left (a^2 C+A b^2\right ) (b \sin (d x)-a \sin (c))}{b (a-b) (a+b) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) (a+b \cos (c+d x))}+\frac {2 b (\sin (c)+i \cos (c)) \left (a^2 (2 A+C)-A b^2\right ) \tan ^{-1}\left (\frac {(\sin (c)+i \cos (c)) \left (\tan \left (\frac {d x}{2}\right ) (b \cos (c)-a)+b \sin (c)\right )}{\sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}}\right )}{\left (a^2-b^2\right ) \sqrt {\left (b^2-a^2\right ) (\cos (c)-i \sin (c))^2}}-A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+A \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{a^2 d (2 A+C \cos (2 (c+d x))+C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Cos[c + d*x]*(C*Cos[c + d*x] + A*Sec[c + d*x])*(-(A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]) + A*Log[Cos[(
c + d*x)/2] + Sin[(c + d*x)/2]] + (2*b*(-(A*b^2) + a^2*(2*A + C))*ArcTan[((I*Cos[c] + Sin[c])*(b*Sin[c] + (-a
+ b*Cos[c])*Tan[(d*x)/2]))/Sqrt[-((a^2 - b^2)*(Cos[c] - I*Sin[c])^2)]]*(I*Cos[c] + Sin[c]))/((a^2 - b^2)*Sqrt[
(-a^2 + b^2)*(Cos[c] - I*Sin[c])^2]) + (a*(A*b^2 + a^2*C)*(-(a*Sin[c]) + b*Sin[d*x]))/((a - b)*b*(a + b)*(a +
b*Cos[c + d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2]))))/(a^2*d*(2*A + C + C*Cos[2*(c + d*x)]))

________________________________________________________________________________________

fricas [B]  time = 5.51, size = 666, normalized size = 4.97 \[ \left [\frac {{\left ({\left (2 \, A + C\right )} a^{3} b - A a b^{3} + {\left ({\left (2 \, A + C\right )} a^{2} b^{2} - A 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 (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4} + {\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4} + {\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{5} + {\left (A - C\right )} a^{3} b^{2} - A 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 ({\left (2 \, A + C\right )} a^{3} b - A a b^{3} + {\left ({\left (2 \, A + C\right )} a^{2} b^{2} - A 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 (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4} + {\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4} + {\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (C a^{5} + {\left (A - C\right )} a^{3} b^{2} - A 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((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 1.22, size = 226, normalized size = 1.69 \[ -\frac {\frac {2 \, {\left (2 \, A a^{2} b + C a^{2} b - A b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}}} - \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} + \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {2 \, {\left (C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(2*A*a^2*b + C*a^2*b - A*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)) - A*log(abs(tan(1/2*d*x +
1/2*c) + 1))/a^2 + A*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 - 2*(C*a^2*tan(1/2*d*x + 1/2*c) + A*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

________________________________________________________________________________________

maple [B]  time = 0.20, size = 342, normalized size = 2.55 \[ \frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A \,b^{2}}{d a \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{d \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}-\frac {4 b \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 b^{3} \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d \,a^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 b \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)/(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)*A*b^2+2/d*a/(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))*A+2/d/a^2*b^3/(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))*A-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))*C-1/d/a^2*A*ln(tan(1/2*d*x+1/2*c)-1)+1/d/a^2*A*ln(tan(1/2*d*x+1/2*c)+1)

________________________________________________________________________________________

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((A+C*cos(d*x+c)^2)*sec(d*x+c)/(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?

________________________________________________________________________________________

mupad [B]  time = 8.46, size = 3850, normalized size = 28.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________