3.908 \(\int \frac{\sec ^4(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=407 \[ -\frac{\tan (c+d x) \left (-a^2 b^2 (6 A-7 C)+9 a^3 b B-12 a^4 C-6 a b^3 B+b^4 (3 A+2 C)\right )}{3 b^4 d \left (a^2-b^2\right )}+\frac{\left (6 a^2 b B-8 a^3 C-2 a b^2 (2 A+C)+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^5 d}+\frac{2 a^2 \left (2 a^2 A b^2-5 a^2 b^2 C-3 a^3 b B+4 a^4 C+4 a b^3 B-3 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}+\frac{\tan (c+d x) \sec (c+d x) \left (3 a^2 b B-4 a^3 C-2 a b^2 (A-C)-b^3 B\right )}{2 b^3 d \left (a^2-b^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.73952, antiderivative size = 407, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.22, Rules used = {4098, 4102, 4092, 4082, 3998, 3770, 3831, 2659, 208} \[ -\frac{\tan (c+d x) \left (-a^2 b^2 (6 A-7 C)+9 a^3 b B-12 a^4 C-6 a b^3 B+b^4 (3 A+2 C)\right )}{3 b^4 d \left (a^2-b^2\right )}+\frac{\left (6 a^2 b B-8 a^3 C-2 a b^2 (2 A+C)+b^3 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^5 d}+\frac{2 a^2 \left (2 a^2 A b^2-5 a^2 b^2 C-3 a^3 b B+4 a^4 C+4 a b^3 B-3 A b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^5 d (a-b)^{3/2} (a+b)^{3/2}}-\frac{\tan (c+d x) \sec ^3(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{\tan (c+d x) \sec ^2(c+d x) \left (4 a^2 C-3 a b B+3 A b^2-b^2 C\right )}{3 b^2 d \left (a^2-b^2\right )}+\frac{\tan (c+d x) \sec (c+d x) \left (3 a^2 b B-4 a^3 C-2 a b^2 (A-C)-b^3 B\right )}{2 b^3 d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^2,x]

[Out]

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

Rule 4098

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

Rule 4102

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[(C*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1)*(d*Csc[e + f*x])^(n - 1))/(b*f*(m + n + 1)), x] + Dist[d/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + (A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C
*n)*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] && GtQ[n, 0]

Rule 4092

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Csc[e + f*x]*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2)
+ A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B*(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m
}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 4082

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

Rule 3998

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x
_Symbol] :> Dist[B/b, Int[Csc[e + f*x], x], x] + Dist[(A*b - a*B)/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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{\sec ^4(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 ) \sec ^3(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\sec ^3(c+d x) \left (3 \left (A b^2-a (b B-a C)\right )+b (b B-a (A+C)) \sec (c+d x)-\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^3(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\sec ^2(c+d x) \left (-2 a \left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right )+b \left (3 A b^2-3 a b B+a^2 C+2 b^2 C\right ) \sec (c+d x)-3 \left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=\frac{\left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^3(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\sec (c+d x) \left (-3 a \left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right )+b \left (3 a^2 b B+3 b^3 B-4 a^3 C-2 a b^2 (3 A+C)\right ) \sec (c+d x)+2 \left (9 a^3 b B-6 a b^3 B-a^2 b^2 (6 A-7 C)-12 a^4 C+b^4 (3 A+2 C)\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (9 a^3 b B-6 a b^3 B-a^2 b^2 (6 A-7 C)-12 a^4 C+b^4 (3 A+2 C)\right ) \tan (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}+\frac{\left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^3(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{\sec (c+d x) \left (-3 a b \left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right )-3 \left (a^2-b^2\right ) \left (6 a^2 b B+b^3 B-8 a^3 C-2 a b^2 (2 A+C)\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 b^4 \left (a^2-b^2\right )}\\ &=-\frac{\left (9 a^3 b B-6 a b^3 B-a^2 b^2 (6 A-7 C)-12 a^4 C+b^4 (3 A+2 C)\right ) \tan (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}+\frac{\left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^3(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (a^2 \left (3 A b^4+3 a^3 b B-4 a b^3 B-a^2 b^2 (2 A-5 C)-4 a^4 C\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b^5 \left (a^2-b^2\right )}+\frac{\left (6 a^2 b B+b^3 B-8 a^3 C-2 a b^2 (2 A+C)\right ) \int \sec (c+d x) \, dx}{2 b^5}\\ &=\frac{\left (6 a^2 b B+b^3 B-8 a^3 C-2 a b^2 (2 A+C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^5 d}-\frac{\left (9 a^3 b B-6 a b^3 B-a^2 b^2 (6 A-7 C)-12 a^4 C+b^4 (3 A+2 C)\right ) \tan (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}+\frac{\left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^3(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (a^2 \left (3 A b^4+3 a^3 b B-4 a b^3 B-a^2 b^2 (2 A-5 C)-4 a^4 C\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{b^6 \left (a^2-b^2\right )}\\ &=\frac{\left (6 a^2 b B+b^3 B-8 a^3 C-2 a b^2 (2 A+C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^5 d}-\frac{\left (9 a^3 b B-6 a b^3 B-a^2 b^2 (6 A-7 C)-12 a^4 C+b^4 (3 A+2 C)\right ) \tan (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}+\frac{\left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^3(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\left (2 a^2 \left (3 A b^4+3 a^3 b B-4 a b^3 B-a^2 b^2 (2 A-5 C)-4 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 )}{b^6 \left (a^2-b^2\right ) d}\\ &=\frac{\left (6 a^2 b B+b^3 B-8 a^3 C-2 a b^2 (2 A+C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^5 d}+\frac{2 a^2 \left (2 a^2 A b^2-3 A b^4-3 a^3 b B+4 a b^3 B+4 a^4 C-5 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-b)^{3/2} b^5 (a+b)^{3/2} d}-\frac{\left (9 a^3 b B-6 a b^3 B-a^2 b^2 (6 A-7 C)-12 a^4 C+b^4 (3 A+2 C)\right ) \tan (c+d x)}{3 b^4 \left (a^2-b^2\right ) d}+\frac{\left (3 a^2 b B-b^3 B-2 a b^2 (A-C)-4 a^3 C\right ) \sec (c+d x) \tan (c+d x)}{2 b^3 \left (a^2-b^2\right ) d}+\frac{\left (3 A b^2-3 a b B+4 a^2 C-b^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{\left (A b^2-a (b B-a C)\right ) \sec ^3(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}

Mathematica [A]  time = 3.81893, size = 605, normalized size = 1.49 \[ \frac{(a \cos (c+d x)+b) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (6 \left (-6 a^2 b B+8 a^3 C+2 a b^2 (2 A+C)-b^3 B\right ) (a \cos (c+d x)+b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 \left (6 a^2 b B-8 a^3 C-2 a b^2 (2 A+C)+b^3 B\right ) (a \cos (c+d x)+b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-\frac{24 a^2 \left (a^2 b^2 (2 A-5 C)-3 a^3 b B+4 a^4 C+4 a b^3 B-3 A b^4\right ) (a \cos (c+d x)+b) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{b \tan (c+d x) \sec ^2(c+d x) \left (\cos (c+d x) \left (a^3 b^2 (29 C-18 A)-24 a^2 b^3 B+27 a^4 b B-36 a^5 C+a b^4 (9 A-2 C)+6 b^5 B\right )+b \left (b^2-a^2\right ) \cos (2 (c+d x)) \left (12 a^2 C-9 a b B+6 A b^2+4 b^2 C\right )-6 a^3 A b^2 \cos (3 (c+d x))-6 a^2 A b^3-6 a^2 b^3 B \cos (3 (c+d x))+9 a^3 b^2 B+7 a^3 b^2 C \cos (3 (c+d x))+4 a^2 b^3 C+9 a^4 b B \cos (3 (c+d x))-12 a^4 b C-12 a^5 C \cos (3 (c+d x))+3 a A b^4 \cos (3 (c+d x))-9 a b^4 B+2 a b^4 C \cos (3 (c+d x))+6 A b^5+8 b^5 C\right )}{b^2-a^2}\right )}{6 b^5 d (a+b \sec (c+d x))^2 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^2,x]

[Out]

((b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-24*a^2*(-3*A*b^4 - 3*a^3*b*B + 4*a*b^3*B + a^
2*b^2*(2*A - 5*C) + 4*a^4*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[c + d*x]))/(a^2 -
b^2)^(3/2) + 6*(-6*a^2*b*B - b^3*B + 8*a^3*C + 2*a*b^2*(2*A + C))*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2] -
Sin[(c + d*x)/2]] + 6*(6*a^2*b*B + b^3*B - 8*a^3*C - 2*a*b^2*(2*A + C))*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x
)/2] + Sin[(c + d*x)/2]] + (b*(-6*a^2*A*b^3 + 6*A*b^5 + 9*a^3*b^2*B - 9*a*b^4*B - 12*a^4*b*C + 4*a^2*b^3*C + 8
*b^5*C + (27*a^4*b*B - 24*a^2*b^3*B + 6*b^5*B + a*b^4*(9*A - 2*C) - 36*a^5*C + a^3*b^2*(-18*A + 29*C))*Cos[c +
d*x] + b*(-a^2 + b^2)*(6*A*b^2 - 9*a*b*B + 12*a^2*C + 4*b^2*C)*Cos[2*(c + d*x)] - 6*a^3*A*b^2*Cos[3*(c + d*x)
] + 3*a*A*b^4*Cos[3*(c + d*x)] + 9*a^4*b*B*Cos[3*(c + d*x)] - 6*a^2*b^3*B*Cos[3*(c + d*x)] - 12*a^5*C*Cos[3*(c
+ d*x)] + 7*a^3*b^2*C*Cos[3*(c + d*x)] + 2*a*b^4*C*Cos[3*(c + d*x)])*Sec[c + d*x]^2*Tan[c + d*x])/(-a^2 + b^2
)))/(6*b^5*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^2)

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Maple [B]  time = 0.117, size = 1254, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x)

[Out]

-6/d*a^2/b/(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+8/d*a^6/b^5
/(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-10/d*a^4/b^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))*C+1/2/d/b^2*ln(tan(1/2*d*x+1/2*c
)+1)*B-1/2/d/b^2*ln(tan(1/2*d*x+1/2*c)-1)*B+8/d*a^3/b^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))*B+4/d*a^4/b^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))*A-1/3/d*C/b^2/(tan(1/2*d*x+1/2*c)-1)^3+1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)^2*B-1/2/d/b^2/(t
an(1/2*d*x+1/2*c)-1)^2*C-1/d/b^2/(tan(1/2*d*x+1/2*c)-1)*A+1/2/d/b^2/(tan(1/2*d*x+1/2*c)-1)*B-1/3/d*C/b^2/(tan(
1/2*d*x+1/2*c)+1)^3-1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)^2*B+1/2/d/b^2/(tan(1/2*d*x+1/2*c)+1)^2*C-6/d*a^5/b^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))*B-1/d/b^3*ln(tan(1/2*d*x+1/2
*c)+1)*a*C+1/d/b^3*ln(tan(1/2*d*x+1/2*c)-1)*a*C-1/d/b^2/(tan(1/2*d*x+1/2*c)+1)*A+1/2/d/b^2/(tan(1/2*d*x+1/2*c)
+1)*B-1/d/b^2/(tan(1/2*d*x+1/2*c)-1)*C-1/d/b^2/(tan(1/2*d*x+1/2*c)+1)*C+2/d*a^4/b^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)*B-1/d/b^3/(tan(1/2*d*x+1/2*c)+1)*a*C-2/d*a^5/b^4/(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-3/d/b^4/(tan(1/2*d*x+1/2*c)+1)*a^
2*C-4/d/b^5*ln(tan(1/2*d*x+1/2*c)+1)*a^3*C-3/d/b^4/(tan(1/2*d*x+1/2*c)-1)*a^2*C-3/d/b^4*ln(tan(1/2*d*x+1/2*c)-
1)*B*a^2+4/d/b^5*ln(tan(1/2*d*x+1/2*c)-1)*a^3*C+2/d/b^3*ln(tan(1/2*d*x+1/2*c)-1)*A*a-1/d/b^3/(tan(1/2*d*x+1/2*
c)-1)*a*C+1/d/b^3/(tan(1/2*d*x+1/2*c)+1)^2*a*C+2/d/b^3/(tan(1/2*d*x+1/2*c)-1)*B*a-2/d/b^3*ln(tan(1/2*d*x+1/2*c
)+1)*A*a+3/d/b^4*ln(tan(1/2*d*x+1/2*c)+1)*B*a^2-1/d/b^3/(tan(1/2*d*x+1/2*c)-1)^2*a*C+2/d/b^3/(tan(1/2*d*x+1/2*
c)+1)*B*a-2/d*a^3/b^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)*A

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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(sec(d*x+c)^4*(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 [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(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**2,x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**4/(a + b*sec(c + d*x))**2, x)

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Giac [A]  time = 1.4407, size = 846, normalized size = 2.08 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(12*(4*C*a^6 - 3*B*a^5*b + 2*A*a^4*b^2 - 5*C*a^4*b^2 + 4*B*a^3*b^3 - 3*A*a^2*b^4)*(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^2
*b^5 - b^7)*sqrt(-a^2 + b^2)) - 12*(C*a^5*tan(1/2*d*x + 1/2*c) - B*a^4*b*tan(1/2*d*x + 1/2*c) + A*a^3*b^2*tan(
1/2*d*x + 1/2*c))/((a^2*b^4 - b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)) - 3*(8*C*a^3
- 6*B*a^2*b + 4*A*a*b^2 + 2*C*a*b^2 - B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^5 + 3*(8*C*a^3 - 6*B*a^2*b
+ 4*A*a*b^2 + 2*C*a*b^2 - B*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^5 - 2*(18*C*a^2*tan(1/2*d*x + 1/2*c)^5 -
12*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*C*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^2*t
an(1/2*d*x + 1/2*c)^5 + 6*C*b^2*tan(1/2*d*x + 1/2*c)^5 - 36*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*B*a*b*tan(1/2*d*
x + 1/2*c)^3 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 - 4*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 18*C*a^2*tan(1/2*d*x + 1/2*c
) - 12*B*a*b*tan(1/2*d*x + 1/2*c) - 6*C*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/2*d*x + 1/2*c) + 3*B*b^2*tan(
1/2*d*x + 1/2*c) + 6*C*b^2*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*b^4))/d