3.13.44 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx\) [1244]

Optimal. Leaf size=276 \[ \frac {(A+8 B-57 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^4 d}+\frac {(4 A+17 B-108 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{42 a^4 d}-\frac {(A+8 B-57 C) \sin (c+d x)}{10 a^4 d \sqrt {\cos (c+d x)}}+\frac {(13 A+29 B-141 C) \sin (c+d x)}{210 a^4 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))^2}+\frac {(4 A+17 B-108 C) \sin (c+d x)}{42 a^4 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^4}+\frac {(3 A+4 B-11 C) \sin (c+d x)}{35 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3} \]

[Out]

1/10*(A+8*B-57*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^4/d+
1/42*(4*A+17*B-108*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^
4/d-1/10*(A+8*B-57*C)*sin(d*x+c)/a^4/d/cos(d*x+c)^(1/2)+1/210*(13*A+29*B-141*C)*sin(d*x+c)/a^4/d/(1+cos(d*x+c)
)^2/cos(d*x+c)^(1/2)+1/42*(4*A+17*B-108*C)*sin(d*x+c)/a^4/d/(1+cos(d*x+c))/cos(d*x+c)^(1/2)-1/7*(A-B+C)*sin(d*
x+c)/d/(a+a*cos(d*x+c))^4/cos(d*x+c)^(1/2)+1/35*(3*A+4*B-11*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^3/cos(d*x+c)^(1
/2)

________________________________________________________________________________________

Rubi [A]
time = 0.57, antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4197, 3120, 3057, 2827, 2716, 2719, 2720} \begin {gather*} \frac {(4 A+17 B-108 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{42 a^4 d}+\frac {(A+8 B-57 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^4 d}-\frac {(A+8 B-57 C) \sin (c+d x)}{10 a^4 d \sqrt {\cos (c+d x)}}+\frac {(4 A+17 B-108 C) \sin (c+d x)}{42 a^4 d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}+\frac {(13 A+29 B-141 C) \sin (c+d x)}{210 a^4 d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)^2}+\frac {(3 A+4 B-11 C) \sin (c+d x)}{35 a d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^4),x]

[Out]

((A + 8*B - 57*C)*EllipticE[(c + d*x)/2, 2])/(10*a^4*d) + ((4*A + 17*B - 108*C)*EllipticF[(c + d*x)/2, 2])/(42
*a^4*d) - ((A + 8*B - 57*C)*Sin[c + d*x])/(10*a^4*d*Sqrt[Cos[c + d*x]]) + ((13*A + 29*B - 141*C)*Sin[c + d*x])
/(210*a^4*d*Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])^2) + ((4*A + 17*B - 108*C)*Sin[c + d*x])/(42*a^4*d*Sqrt[Cos[
c + d*x]]*(1 + Cos[c + d*x])) - ((A - B + C)*Sin[c + d*x])/(7*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^4) + (
(3*A + 4*B - 11*C)*Sin[c + d*x])/(35*a*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3)

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3057

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

Rule 3120

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + a*C)*Cos[e + f*x]*(a
+ b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Dist[1/(b*(b*c - a*d)*(2*m
+ 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(
b*c*m + a*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c*(2*m + 1) - a*d*(m - n -
1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^
2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 4197

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sec[(e_.)
+ (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*
Cos[e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}
, x] &&  !IntegerQ[n] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^4} \, dx &=\int \frac {C+B \cos (c+d x)+A \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^4} \, dx\\ &=-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^4}+\frac {\int \frac {\frac {1}{2} a (A-B+15 C)+\frac {7}{2} a (A+B-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^4}+\frac {(3 A+4 B-11 C) \sin (c+d x)}{35 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {\int \frac {\frac {1}{2} a^2 (2 A-9 B+86 C)+\frac {5}{2} a^2 (3 A+4 B-11 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=\frac {(13 A+29 B-141 C) \sin (c+d x)}{210 a^4 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^4}+\frac {(3 A+4 B-11 C) \sin (c+d x)}{35 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {\int \frac {-\frac {1}{4} a^3 (A+83 B-657 C)+\frac {3}{4} a^3 (13 A+29 B-141 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{105 a^6}\\ &=\frac {(13 A+29 B-141 C) \sin (c+d x)}{210 a^4 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^4}+\frac {(3 A+4 B-11 C) \sin (c+d x)}{35 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(4 A+17 B-108 C) \sin (c+d x)}{42 d \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right )}+\frac {\int \frac {-\frac {21}{4} a^4 (A+8 B-57 C)+\frac {5}{4} a^4 (4 A+17 B-108 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{105 a^8}\\ &=\frac {(13 A+29 B-141 C) \sin (c+d x)}{210 a^4 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^4}+\frac {(3 A+4 B-11 C) \sin (c+d x)}{35 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(4 A+17 B-108 C) \sin (c+d x)}{42 d \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right )}+\frac {(4 A+17 B-108 C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{84 a^4}-\frac {(A+8 B-57 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{20 a^4}\\ &=\frac {(4 A+17 B-108 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{42 a^4 d}-\frac {(A+8 B-57 C) \sin (c+d x)}{10 a^4 d \sqrt {\cos (c+d x)}}+\frac {(13 A+29 B-141 C) \sin (c+d x)}{210 a^4 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^4}+\frac {(3 A+4 B-11 C) \sin (c+d x)}{35 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(4 A+17 B-108 C) \sin (c+d x)}{42 d \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right )}+\frac {(A+8 B-57 C) \int \sqrt {\cos (c+d x)} \, dx}{20 a^4}\\ &=\frac {(A+8 B-57 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^4 d}+\frac {(4 A+17 B-108 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{42 a^4 d}-\frac {(A+8 B-57 C) \sin (c+d x)}{10 a^4 d \sqrt {\cos (c+d x)}}+\frac {(13 A+29 B-141 C) \sin (c+d x)}{210 a^4 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))^2}-\frac {(A-B+C) \sin (c+d x)}{7 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^4}+\frac {(3 A+4 B-11 C) \sin (c+d x)}{35 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(4 A+17 B-108 C) \sin (c+d x)}{42 d \sqrt {\cos (c+d x)} \left (a^4+a^4 \cos (c+d x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 7.60, size = 2316, normalized size = 8.39 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^4),x]

[Out]

(((2*I)/5)*A*Cos[c/2 + (d*x)/2]^8*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2
*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*
d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)
*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4
, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*
d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*
x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/((A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c
+ d*x])^4) + (((16*I)/5)*B*Cos[c/2 + (d*x)/2]^8*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c
+ d*x]^2)*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*
(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I
*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeo
metric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(
-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1
+ E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/((A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])
*(a + a*Sec[c + d*x])^4) - (((114*I)/5)*C*Cos[c/2 + (d*x)/2]^8*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c +
d*x] + C*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c
])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*
x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]
) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*C
os[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*
c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/((A + 2*C + 2*B*Cos[c + d*x] + A*Co
s[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4) - (32*A*Cos[c/2 + (d*x)/2]^8*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5
/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - Ar
cTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]
*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(21*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^
2]*(a + a*Sec[c + d*x])^4) - (136*B*Cos[c/2 + (d*x)/2]^8*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x
- ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]
]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Si
n[d*x - ArcTan[Cot[c]]]])/(21*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Se
c[c + d*x])^4) + (288*C*Cos[c/2 + (d*x)/2]^8*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Co
t[c]]]^2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - S
in[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcT
an[Cot[c]]]])/(7*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^4
) + (Cos[c/2 + (d*x)/2]^8*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((8*(20*C - A*Cos[c] - 8*B*Cos[c] + 37*C*Cos
[c])*Csc[c/2]*Sec[c/2]*Sec[c])/(5*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] + 83*B*Sin[(d*x)/2] -
237*C*Sin[(d*x)/2]))/(105*d) - (16*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + 8*B*Sin[(d*x)/2] - 37*C*Sin[(
d*x)/2]))/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^7*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(7*d) +
(8*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(2*A*Sin[(d*x)/2] - 9*B*Sin[(d*x)/2] + 16*C*Sin[(d*x)/2]))/(35*d) + (64*C*Se
c[c]*Sec[c + d*x]*Sin[d*x])/d - (8*(A + 83*B - 237*C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(105*d) + (8*(2*A - 9*B +
16*C)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(35*d) + (4*(A - B + C)*Sec[c/2 + (d*x)/2]^6*Tan[c/2])/(7*d)))/(Cos[c +
d*x]^(3/2)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^4)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1016\) vs. \(2(304)=608\).
time = 0.28, size = 1017, normalized size = 3.68

method	result	size
default	\(\text {Expression too large to display}\)	\(1017\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/840*(4*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/
2*c)^2)^(1/2)*(20*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-21*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+85*B*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))-168*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-540*C*EllipticF(cos(1/2*d*x+1/2*
c),2^(1/2))+1197*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*(2*sin(1/
2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(20*
A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-21*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+85*B*EllipticF(cos(1/2*d*x+
1/2*c),2^(1/2))-168*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-540*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1197*C
*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+12*(2*sin(1/2*d*x+1/2*c)^2-1)^
(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(20*A*EllipticF(cos(1/
2*d*x+1/2*c),2^(1/2))-21*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+85*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-16
8*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-540*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1197*C*EllipticE(cos(1/2
*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-4*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(20*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))-21*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+85*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-168*B*EllipticE(cos(1
/2*d*x+1/2*c),2^(1/2))-540*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1197*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)
))*cos(1/2*d*x+1/2*c)-168*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+8*B-57*C)*sin(1/2*d*x+1/2*c)
^10+4*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(148*A+1259*B-9036*C)*sin(1/2*d*x+1/2*c)^8-14*(-2*s
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(53*A+499*B-3621*C)*sin(1/2*d*x+1/2*c)^6+2*(-2*sin(1/2*d*x+1/2
*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(181*A+2108*B-15597*C)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2
*d*x+1/2*c)^2)^(1/2)*(59*A+907*B-7053*C)*sin(1/2*d*x+1/2*c)^2)/a^4/cos(1/2*d*x+1/2*c)^7/(-2*sin(1/2*d*x+1/2*c)
^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.68, size = 700, normalized size = 2.54 \begin {gather*} -\frac {2 \, {\left (21 \, {\left (A + 8 \, B - 57 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (64 \, A + 587 \, B - 4248 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (53 \, A + 724 \, B - 5421 \, C\right )} \cos \left (d x + c\right )^{2} - 5 \, {\left (4 \, A - 67 \, B + 564 \, C\right )} \cos \left (d x + c\right ) - 420 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 5 \, {\left (\sqrt {2} {\left (4 i \, A + 17 i \, B - 108 i \, C\right )} \cos \left (d x + c\right )^{5} + 4 \, \sqrt {2} {\left (4 i \, A + 17 i \, B - 108 i \, C\right )} \cos \left (d x + c\right )^{4} + 6 \, \sqrt {2} {\left (4 i \, A + 17 i \, B - 108 i \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, \sqrt {2} {\left (4 i \, A + 17 i \, B - 108 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (4 i \, A + 17 i \, B - 108 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (-4 i \, A - 17 i \, B + 108 i \, C\right )} \cos \left (d x + c\right )^{5} + 4 \, \sqrt {2} {\left (-4 i \, A - 17 i \, B + 108 i \, C\right )} \cos \left (d x + c\right )^{4} + 6 \, \sqrt {2} {\left (-4 i \, A - 17 i \, B + 108 i \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, \sqrt {2} {\left (-4 i \, A - 17 i \, B + 108 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-4 i \, A - 17 i \, B + 108 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, {\left (\sqrt {2} {\left (-i \, A - 8 i \, B + 57 i \, C\right )} \cos \left (d x + c\right )^{5} + 4 \, \sqrt {2} {\left (-i \, A - 8 i \, B + 57 i \, C\right )} \cos \left (d x + c\right )^{4} + 6 \, \sqrt {2} {\left (-i \, A - 8 i \, B + 57 i \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, \sqrt {2} {\left (-i \, A - 8 i \, B + 57 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A - 8 i \, B + 57 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, {\left (\sqrt {2} {\left (i \, A + 8 i \, B - 57 i \, C\right )} \cos \left (d x + c\right )^{5} + 4 \, \sqrt {2} {\left (i \, A + 8 i \, B - 57 i \, C\right )} \cos \left (d x + c\right )^{4} + 6 \, \sqrt {2} {\left (i \, A + 8 i \, B - 57 i \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, \sqrt {2} {\left (i \, A + 8 i \, B - 57 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A + 8 i \, B - 57 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{420 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/420*(2*(21*(A + 8*B - 57*C)*cos(d*x + c)^4 + (64*A + 587*B - 4248*C)*cos(d*x + c)^3 + (53*A + 724*B - 5421*
C)*cos(d*x + c)^2 - 5*(4*A - 67*B + 564*C)*cos(d*x + c) - 420*C)*sqrt(cos(d*x + c))*sin(d*x + c) + 5*(sqrt(2)*
(4*I*A + 17*I*B - 108*I*C)*cos(d*x + c)^5 + 4*sqrt(2)*(4*I*A + 17*I*B - 108*I*C)*cos(d*x + c)^4 + 6*sqrt(2)*(4
*I*A + 17*I*B - 108*I*C)*cos(d*x + c)^3 + 4*sqrt(2)*(4*I*A + 17*I*B - 108*I*C)*cos(d*x + c)^2 + sqrt(2)*(4*I*A
+ 17*I*B - 108*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-4*
I*A - 17*I*B + 108*I*C)*cos(d*x + c)^5 + 4*sqrt(2)*(-4*I*A - 17*I*B + 108*I*C)*cos(d*x + c)^4 + 6*sqrt(2)*(-4*
I*A - 17*I*B + 108*I*C)*cos(d*x + c)^3 + 4*sqrt(2)*(-4*I*A - 17*I*B + 108*I*C)*cos(d*x + c)^2 + sqrt(2)*(-4*I*
A - 17*I*B + 108*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21*(sqrt(2)*(-
I*A - 8*I*B + 57*I*C)*cos(d*x + c)^5 + 4*sqrt(2)*(-I*A - 8*I*B + 57*I*C)*cos(d*x + c)^4 + 6*sqrt(2)*(-I*A - 8*
I*B + 57*I*C)*cos(d*x + c)^3 + 4*sqrt(2)*(-I*A - 8*I*B + 57*I*C)*cos(d*x + c)^2 + sqrt(2)*(-I*A - 8*I*B + 57*I
*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*(sqrt
(2)*(I*A + 8*I*B - 57*I*C)*cos(d*x + c)^5 + 4*sqrt(2)*(I*A + 8*I*B - 57*I*C)*cos(d*x + c)^4 + 6*sqrt(2)*(I*A +
8*I*B - 57*I*C)*cos(d*x + c)^3 + 4*sqrt(2)*(I*A + 8*I*B - 57*I*C)*cos(d*x + c)^2 + sqrt(2)*(I*A + 8*I*B - 57*
I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^4*d*c
os(d*x + c)^5 + 4*a^4*d*cos(d*x + c)^4 + 6*a^4*d*cos(d*x + c)^3 + 4*a^4*d*cos(d*x + c)^2 + a^4*d*cos(d*x + c))

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(7/2)/(a+a*sec(d*x+c))**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^4*cos(d*x + c)^(7/2)), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 )}^{7/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^4} \,d x \end {gather*}

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)^(7/2)*(a + a/cos(c + d*x))^4),x)

[Out]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(7/2)*(a + a/cos(c + d*x))^4), x)

________________________________________________________________________________________