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

Optimal. Leaf size=107 \[ \frac {2 \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d \sqrt {a-b} \sqrt {a+b}}-\frac {(A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {A \tan (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3055, 3001, 3770, 2659, 205} \[ \frac {2 \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d \sqrt {a-b} \sqrt {a+b}}-\frac {(A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {A \tan (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b^2 - a*(b*B - a*C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*Sqrt[a - b]*Sqrt[a + b]*d)
- ((A*b - a*B)*ArcTanh[Sin[c + d*x]])/(a^2*d) + (A*Tan[c + d*x])/(a*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 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 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(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)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[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[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[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+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx &=\frac {A \tan (c+d x)}{a d}+\frac {\int \frac {(-A b+a B+a C \cos (c+d x)) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{a}\\ &=\frac {A \tan (c+d x)}{a d}-\frac {(A b-a B) \int \sec (c+d x) \, dx}{a^2}+\frac {\left (-b (-A b+a B)+a^2 C\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{a^2}\\ &=-\frac {(A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {A \tan (c+d x)}{a d}+\frac {\left (2 \left (A b^2-a (b B-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 d}\\ &=\frac {2 \left (A b^2-a (b B-a C)\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b} d}-\frac {(A b-a B) \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {A \tan (c+d x)}{a d}\\ \end {align*}

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Mathematica [C]  time = 2.77, size = 339, normalized size = 3.17 \[ \frac {2 \cos ^2(c+d x) \left (A \sec ^2(c+d x)+B \sec (c+d x)+C\right ) \left (-\frac {2 i (\cos (c)-i \sin (c)) \left (a (a C-b B)+A b^2\right ) \tan ^{-1}\left (\frac {(\sin (c)+i \cos (c)) \left (\tan \left (\frac {d x}{2}\right ) (b \cos (c)-a)+b \sin (c)\right )}{\sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)-i \sin (c))^2\right )}}\right )}{\sqrt {\left (b^2-a^2\right ) (\cos (c)-i \sin (c))^2}}+(A b-a B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+(a B-A b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a A \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {a A \sin \left (\frac {d x}{2}\right )}{\left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{a^2 d (2 A+2 B \cos (c+d x)+C \cos (2 (c+d x))+C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [B]  time = 5.13, size = 470, normalized size = 4.39 \[ \left [-\frac {{\left (C a^{2} - B a b + A b^{2}\right )} \sqrt {-a^{2} + b^{2}} \cos \left (d x + c\right ) \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A a^{3} - A a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}, \frac {2 \, {\left (C a^{2} - B a b + A b^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) + {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{3} - A a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.26, size = 180, normalized size = 1.68 \[ \frac {\frac {{\left (B a - A b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {{\left (B a - A b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {2 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a} - \frac {2 \, {\left (C a^{2} - B a b + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.22, size = 272, normalized size = 2.54 \[ \frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A \,b^{2}}{d \,a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B b}{d a \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {A}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) B}{a d}-\frac {A}{a d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {A b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) B}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/a^2/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A*b^2-2/d/a/((a-b)*(a+b))^(1/
2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B*b+2/d/((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/a/d*A/(tan(1/2*d*x+1/2*c)-1)+1/d*A*b/a^2*ln(tan(1/2*d*x+1/2*c)-1)-1/a/d*ln(tan(
1/2*d*x+1/2*c)-1)*B-1/a/d*A/(tan(1/2*d*x+1/2*c)+1)-1/d*A*b/a^2*ln(tan(1/2*d*x+1/2*c)+1)+1/a/d*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((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(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 = 8.09, size = 3483, normalized size = 32.55 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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