3.3.49 \(\int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\) [249]

Optimal. Leaf size=290 \[ \frac {7 (33 A+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(63 A+13 C) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}+\frac {7 (33 A+7 C) \sin (c+d x)}{30 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(63 A+13 C) \sin (c+d x)}{6 a^3 d \sqrt {\sec (c+d x)}}-\frac {(A+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {2 (6 A+C) \sin (c+d x)}{15 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2}-\frac {(63 A+13 C) \sin (c+d x)}{10 d \sec ^{\frac {3}{2}}(c+d x) \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

7/30*(33*A+7*C)*sin(d*x+c)/a^3/d/sec(d*x+c)^(3/2)-1/5*(A+C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3-2
/15*(6*A+C)*sin(d*x+c)/a/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2-1/10*(63*A+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+13*C)*sin(d*x+c)/a^3/d/sec(d*x+c)^(1/2)+7/10*(33*A+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)*sec(d*x+c)^(1/2)/a^3/d-1/6*(63
*A+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 = 0.38, antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4170, 4105, 3872, 3854, 3856, 2719, 2720} \begin {gather*} -\frac {(63 A+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+7 C) \sin (c+d x)}{30 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(63 A+13 C) \sin (c+d x)}{6 a^3 d \sqrt {\sec (c+d x)}}-\frac {(63 A+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+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 {2 (6 A+C) \sin (c+d x)}{15 a d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}-\frac {(A+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3854

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 3856

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 3872

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 4105

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 4170

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(-a)*(A + C)*Cot[e + f*x]*(a + b*Csc[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[b*C*n + A*b
*(2*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, C, n}, x
] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx &=-\frac {(A+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+C)+\frac {1}{2} a (9 A-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+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {2 (6 A+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+5 C)+7 a^2 (6 A+C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx}{15 a^4}\\ &=-\frac {(A+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {2 (6 A+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+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+7 C)+\frac {15}{4} a^3 (63 A+13 C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{15 a^6}\\ &=-\frac {(A+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {2 (6 A+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+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+7 C)) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx}{12 a^3}-\frac {(63 A+13 C) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx}{4 a^3}\\ &=\frac {7 (33 A+7 C) \sin (c+d x)}{30 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(63 A+13 C) \sin (c+d x)}{6 a^3 d \sqrt {\sec (c+d x)}}-\frac {(A+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {2 (6 A+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+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+7 C)) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx}{20 a^3}-\frac {(63 A+13 C) \int \sqrt {\sec (c+d x)} \, dx}{12 a^3}\\ &=\frac {7 (33 A+7 C) \sin (c+d x)}{30 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(63 A+13 C) \sin (c+d x)}{6 a^3 d \sqrt {\sec (c+d x)}}-\frac {(A+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {2 (6 A+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+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+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+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+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+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+7 C) \sin (c+d x)}{30 a^3 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(63 A+13 C) \sin (c+d x)}{6 a^3 d \sqrt {\sec (c+d x)}}-\frac {(A+C) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3}-\frac {2 (6 A+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+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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 7.43, size = 1052, normalized size = 3.63 \begin {gather*} -\frac {154 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right ) \sec \left (\frac {c}{2}\right ) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{5 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^3}-\frac {98 \sqrt {2} C e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )\right ) \sec \left (\frac {c}{2}\right ) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{15 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^3}-\frac {84 A \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sec \left (\frac {c}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \sin (c)}{d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^3}-\frac {52 C \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sec \left (\frac {c}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \sin (c)}{3 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^3}+\frac {\cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^{\frac {3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {2 (329 A+78 C+133 A \cos (2 c)+20 C \cos (2 c)) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )}{5 d}-\frac {16 A \cos (2 d x) \sin (2 c)}{d}+\frac {8 A \cos (3 d x) \sin (3 c)}{5 d}+\frac {4 \sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )+C \sin \left (\frac {d x}{2}\right )\right )}{5 d}+\frac {184 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (3 A \sin \left (\frac {d x}{2}\right )+C \sin \left (\frac {d x}{2}\right )\right )}{3 d}-\frac {8 \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (27 A \sin \left (\frac {d x}{2}\right )+17 C \sin \left (\frac {d x}{2}\right )\right )}{15 d}+\frac {8 (133 A+20 C) \cos (c) \sin (d x)}{5 d}-\frac {16 A \cos (2 c) \sin (2 d x)}{d}+\frac {8 A \cos (3 c) \sin (3 d x)}{5 d}+\frac {184 (3 A+C) \tan \left (\frac {c}{2}\right )}{3 d}-\frac {8 (27 A+17 C) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{15 d}+\frac {4 (A+C) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{5 d}\right )}{(A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^3} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + 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 + C*Sec[c + d*x]^2))/(5*d*E^(I*d*x)*(A + 2*C + 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 + C*Sec[c + d*x]^2))/
(15*d*E^(I*d*x)*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) - (84*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 + C*Sec[c + d*x]^2)*Sin[c])/(d*(A +
2*C + 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]*El
lipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2)*Sin[c])/(3*d*(A + 2*C + 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 + C*Sec[c + d*x]^2)*((-2*(329*A
+ 78*C + 133*A*Cos[2*c] + 20*C*Cos[2*c])*Cos[d*x]*Csc[c/2]*Sec[c/2])/(5*d) - (16*A*Cos[2*d*x]*Sin[2*c])/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] + C*Sin[(d*x)/2]))/(5*d) + (
184*Sec[c/2]*Sec[c/2 + (d*x)/2]*(3*A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(3*d) - (8*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*
(27*A*Sin[(d*x)/2] + 17*C*Sin[(d*x)/2]))/(15*d) + (8*(133*A + 20*C)*Cos[c]*Sin[d*x])/(5*d) - (16*A*Cos[2*c]*Si
n[2*d*x])/d + (8*A*Cos[3*c]*Sin[3*d*x])/(5*d) + (184*(3*A + C)*Tan[c/2])/(3*d) - (8*(27*A + 17*C)*Sec[c/2 + (d
*x)/2]^2*Tan[c/2])/(15*d) + (4*(A + C)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d)))/((A + 2*C + A*Cos[2*c + 2*d*x])*
(a + a*Sec[c + d*x])^3)

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Maple [A]
time = 4.62, size = 479, normalized size = 1.65

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 (192 A \left (\cos ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-864 A \left (\cos ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-228 A \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-630 A \left (\cos ^{5}\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}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-1386 A \left (\cos ^{5}\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 )-348 C \left (\cos ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-130 C \left (\cos ^{5}\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}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-294 C \left (\cos ^{5}\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 )+1590 A \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+578 C \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-744 A \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-264 C \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+57 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+37 C \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 A -3 C \right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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}\)	\(479\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)^10-228*A*cos(1/2*d*x+1/2*c)^8-630*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-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))-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*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)*Elliptic
E(cos(1/2*d*x+1/2*c),2^(1/2))+1590*A*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)^
4-264*C*cos(1/2*d*x+1/2*c)^4+57*A*cos(1/2*d*x+1/2*c)^2+37*C*cos(1/2*d*x+1/2*c)^2-3*A-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)] 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+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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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.67, size = 500, normalized size = 1.72 \begin {gather*} -\frac {5 \, {\left (\sqrt {2} {\left (-63 i \, A - 13 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-63 i \, A - 13 i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-63 i \, A - 13 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-63 i \, A - 13 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 (63 i \, A + 13 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (63 i \, A + 13 i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (63 i \, A + 13 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (63 i \, A + 13 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 (-33 i \, A - 7 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-33 i \, A - 7 i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-33 i \, A - 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-33 i \, A - 7 i \, 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 (33 i \, A + 7 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (33 i \, A + 7 i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (33 i \, A + 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (33 i \, A + 7 i \, 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 ) - \frac {2 \, {\left (12 \, A \cos \left (d x + c\right )^{5} - 24 \, A \cos \left (d x + c\right )^{4} - 3 \, {\left (147 \, A + 29 \, C\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (357 \, A + 73 \, C\right )} \cos \left (d x + c\right )^{2} - 5 \, {\left (63 \, A + 13 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(5*(sqrt(2)*(-63*I*A - 13*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-63*I*A - 13*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*
(-63*I*A - 13*I*C)*cos(d*x + c) + sqrt(2)*(-63*I*A - 13*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(
d*x + c)) + 5*(sqrt(2)*(63*I*A + 13*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(63*I*A + 13*I*C)*cos(d*x + c)^2 + 3*sqrt(
2)*(63*I*A + 13*I*C)*cos(d*x + c) + sqrt(2)*(63*I*A + 13*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin
(d*x + c)) + 21*(sqrt(2)*(-33*I*A - 7*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-33*I*A - 7*I*C)*cos(d*x + c)^2 + 3*sqr
t(2)*(-33*I*A - 7*I*C)*cos(d*x + c) + sqrt(2)*(-33*I*A - 7*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4
, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*(sqrt(2)*(33*I*A + 7*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(33*I*A + 7*I*C
)*cos(d*x + c)^2 + 3*sqrt(2)*(33*I*A + 7*I*C)*cos(d*x + c) + sqrt(2)*(33*I*A + 7*I*C))*weierstrassZeta(-4, 0,
weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(12*A*cos(d*x + c)^5 - 24*A*cos(d*x + c)^4 - 3*
(147*A + 29*C)*cos(d*x + c)^3 - 2*(357*A + 73*C)*cos(d*x + c)^2 - 5*(63*A + 13*C)*cos(d*x + c))*sin(d*x + c)/s
qrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*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+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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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+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 + A)/((a*sec(d*x + c) + a)^3*sec(d*x + c)^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((A + 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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