3.1048 \(\int \frac{(a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=419 \[ \frac{\sqrt{\sec (c+d x)} \left (4 a^2 b (4 A+15 C)+10 a^3 B+20 a b^2 B-b^3 (16 A-15 C)\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{15 d \sqrt{a+b \sec (c+d x)}}+\frac{\left (6 a^2 (3 A+5 C)+70 a b B+b^2 (46 A-15 C)\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}-\frac{b \sin (c+d x) \sqrt{\sec (c+d x)} (10 a B+16 A b-15 b C) \sqrt{a+b \sec (c+d x)}}{15 d}+\frac{2 (a B+A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{b^2 (5 a C+2 b B) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{a+b \sec (c+d x)}} \]

[Out]

((10*a^3*B + 20*a*b^2*B - b^3*(16*A - 15*C) + 4*a^2*b*(4*A + 15*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Ellipti
cF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(15*d*Sqrt[a + b*Sec[c + d*x]]) + (b^2*(2*b*B + 5*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]])/(d*Sqrt[a + b*Sec
[c + d*x]]) + ((70*a*b*B + b^2*(46*A - 15*C) + 6*a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a
+ b*Sec[c + d*x]])/(15*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) - (b*(16*A*b + 10*a*B - 15*b*
C)*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(15*d) + (2*(A*b + a*B)*(a + b*Sec[c + d*x])^(3/2
)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/
2))

________________________________________________________________________________________

Rubi [A]  time = 1.64518, antiderivative size = 419, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 13, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.289, Rules used = {4094, 4096, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{\sqrt{\sec (c+d x)} \left (4 a^2 b (4 A+15 C)+10 a^3 B+20 a b^2 B-b^3 (16 A-15 C)\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 d \sqrt{a+b \sec (c+d x)}}+\frac{\left (6 a^2 (3 A+5 C)+70 a b B+b^2 (46 A-15 C)\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}-\frac{b \sin (c+d x) \sqrt{\sec (c+d x)} (10 a B+16 A b-15 b C) \sqrt{a+b \sec (c+d x)}}{15 d}+\frac{2 (a B+A b) \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{b^2 (5 a C+2 b B) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{a+b \sec (c+d x)}} \]

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))/Sec[c + d*x]^(5/2),x]

[Out]

((10*a^3*B + 20*a*b^2*B - b^3*(16*A - 15*C) + 4*a^2*b*(4*A + 15*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Ellipti
cF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(15*d*Sqrt[a + b*Sec[c + d*x]]) + (b^2*(2*b*B + 5*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]])/(d*Sqrt[a + b*Sec
[c + d*x]]) + ((70*a*b*B + b^2*(46*A - 15*C) + 6*a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a
+ b*Sec[c + d*x]])/(15*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) - (b*(16*A*b + 10*a*B - 15*b*
C)*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(15*d) + (2*(A*b + a*B)*(a + b*Sec[c + d*x])^(3/2
)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (2*A*(a + b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/
2))

Rule 4094

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

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 )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2}{5} \int \frac{(a+b \sec (c+d x))^{3/2} \left (\frac{5}{2} (A b+a B)+\frac{1}{2} (3 a A+5 b B+5 a C) \sec (c+d x)-\frac{1}{2} b (2 A-5 C) \sec ^2(c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 (A b+a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4}{15} \int \frac{\sqrt{a+b \sec (c+d x)} \left (\frac{3}{4} \left (5 A b^2+10 a b B+a^2 (3 A+5 C)\right )+\frac{1}{4} \left (8 a A b+5 a^2 B+15 b^2 B+30 a b C\right ) \sec (c+d x)-\frac{1}{4} b (16 A b+10 a B-15 b C) \sec ^2(c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx\\ &=-\frac{b (16 A b+10 a B-15 b C) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 (A b+a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4}{15} \int \frac{\frac{1}{8} a \left (70 a b B+b^2 (46 A-15 C)+6 a^2 (3 A+5 C)\right )+\frac{1}{4} \left (15 A b^3+5 a^3 B+45 a b^2 B+a^2 b (17 A+45 C)\right ) \sec (c+d x)+\frac{15}{8} b^2 (2 b B+5 a C) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx\\ &=-\frac{b (16 A b+10 a B-15 b C) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 (A b+a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{4}{15} \int \frac{\frac{1}{8} a \left (70 a b B+b^2 (46 A-15 C)+6 a^2 (3 A+5 C)\right )+\frac{1}{4} \left (15 A b^3+5 a^3 B+45 a b^2 B+a^2 b (17 A+45 C)\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx+\frac{1}{2} \left (b^2 (2 b B+5 a C)\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=-\frac{b (16 A b+10 a B-15 b C) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 (A b+a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{1}{30} \left (70 a b B+b^2 (46 A-15 C)+6 a^2 (3 A+5 C)\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{30} \left (10 a^3 B+20 a b^2 B-b^3 (16 A-15 C)+4 a^2 b (4 A+15 C)\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx+\frac{\left (b^2 (2 b B+5 a C) \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}{2 \sqrt{a+b \sec (c+d x)}}\\ &=-\frac{b (16 A b+10 a B-15 b C) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 (A b+a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (\left (10 a^3 B+20 a b^2 B-b^3 (16 A-15 C)+4 a^2 b (4 A+15 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}{30 \sqrt{a+b \sec (c+d x)}}+\frac{\left (b^2 (2 b B+5 a C) \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}{2 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (70 a b B+b^2 (46 A-15 C)+6 a^2 (3 A+5 C)\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{30 \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ &=\frac{b^2 (2 b B+5 a C) \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)}}{d \sqrt{a+b \sec (c+d x)}}-\frac{b (16 A b+10 a B-15 b C) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 (A b+a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{\left (\left (10 a^3 B+20 a b^2 B-b^3 (16 A-15 C)+4 a^2 b (4 A+15 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}{30 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (70 a b B+b^2 (46 A-15 C)+6 a^2 (3 A+5 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}{30 \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}\\ &=\frac{\left (10 a^3 B+20 a b^2 B-b^3 (16 A-15 C)+4 a^2 b (4 A+15 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)}}{15 d \sqrt{a+b \sec (c+d x)}}+\frac{b^2 (2 b B+5 a C) \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)}}{d \sqrt{a+b \sec (c+d x)}}+\frac{\left (70 a b B+b^2 (46 A-15 C)+6 a^2 (3 A+5 C)\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{15 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}-\frac{b (16 A b+10 a B-15 b C) \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 (A b+a B) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [C]  time = 7.05177, size = 755, normalized size = 1.8 \[ \frac{(a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{2}{5} a^2 A \sin (2 (c+d x))+\frac{4}{15} a (5 a B+11 A b) \sin (c+d x)+2 b^2 C \tan (c+d x)\right )}{d \sec ^{\frac{9}{2}}(c+d x) (a \cos (c+d x)+b)^2 (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac{(a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{2 \left (68 a^2 A b+180 a^2 b C+20 a^3 B+180 a b^2 B+60 A b^3\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{\sqrt{a \cos (c+d x)+b}}+\frac{2 i \sin (c+d x) \cos (2 (c+d x)) \left (18 a^3 A+70 a^2 b B+30 a^3 C+46 a A b^2-15 a b^2 C\right ) \sqrt{\frac{a-a \cos (c+d x)}{a+b}} \sqrt{\frac{a \cos (c+d x)+a}{a-b}} \left (a \left (2 b \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{\frac{1}{a-b}} \sqrt{a \cos (c+d x)+b}\right ),\frac{b-a}{a+b}\right )+a \Pi \left (1-\frac{a}{b};i \sinh ^{-1}\left (\sqrt{\frac{1}{a-b}} \sqrt{b+a \cos (c+d x)}\right )|\frac{b-a}{a+b}\right )\right )-2 b (a+b) E\left (i \sinh ^{-1}\left (\sqrt{\frac{1}{a-b}} \sqrt{b+a \cos (c+d x)}\right )|\frac{b-a}{a+b}\right )\right )}{b \sqrt{\frac{1}{a-b}} \sqrt{1-\cos ^2(c+d x)} \sqrt{\frac{a^2-a^2 \cos ^2(c+d x)}{a^2}} \left (-a^2+2 (a \cos (c+d x)+b)^2-4 b (a \cos (c+d x)+b)+2 b^2\right )}+\frac{2 \left (18 a^3 A+70 a^2 b B+30 a^3 C+46 a A b^2+135 a b^2 C+60 b^3 B\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{\sqrt{a \cos (c+d x)+b}}\right )}{30 d \sec ^{\frac{9}{2}}(c+d x) (a \cos (c+d x)+b)^{5/2} (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

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))/Sec[c + d*x]^(5/2),x]

[Out]

((a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*(68*a^2*A*b + 60*A*b^3 + 20*a^3*B + 18
0*a*b^2*B + 180*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*(18*a^3*A + 46*a*A*b^2 + 70*a^2*b*B + 60*b^3*B + 30*a^3*C + 135*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)*(18*a^3*A + 46*a
*A*b^2 + 70*a^2*b*B + 30*a^3*C - 15*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*Ell
ipticPi[1 - a/b, I*ArcSinh[Sqrt[(a - b)^(-1)]*Sqrt[b + a*Cos[c + d*x]]], (-a + b)/(a + b)]))*Sin[c + d*x])/(Sq
rt[(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*Co
s[c + d*x]) + 2*(b + a*Cos[c + d*x])^2))))/(30*d*(b + a*Cos[c + d*x])^(5/2)*(A + 2*C + 2*B*Cos[c + d*x] + A*Co
s[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)*((4*
a*(11*A*b + 5*a*B)*Sin[c + d*x])/15 + (2*a^2*A*Sin[2*(c + d*x)])/5 + 2*b^2*C*Tan[c + d*x]))/(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 = 0.646, size = 5634, normalized size = 13.5 \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)^(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))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/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 b^{2} \sec \left (d x + c\right )^{4} +{\left (2 \, C a b + B b^{2}\right )} \sec \left (d x + c\right )^{3} + A a^{2} +{\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )\right )} \sqrt{b \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac{5}{2}}}, x\right ) \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)^(5/2),x, algorithm="fricas")

[Out]

integral((C*b^2*sec(d*x + c)^4 + (2*C*a*b + B*b^2)*sec(d*x + c)^3 + A*a^2 + (C*a^2 + 2*B*a*b + A*b^2)*sec(d*x
+ c)^2 + (B*a^2 + 2*A*a*b)*sec(d*x + c))*sqrt(b*sec(d*x + c) + a)/sec(d*x + c)^(5/2), 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))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/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{{\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}}}{\sec \left (d x + c\right )^{\frac{5}{2}}}\,{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)^(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)^(5/2), x)