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

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

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4034, 4104, 3919, 3831, 2659, 208} \[ -\frac {2 b^2 (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d \sqrt {a-b} \sqrt {a+b}}+\frac {x \left (a^2 A-2 a b B+2 A b^2\right )}{2 a^3}-\frac {(A b-a B) \sin (c+d x)}{a^2 d}+\frac {A \sin (c+d x) \cos (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[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 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 4034

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/(a*f*n), x]
+ Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + A*a*(n +
1)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b
- a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 4104

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rubi steps

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

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Mathematica [A]  time = 0.36, size = 121, normalized size = 0.90 \[ \frac {2 (c+d x) \left (a^2 A-2 a b B+2 A b^2\right )+\frac {8 b^2 (A b-a B) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+a^2 A \sin (2 (c+d x))+4 a (a B-A b) \sin (c+d x)}{4 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 427, normalized size = 3.19 \[ \left [\frac {{\left (A a^{4} - 2 \, B a^{3} b + A a^{2} b^{2} + 2 \, B a b^{3} - 2 \, A b^{4}\right )} d x - {\left (B a b^{2} - A b^{3}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (2 \, B a^{4} - 2 \, A a^{3} b - 2 \, B a^{2} b^{2} + 2 \, A a b^{3} + {\left (A a^{4} - A a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{5} - a^{3} b^{2}\right )} d}, \frac {{\left (A a^{4} - 2 \, B a^{3} b + A a^{2} b^{2} + 2 \, B a b^{3} - 2 \, A b^{4}\right )} d x + 2 \, {\left (B a b^{2} - A b^{3}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (2 \, B a^{4} - 2 \, A a^{3} b - 2 \, B a^{2} b^{2} + 2 \, A a b^{3} + {\left (A a^{4} - A a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{5} - a^{3} b^{2}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.27, size = 227, normalized size = 1.69 \[ \frac {\frac {{\left (A a^{2} - 2 \, B a b + 2 \, A b^{2}\right )} {\left (d x + c\right )}}{a^{3}} + \frac {4 \, {\left (B a b^{2} - A b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{3}} - \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.14, size = 367, normalized size = 2.74 \[ -\frac {2 b^{3} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d \,a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 b^{2} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d \,a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A b}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A}{d a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A b}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{d a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A \,b^{2}}{d \,a^{3}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B b}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d*b^3/a^3/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A+2/d*b^2/a^2/((a-b)*(a
+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B-1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*
x+1/2*c)^3*A-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*A*b+2/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(
1/2*d*x+1/2*c)^3*B+1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)*A-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*ta
n(1/2*d*x+1/2*c)*A*b+2/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)*B+1/d*A/a*arctan(tan(1/2*d*x+1/2*c))+
2/d/a^3*arctan(tan(1/2*d*x+1/2*c))*A*b^2-2/d/a^2*arctan(tan(1/2*d*x+1/2*c))*B*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(cos(d*x+c)^2*(A+B*sec(d*x+c))/(a+b*sec(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*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 6.02, size = 3740, normalized size = 27.91 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((tan(c/2 + (d*x)/2)*(A*a - 2*A*b + 2*B*a))/a^2 - (tan(c/2 + (d*x)/2)^3*(A*a + 2*A*b - 2*B*a))/a^2)/(d*(2*tan(
c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 + 1)) - (atan(((((((8*(2*A*a^10 + 4*A*a^6*b^4 - 6*A*a^7*b^3 + 2*A*a^8*
b^2 - 4*B*a^7*b^3 + 8*B*a^8*b^2 - 2*A*a^9*b - 4*B*a^9*b))/a^6 - (4*tan(c/2 + (d*x)/2)*(A*a^2*1i + A*b^2*2i - B
*a*b*2i)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/a^7)*(A*a^2*1i + A*b^2*2i - B*a*b*2i))/(2*a^3) + (8*tan(c/2 + (d*
x)/2)*(A^2*a^7 - 8*A^2*b^7 + 16*A^2*a*b^6 - 3*A^2*a^6*b - 16*A^2*a^2*b^5 + 16*A^2*a^3*b^4 - 13*A^2*a^4*b^3 + 7
*A^2*a^5*b^2 - 8*B^2*a^2*b^5 + 16*B^2*a^3*b^4 - 12*B^2*a^4*b^3 + 4*B^2*a^5*b^2 + 16*A*B*a*b^6 - 4*A*B*a^6*b -
32*A*B*a^2*b^5 + 28*A*B*a^3*b^4 - 20*A*B*a^4*b^3 + 12*A*B*a^5*b^2))/a^4)*(A*a^2*1i + A*b^2*2i - B*a*b*2i)*1i)/
(2*a^3) - (((((8*(2*A*a^10 + 4*A*a^6*b^4 - 6*A*a^7*b^3 + 2*A*a^8*b^2 - 4*B*a^7*b^3 + 8*B*a^8*b^2 - 2*A*a^9*b -
4*B*a^9*b))/a^6 + (4*tan(c/2 + (d*x)/2)*(A*a^2*1i + A*b^2*2i - B*a*b*2i)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/
a^7)*(A*a^2*1i + A*b^2*2i - B*a*b*2i))/(2*a^3) - (8*tan(c/2 + (d*x)/2)*(A^2*a^7 - 8*A^2*b^7 + 16*A^2*a*b^6 - 3
*A^2*a^6*b - 16*A^2*a^2*b^5 + 16*A^2*a^3*b^4 - 13*A^2*a^4*b^3 + 7*A^2*a^5*b^2 - 8*B^2*a^2*b^5 + 16*B^2*a^3*b^4
- 12*B^2*a^4*b^3 + 4*B^2*a^5*b^2 + 16*A*B*a*b^6 - 4*A*B*a^6*b - 32*A*B*a^2*b^5 + 28*A*B*a^3*b^4 - 20*A*B*a^4*
b^3 + 12*A*B*a^5*b^2))/a^4)*(A*a^2*1i + A*b^2*2i - B*a*b*2i)*1i)/(2*a^3))/((16*(4*A^3*b^8 - 6*A^3*a*b^7 + 6*A^
3*a^2*b^6 - 5*A^3*a^3*b^5 + 2*A^3*a^4*b^4 - A^3*a^5*b^3 - 4*B^3*a^3*b^5 + 4*B^3*a^4*b^4 - 12*A^2*B*a*b^7 + 12*
A*B^2*a^2*b^6 - 14*A*B^2*a^3*b^5 + 6*A*B^2*a^4*b^4 - 4*A*B^2*a^5*b^3 + 16*A^2*B*a^2*b^6 - 12*A^2*B*a^3*b^5 + 9
*A^2*B*a^4*b^4 - 2*A^2*B*a^5*b^3 + A^2*B*a^6*b^2))/a^6 + (((((8*(2*A*a^10 + 4*A*a^6*b^4 - 6*A*a^7*b^3 + 2*A*a^
8*b^2 - 4*B*a^7*b^3 + 8*B*a^8*b^2 - 2*A*a^9*b - 4*B*a^9*b))/a^6 - (4*tan(c/2 + (d*x)/2)*(A*a^2*1i + A*b^2*2i -
B*a*b*2i)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/a^7)*(A*a^2*1i + A*b^2*2i - B*a*b*2i))/(2*a^3) + (8*tan(c/2 + (
d*x)/2)*(A^2*a^7 - 8*A^2*b^7 + 16*A^2*a*b^6 - 3*A^2*a^6*b - 16*A^2*a^2*b^5 + 16*A^2*a^3*b^4 - 13*A^2*a^4*b^3 +
7*A^2*a^5*b^2 - 8*B^2*a^2*b^5 + 16*B^2*a^3*b^4 - 12*B^2*a^4*b^3 + 4*B^2*a^5*b^2 + 16*A*B*a*b^6 - 4*A*B*a^6*b
- 32*A*B*a^2*b^5 + 28*A*B*a^3*b^4 - 20*A*B*a^4*b^3 + 12*A*B*a^5*b^2))/a^4)*(A*a^2*1i + A*b^2*2i - B*a*b*2i))/(
2*a^3) + (((((8*(2*A*a^10 + 4*A*a^6*b^4 - 6*A*a^7*b^3 + 2*A*a^8*b^2 - 4*B*a^7*b^3 + 8*B*a^8*b^2 - 2*A*a^9*b -
4*B*a^9*b))/a^6 + (4*tan(c/2 + (d*x)/2)*(A*a^2*1i + A*b^2*2i - B*a*b*2i)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/a
^7)*(A*a^2*1i + A*b^2*2i - B*a*b*2i))/(2*a^3) - (8*tan(c/2 + (d*x)/2)*(A^2*a^7 - 8*A^2*b^7 + 16*A^2*a*b^6 - 3*
A^2*a^6*b - 16*A^2*a^2*b^5 + 16*A^2*a^3*b^4 - 13*A^2*a^4*b^3 + 7*A^2*a^5*b^2 - 8*B^2*a^2*b^5 + 16*B^2*a^3*b^4
- 12*B^2*a^4*b^3 + 4*B^2*a^5*b^2 + 16*A*B*a*b^6 - 4*A*B*a^6*b - 32*A*B*a^2*b^5 + 28*A*B*a^3*b^4 - 20*A*B*a^4*b
^3 + 12*A*B*a^5*b^2))/a^4)*(A*a^2*1i + A*b^2*2i - B*a*b*2i))/(2*a^3)))*(A*a^2*1i + A*b^2*2i - B*a*b*2i)*1i)/(a
^3*d) - (b^2*atan(((b^2*((a + b)*(a - b))^(1/2)*((8*tan(c/2 + (d*x)/2)*(A^2*a^7 - 8*A^2*b^7 + 16*A^2*a*b^6 - 3
*A^2*a^6*b - 16*A^2*a^2*b^5 + 16*A^2*a^3*b^4 - 13*A^2*a^4*b^3 + 7*A^2*a^5*b^2 - 8*B^2*a^2*b^5 + 16*B^2*a^3*b^4
- 12*B^2*a^4*b^3 + 4*B^2*a^5*b^2 + 16*A*B*a*b^6 - 4*A*B*a^6*b - 32*A*B*a^2*b^5 + 28*A*B*a^3*b^4 - 20*A*B*a^4*
b^3 + 12*A*B*a^5*b^2))/a^4 + (b^2*((a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*(2*A*a^10 + 4*A*a^6*b^4 - 6*A*a^7*b^
3 + 2*A*a^8*b^2 - 4*B*a^7*b^3 + 8*B*a^8*b^2 - 2*A*a^9*b - 4*B*a^9*b))/a^6 - (8*b^2*tan(c/2 + (d*x)/2)*((a + b)
*(a - b))^(1/2)*(A*b - B*a)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/(a^4*(a^5 - a^3*b^2))))/(a^5 - a^3*b^2))*(A*b
- B*a)*1i)/(a^5 - a^3*b^2) + (b^2*((a + b)*(a - b))^(1/2)*((8*tan(c/2 + (d*x)/2)*(A^2*a^7 - 8*A^2*b^7 + 16*A^2
*a*b^6 - 3*A^2*a^6*b - 16*A^2*a^2*b^5 + 16*A^2*a^3*b^4 - 13*A^2*a^4*b^3 + 7*A^2*a^5*b^2 - 8*B^2*a^2*b^5 + 16*B
^2*a^3*b^4 - 12*B^2*a^4*b^3 + 4*B^2*a^5*b^2 + 16*A*B*a*b^6 - 4*A*B*a^6*b - 32*A*B*a^2*b^5 + 28*A*B*a^3*b^4 - 2
0*A*B*a^4*b^3 + 12*A*B*a^5*b^2))/a^4 - (b^2*((a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*(2*A*a^10 + 4*A*a^6*b^4 -
6*A*a^7*b^3 + 2*A*a^8*b^2 - 4*B*a^7*b^3 + 8*B*a^8*b^2 - 2*A*a^9*b - 4*B*a^9*b))/a^6 + (8*b^2*tan(c/2 + (d*x)/2
)*((a + b)*(a - b))^(1/2)*(A*b - B*a)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/(a^4*(a^5 - a^3*b^2))))/(a^5 - a^3*b
^2))*(A*b - B*a)*1i)/(a^5 - a^3*b^2))/((16*(4*A^3*b^8 - 6*A^3*a*b^7 + 6*A^3*a^2*b^6 - 5*A^3*a^3*b^5 + 2*A^3*a^
4*b^4 - A^3*a^5*b^3 - 4*B^3*a^3*b^5 + 4*B^3*a^4*b^4 - 12*A^2*B*a*b^7 + 12*A*B^2*a^2*b^6 - 14*A*B^2*a^3*b^5 + 6
*A*B^2*a^4*b^4 - 4*A*B^2*a^5*b^3 + 16*A^2*B*a^2*b^6 - 12*A^2*B*a^3*b^5 + 9*A^2*B*a^4*b^4 - 2*A^2*B*a^5*b^3 + A
^2*B*a^6*b^2))/a^6 + (b^2*((a + b)*(a - b))^(1/2)*((8*tan(c/2 + (d*x)/2)*(A^2*a^7 - 8*A^2*b^7 + 16*A^2*a*b^6 -
3*A^2*a^6*b - 16*A^2*a^2*b^5 + 16*A^2*a^3*b^4 - 13*A^2*a^4*b^3 + 7*A^2*a^5*b^2 - 8*B^2*a^2*b^5 + 16*B^2*a^3*b
^4 - 12*B^2*a^4*b^3 + 4*B^2*a^5*b^2 + 16*A*B*a*b^6 - 4*A*B*a^6*b - 32*A*B*a^2*b^5 + 28*A*B*a^3*b^4 - 20*A*B*a^
4*b^3 + 12*A*B*a^5*b^2))/a^4 + (b^2*((a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*(2*A*a^10 + 4*A*a^6*b^4 - 6*A*a^7*
b^3 + 2*A*a^8*b^2 - 4*B*a^7*b^3 + 8*B*a^8*b^2 - 2*A*a^9*b - 4*B*a^9*b))/a^6 - (8*b^2*tan(c/2 + (d*x)/2)*((a +
b)*(a - b))^(1/2)*(A*b - B*a)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/(a^4*(a^5 - a^3*b^2))))/(a^5 - a^3*b^2))*(A*
b - B*a))/(a^5 - a^3*b^2) - (b^2*((a + b)*(a - b))^(1/2)*((8*tan(c/2 + (d*x)/2)*(A^2*a^7 - 8*A^2*b^7 + 16*A^2*
a*b^6 - 3*A^2*a^6*b - 16*A^2*a^2*b^5 + 16*A^2*a^3*b^4 - 13*A^2*a^4*b^3 + 7*A^2*a^5*b^2 - 8*B^2*a^2*b^5 + 16*B^
2*a^3*b^4 - 12*B^2*a^4*b^3 + 4*B^2*a^5*b^2 + 16*A*B*a*b^6 - 4*A*B*a^6*b - 32*A*B*a^2*b^5 + 28*A*B*a^3*b^4 - 20
*A*B*a^4*b^3 + 12*A*B*a^5*b^2))/a^4 - (b^2*((a + b)*(a - b))^(1/2)*(A*b - B*a)*((8*(2*A*a^10 + 4*A*a^6*b^4 - 6
*A*a^7*b^3 + 2*A*a^8*b^2 - 4*B*a^7*b^3 + 8*B*a^8*b^2 - 2*A*a^9*b - 4*B*a^9*b))/a^6 + (8*b^2*tan(c/2 + (d*x)/2)
*((a + b)*(a - b))^(1/2)*(A*b - B*a)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2))/(a^4*(a^5 - a^3*b^2))))/(a^5 - a^3*b^
2))*(A*b - B*a))/(a^5 - a^3*b^2)))*((a + b)*(a - b))^(1/2)*(A*b - B*a)*2i)/(d*(a^5 - a^3*b^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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