3.519 \(\int \frac {\sec ^4(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=277 \[ \frac {(11 A-15 B+19 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(245 A-273 B+397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{210 a^2 d}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac {(7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{14 a d \sqrt {a \sec (c+d x)+a}}-\frac {(35 A-63 B+67 C) \tan (c+d x) \sec ^2(c+d x)}{70 a d \sqrt {a \sec (c+d x)+a}}-\frac {(455 A-651 B+799 C) \tan (c+d x)}{105 a d \sqrt {a \sec (c+d x)+a}} \]

[Out]

1/4*(11*A-15*B+19*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/2*(A-B+
C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)-1/105*(455*A-651*B+799*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^
(1/2)-1/70*(35*A-63*B+67*C)*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+1/14*(7*A-7*B+11*C)*sec(d*x+c)^
3*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+1/210*(245*A-273*B+397*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^2/d

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Rubi [A]  time = 0.88, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {4084, 4021, 4010, 4001, 3795, 203} \[ \frac {(11 A-15 B+19 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(245 A-273 B+397 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{210 a^2 d}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac {(7 A-7 B+11 C) \tan (c+d x) \sec ^3(c+d x)}{14 a d \sqrt {a \sec (c+d x)+a}}-\frac {(35 A-63 B+67 C) \tan (c+d x) \sec ^2(c+d x)}{70 a d \sqrt {a \sec (c+d x)+a}}-\frac {(455 A-651 B+799 C) \tan (c+d x)}{105 a d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

((11*A - 15*B + 19*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d)
- ((A - B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((455*A - 651*B + 799*C)*Tan[c
+ d*x])/(105*a*d*Sqrt[a + a*Sec[c + d*x]]) - ((35*A - 63*B + 67*C)*Sec[c + d*x]^2*Tan[c + d*x])/(70*a*d*Sqrt[
a + a*Sec[c + d*x]]) + ((7*A - 7*B + 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(14*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((
245*A - 273*B + 397*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(210*a^2*d)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3795

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

Rule 4001

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[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rule 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(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 + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4021

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

Rule 4084

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*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[e + f*x])^m*(d*Csc[e + f*x])^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)
))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\sec ^4(c+d x) \left (-2 a (A-2 B+2 C)+\frac {1}{2} a (7 A-7 B+11 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(7 A-7 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {\sec ^3(c+d x) \left (\frac {3}{2} a^2 (7 A-7 B+11 C)-\frac {1}{4} a^2 (35 A-63 B+67 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{7 a^3}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(35 A-63 B+67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt {a+a \sec (c+d x)}}+\frac {(7 A-7 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt {a+a \sec (c+d x)}}+\frac {2 \int \frac {\sec ^2(c+d x) \left (-\frac {1}{2} a^3 (35 A-63 B+67 C)+\frac {1}{8} a^3 (245 A-273 B+397 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{35 a^4}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(35 A-63 B+67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt {a+a \sec (c+d x)}}+\frac {(7 A-7 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt {a+a \sec (c+d x)}}+\frac {(245 A-273 B+397 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{210 a^2 d}+\frac {4 \int \frac {\sec (c+d x) \left (\frac {1}{16} a^4 (245 A-273 B+397 C)-\frac {1}{8} a^4 (455 A-651 B+799 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{105 a^5}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(455 A-651 B+799 C) \tan (c+d x)}{105 a d \sqrt {a+a \sec (c+d x)}}-\frac {(35 A-63 B+67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt {a+a \sec (c+d x)}}+\frac {(7 A-7 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt {a+a \sec (c+d x)}}+\frac {(245 A-273 B+397 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{210 a^2 d}+\frac {(11 A-15 B+19 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(455 A-651 B+799 C) \tan (c+d x)}{105 a d \sqrt {a+a \sec (c+d x)}}-\frac {(35 A-63 B+67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt {a+a \sec (c+d x)}}+\frac {(7 A-7 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt {a+a \sec (c+d x)}}+\frac {(245 A-273 B+397 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{210 a^2 d}-\frac {(11 A-15 B+19 C) \operatorname {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{2 a d}\\ &=\frac {(11 A-15 B+19 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(455 A-651 B+799 C) \tan (c+d x)}{105 a d \sqrt {a+a \sec (c+d x)}}-\frac {(35 A-63 B+67 C) \sec ^2(c+d x) \tan (c+d x)}{70 a d \sqrt {a+a \sec (c+d x)}}+\frac {(7 A-7 B+11 C) \sec ^3(c+d x) \tan (c+d x)}{14 a d \sqrt {a+a \sec (c+d x)}}+\frac {(245 A-273 B+397 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{210 a^2 d}\\ \end {align*}

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Mathematica [C]  time = 11.52, size = 2746, normalized size = 9.91 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

(4*Cos[(c + d*x)/2]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - 2*Sin[(c + d*x)/2]^2)^(-1)]*Sqrt[1 - 2
*Sin[(c + d*x)/2]^2]*((4*A*Sin[(c + d*x)/2])/(7*(1 - 2*Sin[(c + d*x)/2]^2)^(7/2)) - ((A - B + C)*(1 - 2*Sin[(c
+ d*x)/2]))/(28*(1 + Sin[(c + d*x)/2])*(1 - 2*Sin[(c + d*x)/2]^2)^(7/2)) + ((A - B + C)*(1 + 2*Sin[(c + d*x)/
2]))/(28*(1 - Sin[(c + d*x)/2])*(1 - 2*Sin[(c + d*x)/2]^2)^(7/2)) - ((A - B + C)*(315*ArcTan[(1 - 2*Sin[(c + d
*x)/2])/Sqrt[1 - 2*Sin[(c + d*x)/2]^2]] + (5 + 3*Sin[(c + d*x)/2])/((1 - Sin[(c + d*x)/2])*(1 - 2*Sin[(c + d*x
)/2]^2)^(5/2)) - (11 + 17*Sin[(c + d*x)/2])/((1 - Sin[(c + d*x)/2])*(1 - 2*Sin[(c + d*x)/2]^2)^(3/2)) + (61 +
71*Sin[(c + d*x)/2])/((1 - Sin[(c + d*x)/2])*Sqrt[1 - 2*Sin[(c + d*x)/2]^2]) + (193*Sqrt[1 - 2*Sin[(c + d*x)/2
]^2])/(1 - Sin[(c + d*x)/2])))/70 + ((A - B + C)*(315*ArcTan[(1 + 2*Sin[(c + d*x)/2])/Sqrt[1 - 2*Sin[(c + d*x)
/2]^2]] + (5 - 3*Sin[(c + d*x)/2])/((1 + Sin[(c + d*x)/2])*(1 - 2*Sin[(c + d*x)/2]^2)^(5/2)) - (11 - 17*Sin[(c
+ d*x)/2])/((1 + Sin[(c + d*x)/2])*(1 - 2*Sin[(c + d*x)/2]^2)^(3/2)) + (61 - 71*Sin[(c + d*x)/2])/((1 + Sin[(
c + d*x)/2])*Sqrt[1 - 2*Sin[(c + d*x)/2]^2]) + (193*Sqrt[1 - 2*Sin[(c + d*x)/2]^2])/(1 + Sin[(c + d*x)/2])))/7
0 - ((7*A - 3*B - C)*Csc[(c + d*x)/2]^9*(363825*Sin[(c + d*x)/2]^2 - 4729725*Sin[(c + d*x)/2]^4 + 26785605*Sin
[(c + d*x)/2]^6 - 86790165*Sin[(c + d*x)/2]^8 + 177677808*Sin[(c + d*x)/2]^10 - 239283044*Sin[(c + d*x)/2]^12
+ 52080*Hypergeometric2F1[2, 11/2, 13/2, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^12 +
560*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Si
n[(c + d*x)/2]^12 + 213120160*Sin[(c + d*x)/2]^14 - 168280*Hypergeometric2F1[2, 11/2, 13/2, Sin[(c + d*x)/2]^2
/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^14 - 2240*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}
, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^14 - 121497024*Sin[(c + d*x)/2]^16 + 212520
*Hypergeometric2F1[2, 11/2, 13/2, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^16 + 3360*H
ypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c +
d*x)/2]^16 + 40125184*Sin[(c + d*x)/2]^18 - 124320*Hypergeometric2F1[2, 11/2, 13/2, Sin[(c + d*x)/2]^2/(-1 +
2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^18 - 2240*HypergeometricPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[(
c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^18 - 5840384*Sin[(c + d*x)/2]^20 + 28000*Hypergeom
etric2F1[2, 11/2, 13/2, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^20 + 560*Hypergeometr
icPFQ[{2, 2, 2, 2, 11/2}, {1, 1, 1, 13/2}, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^20
+ 363825*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c
+ d*x)/2]^2)] - 5336100*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^2*Sqrt
[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] + 34636140*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d
*x)/2]^2)]]*Sin[(c + d*x)/2]^4*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] - 131060160*ArcTanh[Sqrt[S
in[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^6*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*
x)/2]^2)] + 320535600*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^8*Sqrt[Si
n[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] - 530671680*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x
)/2]^2)]]*Sin[(c + d*x)/2]^10*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] + 604296000*ArcTanh[Sqrt[Si
n[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^12*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*
x)/2]^2)] - 468948480*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^14*Sqrt[S
in[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] + 237726720*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*
x)/2]^2)]]*Sin[(c + d*x)/2]^16*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] - 70963200*ArcTanh[Sqrt[Si
n[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^18*Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*
x)/2]^2)] + 9461760*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Sin[(c + d*x)/2]^20*Sqrt[Sin
[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)] - 1120*Cos[(c + d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 11/2}, {1,
1, 13/2}, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^12*(-6 + 5*Sin[(c + d*x)/2]^2) + 28
0*Cos[(c + d*x)/2]^4*HypergeometricPFQ[{2, 2, 11/2}, {1, 13/2}, Sin[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)
]*Sin[(c + d*x)/2]^12*(103 - 164*Sin[(c + d*x)/2]^2 + 70*Sin[(c + d*x)/2]^4)))/(80850*(1 - 2*Sin[(c + d*x)/2]^
2)^(9/2)*(-1 + 2*Sin[(c + d*x)/2]^2)) + (8*A*((3*Sin[(c + d*x)/2])/(1 - 2*Sin[(c + d*x)/2]^2)^(5/2) + 4*(Sin[(
c + d*x)/2]/(1 - 2*Sin[(c + d*x)/2]^2)^(3/2) + (2*Sin[(c + d*x)/2])/Sqrt[1 - 2*Sin[(c + d*x)/2]^2])))/35))/(d*
(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))

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fricas [A]  time = 0.57, size = 575, normalized size = 2.08 \[ \left [-\frac {105 \, \sqrt {2} {\left ({\left (11 \, A - 15 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (11 \, A - 15 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (11 \, A - 15 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (665 \, A - 1029 \, B + 1201 \, C\right )} \cos \left (d x + c\right )^{4} + 12 \, {\left (35 \, A - 63 \, B + 67 \, C\right )} \cos \left (d x + c\right )^{3} - 28 \, {\left (5 \, A - 3 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2} - 12 \, {\left (7 \, B - 3 \, C\right )} \cos \left (d x + c\right ) - 60 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{840 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}}, -\frac {105 \, \sqrt {2} {\left ({\left (11 \, A - 15 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (11 \, A - 15 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (11 \, A - 15 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (665 \, A - 1029 \, B + 1201 \, C\right )} \cos \left (d x + c\right )^{4} + 12 \, {\left (35 \, A - 63 \, B + 67 \, C\right )} \cos \left (d x + c\right )^{3} - 28 \, {\left (5 \, A - 3 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2} - 12 \, {\left (7 \, B - 3 \, C\right )} \cos \left (d x + c\right ) - 60 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{420 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[-1/840*(105*sqrt(2)*((11*A - 15*B + 19*C)*cos(d*x + c)^5 + 2*(11*A - 15*B + 19*C)*cos(d*x + c)^4 + (11*A - 15
*B + 19*C)*cos(d*x + c)^3)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x +
c)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((665*
A - 1029*B + 1201*C)*cos(d*x + c)^4 + 12*(35*A - 63*B + 67*C)*cos(d*x + c)^3 - 28*(5*A - 3*B + 7*C)*cos(d*x +
c)^2 - 12*(7*B - 3*C)*cos(d*x + c) - 60*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*
x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3), -1/420*(105*sqrt(2)*((11*A - 15*B + 19*C)*cos(d*x +
c)^5 + 2*(11*A - 15*B + 19*C)*cos(d*x + c)^4 + (11*A - 15*B + 19*C)*cos(d*x + c)^3)*sqrt(a)*arctan(sqrt(2)*sq
rt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 2*((665*A - 1029*B + 1201*C)*cos(
d*x + c)^4 + 12*(35*A - 63*B + 67*C)*cos(d*x + c)^3 - 28*(5*A - 3*B + 7*C)*cos(d*x + c)^2 - 12*(7*B - 3*C)*cos
(d*x + c) - 60*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*
x + c)^4 + a^2*d*cos(d*x + c)^3)]

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giac [B]  time = 3.02, size = 560, normalized size = 2.02 \[ \frac {\frac {105 \, {\left (11 \, \sqrt {2} A - 15 \, \sqrt {2} B + 19 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {{\left ({\left ({\left ({\left (\frac {105 \, {\left (\sqrt {2} A a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - \sqrt {2} B a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) + \sqrt {2} C a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{3}} - \frac {4 \, {\left (455 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - 693 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) + 877 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {14 \, {\left (305 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - 453 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) + 517 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {140 \, {\left (25 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - 39 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) + 47 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {105 \, {\left (9 \, \sqrt {2} A a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) - 17 \, \sqrt {2} B a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) + 17 \, \sqrt {2} C a^{5} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{420 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

1/420*(105*(11*sqrt(2)*A - 15*sqrt(2)*B + 19*sqrt(2)*C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1
/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - ((((105*(sqrt(2)*A*a^5*sgn(tan(1/2*d*x
+ 1/2*c)^2 - 1) - sqrt(2)*B*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + sqrt(2)*C*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 -
1))*tan(1/2*d*x + 1/2*c)^2/a^3 - 4*(455*sqrt(2)*A*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 693*sqrt(2)*B*a^5*sgn(
tan(1/2*d*x + 1/2*c)^2 - 1) + 877*sqrt(2)*C*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^3)*tan(1/2*d*x + 1/2*c)^2 +
14*(305*sqrt(2)*A*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 453*sqrt(2)*B*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 5
17*sqrt(2)*C*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^3)*tan(1/2*d*x + 1/2*c)^2 - 140*(25*sqrt(2)*A*a^5*sgn(tan(
1/2*d*x + 1/2*c)^2 - 1) - 39*sqrt(2)*B*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 47*sqrt(2)*C*a^5*sgn(tan(1/2*d*x
+ 1/2*c)^2 - 1))/a^3)*tan(1/2*d*x + 1/2*c)^2 + 105*(9*sqrt(2)*A*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 17*sqrt(
2)*B*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 17*sqrt(2)*C*a^5*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^3)*tan(1/2*d*x
+ 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d

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maple [B]  time = 2.63, size = 1437, normalized size = 5.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x)

[Out]

1/3360/d*(-1+cos(d*x+c))*(-960*C+2688*B*cos(d*x+c)^2-2240*A*cos(d*x+c)^2+1536*C*cos(d*x+c)+6352*C*cos(d*x+c)^4
-3712*C*cos(d*x+c)^2+3920*A*cos(d*x+c)^4-4368*B*cos(d*x+c)^4-1344*B*cos(d*x+c)+1155*A*ln(((-2*cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*sin(d*x+c)+1995*C*l
n(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7
/2)*sin(d*x+c)+16000*C*cos(d*x+c)^3+8960*A*cos(d*x+c)^3-13440*B*cos(d*x+c)^3-10640*A*cos(d*x+c)^5+16464*B*cos(
d*x+c)^5-6300*B*sin(d*x+c)*cos(d*x+c)^3*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(
d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+7980*C*sin(d*x+c)*cos(d*x+c)^3*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-9450*B*sin(d*x+c)*cos(d*x+c)^
2*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(7/2)+11970*C*sin(d*x+c)*cos(d*x+c)^2*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d
*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+4620*A*cos(d*x+c)^3*sin(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-19216*C*cos(d*x+c)^5-1575*B*ln
(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/
2)*sin(d*x+c)+6930*A*cos(d*x+c)^2*sin(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)
/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+1155*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*ln(((-2*cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^4-1575*B*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(7/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*sin(d*x+c)*c
os(d*x+c)^4+1995*C*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-co
s(d*x+c)+1)/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^4+7980*C*sin(d*x+c)*cos(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c))
)^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-6300*B*sin(d*x+c)*cos(d*x+c)
*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^
(7/2)+4620*A*cos(d*x+c)*sin(d*x+c)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c
))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2))*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)^3/cos(d*x+c)^3/a^2

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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)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + d*x))^(3/2)),x)

[Out]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + d*x))^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**4/(a*(sec(c + d*x) + 1))**(3/2), x)

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