3.634 \(\int \frac{A+B \sec (c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=526 \[ -\frac{\left (-5 a^2 B+2 a A b+3 b^2 B\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{3 b^2 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 a \left (2 a^2 A b-5 a^3 B+9 a b^2 B-6 A b^3\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right )^2 \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}+\frac{2 a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}-\frac{\left (6 a^3 A b+26 a^2 b^2 B-15 a^4 B-14 a A b^3-3 b^4 B\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)}}+\frac{\left (6 a^3 A b+26 a^2 b^2 B-15 a^4 B-14 a A b^3-3 b^4 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{(2 A b-5 a B) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{b^3 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}} \]

[Out]

-((2*a*A*b - 5*a^2*B + 3*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(3*b
^2*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + ((2*A*b - 5*a*B)*Sqrt[(b + a*Cos[c + d*x])/(a
+ b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(b^3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + ((6*a^3
*A*b - 14*a*A*b^3 - 15*a^4*B + 26*a^2*b^2*B - 3*b^4*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)
]*Sqrt[a + b*Sec[c + d*x]])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (2*a*(A*b - a*B)*Sin[
c + d*x])/(3*b*(a^2 - b^2)*d*Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(3/2)) + (2*a*(2*a^2*A*b - 6*A*b^3 - 5*a^
3*B + 9*a*b^2*B)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)^2*d*Cos[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]) - ((6*a^3*A
*b - 14*a*A*b^3 - 15*a^4*B + 26*a^2*b^2*B - 3*b^4*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*b^3*(a^2 - b^2)
^2*d*Sqrt[Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.99028, antiderivative size = 526, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 15, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {2955, 4029, 4098, 4102, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 a \left (2 a^2 A b-5 a^3 B+9 a b^2 B-6 A b^3\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right )^2 \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}+\frac{2 a (A b-a B) \sin (c+d x)}{3 b d \left (a^2-b^2\right ) \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}-\frac{\left (6 a^3 A b+26 a^2 b^2 B-15 a^4 B-14 a A b^3-3 b^4 B\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)}}-\frac{\left (-5 a^2 B+2 a A b+3 b^2 B\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 b^2 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (6 a^3 A b+26 a^2 b^2 B-15 a^4 B-14 a A b^3-3 b^4 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{(2 A b-5 a B) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{b^3 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(7/2)*(a + b*Sec[c + d*x])^(5/2)),x]

[Out]

-((2*a*A*b - 5*a^2*B + 3*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(3*b
^2*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + ((2*A*b - 5*a*B)*Sqrt[(b + a*Cos[c + d*x])/(a
+ b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(b^3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + ((6*a^3
*A*b - 14*a*A*b^3 - 15*a^4*B + 26*a^2*b^2*B - 3*b^4*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)
]*Sqrt[a + b*Sec[c + d*x]])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (2*a*(A*b - a*B)*Sin[
c + d*x])/(3*b*(a^2 - b^2)*d*Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(3/2)) + (2*a*(2*a^2*A*b - 6*A*b^3 - 5*a^
3*B + 9*a*b^2*B)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)^2*d*Cos[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]) - ((6*a^3*A
*b - 14*a*A*b^3 - 15*a^4*B + 26*a^2*b^2*B - 3*b^4*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*b^3*(a^2 - b^2)
^2*d*Sqrt[Cos[c + d*x]])

Rule 2955

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Csc[e + f*x])^m*(
c + d*Csc[e + f*x])^n)/(g*Csc[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 4029

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

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 4108

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Dist[C/d^2, Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a +
b*Csc[e + f*x]], x], x] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]]), x] /; Fre
eQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]

Rule 3859

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2807

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

Rule 2805

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

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)}{\cos ^{\frac{7}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\\ &=\frac{2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (\frac{3}{2} a (A b-a B)-\frac{3}{2} b (A b-a B) \sec (c+d x)-\frac{1}{2} \left (2 a A b-5 a^2 B+3 b^2 B\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )}\\ &=\frac{2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac{2 a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}-\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)} \left (-\frac{1}{4} a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right )-\frac{1}{4} b \left (a^2 A b+3 A b^3+2 a^3 B-6 a b^2 B\right ) \sec (c+d x)+\frac{1}{4} \left (6 a^3 A b-14 a A b^3-15 a^4 B+26 a^2 b^2 B-3 b^4 B\right ) \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac{2 a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}-\frac{\left (6 a^3 A b-14 a A b^3-15 a^4 B+26 a^2 b^2 B-3 b^4 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}-\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{8} a \left (6 a^3 A b-14 a A b^3-15 a^4 B+26 a^2 b^2 B-3 b^4 B\right )-\frac{1}{4} a b \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \sec (c+d x)-\frac{3}{8} \left (a^2-b^2\right )^2 (2 A b-5 a B) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=\frac{2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac{2 a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}-\frac{\left (6 a^3 A b-14 a A b^3-15 a^4 B+26 a^2 b^2 B-3 b^4 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}-\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{8} a \left (6 a^3 A b-14 a A b^3-15 a^4 B+26 a^2 b^2 B-3 b^4 B\right )-\frac{1}{4} a b \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}+\frac{\left ((2 A b-5 a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{2 b^3}\\ &=\frac{2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac{2 a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}-\frac{\left (6 a^3 A b-14 a A b^3-15 a^4 B+26 a^2 b^2 B-3 b^4 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}-\frac{\left (\left (2 a A b-5 a^2 B+3 b^2 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx}{6 b^2 \left (a^2-b^2\right )}-\frac{\left (\left (-6 a^3 A b+14 a A b^3+15 a^4 B-26 a^2 b^2 B+3 b^4 B\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{6 b^3 \left (a^2-b^2\right )^2}+\frac{\left ((2 A b-5 a B) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt{b+a \cos (c+d x)}} \, dx}{2 b^3 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}\\ &=\frac{2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac{2 a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}-\frac{\left (6 a^3 A b-14 a A b^3-15 a^4 B+26 a^2 b^2 B-3 b^4 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}-\frac{\left (\left (2 a A b-5 a^2 B+3 b^2 B\right ) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{6 b^2 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left ((2 A b-5 a B) \sqrt{\frac{b+a \cos (c+d x)}{a+b}}\right ) \int \frac{\sec (c+d x)}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{2 b^3 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (-6 a^3 A b+14 a A b^3+15 a^4 B-26 a^2 b^2 B+3 b^4 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{6 b^3 \left (a^2-b^2\right )^2 \sqrt{b+a \cos (c+d x)}}\\ &=\frac{(2 A b-5 a B) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{b^3 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac{2 a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}-\frac{\left (6 a^3 A b-14 a A b^3-15 a^4 B+26 a^2 b^2 B-3 b^4 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}-\frac{\left (\left (2 a A b-5 a^2 B+3 b^2 B\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{6 b^2 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (-6 a^3 A b+14 a A b^3+15 a^4 B-26 a^2 b^2 B+3 b^4 B\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}} \, dx}{6 b^3 \left (a^2-b^2\right )^2 \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}\\ &=-\frac{\left (2 a A b-5 a^2 B+3 b^2 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 )}{3 b^2 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{(2 A b-5 a B) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{b^3 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (6 a^3 A b-14 a A b^3-15 a^4 B+26 a^2 b^2 B-3 b^4 B\right ) \sqrt{\cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}+\frac{2 a (A b-a B) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}+\frac{2 a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}-\frac{\left (6 a^3 A b-14 a A b^3-15 a^4 B+26 a^2 b^2 B-3 b^4 B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 35.9732, size = 184379, normalized size = 350.53 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(7/2)*(a + b*Sec[c + d*x])^(5/2)),x]

[Out]

Result too large to show

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Maple [C]  time = 0.714, size = 5358, normalized size = 10.2 \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))/cos(d*x+c)^(7/2)/(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))/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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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((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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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))/cos(d*x+c)**(7/2)/(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}} \cos \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(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)*cos(d*x + c)^(7/2)), x)