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

Optimal. Leaf size=610 \[ \frac{2 (a-b) \sqrt{a+b} \cot (c+d x) \left (-15 a^2 b^2 (33 A-121 B+19 C)+10 a^3 b (11 B-3 C)-40 a^4 C+6 a b^3 (660 A-209 B+505 C)-3 b^4 (275 A-539 B+225 C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{3465 b^3 d}+\frac{2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{693 b^2 d}-\frac{2 \tan (c+d x) \left (110 a^2 b B-40 a^3 C-5 a b^2 (99 A+67 C)-539 b^3 B\right ) (a+b \sec (c+d x))^{3/2}}{3465 b^2 d}-\frac{2 \tan (c+d x) \left (-15 a^2 b^2 (33 A+19 C)+110 a^3 b B-40 a^4 C-1254 a b^3 B-75 b^4 (11 A+9 C)\right ) \sqrt{a+b \sec (c+d x)}}{3465 b^2 d}+\frac{2 (a-b) \sqrt{a+b} \cot (c+d x) \left (-15 a^3 b^2 (33 A+17 C)-3069 a^2 b^3 B+110 a^4 b B-40 a^5 C-15 a b^4 (319 A+247 C)-1617 b^5 B\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3465 b^4 d}+\frac{2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{99 b^2 d}+\frac{2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d} \]

[Out]

(2*(a - b)*Sqrt[a + b]*(110*a^4*b*B - 3069*a^2*b^3*B - 1617*b^5*B - 40*a^5*C - 15*a^3*b^2*(33*A + 17*C) - 15*a
*b^4*(319*A + 247*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sq
rt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3465*b^4*d) + (2*(a - b)*Sqrt[a +
b]*(10*a^3*b*(11*B - 3*C) - 40*a^4*C - 15*a^2*b^2*(33*A - 121*B + 19*C) - 3*b^4*(275*A - 539*B + 225*C) + 6*a
*b^3*(660*A - 209*B + 505*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a
- b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3465*b^3*d) - (2*(110*a^3
*b*B - 1254*a*b^3*B - 40*a^4*C - 75*b^4*(11*A + 9*C) - 15*a^2*b^2*(33*A + 19*C))*Sqrt[a + b*Sec[c + d*x]]*Tan[
c + d*x])/(3465*b^2*d) - (2*(110*a^2*b*B - 539*b^3*B - 40*a^3*C - 5*a*b^2*(99*A + 67*C))*(a + b*Sec[c + d*x])^
(3/2)*Tan[c + d*x])/(3465*b^2*d) + (2*(99*A*b^2 - 22*a*b*B + 8*a^2*C + 81*b^2*C)*(a + b*Sec[c + d*x])^(5/2)*Ta
n[c + d*x])/(693*b^2*d) + (2*(11*b*B - 4*a*C)*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(99*b^2*d) + (2*C*Sec[c
+ d*x]*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(11*b*d)

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Rubi [A]  time = 2.10959, antiderivative size = 610, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {4092, 4082, 4002, 4005, 3832, 4004} \[ \frac{2 \tan (c+d x) \left (8 a^2 C-22 a b B+99 A b^2+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2}}{693 b^2 d}-\frac{2 \tan (c+d x) \left (110 a^2 b B-40 a^3 C-5 a b^2 (99 A+67 C)-539 b^3 B\right ) (a+b \sec (c+d x))^{3/2}}{3465 b^2 d}-\frac{2 \tan (c+d x) \left (-15 a^2 b^2 (33 A+19 C)+110 a^3 b B-40 a^4 C-1254 a b^3 B-75 b^4 (11 A+9 C)\right ) \sqrt{a+b \sec (c+d x)}}{3465 b^2 d}+\frac{2 (a-b) \sqrt{a+b} \cot (c+d x) \left (-15 a^2 b^2 (33 A-121 B+19 C)+10 a^3 b (11 B-3 C)-40 a^4 C+6 a b^3 (660 A-209 B+505 C)-3 b^4 (275 A-539 B+225 C)\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3465 b^3 d}+\frac{2 (a-b) \sqrt{a+b} \cot (c+d x) \left (-15 a^3 b^2 (33 A+17 C)-3069 a^2 b^3 B+110 a^4 b B-40 a^5 C-15 a b^4 (319 A+247 C)-1617 b^5 B\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3465 b^4 d}+\frac{2 (11 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{99 b^2 d}+\frac{2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{7/2}}{11 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a - b)*Sqrt[a + b]*(110*a^4*b*B - 3069*a^2*b^3*B - 1617*b^5*B - 40*a^5*C - 15*a^3*b^2*(33*A + 17*C) - 15*a
*b^4*(319*A + 247*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sq
rt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3465*b^4*d) + (2*(a - b)*Sqrt[a +
b]*(10*a^3*b*(11*B - 3*C) - 40*a^4*C - 15*a^2*b^2*(33*A - 121*B + 19*C) - 3*b^4*(275*A - 539*B + 225*C) + 6*a
*b^3*(660*A - 209*B + 505*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a
- b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3465*b^3*d) - (2*(110*a^3
*b*B - 1254*a*b^3*B - 40*a^4*C - 75*b^4*(11*A + 9*C) - 15*a^2*b^2*(33*A + 19*C))*Sqrt[a + b*Sec[c + d*x]]*Tan[
c + d*x])/(3465*b^2*d) - (2*(110*a^2*b*B - 539*b^3*B - 40*a^3*C - 5*a*b^2*(99*A + 67*C))*(a + b*Sec[c + d*x])^
(3/2)*Tan[c + d*x])/(3465*b^2*d) + (2*(99*A*b^2 - 22*a*b*B + 8*a^2*C + 81*b^2*C)*(a + b*Sec[c + d*x])^(5/2)*Ta
n[c + d*x])/(693*b^2*d) + (2*(11*b*B - 4*a*C)*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(99*b^2*d) + (2*C*Sec[c
+ d*x]*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(11*b*d)

Rule 4092

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Csc[e + f*x]*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m
+ 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2)
+ A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B*(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m
}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 4082

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

Rule 4002

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

Rule 4005

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, Int[(Csc[e + f*x]*(1 +
Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac{2 \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (a C+\frac{1}{2} b (11 A+9 C) \sec (c+d x)+\frac{1}{2} (11 b B-4 a C) \sec ^2(c+d x)\right ) \, dx}{11 b}\\ &=\frac{2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac{2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac{4 \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (\frac{1}{4} b (77 b B-10 a C)+\frac{1}{4} \left (99 A b^2-22 a b B+8 a^2 C+81 b^2 C\right ) \sec (c+d x)\right ) \, dx}{99 b^2}\\ &=\frac{2 \left (99 A b^2-22 a b B+8 a^2 C+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}+\frac{2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac{2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac{8 \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (\frac{3}{8} b \left (165 A b^2+143 a b B-10 a^2 C+135 b^2 C\right )-\frac{1}{8} \left (110 a^2 b B-539 b^3 B-40 a^3 C-5 a b^2 (99 A+67 C)\right ) \sec (c+d x)\right ) \, dx}{693 b^2}\\ &=-\frac{2 \left (110 a^2 b B-539 b^3 B-40 a^3 C-5 a b^2 (99 A+67 C)\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^2 d}+\frac{2 \left (99 A b^2-22 a b B+8 a^2 C+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}+\frac{2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac{2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac{16 \int \sec (c+d x) \sqrt{a+b \sec (c+d x)} \left (\frac{3}{16} b \left (605 a^2 b B+539 b^3 B-10 a^3 C+10 a b^2 (132 A+101 C)\right )-\frac{3}{16} \left (110 a^3 b B-1254 a b^3 B-40 a^4 C-75 b^4 (11 A+9 C)-15 a^2 b^2 (33 A+19 C)\right ) \sec (c+d x)\right ) \, dx}{3465 b^2}\\ &=-\frac{2 \left (110 a^3 b B-1254 a b^3 B-40 a^4 C-75 b^4 (11 A+9 C)-15 a^2 b^2 (33 A+19 C)\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{3465 b^2 d}-\frac{2 \left (110 a^2 b B-539 b^3 B-40 a^3 C-5 a b^2 (99 A+67 C)\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^2 d}+\frac{2 \left (99 A b^2-22 a b B+8 a^2 C+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}+\frac{2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac{2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}+\frac{32 \int \frac{\sec (c+d x) \left (\frac{3}{32} b \left (1705 a^3 b B+2871 a b^3 B+10 a^4 C+75 b^4 (11 A+9 C)+15 a^2 b^2 (297 A+221 C)\right )-\frac{3}{32} \left (110 a^4 b B-3069 a^2 b^3 B-1617 b^5 B-40 a^5 C-15 a^3 b^2 (33 A+17 C)-15 a b^4 (319 A+247 C)\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{10395 b^2}\\ &=-\frac{2 \left (110 a^3 b B-1254 a b^3 B-40 a^4 C-75 b^4 (11 A+9 C)-15 a^2 b^2 (33 A+19 C)\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{3465 b^2 d}-\frac{2 \left (110 a^2 b B-539 b^3 B-40 a^3 C-5 a b^2 (99 A+67 C)\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^2 d}+\frac{2 \left (99 A b^2-22 a b B+8 a^2 C+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}+\frac{2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac{2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}-\frac{\left (110 a^4 b B-3069 a^2 b^3 B-1617 b^5 B-40 a^5 C-15 a^3 b^2 (33 A+17 C)-15 a b^4 (319 A+247 C)\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{3465 b^2}+\frac{\left (32 \left (\frac{3}{32} b \left (1705 a^3 b B+2871 a b^3 B+10 a^4 C+75 b^4 (11 A+9 C)+15 a^2 b^2 (297 A+221 C)\right )+\frac{3}{32} \left (110 a^4 b B-3069 a^2 b^3 B-1617 b^5 B-40 a^5 C-15 a^3 b^2 (33 A+17 C)-15 a b^4 (319 A+247 C)\right )\right )\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{10395 b^2}\\ &=\frac{2 (a-b) \sqrt{a+b} \left (110 a^4 b B-3069 a^2 b^3 B-1617 b^5 B-40 a^5 C-15 a^3 b^2 (33 A+17 C)-15 a b^4 (319 A+247 C)\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{3465 b^4 d}+\frac{2 (a-b) \sqrt{a+b} \left (a^3 b (110 B-30 C)-40 a^4 C-15 a^2 b^2 (33 A-121 B+19 C)-3 b^4 (275 A-539 B+225 C)+6 a b^3 (660 A-209 B+505 C)\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{3465 b^3 d}-\frac{2 \left (110 a^3 b B-1254 a b^3 B-40 a^4 C-75 b^4 (11 A+9 C)-15 a^2 b^2 (33 A+19 C)\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{3465 b^2 d}-\frac{2 \left (110 a^2 b B-539 b^3 B-40 a^3 C-5 a b^2 (99 A+67 C)\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{3465 b^2 d}+\frac{2 \left (99 A b^2-22 a b B+8 a^2 C+81 b^2 C\right ) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{693 b^2 d}+\frac{2 (11 b B-4 a C) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{99 b^2 d}+\frac{2 C \sec (c+d x) (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{11 b d}\\ \end{align*}

Mathematica [A]  time = 21.8605, size = 1090, normalized size = 1.79 \[ \frac{\cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \left (\frac{4}{99} \left (11 B \sin (c+d x) b^2+23 a C \sin (c+d x) b\right ) \sec ^4(c+d x)+\frac{4}{11} b^2 C \tan (c+d x) \sec ^4(c+d x)+\frac{4}{693} \left (113 C \sin (c+d x) a^2+209 b B \sin (c+d x) a+99 A b^2 \sin (c+d x)+81 b^2 C \sin (c+d x)\right ) \sec ^3(c+d x)+\frac{4 \left (15 C \sin (c+d x) a^3+825 b B \sin (c+d x) a^2+1485 A b^2 \sin (c+d x) a+1145 b^2 C \sin (c+d x) a+539 b^3 B \sin (c+d x)\right ) \sec ^2(c+d x)}{3465 b}+\frac{4 \left (-20 C \sin (c+d x) a^4+55 b B \sin (c+d x) a^3+1485 A b^2 \sin (c+d x) a^2+1025 b^2 C \sin (c+d x) a^2+1793 b^3 B \sin (c+d x) a+825 A b^4 \sin (c+d x)+675 b^4 C \sin (c+d x)\right ) \sec (c+d x)}{3465 b^2}+\frac{4 \left (40 C a^5-110 b B a^4+495 A b^2 a^3+255 b^2 C a^3+3069 b^3 B a^2+4785 A b^4 a+3705 b^4 C a+1617 b^5 B\right ) \sin (c+d x)}{3465 b^3}\right )}{d (b+a \cos (c+d x))^2 (\cos (2 c+2 d x) A+A+2 C+2 B \cos (c+d x))}-\frac{4 (a+b \sec (c+d x))^{5/2} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right ) \sqrt{\frac{1}{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )}} \left ((a+b) \left (40 C a^5-110 b B a^4+15 b^2 (33 A+17 C) a^3+3069 b^3 B a^2+15 b^4 (319 A+247 C) a+1617 b^5 B\right ) E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right ) \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b}{a+b}} \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )-b (a+b) \left (40 C a^4-10 b (11 B+3 C) a^3+15 b^2 (33 A+121 B+19 C) a^2+6 b^3 (660 A+209 B+505 C) a+3 b^4 (275 A+539 B+225 C)\right ) \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right ) \sqrt{1-\tan ^2\left (\frac{1}{2} (c+d x)\right )} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b}{a+b}} \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )+\left (40 C a^5-110 b B a^4+15 b^2 (33 A+17 C) a^3+3069 b^3 B a^2+15 b^4 (319 A+247 C) a+1617 b^5 B\right ) \tan \left (\frac{1}{2} (c+d x)\right ) \left (-b \tan ^4\left (\frac{1}{2} (c+d x)\right )+a \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )-1\right )^2+b\right )\right )}{3465 b^3 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) \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )+1\right )^{3/2} \sqrt{\frac{-a \tan ^2\left (\frac{1}{2} (c+d x)\right )+b \tan ^2\left (\frac{1}{2} (c+d x)\right )+a+b}{\tan ^2\left (\frac{1}{2} (c+d x)\right )+1}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*((a
+ b)*(-110*a^4*b*B + 3069*a^2*b^3*B + 1617*b^5*B + 40*a^5*C + 15*a^3*b^2*(33*A + 17*C) + 15*a*b^4*(319*A + 24
7*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^
2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - b*(a + b)*(40*a^4*C - 10*a^3*b*(11*B
+ 3*C) + 15*a^2*b^2*(33*A + 121*B + 19*C) + 3*b^4*(275*A + 539*B + 225*C) + 6*a*b^3*(660*A + 209*B + 505*C))*E
llipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt
[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (-110*a^4*b*B + 3069*a^2*b^3*B + 1617*b^5*B
+ 40*a^5*C + 15*a^3*b^2*(33*A + 17*C) + 15*a*b^4*(319*A + 247*C))*Tan[(c + d*x)/2]*(b - b*Tan[(c + d*x)/2]^4 +
a*(-1 + Tan[(c + d*x)/2]^2)^2)))/(3465*b^3*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)*(1 + Tan[(c + d*x)/2]^2)^(3/2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c +
d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]) + (Cos[c + d*x]^4*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Se
c[c + d*x]^2)*((4*(495*a^3*A*b^2 + 4785*a*A*b^4 - 110*a^4*b*B + 3069*a^2*b^3*B + 1617*b^5*B + 40*a^5*C + 255*a
^3*b^2*C + 3705*a*b^4*C)*Sin[c + d*x])/(3465*b^3) + (4*Sec[c + d*x]^4*(11*b^2*B*Sin[c + d*x] + 23*a*b*C*Sin[c
+ d*x]))/99 + (4*Sec[c + d*x]^3*(99*A*b^2*Sin[c + d*x] + 209*a*b*B*Sin[c + d*x] + 113*a^2*C*Sin[c + d*x] + 81*
b^2*C*Sin[c + d*x]))/693 + (4*Sec[c + d*x]^2*(1485*a*A*b^2*Sin[c + d*x] + 825*a^2*b*B*Sin[c + d*x] + 539*b^3*B
*Sin[c + d*x] + 15*a^3*C*Sin[c + d*x] + 1145*a*b^2*C*Sin[c + d*x]))/(3465*b) + (4*Sec[c + d*x]*(1485*a^2*A*b^2
*Sin[c + d*x] + 825*A*b^4*Sin[c + d*x] + 55*a^3*b*B*Sin[c + d*x] + 1793*a*b^3*B*Sin[c + d*x] - 20*a^4*C*Sin[c
+ d*x] + 1025*a^2*b^2*C*Sin[c + d*x] + 675*b^4*C*Sin[c + d*x]))/(3465*b^2) + (4*b^2*C*Sec[c + d*x]^4*Tan[c + d
*x])/11))/(d*(b + a*Cos[c + d*x])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]))

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Maple [B]  time = 3.07, size = 7208, normalized size = 11.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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(sec(d*x+c)^2*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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 ({\left (C b^{2} \sec \left (d x + c\right )^{6} +{\left (2 \, C a b + B b^{2}\right )} \sec \left (d x + c\right )^{5} + A a^{2} \sec \left (d x + c\right )^{2} +{\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{4} +{\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )^{3}\right )} \sqrt{b \sec \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b^2*sec(d*x + c)^6 + (2*C*a*b + B*b^2)*sec(d*x + c)^5 + A*a^2*sec(d*x + c)^2 + (C*a^2 + 2*B*a*b +
A*b^2)*sec(d*x + c)^4 + (B*a^2 + 2*A*a*b)*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a), 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(sec(d*x+c)**2*(a+b*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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