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

Optimal. Leaf size=502 \[ \frac {2 \tan (c+d x) \left (-10 a^2 C+45 a b B+63 A b^2+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2}}{315 b d}+\frac {2 \tan (c+d x) \left (-10 a^3 C+45 a^2 b B+6 a b^2 (28 A+19 C)+75 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{315 b d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (10 a^3 C+15 a^2 b (21 A-3 B+11 C)-6 a b^2 (28 A-60 B+19 C)+3 b^3 (63 A-25 B+49 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^2 d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-10 a^4 C+45 a^3 b B+3 a^2 b^2 (161 A+93 C)+435 a b^3 B+21 b^4 (9 A+7 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^3 d}+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{63 b d}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d} \]

[Out]

-2/315*(a-b)*(45*a^3*b*B+435*a*b^3*B-10*a^4*C+21*b^4*(9*A+7*C)+3*a^2*b^2*(161*A+93*C))*cot(d*x+c)*EllipticE((a
+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*
x+c))/(a-b))^(1/2)/b^3/d+2/315*(a-b)*(10*a^3*C+15*a^2*b*(21*A-3*B+11*C)-6*a*b^2*(28*A-60*B+19*C)+3*b^3*(63*A-2
5*B+49*C))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(
d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^2/d+2/315*(63*A*b^2+45*B*a*b-10*C*a^2+49*C*b^2)*(a+b*se
c(d*x+c))^(3/2)*tan(d*x+c)/b/d+2/63*(9*B*b-2*C*a)*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b/d+2/9*C*(a+b*sec(d*x+c))
^(7/2)*tan(d*x+c)/b/d+2/315*(45*a^2*b*B+75*b^3*B-10*a^3*C+6*a*b^2*(28*A+19*C))*(a+b*sec(d*x+c))^(1/2)*tan(d*x+
c)/b/d

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Rubi [A]  time = 1.26, antiderivative size = 502, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {4082, 4002, 4005, 3832, 4004} \[ \frac {2 \tan (c+d x) \left (-10 a^2 C+45 a b B+63 A b^2+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2}}{315 b d}+\frac {2 \tan (c+d x) \left (45 a^2 b B-10 a^3 C+6 a b^2 (28 A+19 C)+75 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{315 b d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (15 a^2 b (21 A-3 B+11 C)+10 a^3 C-6 a b^2 (28 A-60 B+19 C)+3 b^3 (63 A-25 B+49 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^2 d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (3 a^2 b^2 (161 A+93 C)+45 a^3 b B-10 a^4 C+435 a b^3 B+21 b^4 (9 A+7 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{315 b^3 d}+\frac {2 (9 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{63 b d}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(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]*(45*a^3*b*B + 435*a*b^3*B - 10*a^4*C + 21*b^4*(9*A + 7*C) + 3*a^2*b^2*(161*A + 93*C))*
Cot[c + d*x]*EllipticE[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))])/(315*b^3*d) + (2*(a - b)*Sqrt[a + b]*(10*a^3*C + 15*a^2*
b*(21*A - 3*B + 11*C) - 6*a*b^2*(28*A - 60*B + 19*C) + 3*b^3*(63*A - 25*B + 49*C))*Cot[c + d*x]*EllipticF[ArcS
in[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))])/(315*b^2*d) + (2*(45*a^2*b*B + 75*b^3*B - 10*a^3*C + 6*a*b^2*(28*A + 19*C))*Sqrt[a
+ b*Sec[c + d*x]]*Tan[c + d*x])/(315*b*d) + (2*(63*A*b^2 + 45*a*b*B - 10*a^2*C + 49*b^2*C)*(a + b*Sec[c + d*x]
)^(3/2)*Tan[c + d*x])/(315*b*d) + (2*(9*b*B - 2*a*C)*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(63*b*d) + (2*C*
(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*b*d)

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 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 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]

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 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]

Rubi steps

\begin {align*} \int \sec (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 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {2 \int \sec (c+d x) (a+b \sec (c+d x))^{5/2} \left (\frac {1}{2} b (9 A+7 C)+\frac {1}{2} (9 b B-2 a C) \sec (c+d x)\right ) \, dx}{9 b}\\ &=\frac {2 (9 b B-2 a C) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 C (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {4 \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (\frac {3}{4} b (21 a A+15 b B+13 a C)+\frac {1}{4} \left (63 A b^2+45 a b B-10 a^2 C+49 b^2 C\right ) \sec (c+d x)\right ) \, dx}{63 b}\\ &=\frac {2 \left (63 A b^2+45 a b B-10 a^2 C+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}+\frac {2 (9 b B-2 a C) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 C (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {8 \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (\frac {3}{8} b \left (120 a b B+7 b^2 (9 A+7 C)+5 a^2 (21 A+11 C)\right )+\frac {3}{8} \left (45 a^2 b B+75 b^3 B-10 a^3 C+6 a b^2 (28 A+19 C)\right ) \sec (c+d x)\right ) \, dx}{315 b}\\ &=\frac {2 \left (45 a^2 b B+75 b^3 B-10 a^3 C+6 a b^2 (28 A+19 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}+\frac {2 \left (63 A b^2+45 a b B-10 a^2 C+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}+\frac {2 (9 b B-2 a C) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 C (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {16 \int \frac {\sec (c+d x) \left (\frac {3}{16} b \left (405 a^2 b B+75 b^3 B+5 a^3 (63 A+31 C)+3 a b^2 (119 A+87 C)\right )+\frac {3}{16} \left (45 a^3 b B+435 a b^3 B-10 a^4 C+21 b^4 (9 A+7 C)+3 a^2 b^2 (161 A+93 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{945 b}\\ &=\frac {2 \left (45 a^2 b B+75 b^3 B-10 a^3 C+6 a b^2 (28 A+19 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}+\frac {2 \left (63 A b^2+45 a b B-10 a^2 C+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}+\frac {2 (9 b B-2 a C) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 C (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac {\left ((a-b) \left (10 a^3 C+15 a^2 b (21 A-3 B+11 C)-6 a b^2 (28 A-60 B+19 C)+3 b^3 (63 A-25 B+49 C)\right )\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{315 b}+\frac {\left (45 a^3 b B+435 a b^3 B-10 a^4 C+21 b^4 (9 A+7 C)+3 a^2 b^2 (161 A+93 C)\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{315 b}\\ &=-\frac {2 (a-b) \sqrt {a+b} \left (45 a^3 b B+435 a b^3 B-10 a^4 C+21 b^4 (9 A+7 C)+3 a^2 b^2 (161 A+93 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}}}{315 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (10 a^3 C+15 a^2 b (21 A-3 B+11 C)-6 a b^2 (28 A-60 B+19 C)+3 b^3 (63 A-25 B+49 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}}}{315 b^2 d}+\frac {2 \left (45 a^2 b B+75 b^3 B-10 a^3 C+6 a b^2 (28 A+19 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b d}+\frac {2 \left (63 A b^2+45 a b B-10 a^2 C+49 b^2 C\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}+\frac {2 (9 b B-2 a C) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac {2 C (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}\\ \end {align*}

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Mathematica [B]  time = 26.70, size = 4220, normalized size = 8.41 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[c + d*x]^4*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*(483*a^2*A*b^2 + 189*A*
b^4 + 45*a^3*b*B + 435*a*b^3*B - 10*a^4*C + 279*a^2*b^2*C + 147*b^4*C)*Sin[c + d*x])/(315*b^2) + (4*Sec[c + d*
x]^3*(9*b^2*B*Sin[c + d*x] + 19*a*b*C*Sin[c + d*x]))/63 + (4*Sec[c + d*x]^2*(63*A*b^2*Sin[c + d*x] + 135*a*b*B
*Sin[c + d*x] + 75*a^2*C*Sin[c + d*x] + 49*b^2*C*Sin[c + d*x]))/315 + (4*Sec[c + d*x]*(231*a*A*b^2*Sin[c + d*x
] + 135*a^2*b*B*Sin[c + d*x] + 75*b^3*B*Sin[c + d*x] + 5*a^3*C*Sin[c + d*x] + 163*a*b^2*C*Sin[c + d*x]))/(315*
b) + (4*b^2*C*Sec[c + d*x]^3*Tan[c + d*x])/9))/(d*(b + a*Cos[c + d*x])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2
*c + 2*d*x])) + (4*((-46*a^2*A*b)/(15*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (6*A*b^3)/(5*Sqrt[b + a*C
os[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*a^3*B)/(7*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (58*a*b^2*B)/(2
1*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (4*a^4*C)/(63*b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]])
- (62*a^2*b*C)/(35*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (14*b^3*C)/(15*Sqrt[b + a*Cos[c + d*x]]*Sqrt
[Sec[c + d*x]]) - (16*a^3*A*Sqrt[Sec[c + d*x]])/(15*Sqrt[b + a*Cos[c + d*x]]) + (16*a*A*b^2*Sqrt[Sec[c + d*x]]
)/(15*Sqrt[b + a*Cos[c + d*x]]) - (2*a^4*B*Sqrt[Sec[c + d*x]])/(7*b*Sqrt[b + a*Cos[c + d*x]]) - (4*a^2*b*B*Sqr
t[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) + (10*b^3*B*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) -
 (248*a^3*C*Sqrt[Sec[c + d*x]])/(315*Sqrt[b + a*Cos[c + d*x]]) + (4*a^5*C*Sqrt[Sec[c + d*x]])/(63*b^2*Sqrt[b +
 a*Cos[c + d*x]]) + (76*a*b^2*C*Sqrt[Sec[c + d*x]])/(105*Sqrt[b + a*Cos[c + d*x]]) - (46*a^3*A*Cos[2*(c + d*x)
]*Sqrt[Sec[c + d*x]])/(15*Sqrt[b + a*Cos[c + d*x]]) - (6*a*A*b^2*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*Sqrt[
b + a*Cos[c + d*x]]) - (2*a^4*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(7*b*Sqrt[b + a*Cos[c + d*x]]) - (58*a^2*
b*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) - (62*a^3*C*Cos[2*(c + d*x)]*Sqrt[Sec[c
 + d*x]])/(35*Sqrt[b + a*Cos[c + d*x]]) + (4*a^5*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(63*b^2*Sqrt[b + a*Cos
[c + d*x]]) - (14*a*b^2*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(15*Sqrt[b + a*Cos[c + d*x]]))*Sqrt[Cos[(c + d*
x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((a + b)*((-45*a^3*b*
B - 435*a*b^3*B + 10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]]
, (a - b)/(a + b)] + b*(-10*a^3*C + 15*a^2*b*(21*A + 3*B + 11*C) + 6*a*b^2*(28*A + 60*B + 19*C) + 3*b^3*(63*A
+ 25*B + 49*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*
Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)]*Sec[c + d*x] + (-45*a^3*b*B - 435*a*b^3*B + 10*a^4*C -
 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Tan[(c +
d*x)/2]))/(315*b^2*d*(b + a*Cos[c + d*x])^3*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(Sec[(c + d*x)/2
]^2)^(3/2)*Sec[c + d*x]^(9/2)*((2*a*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sin[c + d*x]*((a + b)*((-45*a^3*b*B
- 435*a*b^3*B + 10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]],
(a - b)/(a + b)] + b*(-10*a^3*C + 15*a^2*b*(21*A + 3*B + 11*C) + 6*a*b^2*(28*A + 60*B + 19*C) + 3*b^3*(63*A +
25*B + 49*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sq
rt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)]*Sec[c + d*x] + (-45*a^3*b*B - 435*a*b^3*B + 10*a^4*C - 2
1*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Tan[(c + d*
x)/2]))/(315*b^2*(b + a*Cos[c + d*x])^(3/2)*(Sec[(c + d*x)/2]^2)^(3/2)) - (2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d
*x]]*Tan[(c + d*x)/2]*((a + b)*((-45*a^3*b*B - 435*a*b^3*B + 10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A
+ 93*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(-10*a^3*C + 15*a^2*b*(21*A + 3*B + 11*C) +
6*a*b^2*(28*A + 60*B + 19*C) + 3*b^3*(63*A + 25*B + 49*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)
])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)]*Sec[c + d*x
] + (-45*a^3*b*B - 435*a*b^3*B + 10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*C))*Cos[c + d*x]*(b + a
*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2]))/(105*b^2*Sqrt[b + a*Cos[c + d*x]]*(Sec[(c + d*x)/2]^2)^(3
/2)) + (2*((a + b)*((-45*a^3*b*B - 435*a*b^3*B + 10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*C))*Ell
ipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(-10*a^3*C + 15*a^2*b*(21*A + 3*B + 11*C) + 6*a*b^2*(28*
A + 60*B + 19*C) + 3*b^3*(63*A + 25*B + 49*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*(Cos[c +
d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)]*Sec[c + d*x] + (-45*a^3
*b*B - 435*a*b^3*B + 10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*C))*Cos[c + d*x]*(b + a*Cos[c + d*x
])*Sec[(c + d*x)/2]^4*Tan[(c + d*x)/2])*(-(Cos[(c + d*x)/2]*Sec[c + d*x]*Sin[(c + d*x)/2]) + Cos[(c + d*x)/2]^
2*Sec[c + d*x]*Tan[c + d*x]))/(315*b^2*Sqrt[b + a*Cos[c + d*x]]*(Sec[(c + d*x)/2]^2)^(3/2)*Sqrt[Cos[(c + d*x)/
2]^2*Sec[c + d*x]]) + (4*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(((-45*a^3*b*B - 435*a*b^3*B + 10*a^4*C - 21*b^
4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^6)/2 - a*(-45*a^3
*b*B - 435*a*b^3*B + 10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*C))*Cos[c + d*x]*Sec[(c + d*x)/2]^4
*Sin[c + d*x]*Tan[(c + d*x)/2] - (-45*a^3*b*B - 435*a*b^3*B + 10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A
 + 93*C))*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Sin[c + d*x]*Tan[(c + d*x)/2] + 2*(-45*a^3*b*B - 435*a*b^3*B
 + 10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*C))*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2
]^4*Tan[(c + d*x)/2]^2 + (3*(a + b)*((-45*a^3*b*B - 435*a*b^3*B + 10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(1
61*A + 93*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(-10*a^3*C + 15*a^2*b*(21*A + 3*B + 11*
C) + 6*a*b^2*(28*A + 60*B + 19*C) + 3*b^3*(63*A + 25*B + 49*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a
 + b)])*Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^2]*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)]*Sec[c +
d*x]*(-(Sec[(c + d*x)/2]^2*Sin[c + d*x]) + Cos[c + d*x]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/2 + ((a + b)*((-
45*a^3*b*B - 435*a*b^3*B + 10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*C))*EllipticE[ArcSin[Tan[(c +
 d*x)/2]], (a - b)/(a + b)] + b*(-10*a^3*C + 15*a^2*b*(21*A + 3*B + 11*C) + 6*a*b^2*(28*A + 60*B + 19*C) + 3*b
^3*(63*A + 25*B + 49*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)])*(Cos[c + d*x]*Sec[(c + d*x)/2]^
2)^(3/2)*Sec[c + d*x]*(-((a*Sec[(c + d*x)/2]^2*Sin[c + d*x])/(a + b)) + ((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]
^2*Tan[(c + d*x)/2])/(a + b)))/(2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)]) + (a + b)*(Cos[c +
d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)]*Sec[c + d*x]*((b*(-10*a
^3*C + 15*a^2*b*(21*A + 3*B + 11*C) + 6*a*b^2*(28*A + 60*B + 19*C) + 3*b^3*(63*A + 25*B + 49*C))*Sec[(c + d*x)
/2]^2)/(2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[1 - ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]) + ((-45*a^3*b*B - 435*a
*b^3*B + 10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*C))*Sec[(c + d*x)/2]^2*Sqrt[1 - ((a - b)*Tan[(c
 + d*x)/2]^2)/(a + b)])/(2*Sqrt[1 - Tan[(c + d*x)/2]^2])) + (a + b)*((-45*a^3*b*B - 435*a*b^3*B + 10*a^4*C - 2
1*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*C))*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + b*(-10*a^
3*C + 15*a^2*b*(21*A + 3*B + 11*C) + 6*a*b^2*(28*A + 60*B + 19*C) + 3*b^3*(63*A + 25*B + 49*C))*EllipticF[ArcS
in[Tan[(c + d*x)/2]], (a - b)/(a + b)])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sqrt[((b + a*Cos[c + d*x])*Sec
[(c + d*x)/2]^2)/(a + b)]*Sec[c + d*x]*Tan[c + d*x]))/(315*b^2*Sqrt[b + a*Cos[c + d*x]]*(Sec[(c + d*x)/2]^2)^(
3/2))))

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \sec \left (d x + c\right )^{5} + {\left (2 \, C a b + B b^{2}\right )} \sec \left (d x + c\right )^{4} + A a^{2} \sec \left (d x + c\right ) + {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{3} + {\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )^{2}\right )} \sqrt {b \sec \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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), x)

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maple [B]  time = 3.65, size = 6163, normalized size = 12.28 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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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