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

Optimal. Leaf size=472 \[ \frac{2 \left (16 a^2 A b^2+a^4 A-9 a^3 b B+8 a b^3 B-16 A b^4\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{3 a^4 d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (-13 a^2 A b^2+a^4 A+8 a^3 b B-4 a b^3 B+8 A b^4\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)}}+\frac{2 b \left (10 a^2 A b-7 a^3 B+3 a b^2 B-6 A b^3\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (-28 a^2 A b^3+8 a^4 A b+15 a^3 b^2 B-3 a^5 B-8 a b^4 B+16 A b^5\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^4 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

[Out]

(2*(a^4*A + 16*a^2*A*b^2 - 16*A*b^4 - 9*a^3*b*B + 8*a*b^3*B)*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*(8*a^4*A*b - 2
8*a^2*A*b^3 + 16*A*b^5 - 3*a^5*B + 15*a^3*b^2*B - 8*a*b^4*B)*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)]*Sqrt[Sec[c + d*x]]) + (2*b*(A*b - a*B
)*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*b*(10*a^2*A*b - 6*A*b^3
- 7*a^3*B + 3*a*b^2*B)*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
*(a^4*A - 13*a^2*A*b^2 + 8*A*b^4 + 8*a^3*b*B - 4*a*b^3*B)*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 = 1.40857, antiderivative size = 472, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4030, 4100, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 \left (-13 a^2 A b^2+a^4 A+8 a^3 b B-4 a b^3 B+8 A b^4\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)}}+\frac{2 b \left (10 a^2 A b-7 a^3 B+3 a b^2 B-6 A b^3\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 b (A b-a B) \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (16 a^2 A b^2+a^4 A-9 a^3 b B+8 a b^3 B-16 A b^4\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^4 d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (-28 a^2 A b^3+8 a^4 A b+15 a^3 b^2 B-3 a^5 B-8 a b^4 B+16 A b^5\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^4 d \left (a^2-b^2\right )^2 \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a^4*A + 16*a^2*A*b^2 - 16*A*b^4 - 9*a^3*b*B + 8*a*b^3*B)*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*(8*a^4*A*b - 2
8*a^2*A*b^3 + 16*A*b^5 - 3*a^5*B + 15*a^3*b^2*B - 8*a*b^4*B)*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)]*Sqrt[Sec[c + d*x]]) + (2*b*(A*b - a*B
)*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*b*(10*a^2*A*b - 6*A*b^3
- 7*a^3*B + 3*a*b^2*B)*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
*(a^4*A - 13*a^2*A*b^2 + 8*A*b^4 + 8*a^3*b*B - 4*a*b^3*B)*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 4030

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

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)}{\sec ^{\frac{3}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx &=\frac{2 b (A b-a B) \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 (a^2 A-2 A b^2+a b B\right )+\frac{3}{2} a (A b-a B) \sec (c+d x)-2 b (A b-a B) \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 b (A b-a B) \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 b \left (10 a^2 A b-6 A b^3-7 a^3 B+3 a b^2 B\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 (a^4 A-13 a^2 A b^2+8 A b^4+8 a^3 b B-4 a b^3 B\right )-\frac{1}{4} a \left (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\right ) \sec (c+d x)+\frac{1}{2} b \left (10 a^2 A b-6 A b^3-7 a^3 B+3 a b^2 B\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 b (A b-a B) \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 b \left (10 a^2 A b-6 A b^3-7 a^3 B+3 a b^2 B\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 (a^4 A-13 a^2 A b^2+8 A b^4+8 a^3 b B-4 a b^3 B\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 (8 a^4 A b-28 a^2 A b^3+16 A b^5-3 a^5 B+15 a^3 b^2 B-8 a b^4 B\right )-\frac{3}{8} a \left (a^4 A+7 a^2 A b^2-4 A b^4-6 a^3 b B+2 a b^3 B\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 b (A b-a B) \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 b \left (10 a^2 A b-6 A b^3-7 a^3 B+3 a b^2 B\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 (a^4 A-13 a^2 A b^2+8 A b^4+8 a^3 b B-4 a b^3 B\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 (a^4 A+16 a^2 A b^2-16 A b^4-9 a^3 b B+8 a b^3 B\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 (8 a^4 A b-28 a^2 A b^3+16 A b^5-3 a^5 B+15 a^3 b^2 B-8 a b^4 B\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 b (A b-a B) \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 b \left (10 a^2 A b-6 A b^3-7 a^3 B+3 a b^2 B\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 (a^4 A-13 a^2 A b^2+8 A b^4+8 a^3 b B-4 a b^3 B\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 (a^4 A+16 a^2 A b^2-16 A b^4-9 a^3 b B+8 a b^3 B\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 (8 a^4 A b-28 a^2 A b^3+16 A b^5-3 a^5 B+15 a^3 b^2 B-8 a b^4 B\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 b (A b-a B) \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 b \left (10 a^2 A b-6 A b^3-7 a^3 B+3 a b^2 B\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 (a^4 A-13 a^2 A b^2+8 A b^4+8 a^3 b B-4 a b^3 B\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 (a^4 A+16 a^2 A b^2-16 A b^4-9 a^3 b B+8 a b^3 B\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 (8 a^4 A b-28 a^2 A b^3+16 A b^5-3 a^5 B+15 a^3 b^2 B-8 a b^4 B\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 (a^4 A+16 a^2 A b^2-16 A b^4-9 a^3 b B+8 a b^3 B\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 (8 a^4 A b-28 a^2 A b^3+16 A b^5-3 a^5 B+15 a^3 b^2 B-8 a b^4 B\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 b (A b-a B) \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 b \left (10 a^2 A b-6 A b^3-7 a^3 B+3 a b^2 B\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 (a^4 A-13 a^2 A b^2+8 A b^4+8 a^3 b B-4 a b^3 B\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 [A]  time = 2.90866, size = 353, normalized size = 0.75 \[ \frac{2 \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b) \left (\frac{\left (\frac{a \cos (c+d x)+b}{a+b}\right )^{3/2} \left (a^2 \left (7 a^2 A b^2+a^4 A-6 a^3 b B+2 a b^3 B-4 A b^4\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )+\left (28 a^2 A b^3-8 a^4 A b-15 a^3 b^2 B+3 a^5 B+8 a b^4 B-16 A b^5\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )-b \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )\right )\right )}{(a-b)^2 (a+b)}+\frac{a \sin (c+d x) \left (2 a b \left (-16 a^2 A b^2+2 a^4 A+9 a^3 b B-5 a b^3 B+10 A b^4\right ) \cos (c+d x)+A \left (a^3-a b^2\right )^2 \cos (2 (c+d x))-25 a^2 A b^4+a^6 A+16 a^3 b^3 B-8 a b^5 B+16 A b^6\right )}{2 \left (a^2-b^2\right )^2}\right )}{3 a^4 d (a+b \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b + a*Cos[c + d*x])*Sec[c + d*x]^(5/2)*((((b + a*Cos[c + d*x])/(a + b))^(3/2)*(a^2*(a^4*A + 7*a^2*A*b^2 -
4*A*b^4 - 6*a^3*b*B + 2*a*b^3*B)*EllipticF[(c + d*x)/2, (2*a)/(a + b)] + (-8*a^4*A*b + 28*a^2*A*b^3 - 16*A*b^5
+ 3*a^5*B - 15*a^3*b^2*B + 8*a*b^4*B)*((a + b)*EllipticE[(c + d*x)/2, (2*a)/(a + b)] - b*EllipticF[(c + d*x)/
2, (2*a)/(a + b)])))/((a - b)^2*(a + b)) + (a*(a^6*A - 25*a^2*A*b^4 + 16*A*b^6 + 16*a^3*b^3*B - 8*a*b^5*B + 2*
a*b*(2*a^4*A - 16*a^2*A*b^2 + 10*A*b^4 + 9*a^3*b*B - 5*a*b^3*B)*Cos[c + d*x] + A*(a^3 - a*b^2)^2*Cos[2*(c + d*
x)])*Sin[c + d*x])/(2*(a^2 - b^2)^2)))/(3*a^4*d*(a + b*Sec[c + d*x])^(5/2))

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Maple [B]  time = 0.541, size = 6745, normalized size = 14.3 \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))/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))/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 (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))/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((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))/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{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))/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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