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

Optimal. Leaf size=298 \[ \frac{\sin (c+d x) \left (-a^2 b (2 A-C)+a^3 B-2 a b^2 B+3 A b^3\right )}{a^3 d \left (a^2-b^2\right )}-\frac{\sin (c+d x) \cos (c+d x) \left (a^2 (-(A-2 C))-2 a b B+3 A b^2\right )}{2 a^2 d \left (a^2-b^2\right )}-\frac{2 b \left (4 a^2 A b^2-a^2 b^2 C-3 a^3 b B+2 a^4 C+2 a b^3 B-3 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\sin (c+d x) \cos (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{x \left (a^2 (A+2 C)-4 a b B+6 A b^2\right )}{2 a^4} \]

[Out]

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

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Rubi [A]  time = 1.23513, antiderivative size = 298, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4100, 4104, 3919, 3831, 2659, 208} \[ \frac{\sin (c+d x) \left (-a^2 b (2 A-C)+a^3 B-2 a b^2 B+3 A b^3\right )}{a^3 d \left (a^2-b^2\right )}-\frac{\sin (c+d x) \cos (c+d x) \left (a^2 (-(A-2 C))-2 a b B+3 A b^2\right )}{2 a^2 d \left (a^2-b^2\right )}-\frac{2 b \left (4 a^2 A b^2-a^2 b^2 C-3 a^3 b B+2 a^4 C+2 a b^3 B-3 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\sin (c+d x) \cos (c+d x) \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{x \left (a^2 (A+2 C)-4 a b B+6 A b^2\right )}{2 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4100

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*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a +
b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), I
nt[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*
(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] &
& ILtQ[n, 0])

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]

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx &=\frac{\left (A b^2-a (b B-a C)\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\cos ^2(c+d x) \left (3 A b^2-2 a b B-a^2 (A-2 C)+a (A b-a B+b C) \sec (c+d x)-2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{\left (3 A b^2-2 a b B-a^2 (A-2 C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\int \frac{\cos (c+d x) \left (2 \left (3 A b^3+a^3 B-2 a b^2 B-a^2 b (2 A-C)\right )+a \left (A b^2-2 a b B+a^2 (A+2 C)\right ) \sec (c+d x)-b \left (3 A b^2-2 a b B-a^2 (A-2 C)\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac{\left (3 A b^3+a^3 B-2 a b^2 B-a^2 b (2 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (3 A b^2-2 a b B-a^2 (A-2 C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{-\left (a^2-b^2\right ) \left (6 A b^2-4 a b B+a^2 (A+2 C)\right )+a b \left (3 A b^2-2 a b B-a^2 (A-2 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac{\left (6 A b^2-4 a b B+a^2 (A+2 C)\right ) x}{2 a^4}+\frac{\left (3 A b^3+a^3 B-2 a b^2 B-a^2 b (2 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (3 A b^2-2 a b B-a^2 (A-2 C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (b \left (3 A b^4+3 a^3 b B-2 a b^3 B-a^2 b^2 (4 A-C)-2 a^4 C\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^4 \left (a^2-b^2\right )}\\ &=\frac{\left (6 A b^2-4 a b B+a^2 (A+2 C)\right ) x}{2 a^4}+\frac{\left (3 A b^3+a^3 B-2 a b^2 B-a^2 b (2 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (3 A b^2-2 a b B-a^2 (A-2 C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (3 A b^4+3 a^3 b B-2 a b^3 B-a^2 b^2 (4 A-C)-2 a^4 C\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^4 \left (a^2-b^2\right )}\\ &=\frac{\left (6 A b^2-4 a b B+a^2 (A+2 C)\right ) x}{2 a^4}+\frac{\left (3 A b^3+a^3 B-2 a b^2 B-a^2 b (2 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (3 A b^2-2 a b B-a^2 (A-2 C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (2 \left (3 A b^4+3 a^3 b B-2 a b^3 B-a^2 b^2 (4 A-C)-2 a^4 C\right )\right ) \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^4 \left (a^2-b^2\right ) d}\\ &=\frac{\left (6 A b^2-4 a b B+a^2 (A+2 C)\right ) x}{2 a^4}-\frac{2 b \left (4 a^2 A b^2-3 A b^4-3 a^3 b B+2 a b^3 B+2 a^4 C-a^2 b^2 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}+\frac{\left (3 A b^3+a^3 B-2 a b^2 B-a^2 b (2 A-C)\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}-\frac{\left (3 A b^2-2 a b B-a^2 (A-2 C)\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac{\left (A b^2-a (b B-a C)\right ) \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.40944, size = 206, normalized size = 0.69 \[ \frac{2 (c+d x) \left (a^2 (A+2 C)-4 a b B+6 A b^2\right )-\frac{8 b \left (a^2 b^2 (C-4 A)+3 a^3 b B-2 a^4 C-2 a b^3 B+3 A b^4\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+a^2 A \sin (2 (c+d x))+\frac{4 a b^2 \sin (c+d x) \left (a (a C-b B)+A b^2\right )}{(a-b) (a+b) (a \cos (c+d x)+b)}+4 a (a B-2 A b) \sin (c+d x)}{4 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(6*A*b^2 - 4*a*b*B + a^2*(A + 2*C))*(c + d*x) - (8*b*(3*A*b^4 + 3*a^3*b*B - 2*a*b^3*B - 2*a^4*C + a^2*b^2*(
-4*A + C))*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) + 4*a*(-2*A*b + a*B)*Sin[c
+ d*x] + (4*a*b^2*(A*b^2 + a*(-(b*B) + a*C))*Sin[c + d*x])/((a - b)*(a + b)*(b + a*Cos[c + d*x])) + a^2*A*Sin[
2*(c + d*x)])/(4*a^4*d)

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Maple [B]  time = 0.138, size = 857, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*A-4/d/a^3/(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^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*B+1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*
d*x+1/2*c)*A-4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)*A*b+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(
1/2*d*x+1/2*c)*B+1/d*A/a^2*arctan(tan(1/2*d*x+1/2*c))+6/d/a^4*arctan(tan(1/2*d*x+1/2*c))*A*b^2-4/d/a^3*arctan(
tan(1/2*d*x+1/2*c))*B*b+2/d/a^2*arctan(tan(1/2*d*x+1/2*c))*C-2/d*b^4/a^3/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2
*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)*A+2/d*b^3/a^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a
-tan(1/2*d*x+1/2*c)^2*b-a-b)*B-2/d*b^2/a/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*
c)^2*b-a-b)*C-8/d/a^2/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A*
b^3+6/d*b^5/a^4/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*A+6/d*b^
2/a/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B-4/d*b^4/a^3/(a+b)/
(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B-4/d/(a+b)/(a-b)/((a+b)*(a-b)
)^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*b*C+2/d*b^3/a^2/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*
arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*C

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.819258, size = 2398, normalized size = 8.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(((A + 2*C)*a^7 - 4*B*a^6*b + 4*(A - C)*a^5*b^2 + 8*B*a^4*b^3 - (11*A - 2*C)*a^3*b^4 - 4*B*a^2*b^5 + 6*A*
a*b^6)*d*x*cos(d*x + c) + ((A + 2*C)*a^6*b - 4*B*a^5*b^2 + 4*(A - C)*a^4*b^3 + 8*B*a^3*b^4 - (11*A - 2*C)*a^2*
b^5 - 4*B*a*b^6 + 6*A*b^7)*d*x + (2*C*a^4*b^2 - 3*B*a^3*b^3 + (4*A - C)*a^2*b^4 + 2*B*a*b^5 - 3*A*b^6 + (2*C*a
^5*b - 3*B*a^4*b^2 + (4*A - C)*a^3*b^3 + 2*B*a^2*b^4 - 3*A*a*b^5)*cos(d*x + c))*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^6*b - 2*(2*A - C)*a^5*b^2 - 6*B*a^4*b^3 + 2*(5*A - C
)*a^3*b^4 + 4*B*a^2*b^5 - 6*A*a*b^6 + (A*a^7 - 2*A*a^5*b^2 + A*a^3*b^4)*cos(d*x + c)^2 + (2*B*a^7 - 3*A*a^6*b
- 4*B*a^5*b^2 + 6*A*a^4*b^3 + 2*B*a^3*b^4 - 3*A*a^2*b^5)*cos(d*x + c))*sin(d*x + c))/((a^9 - 2*a^7*b^2 + a^5*b
^4)*d*cos(d*x + c) + (a^8*b - 2*a^6*b^3 + a^4*b^5)*d), 1/2*(((A + 2*C)*a^7 - 4*B*a^6*b + 4*(A - C)*a^5*b^2 + 8
*B*a^4*b^3 - (11*A - 2*C)*a^3*b^4 - 4*B*a^2*b^5 + 6*A*a*b^6)*d*x*cos(d*x + c) + ((A + 2*C)*a^6*b - 4*B*a^5*b^2
+ 4*(A - C)*a^4*b^3 + 8*B*a^3*b^4 - (11*A - 2*C)*a^2*b^5 - 4*B*a*b^6 + 6*A*b^7)*d*x - 2*(2*C*a^4*b^2 - 3*B*a^
3*b^3 + (4*A - C)*a^2*b^4 + 2*B*a*b^5 - 3*A*b^6 + (2*C*a^5*b - 3*B*a^4*b^2 + (4*A - C)*a^3*b^3 + 2*B*a^2*b^4 -
3*A*a*b^5)*cos(d*x + c))*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^6*b - 2*(2*A - C)*a^5*b^2 - 6*B*a^4*b^3 + 2*(5*A - C)*a^3*b^4 + 4*B*a^2*b^5 - 6*A*a*b^6 + (A*a
^7 - 2*A*a^5*b^2 + A*a^3*b^4)*cos(d*x + c)^2 + (2*B*a^7 - 3*A*a^6*b - 4*B*a^5*b^2 + 6*A*a^4*b^3 + 2*B*a^3*b^4
- 3*A*a^2*b^5)*cos(d*x + c))*sin(d*x + c))/((a^9 - 2*a^7*b^2 + a^5*b^4)*d*cos(d*x + c) + (a^8*b - 2*a^6*b^3 +
a^4*b^5)*d)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.27657, size = 513, normalized size = 1.72 \begin{align*} -\frac{\frac{4 \,{\left (2 \, C a^{4} b - 3 \, B a^{3} b^{2} + 4 \, A a^{2} b^{3} - C a^{2} b^{3} + 2 \, B a b^{4} - 3 \, A b^{5}\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^{6} - a^{4} b^{2}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{4 \,{\left (C a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{5} - a^{3} 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 )}} - \frac{{\left (A a^{2} + 2 \, C a^{2} - 4 \, B a b + 6 \, A b^{2}\right )}{\left (d x + c\right )}}{a^{4}} + \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} + 4 \, 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 ) + 4 \, 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^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*(2*C*a^4*b - 3*B*a^3*b^2 + 4*A*a^2*b^3 - C*a^2*b^3 + 2*B*a*b^4 - 3*A*b^5)*(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^6 - a
^4*b^2)*sqrt(-a^2 + b^2)) + 4*(C*a^2*b^2*tan(1/2*d*x + 1/2*c) - B*a*b^3*tan(1/2*d*x + 1/2*c) + A*b^4*tan(1/2*d
*x + 1/2*c))/((a^5 - a^3*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)) - (A*a^2 + 2*C*a^
2 - 4*B*a*b + 6*A*b^2)*(d*x + c)/a^4 + 2*(A*a*tan(1/2*d*x + 1/2*c)^3 - 2*B*a*tan(1/2*d*x + 1/2*c)^3 + 4*A*b*ta
n(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) + 4*A*b*tan(1/2*d*x + 1/2*c))/((t
an(1/2*d*x + 1/2*c)^2 + 1)^2*a^3))/d