3.921 \(\int \frac{\cos (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=330 \[ -\frac{\sin (c+d x) \left (11 a^2 A b^2+a^4 (-(2 A-3 C))-5 a^3 b B+2 a b^3 B-6 A b^4\right )}{2 a^3 d \left (a^2-b^2\right )^2}-\frac{\left (-a^4 b^2 (12 A+C)+15 a^2 A b^4-5 a^3 b^3 B+6 a^5 b B-2 a^6 C+2 a b^5 B-6 A b^6\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\sin (c+d x) \left (-a^2 b^2 (6 A+C)+4 a^3 b B-2 a^4 C-a b^3 B+3 A b^4\right )}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{x (3 A b-a B)}{a^4} \]

[Out]

-(((3*A*b - a*B)*x)/a^4) - ((15*a^2*A*b^4 - 6*A*b^6 + 6*a^5*b*B - 5*a^3*b^3*B + 2*a*b^5*B - 2*a^6*C - a^4*b^2*
(12*A + C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*(a - b)^(5/2)*(a + b)^(5/2)*d) - ((11*a^
2*A*b^2 - 6*A*b^4 - 5*a^3*b*B + 2*a*b^3*B - a^4*(2*A - 3*C))*Sin[c + d*x])/(2*a^3*(a^2 - b^2)^2*d) + ((A*b^2 -
a*(b*B - a*C))*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) - ((3*A*b^4 + 4*a^3*b*B - a*b^3*B - 2
*a^4*C - a^2*b^2*(6*A + C))*Sin[c + d*x])/(2*a^2*(a^2 - b^2)^2*d*(a + b*Sec[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 3.14895, antiderivative size = 330, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4100, 4104, 3919, 3831, 2659, 208} \[ -\frac{\sin (c+d x) \left (11 a^2 A b^2+a^4 (-(2 A-3 C))-5 a^3 b B+2 a b^3 B-6 A b^4\right )}{2 a^3 d \left (a^2-b^2\right )^2}-\frac{\left (-a^4 b^2 (12 A+C)+15 a^2 A b^4-5 a^3 b^3 B+6 a^5 b B-2 a^6 C+2 a b^5 B-6 A b^6\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\sin (c+d x) \left (-a^2 b^2 (6 A+C)+4 a^3 b B-2 a^4 C-a b^3 B+3 A b^4\right )}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}+\frac{\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac{x (3 A b-a B)}{a^4} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^3,x]

[Out]

-(((3*A*b - a*B)*x)/a^4) - ((15*a^2*A*b^4 - 6*A*b^6 + 6*a^5*b*B - 5*a^3*b^3*B + 2*a*b^5*B - 2*a^6*C - a^4*b^2*
(12*A + C))*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*(a - b)^(5/2)*(a + b)^(5/2)*d) - ((11*a^
2*A*b^2 - 6*A*b^4 - 5*a^3*b*B + 2*a*b^3*B - a^4*(2*A - 3*C))*Sin[c + d*x])/(2*a^3*(a^2 - b^2)^2*d) + ((A*b^2 -
a*(b*B - a*C))*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) - ((3*A*b^4 + 4*a^3*b*B - a*b^3*B - 2
*a^4*C - a^2*b^2*(6*A + C))*Sin[c + d*x])/(2*a^2*(a^2 - b^2)^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 (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx &=\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\int \frac{\cos (c+d x) \left (3 A b^2-a b B-a^2 (2 A-C)+2 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))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}+\frac{\int \frac{\cos (c+d x) \left (-11 a^2 A b^2+6 A b^4+5 a^3 b B-2 a b^3 B+a^4 (2 A-3 C)+a \left (A b^3+2 a^3 B+a b^2 B-a^2 b (4 A+3 C)\right ) \sec (c+d x)-\left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (11 a^2 A b^2-6 A b^4-5 a^3 b B+2 a b^3 B-a^4 (2 A-3 C)\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\int \frac{2 \left (a^2-b^2\right )^2 (3 A b-a B)+a \left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac{(3 A b-a B) x}{a^4}-\frac{\left (11 a^2 A b^2-6 A b^4-5 a^3 b B+2 a b^3 B-a^4 (2 A-3 C)\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (15 a^2 A b^4-6 A b^6+6 a^5 b B-5 a^3 b^3 B+2 a b^5 B-2 a^6 C-a^4 b^2 (12 A+C)\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^4 \left (a^2-b^2\right )^2}\\ &=-\frac{(3 A b-a B) x}{a^4}-\frac{\left (11 a^2 A b^2-6 A b^4-5 a^3 b B+2 a b^3 B-a^4 (2 A-3 C)\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (15 a^2 A b^4-6 A b^6+6 a^5 b B-5 a^3 b^3 B+2 a b^5 B-2 a^6 C-a^4 b^2 (12 A+C)\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{2 a^4 b \left (a^2-b^2\right )^2}\\ &=-\frac{(3 A b-a B) x}{a^4}-\frac{\left (11 a^2 A b^2-6 A b^4-5 a^3 b B+2 a b^3 B-a^4 (2 A-3 C)\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}-\frac{\left (15 a^2 A b^4-6 A b^6+6 a^5 b B-5 a^3 b^3 B+2 a b^5 B-2 a^6 C-a^4 b^2 (12 A+C)\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 b \left (a^2-b^2\right )^2 d}\\ &=-\frac{(3 A b-a B) x}{a^4}+\frac{\left (12 a^4 A b^2-15 a^2 A b^4+6 A b^6-6 a^5 b B+5 a^3 b^3 B-2 a b^5 B+2 a^6 C+a^4 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)^{5/2} (a+b)^{5/2} d}-\frac{\left (11 a^2 A b^2-6 A b^4-5 a^3 b B+2 a b^3 B-a^4 (2 A-3 C)\right ) \sin (c+d x)}{2 a^3 \left (a^2-b^2\right )^2 d}+\frac{\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac{\left (3 A b^4+4 a^3 b B-a b^3 B-2 a^4 C-a^2 b^2 (6 A+C)\right ) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [C]  time = 7.12364, size = 1015, normalized size = 3.08 \[ -\frac{2 (3 A b-a B) x \sec (c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) (b+a \cos (c+d x))^3}{a^4 (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^3}+\frac{\left (2 C a^6-6 b B a^5+12 A b^2 a^4+b^2 C a^4+5 b^3 B a^3-15 A b^4 a^2-2 b^5 B a+6 A b^6\right ) \sec (c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (-\frac{2 i \tan ^{-1}\left (\sec \left (\frac{d x}{2}\right ) \left (\frac{\cos (c)}{\sqrt{a^2-b^2} \sqrt{\cos (2 c)-i \sin (2 c)}}-\frac{i \sin (c)}{\sqrt{a^2-b^2} \sqrt{\cos (2 c)-i \sin (2 c)}}\right ) \left (i a \sin \left (c+\frac{d x}{2}\right )-i b \sin \left (\frac{d x}{2}\right )\right )\right ) \cos (c)}{a^4 \sqrt{a^2-b^2} d \sqrt{\cos (2 c)-i \sin (2 c)}}-\frac{2 \tan ^{-1}\left (\sec \left (\frac{d x}{2}\right ) \left (\frac{\cos (c)}{\sqrt{a^2-b^2} \sqrt{\cos (2 c)-i \sin (2 c)}}-\frac{i \sin (c)}{\sqrt{a^2-b^2} \sqrt{\cos (2 c)-i \sin (2 c)}}\right ) \left (i a \sin \left (c+\frac{d x}{2}\right )-i b \sin \left (\frac{d x}{2}\right )\right )\right ) \sin (c)}{a^4 \sqrt{a^2-b^2} d \sqrt{\cos (2 c)-i \sin (2 c)}}\right ) (b+a \cos (c+d x))^3}{\left (b^2-a^2\right )^2 (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^3}+\frac{2 A \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \tan (c+d x) (b+a \cos (c+d x))^3}{a^3 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^3}+\frac{\sec (c) \sec (c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (-6 A \sin (c) b^6+4 a B \sin (c) b^5+5 a A \sin (d x) b^5+9 a^2 A \sin (c) b^4-2 a^2 C \sin (c) b^4-3 a^2 B \sin (d x) b^4-7 a^3 B \sin (c) b^3-8 a^3 A \sin (d x) b^3+a^3 C \sin (d x) b^3+5 a^4 C \sin (c) b^2+6 a^4 B \sin (d x) b^2-4 a^5 C \sin (d x) b\right ) (b+a \cos (c+d x))^2}{a^4 \left (a^2-b^2\right )^2 d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^3}+\frac{\sec (c) \sec (c+d x) \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (-A \sin (c) b^5+a B \sin (c) b^4+a A \sin (d x) b^4-a^2 C \sin (c) b^3-a^2 B \sin (d x) b^3+a^3 C \sin (d x) b^2\right ) (b+a \cos (c+d x))}{a^4 \left (a^2-b^2\right ) d (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) (a+b \sec (c+d x))^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^3,x]

[Out]

(-2*(3*A*b - a*B)*x*(b + a*Cos[c + d*x])^3*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a^4*(A + 2*C
+ 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^3) + ((12*a^4*A*b^2 - 15*a^2*A*b^4 + 6*A*b^6 -
6*a^5*b*B + 5*a^3*b^3*B - 2*a*b^5*B + 2*a^6*C + a^4*b^2*C)*(b + a*Cos[c + d*x])^3*Sec[c + d*x]*(A + B*Sec[c +
d*x] + C*Sec[c + d*x]^2)*(((-2*I)*ArcTan[Sec[(d*x)/2]*(Cos[c]/(Sqrt[a^2 - b^2]*Sqrt[Cos[2*c] - I*Sin[2*c]]) -
(I*Sin[c])/(Sqrt[a^2 - b^2]*Sqrt[Cos[2*c] - I*Sin[2*c]]))*((-I)*b*Sin[(d*x)/2] + I*a*Sin[c + (d*x)/2])]*Cos[c]
)/(a^4*Sqrt[a^2 - b^2]*d*Sqrt[Cos[2*c] - I*Sin[2*c]]) - (2*ArcTan[Sec[(d*x)/2]*(Cos[c]/(Sqrt[a^2 - b^2]*Sqrt[C
os[2*c] - I*Sin[2*c]]) - (I*Sin[c])/(Sqrt[a^2 - b^2]*Sqrt[Cos[2*c] - I*Sin[2*c]]))*((-I)*b*Sin[(d*x)/2] + I*a*
Sin[c + (d*x)/2])]*Sin[c])/(a^4*Sqrt[a^2 - b^2]*d*Sqrt[Cos[2*c] - I*Sin[2*c]])))/((-a^2 + b^2)^2*(A + 2*C + 2*
B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^3) + ((b + a*Cos[c + d*x])*Sec[c]*Sec[c + d*x]*(A +
B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-(A*b^5*Sin[c]) + a*b^4*B*Sin[c] - a^2*b^3*C*Sin[c] + a*A*b^4*Sin[d*x] - a
^2*b^3*B*Sin[d*x] + a^3*b^2*C*Sin[d*x]))/(a^4*(a^2 - b^2)*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*
(a + b*Sec[c + d*x])^3) + ((b + a*Cos[c + d*x])^2*Sec[c]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*
(9*a^2*A*b^4*Sin[c] - 6*A*b^6*Sin[c] - 7*a^3*b^3*B*Sin[c] + 4*a*b^5*B*Sin[c] + 5*a^4*b^2*C*Sin[c] - 2*a^2*b^4*
C*Sin[c] - 8*a^3*A*b^3*Sin[d*x] + 5*a*A*b^5*Sin[d*x] + 6*a^4*b^2*B*Sin[d*x] - 3*a^2*b^4*B*Sin[d*x] - 4*a^5*b*C
*Sin[d*x] + a^3*b^3*C*Sin[d*x]))/(a^4*(a^2 - b^2)^2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b
*Sec[c + d*x])^3) + (2*A*(b + a*Cos[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Tan[c + d*x])/(a^3*d*(
A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^3)

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Maple [B]  time = 0.139, size = 1756, normalized size = 5.3 \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)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x)

[Out]

-4/d*a/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*b*C-6/d/(tan(1/2
*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*b^2*B+4/d*a/(tan(1/2*
d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*b*C-8/d/a/(tan(1/2*d*x
+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*A*b^3+1/d/a^2/(tan(1/2*d*x+1/2*c)^2
*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*A*b^4+8/d/a/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*
d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A*b^3+1/d/a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2
*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A*b^4-2/d/a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/
2*d*x+1/2*c)^2*b-a-b)^2*b^4/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B-1/d/a/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c
)^2*b-a-b)^2*b^3/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*B-4/d/a^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b
)^2*b^5/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*A-1/d/a/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)
^2*b^3/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*B+2/d/a^2/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b
)^2*b^4/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*B+4/d/a^3/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-
b)^2*b^5/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c)*A+1/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a
-b)^2*tan(1/2*d*x+1/2*c)*C*b^2-6/d*b/(a^4-2*a^2*b^2+b^4)/((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+1/d/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(
a-b))^(1/2))*b^2*C+2/d/a^3*B*arctan(tan(1/2*d*x+1/2*c))+5/d/a/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh(
(a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*B*b^3-2/d/a^3/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a
-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*b^5*B-15/d/a^2/(a^4-2*a^2*b^2+b^4)/((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^4+6/d/a^4/(a^4-2*a^2*b^2+b^4)/((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^6+2/d/a^3*A*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-6/d/a^4*A
*arctan(tan(1/2*d*x+1/2*c))*b+6/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a+b)/(a-b)^2*tan(1/2*
d*x+1/2*c)*b^2*B+12/d*b^2/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b
))^(1/2))*A+2/d/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))*
C*a^2+1/d/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3*C*b
^2

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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)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.961969, size = 3634, normalized size = 11.01 \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)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(4*(B*a^9 - 3*A*a^8*b - 3*B*a^7*b^2 + 9*A*a^6*b^3 + 3*B*a^5*b^4 - 9*A*a^4*b^5 - B*a^3*b^6 + 3*A*a^2*b^7)*
d*x*cos(d*x + c)^2 + 8*(B*a^8*b - 3*A*a^7*b^2 - 3*B*a^6*b^3 + 9*A*a^5*b^4 + 3*B*a^4*b^5 - 9*A*a^3*b^6 - B*a^2*
b^7 + 3*A*a*b^8)*d*x*cos(d*x + c) + 4*(B*a^7*b^2 - 3*A*a^6*b^3 - 3*B*a^5*b^4 + 9*A*a^4*b^5 + 3*B*a^3*b^6 - 9*A
*a^2*b^7 - B*a*b^8 + 3*A*b^9)*d*x + (2*C*a^6*b^2 - 6*B*a^5*b^3 + (12*A + C)*a^4*b^4 + 5*B*a^3*b^5 - 15*A*a^2*b
^6 - 2*B*a*b^7 + 6*A*b^8 + (2*C*a^8 - 6*B*a^7*b + (12*A + C)*a^6*b^2 + 5*B*a^5*b^3 - 15*A*a^4*b^4 - 2*B*a^3*b^
5 + 6*A*a^2*b^6)*cos(d*x + c)^2 + 2*(2*C*a^7*b - 6*B*a^6*b^2 + (12*A + C)*a^5*b^3 + 5*B*a^4*b^4 - 15*A*a^3*b^5
- 2*B*a^2*b^6 + 6*A*a*b^7)*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*((2*A - 3*C)*a^7*b^2 + 5*B*a^6*b^3 - (13*A - 3*C)*a^5*b^4 - 7*B*a^4*b^5 + 17*A*a^3*b^6 + 2*B*a^
2*b^7 - 6*A*a*b^8 + 2*(A*a^9 - 3*A*a^7*b^2 + 3*A*a^5*b^4 - A*a^3*b^6)*cos(d*x + c)^2 + (4*(A - C)*a^8*b + 6*B*
a^7*b^2 - 5*(4*A - C)*a^6*b^3 - 9*B*a^5*b^4 + (25*A - C)*a^4*b^5 + 3*B*a^3*b^6 - 9*A*a^2*b^7)*cos(d*x + c))*si
n(d*x + c))/((a^12 - 3*a^10*b^2 + 3*a^8*b^4 - a^6*b^6)*d*cos(d*x + c)^2 + 2*(a^11*b - 3*a^9*b^3 + 3*a^7*b^5 -
a^5*b^7)*d*cos(d*x + c) + (a^10*b^2 - 3*a^8*b^4 + 3*a^6*b^6 - a^4*b^8)*d), 1/2*(2*(B*a^9 - 3*A*a^8*b - 3*B*a^7
*b^2 + 9*A*a^6*b^3 + 3*B*a^5*b^4 - 9*A*a^4*b^5 - B*a^3*b^6 + 3*A*a^2*b^7)*d*x*cos(d*x + c)^2 + 4*(B*a^8*b - 3*
A*a^7*b^2 - 3*B*a^6*b^3 + 9*A*a^5*b^4 + 3*B*a^4*b^5 - 9*A*a^3*b^6 - B*a^2*b^7 + 3*A*a*b^8)*d*x*cos(d*x + c) +
2*(B*a^7*b^2 - 3*A*a^6*b^3 - 3*B*a^5*b^4 + 9*A*a^4*b^5 + 3*B*a^3*b^6 - 9*A*a^2*b^7 - B*a*b^8 + 3*A*b^9)*d*x +
(2*C*a^6*b^2 - 6*B*a^5*b^3 + (12*A + C)*a^4*b^4 + 5*B*a^3*b^5 - 15*A*a^2*b^6 - 2*B*a*b^7 + 6*A*b^8 + (2*C*a^8
- 6*B*a^7*b + (12*A + C)*a^6*b^2 + 5*B*a^5*b^3 - 15*A*a^4*b^4 - 2*B*a^3*b^5 + 6*A*a^2*b^6)*cos(d*x + c)^2 + 2*
(2*C*a^7*b - 6*B*a^6*b^2 + (12*A + C)*a^5*b^3 + 5*B*a^4*b^4 - 15*A*a^3*b^5 - 2*B*a^2*b^6 + 6*A*a*b^7)*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*A - 3*C
)*a^7*b^2 + 5*B*a^6*b^3 - (13*A - 3*C)*a^5*b^4 - 7*B*a^4*b^5 + 17*A*a^3*b^6 + 2*B*a^2*b^7 - 6*A*a*b^8 + 2*(A*a
^9 - 3*A*a^7*b^2 + 3*A*a^5*b^4 - A*a^3*b^6)*cos(d*x + c)^2 + (4*(A - C)*a^8*b + 6*B*a^7*b^2 - 5*(4*A - C)*a^6*
b^3 - 9*B*a^5*b^4 + (25*A - C)*a^4*b^5 + 3*B*a^3*b^6 - 9*A*a^2*b^7)*cos(d*x + c))*sin(d*x + c))/((a^12 - 3*a^1
0*b^2 + 3*a^8*b^4 - a^6*b^6)*d*cos(d*x + c)^2 + 2*(a^11*b - 3*a^9*b^3 + 3*a^7*b^5 - a^5*b^7)*d*cos(d*x + c) +
(a^10*b^2 - 3*a^8*b^4 + 3*a^6*b^6 - a^4*b^8)*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)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.52182, size = 900, normalized size = 2.73 \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)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

((2*C*a^6 - 6*B*a^5*b + 12*A*a^4*b^2 + C*a^4*b^2 + 5*B*a^3*b^3 - 15*A*a^2*b^4 - 2*B*a*b^5 + 6*A*b^6)*(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^8 - 2*a^6*b^2 + a^4*b^4)*sqrt(-a^2 + b^2)) + (4*C*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 6*B*a^4*b^2*tan
(1/2*d*x + 1/2*c)^3 - 3*C*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 + 8*A*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 + 5*B*a^3*b^3*ta
n(1/2*d*x + 1/2*c)^3 - C*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 7*A*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 + 3*B*a^2*b^4*tan
(1/2*d*x + 1/2*c)^3 - 5*A*a*b^5*tan(1/2*d*x + 1/2*c)^3 - 2*B*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 4*A*b^6*tan(1/2*d*
x + 1/2*c)^3 - 4*C*a^5*b*tan(1/2*d*x + 1/2*c) + 6*B*a^4*b^2*tan(1/2*d*x + 1/2*c) - 3*C*a^4*b^2*tan(1/2*d*x + 1
/2*c) - 8*A*a^3*b^3*tan(1/2*d*x + 1/2*c) + 5*B*a^3*b^3*tan(1/2*d*x + 1/2*c) + C*a^3*b^3*tan(1/2*d*x + 1/2*c) -
7*A*a^2*b^4*tan(1/2*d*x + 1/2*c) - 3*B*a^2*b^4*tan(1/2*d*x + 1/2*c) + 5*A*a*b^5*tan(1/2*d*x + 1/2*c) - 2*B*a*
b^5*tan(1/2*d*x + 1/2*c) + 4*A*b^6*tan(1/2*d*x + 1/2*c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*(a*tan(1/2*d*x + 1/2*c)^
2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^2) + (B*a - 3*A*b)*(d*x + c)/a^4 + 2*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*
x + 1/2*c)^2 + 1)*a^3))/d