3.1068 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=521 \[ -\frac{2 \sqrt{\sec (c+d x)} \left (-2 a^2 b^2 (8 A-C)+a^4 (-(A+3 C))+9 a^3 b B-8 a b^3 B+16 A b^4\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{3 a^4 d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 \sin (c+d x) \left (-a^2 b^2 (13 A-C)+a^4 (A-5 C)+8 a^3 b B-4 a b^3 B+8 A b^4\right ) \sqrt{a+b \sec (c+d x)}}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)}}+\frac{2 \sin (c+d x) \left (10 a^2 A b^2-7 a^3 b B+4 a^4 C+3 a b^3 B-6 A b^4\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (-2 a^2 b^3 (14 A-C)+a^4 (8 A b-6 b C)+15 a^3 b^2 B-3 a^5 B-8 a b^4 B+16 A b^5\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^4 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

[Out]

(-2*(16*A*b^4 + 9*a^3*b*B - 8*a*b^3*B - 2*a^2*b^2*(8*A - C) - a^4*(A + 3*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)
]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(3*a^4*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) - (
2*(16*A*b^5 - 3*a^5*B + 15*a^3*b^2*B - 8*a*b^4*B - 2*a^2*b^3*(14*A - C) + a^4*(8*A*b - 6*b*C))*EllipticE[(c +
d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a^4*(a^2 - b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqr
t[Sec[c + d*x]]) + (2*(A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]*(a + b*Sec[c
+ d*x])^(3/2)) + (2*(10*a^2*A*b^2 - 6*A*b^4 - 7*a^3*b*B + 3*a*b^3*B + 4*a^4*C)*Sin[c + d*x])/(3*a^2*(a^2 - b^
2)^2*d*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(8*A*b^4 + 8*a^3*b*B - 4*a*b^3*B + a^4*(A - 5*C) - a^
2*b^2*(13*A - C))*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*a^3*(a^2 - b^2)^2*d*Sqrt[Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.60248, antiderivative size = 521, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4100, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 \sin (c+d x) \left (-a^2 b^2 (13 A-C)+a^4 (A-5 C)+8 a^3 b B-4 a b^3 B+8 A b^4\right ) \sqrt{a+b \sec (c+d x)}}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)}}+\frac{2 \sin (c+d x) \left (10 a^2 A b^2-7 a^3 b B+4 a^4 C+3 a b^3 B-6 A b^4\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac{2 \sqrt{\sec (c+d x)} \left (-2 a^2 b^2 (8 A-C)+a^4 (-(A+3 C))+9 a^3 b B-8 a b^3 B+16 A b^4\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^4 d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (-2 a^2 b^3 (14 A-C)+a^4 (8 A b-6 b C)+15 a^3 b^2 B-3 a^5 B-8 a b^4 B+16 A b^5\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^4 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(5/2)),x]

[Out]

(-2*(16*A*b^4 + 9*a^3*b*B - 8*a*b^3*B - 2*a^2*b^2*(8*A - C) - a^4*(A + 3*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)
]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(3*a^4*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) - (
2*(16*A*b^5 - 3*a^5*B + 15*a^3*b^2*B - 8*a*b^4*B - 2*a^2*b^3*(14*A - C) + a^4*(8*A*b - 6*b*C))*EllipticE[(c +
d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a^4*(a^2 - b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqr
t[Sec[c + d*x]]) + (2*(A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]]*(a + b*Sec[c
+ d*x])^(3/2)) + (2*(10*a^2*A*b^2 - 6*A*b^4 - 7*a^3*b*B + 3*a*b^3*B + 4*a^4*C)*Sin[c + d*x])/(3*a^2*(a^2 - b^
2)^2*d*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(8*A*b^4 + 8*a^3*b*B - 4*a*b^3*B + a^4*(A - 5*C) - a^
2*b^2*(13*A - C))*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*a^3*(a^2 - b^2)^2*d*Sqrt[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 4035

Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]

Rule 3856

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

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 3858

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

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx &=\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac{2 \int \frac{\frac{3}{2} \left (2 A b^2-a b B-a^2 (A-C)\right )+\frac{3}{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)}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )}\\ &=\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (10 a^2 A b^2-6 A b^4-7 a^3 b B+3 a b^3 B+4 a^4 C\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{4 \int \frac{\frac{3}{4} \left (8 A b^4+8 a^3 b B-4 a b^3 B+a^4 (A-5 C)-a^2 b^2 (13 A-C)\right )+\frac{1}{4} a \left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \sec (c+d x)+\frac{1}{2} \left (10 a^2 A b^2-6 A b^4-7 a^3 b B+3 a b^3 B+4 a^4 C\right ) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (10 a^2 A b^2-6 A b^4-7 a^3 b B+3 a b^3 B+4 a^4 C\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (8 A b^4+8 a^3 b B-4 a b^3 B+a^4 (A-5 C)-a^2 b^2 (13 A-C)\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)}}-\frac{8 \int \frac{\frac{3}{8} \left (16 A b^5-3 a^5 B+15 a^3 b^2 B-8 a b^4 B-2 a^2 b^3 (14 A-C)+a^4 (8 A b-6 b C)\right )+\frac{3}{8} a \left (4 A b^4+6 a^3 b B-2 a b^3 B-a^2 b^2 (7 A+C)-a^4 (A+3 C)\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{9 a^3 \left (a^2-b^2\right )^2}\\ &=\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (10 a^2 A b^2-6 A b^4-7 a^3 b B+3 a b^3 B+4 a^4 C\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (8 A b^4+8 a^3 b B-4 a b^3 B+a^4 (A-5 C)-a^2 b^2 (13 A-C)\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)}}-\frac{\left (16 A b^4+9 a^3 b B-8 a b^3 B-2 a^2 b^2 (8 A-C)-a^4 (A+3 C)\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )}-\frac{\left (16 A b^5-3 a^5 B+15 a^3 b^2 B-8 a b^4 B-2 a^2 b^3 (14 A-C)+a^4 (8 A b-6 b C)\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right )^2}\\ &=\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (10 a^2 A b^2-6 A b^4-7 a^3 b B+3 a b^3 B+4 a^4 C\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (8 A b^4+8 a^3 b B-4 a b^3 B+a^4 (A-5 C)-a^2 b^2 (13 A-C)\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)}}-\frac{\left (\left (16 A b^4+9 a^3 b B-8 a b^3 B-2 a^2 b^2 (8 A-C)-a^4 (A+3 C)\right ) \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{3 a^4 \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (16 A b^5-3 a^5 B+15 a^3 b^2 B-8 a b^4 B-2 a^2 b^3 (14 A-C)+a^4 (8 A b-6 b C)\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{3 a^4 \left (a^2-b^2\right )^2 \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ &=\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (10 a^2 A b^2-6 A b^4-7 a^3 b B+3 a b^3 B+4 a^4 C\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (8 A b^4+8 a^3 b B-4 a b^3 B+a^4 (A-5 C)-a^2 b^2 (13 A-C)\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)}}-\frac{\left (\left (16 A b^4+9 a^3 b B-8 a b^3 B-2 a^2 b^2 (8 A-C)-a^4 (A+3 C)\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{3 a^4 \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (16 A b^5-3 a^5 B+15 a^3 b^2 B-8 a b^4 B-2 a^2 b^3 (14 A-C)+a^4 (8 A b-6 b C)\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}} \, dx}{3 a^4 \left (a^2-b^2\right )^2 \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}\\ &=-\frac{2 \left (16 A b^4+9 a^3 b B-8 a b^3 B-2 a^2 b^2 (8 A-C)-a^4 (A+3 C)\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{\sec (c+d x)}}{3 a^4 \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (16 A b^5-3 a^5 B+15 a^3 b^2 B-8 a b^4 B-2 a^2 b^3 (14 A-C)+a^4 (8 A b-6 b C)\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{3 a^4 \left (a^2-b^2\right )^2 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}+\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (10 a^2 A b^2-6 A b^4-7 a^3 b B+3 a b^3 B+4 a^4 C\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (8 A b^4+8 a^3 b B-4 a b^3 B+a^4 (A-5 C)-a^2 b^2 (13 A-C)\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt{\sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 9.23639, size = 7608, normalized size = 14.6 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(5/2)),x]

[Out]

Result too large to show

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Maple [B]  time = 0.599, size = 8777, normalized size = 16.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(5/2),x)

[Out]

result too large to display

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Maxima [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((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt{b \sec \left (d x + c\right ) + a} \sqrt{\sec \left (d x + c\right )}}{b^{3} \sec \left (d x + c\right )^{5} + 3 \, a b^{2} \sec \left (d x + c\right )^{4} + 3 \, a^{2} b \sec \left (d x + c\right )^{3} + a^{3} \sec \left (d x + c\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a)*sqrt(sec(d*x + c))/(b^3*sec(d*x + c)
^5 + 3*a*b^2*sec(d*x + c)^4 + 3*a^2*b*sec(d*x + c)^3 + a^3*sec(d*x + c)^2), x)

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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((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(3/2)/(a+b*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^(5/2)*sec(d*x + c)^(3/2)), x)