3.13.41 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^4} \, dx\) [1241]

Optimal. Leaf size=232 \[ -\frac {(8 A+B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^4 d}+\frac {(17 A+4 B+3 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{42 a^4 d}-\frac {(83 A+B-15 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{210 a^4 d (1+\cos (c+d x))^2}+\frac {(8 A+B) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 a^4 d (1+\cos (c+d x))}-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(9 A-2 B-5 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3} \]

[Out]

-1/10*(8*A+B)*(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
*(17*A+4*B+3*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/
7*(A-B+C)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos(d*x+c))^4-1/35*(9*A-2*B-5*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d/
(a+a*cos(d*x+c))^3-1/210*(83*A+B-15*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/a^4/d/(1+cos(d*x+c))^2+1/10*(8*A+B)*sin(d*x
+c)*cos(d*x+c)^(1/2)/a^4/d/(1+cos(d*x+c))

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Rubi [A]
time = 0.53, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4197, 3120, 3056, 3057, 2827, 2720, 2719} \begin {gather*} \frac {(17 A+4 B+3 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{42 a^4 d}-\frac {(83 A+B-15 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{210 a^4 d (\cos (c+d x)+1)^2}-\frac {(8 A+B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^4 d}+\frac {(8 A+B) \sin (c+d x) \sqrt {\cos (c+d x)}}{10 a^4 d (\cos (c+d x)+1)}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d (a \cos (c+d x)+a)^4}-\frac {(9 A-2 B-5 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{35 a d (a \cos (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/10*((8*A + B)*EllipticE[(c + d*x)/2, 2])/(a^4*d) + ((17*A + 4*B + 3*C)*EllipticF[(c + d*x)/2, 2])/(42*a^4*d
) - ((83*A + B - 15*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(210*a^4*d*(1 + Cos[c + d*x])^2) + ((8*A + B)*Sqrt[Cos
[c + d*x]]*Sin[c + d*x])/(10*a^4*d*(1 + Cos[c + d*x])) - ((A - B + C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(7*d*(a
+ a*Cos[c + d*x])^4) - ((9*A - 2*B - 5*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3)

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 3056

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^n/(a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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 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)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^4} \, dx &=\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^4} \, dx\\ &=-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (-\frac {1}{2} a (5 A-5 B-9 C)+\frac {1}{2} a (13 A+B-C) \cos (c+d x)\right )}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(9 A-2 B-5 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\sqrt {\cos (c+d x)} \left (-\frac {3}{2} a^2 (9 A-2 B-5 C)+\frac {7}{2} a^2 (8 A+B) \cos (c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(83 A+B-15 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{210 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(9 A-2 B-5 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {-\frac {1}{4} a^3 (83 A+B-15 C)+\frac {1}{4} a^3 (253 A+41 B+15 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{105 a^6}\\ &=-\frac {(83 A+B-15 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{210 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(9 A-2 B-5 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(8 A+B) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {\int \frac {\frac {5}{4} a^4 (17 A+4 B+3 C)-\frac {21}{4} a^4 (8 A+B) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{105 a^8}\\ &=-\frac {(83 A+B-15 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{210 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(9 A-2 B-5 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(8 A+B) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^4+a^4 \cos (c+d x)\right )}-\frac {(8 A+B) \int \sqrt {\cos (c+d x)} \, dx}{20 a^4}+\frac {(17 A+4 B+3 C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{84 a^4}\\ &=-\frac {(8 A+B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^4 d}+\frac {(17 A+4 B+3 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{42 a^4 d}-\frac {(83 A+B-15 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{210 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(9 A-2 B-5 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {(8 A+B) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 7.31, size = 1862, normalized size = 8.03 \begin {gather*} \text {Too large to display} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((-16*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) - (((2*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*Hyperge
ometric2F1[-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) - (136*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 - 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 + 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) - (32*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 + 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)
- (8*C*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 + Sin[d*x - ArcTan[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)*((16*(8*A + B)*Csc[c])/(5*d) + (16*Sec[c/2]*Sec[c/2 + (d*x)/
2]*(8*A*Sin[(d*x)/2] + B*Sin[(d*x)/2]))/(5*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(167*A*Sin[(d*x)/2] - 41*B*Si
n[(d*x)/2] - 15*C*Sin[(d*x)/2]))/(105*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*(19*A*Sin[(d*x)/2] - 12*B*Sin[(d*x)/2] + 5*C*Sin[(d
*x)/2]))/(35*d) - (8*(167*A - 41*B - 15*C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(105*d) + (8*(19*A - 12*B + 5*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)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(594\) vs. \(2(264)=528\).
time = 0.21, size = 595, normalized size = 2.56

method	result	size
default	\(-\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (1344 A \left (\cos ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+340 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+672 A \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+168 B \left (\cos ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+80 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+84 B \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+60 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2684 A \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-88 B \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+60 C \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1902 A \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-306 B \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-30 C \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-706 A \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+328 B \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-90 C \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+159 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-117 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+75 C \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 A +15 B -15 C \right )}{840 a^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(595\)

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)^(1/2)/(a+a*sec(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

-1/840*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(1344*A*cos(1/2*d*x+1/2*c)^10+340*A*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^7
+672*A*cos(1/2*d*x+1/2*c)^7*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d
*x+1/2*c),2^(1/2))+168*B*cos(1/2*d*x+1/2*c)^10+80*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(
1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^7+84*B*cos(1/2*d*x+1/2*c)^7*(sin(1/2*d*x+1/2*c)^
2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+60*C*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^7-2684*A*cos(1
/2*d*x+1/2*c)^8-88*B*cos(1/2*d*x+1/2*c)^8+60*C*cos(1/2*d*x+1/2*c)^8+1902*A*cos(1/2*d*x+1/2*c)^6-306*B*cos(1/2*
d*x+1/2*c)^6-30*C*cos(1/2*d*x+1/2*c)^6-706*A*cos(1/2*d*x+1/2*c)^4+328*B*cos(1/2*d*x+1/2*c)^4-90*C*cos(1/2*d*x+
1/2*c)^4+159*A*cos(1/2*d*x+1/2*c)^2-117*B*cos(1/2*d*x+1/2*c)^2+75*C*cos(1/2*d*x+1/2*c)^2-15*A+15*B-15*C)/a^4/c
os(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

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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)^(1/2)/(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.64, size = 613, normalized size = 2.64 \begin {gather*} \frac {2 \, {\left (21 \, {\left (8 \, A + B\right )} \cos \left (d x + c\right )^{3} + {\left (337 \, A + 104 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (284 \, A + 73 \, B + 60 \, C\right )} \cos \left (d x + c\right ) + 85 \, A + 20 \, B + 15 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, {\left (\sqrt {2} {\left (17 i \, A + 4 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, \sqrt {2} {\left (17 i \, A + 4 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{3} + 6 \, \sqrt {2} {\left (17 i \, A + 4 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, \sqrt {2} {\left (17 i \, A + 4 i \, B + 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (17 i \, A + 4 i \, B + 3 i \, 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 (-17 i \, A - 4 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, \sqrt {2} {\left (-17 i \, A - 4 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{3} + 6 \, \sqrt {2} {\left (-17 i \, A - 4 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, \sqrt {2} {\left (-17 i \, A - 4 i \, B - 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-17 i \, A - 4 i \, B - 3 i \, 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 (8 i \, A + i \, B\right )} \cos \left (d x + c\right )^{4} + 4 \, \sqrt {2} {\left (8 i \, A + i \, B\right )} \cos \left (d x + c\right )^{3} + 6 \, \sqrt {2} {\left (8 i \, A + i \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, \sqrt {2} {\left (8 i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (8 i \, A + i \, B\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 (-8 i \, A - i \, B\right )} \cos \left (d x + c\right )^{4} + 4 \, \sqrt {2} {\left (-8 i \, A - i \, B\right )} \cos \left (d x + c\right )^{3} + 6 \, \sqrt {2} {\left (-8 i \, A - i \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, \sqrt {2} {\left (-8 i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-8 i \, A - i \, B\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 )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\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)^(1/2)/(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/420*(2*(21*(8*A + B)*cos(d*x + c)^3 + (337*A + 104*B + 15*C)*cos(d*x + c)^2 + (284*A + 73*B + 60*C)*cos(d*x
+ c) + 85*A + 20*B + 15*C)*sqrt(cos(d*x + c))*sin(d*x + c) - 5*(sqrt(2)*(17*I*A + 4*I*B + 3*I*C)*cos(d*x + c)^
4 + 4*sqrt(2)*(17*I*A + 4*I*B + 3*I*C)*cos(d*x + c)^3 + 6*sqrt(2)*(17*I*A + 4*I*B + 3*I*C)*cos(d*x + c)^2 + 4*
sqrt(2)*(17*I*A + 4*I*B + 3*I*C)*cos(d*x + c) + sqrt(2)*(17*I*A + 4*I*B + 3*I*C))*weierstrassPInverse(-4, 0, c
os(d*x + c) + I*sin(d*x + c)) - 5*(sqrt(2)*(-17*I*A - 4*I*B - 3*I*C)*cos(d*x + c)^4 + 4*sqrt(2)*(-17*I*A - 4*I
*B - 3*I*C)*cos(d*x + c)^3 + 6*sqrt(2)*(-17*I*A - 4*I*B - 3*I*C)*cos(d*x + c)^2 + 4*sqrt(2)*(-17*I*A - 4*I*B -
3*I*C)*cos(d*x + c) + sqrt(2)*(-17*I*A - 4*I*B - 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x
+ c)) - 21*(sqrt(2)*(8*I*A + I*B)*cos(d*x + c)^4 + 4*sqrt(2)*(8*I*A + I*B)*cos(d*x + c)^3 + 6*sqrt(2)*(8*I*A +
I*B)*cos(d*x + c)^2 + 4*sqrt(2)*(8*I*A + I*B)*cos(d*x + c) + sqrt(2)*(8*I*A + I*B))*weierstrassZeta(-4, 0, we
ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 21*(sqrt(2)*(-8*I*A - I*B)*cos(d*x + c)^4 + 4*sqrt(
2)*(-8*I*A - I*B)*cos(d*x + c)^3 + 6*sqrt(2)*(-8*I*A - I*B)*cos(d*x + c)^2 + 4*sqrt(2)*(-8*I*A - I*B)*cos(d*x
+ c) + sqrt(2)*(-8*I*A - I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)
)))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

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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)**(1/2)/(a+a*sec(d*x+c))**4,x)

[Out]

Timed out

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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)^(1/2)/(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*sqrt(cos(d*x + c))), x)

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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}}{\sqrt {\cos \left (c+d\,x\right )}\,{\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)^(1/2)*(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)^(1/2)*(a + a/cos(c + d*x))^4), x)

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