3.1046 \(\int \frac{(a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=453 \[ \frac{\sqrt{\sec (c+d x)} \left (a^2 b (96 A+59 C)+48 a^3 B+66 a b^2 B+8 b^3 (3 A+2 C)\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{24 d \sqrt{a+b \sec (c+d x)}}+\frac{\sin (c+d x) \sqrt{\sec (c+d x)} \left (15 a^2 C+42 a b B+24 A b^2+16 b^2 C\right ) \sqrt{a+b \sec (c+d x)}}{24 d}-\frac{\left (a^2 (-(48 A-33 C))+54 a b B+8 b^2 (3 A+2 C)\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{24 d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{\sqrt{\sec (c+d x)} \left (30 a^2 b B+5 a^3 C+20 a b^2 (2 A+C)+8 b^3 B\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{8 d \sqrt{a+b \sec (c+d x)}}+\frac{(5 a C+6 b B) \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}{12 d}+\frac{C \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{5/2}}{3 d} \]

[Out]

((48*a^3*B + 66*a*b^2*B + 8*b^3*(3*A + 2*C) + a^2*b*(96*A + 59*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Elliptic
F[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(24*d*Sqrt[a + b*Sec[c + d*x]]) + ((30*a^2*b*B + 8*b^3*B + 5
*a^3*C + 20*a*b^2*(2*A + C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)]*Sqrt
[Sec[c + d*x]])/(8*d*Sqrt[a + b*Sec[c + d*x]]) - ((54*a*b*B - a^2*(48*A - 33*C) + 8*b^2*(3*A + 2*C))*EllipticE
[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(24*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d
*x]]) + ((24*A*b^2 + 42*a*b*B + 15*a^2*C + 16*b^2*C)*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])
/(24*d) + ((6*b*B + 5*a*C)*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(12*d) + (C*Sqrt[Sec[c
+ d*x]]*(a + b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(3*d)

________________________________________________________________________________________

Rubi [A]  time = 1.68136, antiderivative size = 453, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 12, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {4096, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{\sin (c+d x) \sqrt{\sec (c+d x)} \left (15 a^2 C+42 a b B+24 A b^2+16 b^2 C\right ) \sqrt{a+b \sec (c+d x)}}{24 d}+\frac{\sqrt{\sec (c+d x)} \left (a^2 b (96 A+59 C)+48 a^3 B+66 a b^2 B+8 b^3 (3 A+2 C)\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{24 d \sqrt{a+b \sec (c+d x)}}-\frac{\left (a^2 (-(48 A-33 C))+54 a b B+8 b^2 (3 A+2 C)\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{24 d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{\sqrt{\sec (c+d x)} \left (30 a^2 b B+5 a^3 C+20 a b^2 (2 A+C)+8 b^3 B\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{8 d \sqrt{a+b \sec (c+d x)}}+\frac{(5 a C+6 b B) \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}}{12 d}+\frac{C \sin (c+d x) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{5/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((48*a^3*B + 66*a*b^2*B + 8*b^3*(3*A + 2*C) + a^2*b*(96*A + 59*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Elliptic
F[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(24*d*Sqrt[a + b*Sec[c + d*x]]) + ((30*a^2*b*B + 8*b^3*B + 5
*a^3*C + 20*a*b^2*(2*A + C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)]*Sqrt
[Sec[c + d*x]])/(8*d*Sqrt[a + b*Sec[c + d*x]]) - ((54*a*b*B - a^2*(48*A - 33*C) + 8*b^2*(3*A + 2*C))*EllipticE
[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(24*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d
*x]]) + ((24*A*b^2 + 42*a*b*B + 15*a^2*C + 16*b^2*C)*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])
/(24*d) + ((6*b*B + 5*a*C)*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(12*d) + (C*Sqrt[Sec[c
+ d*x]]*(a + b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(3*d)

Rule 4096

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

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))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx &=\frac{C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{1}{3} \int \frac{(a+b \sec (c+d x))^{3/2} \left (\frac{1}{2} a (6 A-C)+(3 A b+3 a B+2 b C) \sec (c+d x)+\frac{1}{2} (6 b B+5 a C) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{(6 b B+5 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{1}{6} \int \frac{\sqrt{a+b \sec (c+d x)} \left (\frac{3}{4} a (8 a A-2 b B-3 a C)+\frac{1}{2} \left (12 a^2 B+6 b^2 B+a b (24 A+11 C)\right ) \sec (c+d x)+\frac{1}{4} \left (24 A b^2+42 a b B+15 a^2 C+16 b^2 C\right ) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{\left (24 A b^2+42 a b B+15 a^2 C+16 b^2 C\right ) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{(6 b B+5 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{1}{6} \int \frac{-\frac{1}{8} a \left (54 a b B-a^2 (48 A-33 C)+8 b^2 (3 A+2 C)\right )+\frac{1}{4} a \left (24 a^2 B+6 b^2 B+a b (72 A+13 C)\right ) \sec (c+d x)+\frac{3}{8} \left (30 a^2 b B+8 b^3 B+5 a^3 C+20 a b^2 (2 A+C)\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{\left (24 A b^2+42 a b B+15 a^2 C+16 b^2 C\right ) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{(6 b B+5 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{1}{6} \int \frac{-\frac{1}{8} a \left (54 a b B-a^2 (48 A-33 C)+8 b^2 (3 A+2 C)\right )+\frac{1}{4} a \left (24 a^2 B+6 b^2 B+a b (72 A+13 C)\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx+\frac{1}{16} \left (30 a^2 b B+8 b^3 B+5 a^3 C+20 a b^2 (2 A+C)\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{\left (24 A b^2+42 a b B+15 a^2 C+16 b^2 C\right ) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{(6 b B+5 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{1}{48} \left (-54 a b B+a^2 (48 A-33 C)-8 b^2 (3 A+2 C)\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{48} \left (48 a^3 B+66 a b^2 B+8 b^3 (3 A+2 C)+a^2 b (96 A+59 C)\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx+\frac{\left (\left (30 a^2 b B+8 b^3 B+5 a^3 C+20 a b^2 (2 A+C)\right ) \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt{b+a \cos (c+d x)}} \, dx}{16 \sqrt{a+b \sec (c+d x)}}\\ &=\frac{\left (24 A b^2+42 a b B+15 a^2 C+16 b^2 C\right ) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{(6 b B+5 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{\left (\left (48 a^3 B+66 a b^2 B+8 b^3 (3 A+2 C)+a^2 b (96 A+59 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}{48 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (30 a^2 b B+8 b^3 B+5 a^3 C+20 a b^2 (2 A+C)\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{16 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (-54 a b B+a^2 (48 A-33 C)-8 b^2 (3 A+2 C)\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{48 \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ &=\frac{\left (30 a^2 b B+8 b^3 B+5 a^3 C+20 a b^2 (2 A+C)\right ) \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 ) \sqrt{\sec (c+d x)}}{8 d \sqrt{a+b \sec (c+d x)}}+\frac{\left (24 A b^2+42 a b B+15 a^2 C+16 b^2 C\right ) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{(6 b B+5 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}+\frac{\left (\left (48 a^3 B+66 a b^2 B+8 b^3 (3 A+2 C)+a^2 b (96 A+59 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}{48 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (-54 a b B+a^2 (48 A-33 C)-8 b^2 (3 A+2 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}{48 \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}\\ &=\frac{\left (48 a^3 B+66 a b^2 B+8 b^3 (3 A+2 C)+a^2 b (96 A+59 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)}}{24 d \sqrt{a+b \sec (c+d x)}}+\frac{\left (30 a^2 b B+8 b^3 B+5 a^3 C+20 a b^2 (2 A+C)\right ) \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 ) \sqrt{\sec (c+d x)}}{8 d \sqrt{a+b \sec (c+d x)}}-\frac{\left (54 a b B-a^2 (48 A-33 C)+8 b^2 (3 A+2 C)\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{24 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}+\frac{\left (24 A b^2+42 a b B+15 a^2 C+16 b^2 C\right ) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{24 d}+\frac{(6 b B+5 a C) \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{12 d}+\frac{C \sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [C]  time = 7.30106, size = 817, normalized size = 1.8 \[ \frac{\left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (\frac{2 \left (96 B a^3+288 A b a^2+52 b C a^2+24 b^2 B a\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{\sqrt{b+a \cos (c+d x)}}+\frac{2 \left (48 A a^3-3 C a^3+126 b B a^2+216 A b^2 a+104 b^2 C a+48 b^3 B\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{\sqrt{b+a \cos (c+d x)}}+\frac{2 i \left (48 A a^3-33 C a^3-54 b B a^2-24 A b^2 a-16 b^2 C a\right ) \sqrt{\frac{a-a \cos (c+d x)}{a+b}} \sqrt{\frac{\cos (c+d x) a+a}{a-b}} \cos (2 (c+d x)) \left (a \left (2 b \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{\frac{1}{a-b}} \sqrt{b+a \cos (c+d x)}\right ),\frac{b-a}{a+b}\right )+a \Pi \left (1-\frac{a}{b};i \sinh ^{-1}\left (\sqrt{\frac{1}{a-b}} \sqrt{b+a \cos (c+d x)}\right )|\frac{b-a}{a+b}\right )\right )-2 b (a+b) E\left (i \sinh ^{-1}\left (\sqrt{\frac{1}{a-b}} \sqrt{b+a \cos (c+d x)}\right )|\frac{b-a}{a+b}\right )\right ) \sin (c+d x)}{\sqrt{\frac{1}{a-b}} b \sqrt{1-\cos ^2(c+d x)} \sqrt{\frac{a^2-a^2 \cos ^2(c+d x)}{a^2}} \left (-a^2+2 b^2+2 (b+a \cos (c+d x))^2-4 b (b+a \cos (c+d x))\right )}\right ) (a+b \sec (c+d x))^{5/2}}{48 d (b+a \cos (c+d x))^{5/2} (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) \sec ^{\frac{9}{2}}(c+d x)}+\frac{\left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (\frac{1}{6} \left (6 B \sin (c+d x) b^2+13 a C \sin (c+d x) b\right ) \sec ^2(c+d x)+\frac{2}{3} b^2 C \tan (c+d x) \sec ^2(c+d x)+\frac{1}{12} \left (33 C \sin (c+d x) a^2+54 b B \sin (c+d x) a+24 A b^2 \sin (c+d x)+16 b^2 C \sin (c+d x)\right ) \sec (c+d x)\right ) (a+b \sec (c+d x))^{5/2}}{d (b+a \cos (c+d x))^2 (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x)) \sec ^{\frac{9}{2}}(c+d x)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*(288*a^2*A*b + 96*a^3*B + 24*a*b^2*B +
52*a^2*b*C)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/Sqrt[b + a*Cos[c + d*x]
] + (2*(48*a^3*A + 216*a*A*b^2 + 126*a^2*b*B + 48*b^3*B - 3*a^3*C + 104*a*b^2*C)*Sqrt[(b + a*Cos[c + d*x])/(a
+ b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/Sqrt[b + a*Cos[c + d*x]] + ((2*I)*(48*a^3*A - 24*a*A*b^2 - 54
*a^2*b*B - 33*a^3*C - 16*a*b^2*C)*Sqrt[(a - a*Cos[c + d*x])/(a + b)]*Sqrt[(a + a*Cos[c + d*x])/(a - b)]*Cos[2*
(c + d*x)]*(-2*b*(a + b)*EllipticE[I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)] +
a*(2*b*EllipticF[I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)] + a*EllipticPi[1 -
a/b, I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)]))*Sin[c + d*x])/(Sqrt[(a - b)^
(-1)]*b*Sqrt[1 - Cos[c + d*x]^2]*Sqrt[(a^2 - a^2*Cos[c + d*x]^2)/a^2]*(-a^2 + 2*b^2 - 4*b*(b + a*Cos[c + d*x])
+ 2*(b + a*Cos[c + d*x])^2))))/(48*d*(b + a*Cos[c + d*x])^(5/2)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d
*x])*Sec[c + d*x]^(9/2)) + ((a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((Sec[c + d*x]^
2*(6*b^2*B*Sin[c + d*x] + 13*a*b*C*Sin[c + d*x]))/6 + (Sec[c + d*x]*(24*A*b^2*Sin[c + d*x] + 54*a*b*B*Sin[c +
d*x] + 33*a^2*C*Sin[c + d*x] + 16*b^2*C*Sin[c + d*x]))/12 + (2*b^2*C*Sec[c + d*x]^2*Tan[c + d*x])/3))/(d*(b +
a*Cos[c + d*x])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(9/2))

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Maple [C]  time = 0.672, size = 6194, normalized size = 13.7 \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))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),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))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),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))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),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))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(1/2),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)/sqrt(sec(d*x + c)), x)