3.577 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=313 \[ -\frac {(63 A-33 B+13 C) \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}+\frac {7 (33 A-17 B+7 C) \sin (c+d x)}{30 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(63 A-33 B+13 C) \sin (c+d x)}{6 a^3 d \sqrt {\sec (c+d x)}}-\frac {(63 A-33 B+13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac {7 (33 A-17 B+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(12 A-7 B+2 C) \sin (c+d x)}{15 a d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3} \]

[Out]

7/30*(33*A-17*B+7*C)*sin(d*x+c)/a^3/d/sec(d*x+c)^(3/2)-1/5*(A-B+C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+
c))^3-1/15*(12*A-7*B+2*C)*sin(d*x+c)/a/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2-1/10*(63*A-33*B+13*C)*sin(d*x+c)/
d/sec(d*x+c)^(3/2)/(a^3+a^3*sec(d*x+c))-1/6*(63*A-33*B+13*C)*sin(d*x+c)/a^3/d/sec(d*x+c)^(1/2)+7/10*(33*A-17*B
+7*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))*cos(d*x+c)^(1/2)*s
ec(d*x+c)^(1/2)/a^3/d-1/6*(63*A-33*B+13*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^3/d

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Rubi [A]  time = 0.64, antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4084, 4020, 3787, 3769, 3771, 2639, 2641} \[ -\frac {(63 A-33 B+13 C) \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}+\frac {7 (33 A-17 B+7 C) \sin (c+d x)}{30 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(63 A-33 B+13 C) \sin (c+d x)}{6 a^3 d \sqrt {\sec (c+d x)}}-\frac {(63 A-33 B+13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac {7 (33 A-17 B+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(12 A-7 B+2 C) \sin (c+d x)}{15 a d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3),x]

[Out]

(7*(33*A - 17*B + 7*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(10*a^3*d) - ((63*A -
33*B + 13*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(6*a^3*d) + (7*(33*A - 17*B + 7*
C)*Sin[c + d*x])/(30*a^3*d*Sec[c + d*x]^(3/2)) - ((63*A - 33*B + 13*C)*Sin[c + d*x])/(6*a^3*d*Sqrt[Sec[c + d*x
]]) - ((A - B + C)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^3) - ((12*A - 7*B + 2*C)*Sin[c +
d*x])/(15*a*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2) - ((63*A - 33*B + 13*C)*Sin[c + d*x])/(10*d*Sec[c +
d*x]^(3/2)*(a^3 + a^3*Sec[c + d*x]))

Rule 2639

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

Rule 2641

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

Rule 3769

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

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4020

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

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 {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx &=-\frac {(A-B+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}+\frac {\int \frac {\frac {5}{2} a (3 A-B+C)-\frac {1}{2} a (9 A-9 B-C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {(12 A-7 B+2 C) \sin (c+d x)}{15 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}+\frac {\int \frac {\frac {5}{2} a^2 (21 A-10 B+5 C)-\frac {7}{2} a^2 (12 A-7 B+2 C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{15 a^4}\\ &=-\frac {(A-B+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {(12 A-7 B+2 C) \sin (c+d x)}{15 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {(63 A-33 B+13 C) \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \frac {\frac {35}{4} a^3 (33 A-17 B+7 C)-\frac {15}{4} a^3 (63 A-33 B+13 C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{15 a^6}\\ &=-\frac {(A-B+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {(12 A-7 B+2 C) \sin (c+d x)}{15 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {(63 A-33 B+13 C) \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(7 (33 A-17 B+7 C)) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{12 a^3}-\frac {(63 A-33 B+13 C) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{4 a^3}\\ &=\frac {7 (33 A-17 B+7 C) \sin (c+d x)}{30 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(63 A-33 B+13 C) \sin (c+d x)}{6 a^3 d \sqrt {\sec (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {(12 A-7 B+2 C) \sin (c+d x)}{15 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {(63 A-33 B+13 C) \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(7 (33 A-17 B+7 C)) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{20 a^3}-\frac {(63 A-33 B+13 C) \int \sqrt {\sec (c+d x)} \, dx}{12 a^3}\\ &=\frac {7 (33 A-17 B+7 C) \sin (c+d x)}{30 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(63 A-33 B+13 C) \sin (c+d x)}{6 a^3 d \sqrt {\sec (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {(12 A-7 B+2 C) \sin (c+d x)}{15 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {(63 A-33 B+13 C) \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\left (7 (33 A-17 B+7 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3}-\frac {\left ((63 A-33 B+13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3}\\ &=\frac {7 (33 A-17 B+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(63 A-33 B+13 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}+\frac {7 (33 A-17 B+7 C) \sin (c+d x)}{30 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(63 A-33 B+13 C) \sin (c+d x)}{6 a^3 d \sqrt {\sec (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {(12 A-7 B+2 C) \sin (c+d x)}{15 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {(63 A-33 B+13 C) \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [C]  time = 7.85, size = 1555, normalized size = 4.97 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^3),x]

[Out]

(-154*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/
2]^6*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4,
7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(5*d*E^(I*d*x)*(A
+ 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) + (238*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(
1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c
+ d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*
Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(15*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c
+ 2*d*x])*(a + a*Sec[c + d*x])^3) - (98*Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^(
(2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*
I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*S
ec[c + d*x]^2))/(15*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) - (8
4*A*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(3/2)*(A
+ B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[c])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c
+ d*x])^3) + (44*B*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c
+ d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[c])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*
x])*(a + a*Sec[c + d*x])^3) - (52*C*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]
*Sec[c/2]*Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[c])/(3*d*(A + 2*C + 2*B*Cos[c + d*x]
+ A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) + (Cos[c/2 + (d*x)/2]^6*Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x] +
C*Sec[c + d*x]^2)*((-2*(329*A - 178*B + 78*C + 133*A*Cos[2*c] - 60*B*Cos[2*c] + 20*C*Cos[2*c])*Cos[d*x]*Csc[c
/2]*Sec[c/2])/(5*d) - (16*(3*A - B)*Cos[2*d*x]*Sin[2*c])/(3*d) + (8*A*Cos[3*d*x]*Sin[3*c])/(5*d) + (4*Sec[c/2]
*Sec[c/2 + (d*x)/2]^5*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(5*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/
2]^3*(27*A*Sin[(d*x)/2] - 22*B*Sin[(d*x)/2] + 17*C*Sin[(d*x)/2]))/(15*d) + (8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(69*
A*Sin[(d*x)/2] - 43*B*Sin[(d*x)/2] + 23*C*Sin[(d*x)/2]))/(3*d) + (8*(133*A - 60*B + 20*C)*Cos[c]*Sin[d*x])/(5*
d) - (16*(3*A - B)*Cos[2*c]*Sin[2*d*x])/(3*d) + (8*A*Cos[3*c]*Sin[3*d*x])/(5*d) + (8*(69*A - 43*B + 23*C)*Tan[
c/2])/(3*d) - (8*(27*A - 22*B + 17*C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(15*d) + (4*(A - B + C)*Sec[c/2 + (d*x)/2
]^4*Tan[c/2])/(5*d)))/((A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3)

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{a^{3} \sec \left (d x + c\right )^{6} + 3 \, a^{3} \sec \left (d x + c\right )^{5} + 3 \, a^{3} \sec \left (d x + c\right )^{4} + a^{3} \sec \left (d x + c\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 6.02, size = 666, normalized size = 2.13 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x)

[Out]

-1/60/a^3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(192*A*cos(1/2*d*x+1/2*c)^12-864*A*cos(1/2*d
*x+1/2*c)^10+160*B*cos(1/2*d*x+1/2*c)^10-228*A*cos(1/2*d*x+1/2*c)^8-630*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)^5-1386*A*cos(1/2*d*x+1/2*c
)^5*(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))+468*B
*cos(1/2*d*x+1/2*c)^8+330*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)^5+714*B*cos(1/2*d*x+1/2*c)^5*(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))-348*C*cos(1/2*d*x+1/2*c)^8-130*C*cos(1/2*d*x+1/2*c)^
5*(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))-294*C*c
os(1/2*d*x+1/2*c)^5*(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))+1590*A*cos(1/2*d*x+1/2*c)^6-1058*B*cos(1/2*d*x+1/2*c)^6+578*C*cos(1/2*d*x+1/2*c)^6-744*A*cos(1/2*d*
x+1/2*c)^4+474*B*cos(1/2*d*x+1/2*c)^4-264*C*cos(1/2*d*x+1/2*c)^4+57*A*cos(1/2*d*x+1/2*c)^2-47*B*cos(1/2*d*x+1/
2*c)^2+37*C*cos(1/2*d*x+1/2*c)^2-3*A+3*B-3*C)/cos(1/2*d*x+1/2*c)^5/(-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)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))^3*(1/cos(c + d*x))^(5/2)),x)

[Out]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))^3*(1/cos(c + d*x))^(5/2)), 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((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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