\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\) [1237]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (warning: unable to verify)
   Maple [B] (warning: unable to verify)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F(-1)]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 43, antiderivative size = 229 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\frac {(A+9 B-49 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(A+3 B-13 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A+9 B-49 C) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(2 A+3 B-8 C) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {(A+3 B-13 C) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

1/10*(A+9*B-49*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^3/d+
1/6*(A+3*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))/a^3/d-1
/10*(A+9*B-49*C)*sin(d*x+c)/a^3/d/cos(d*x+c)^(1/2)-1/5*(A-B+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3/cos(d*x+c)^(1/2
)+1/15*(2*A+3*B-8*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2/cos(d*x+c)^(1/2)+1/6*(A+3*B-13*C)*sin(d*x+c)/d/(a^3+a^3
*cos(d*x+c))/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4197, 3120, 3057, 2827, 2716, 2719, 2720} \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\frac {(A+3 B-13 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(A+9 B-49 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac {(A+9 B-49 C) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}+\frac {(A+3 B-13 C) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3 \cos (c+d x)+a^3\right )}+\frac {(2 A+3 B-8 C) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3} \]

[In]

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

[Out]

((A + 9*B - 49*C)*EllipticE[(c + d*x)/2, 2])/(10*a^3*d) + ((A + 3*B - 13*C)*EllipticF[(c + d*x)/2, 2])/(6*a^3*
d) - ((A + 9*B - 49*C)*Sin[c + d*x])/(10*a^3*d*Sqrt[Cos[c + d*x]]) - ((A - B + C)*Sin[c + d*x])/(5*d*Sqrt[Cos[
c + d*x]]*(a + a*Cos[c + d*x])^3) + ((2*A + 3*B - 8*C)*Sin[c + d*x])/(15*a*d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c +
 d*x])^2) + ((A + 3*B - 13*C)*Sin[c + d*x])/(6*d*Sqrt[Cos[c + d*x]]*(a^3 + a^3*Cos[c + d*x]))

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*} \text {integral}& = \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))^3} \, dx \\ & = -\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {\int \frac {\frac {1}{2} a (A-B+11 C)+\frac {5}{2} a (A+B-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(2 A+3 B-8 C) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a^2 (A-6 B+41 C)+\frac {3}{2} a^2 (2 A+3 B-8 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{15 a^4} \\ & = -\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(2 A+3 B-8 C) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {(A+3 B-13 C) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \frac {-\frac {3}{4} a^3 (A+9 B-49 C)+\frac {5}{4} a^3 (A+3 B-13 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a^6} \\ & = -\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(2 A+3 B-8 C) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {(A+3 B-13 C) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )}-\frac {(A+9 B-49 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{20 a^3}+\frac {(A+3 B-13 C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{12 a^3} \\ & = \frac {(A+3 B-13 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A+9 B-49 C) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(2 A+3 B-8 C) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {(A+3 B-13 C) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(A+9 B-49 C) \int \sqrt {\cos (c+d x)} \, dx}{20 a^3} \\ & = \frac {(A+9 B-49 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(A+3 B-13 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A+9 B-49 C) \sin (c+d x)}{10 a^3 d \sqrt {\cos (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3}+\frac {(2 A+3 B-8 C) \sin (c+d x)}{15 a d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {(A+3 B-13 C) \sin (c+d x)}{6 d \sqrt {\cos (c+d x)} \left (a^3+a^3 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 14.14 (sec) , antiderivative size = 1851, normalized size of antiderivative = 8.08 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=-\frac {4 A \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))^3}-\frac {4 B \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))^3}+\frac {52 C \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))^3}+\frac {\cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {4 (20 C-A \cos (c)-9 B \cos (c)+29 C \cos (c)) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec (c)}{5 d}-\frac {8 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )+9 B \sin \left (\frac {d x}{2}\right )-29 C \sin \left (\frac {d x}{2}\right )\right )}{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 )-B \sin \left (\frac {d x}{2}\right )+C \sin \left (\frac {d x}{2}\right )\right )}{5 d}+\frac {8 \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )-6 B \sin \left (\frac {d x}{2}\right )+11 C \sin \left (\frac {d x}{2}\right )\right )}{15 d}+\frac {32 C \sec (c) \sec (c+d x) \sin (d x)}{d}+\frac {8 (A-6 B+11 C) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{15 d}+\frac {4 (A-B+C) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{5 d}\right )}{\sqrt {\cos (c+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}-\frac {2 A \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{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}-\frac {18 B \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{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}+\frac {98 C \cos ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{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} \]

[In]

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

[Out]

(-4*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]
*Sec[c + d*x]*(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]]]])/(3*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])^3) - (4*B*Cos[c/2 + (
d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(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]]]])/(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])^3) + (52*C*Cos[c/2 + (d*x)/2]^6*Csc[c/2]
*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(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]]]])/(3*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])^3) + (Cos[c/2 + (d*x)/2]^6*(A + B*Sec[c + d*x] + C*Sec[c
 + d*x]^2)*((4*(20*C - A*Cos[c] - 9*B*Cos[c] + 29*C*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/(5*d) - (8*Sec[c/2]*Sec[
c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + 9*B*Sin[(d*x)/2] - 29*C*Sin[(d*x)/2]))/(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*(A*Sin[(d*x)/2
] - 6*B*Sin[(d*x)/2] + 11*C*Sin[(d*x)/2]))/(15*d) + (32*C*Sec[c]*Sec[c + d*x]*Sin[d*x])/d + (8*(A - 6*B + 11*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)))/(Sqrt[Cos[c +
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) - (2*A*Cos[c/2 + (d*x)/2]^6*Cs
c[c/2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4},
Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + C
os[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Si
n[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])
/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + 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) - (18*B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*
x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^
2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*
Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Ta
n[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sq
rt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + 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) + (98*C*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C
*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]
]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcT
an[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] +
 (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan
[Tan[c]]]*Sqrt[1 + 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)

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(788\) vs. \(2(261)=522\).

Time = 3.39 (sec) , antiderivative size = 789, normalized size of antiderivative = 3.45

method result size
default \(\text {Expression too large to display}\) \(789\)

[In]

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

[Out]

1/60*(-2*(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)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*B*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))-27*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-65*C*EllipticF(cos(1/2*d*x+1/2*c),2
^(1/2))+147*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)^4+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)*(5*A*Ellip
ticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*B*EllipticF(cos(1/2*d*x+1/2*c),2
^(1/2))-27*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-65*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*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)-2*(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)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),
2^(1/2))-3*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-27*B*EllipticE(c
os(1/2*d*x+1/2*c),2^(1/2))-65*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/
2)))*cos(1/2*d*x+1/2*c)+12*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+9*B-49*C)*sin(1/2*d*x+1/2*c
)^8-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(13*A+147*B-817*C)*sin(1/2*d*x+1/2*c)^6+6*(-2*sin(1
/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A+43*B-248*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)*(A+69*B-439*C)*sin(1/2*d*x+1/2*c)^2)/a^3/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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 574, normalized size of antiderivative = 2.51 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=-\frac {2 \, {\left (3 \, {\left (A + 9 \, B - 49 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (2 \, A + 33 \, B - 188 \, C\right )} \cos \left (d x + c\right )^{2} - 5 \, {\left (A - 9 \, B + 59 \, C\right )} \cos \left (d x + c\right ) - 60 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 5 \, {\left (\sqrt {2} {\left (i \, A + 3 i \, B - 13 i \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (i \, A + 3 i \, B - 13 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (i \, A + 3 i \, B - 13 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A + 3 i \, B - 13 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 (-i \, A - 3 i \, B + 13 i \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (-i \, A - 3 i \, B + 13 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-i \, A - 3 i \, B + 13 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A - 3 i \, B + 13 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 ) + 3 \, {\left (\sqrt {2} {\left (-i \, A - 9 i \, B + 49 i \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (-i \, A - 9 i \, B + 49 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-i \, A - 9 i \, B + 49 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A - 9 i \, B + 49 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 ) + 3 \, {\left (\sqrt {2} {\left (i \, A + 9 i \, B - 49 i \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, \sqrt {2} {\left (i \, A + 9 i \, B - 49 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (i \, A + 9 i \, B - 49 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A + 9 i \, B - 49 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 )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]

[In]

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

[Out]

-1/60*(2*(3*(A + 9*B - 49*C)*cos(d*x + c)^3 + 2*(2*A + 33*B - 188*C)*cos(d*x + c)^2 - 5*(A - 9*B + 59*C)*cos(d
*x + c) - 60*C)*sqrt(cos(d*x + c))*sin(d*x + c) + 5*(sqrt(2)*(I*A + 3*I*B - 13*I*C)*cos(d*x + c)^4 + 3*sqrt(2)
*(I*A + 3*I*B - 13*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A + 3*I*B - 13*I*C)*cos(d*x + c)^2 + sqrt(2)*(I*A + 3*I*
B - 13*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-I*A - 3*I*B
 + 13*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(-I*A - 3*I*B + 13*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A - 3*I*B + 13*I*
C)*cos(d*x + c)^2 + sqrt(2)*(-I*A - 3*I*B + 13*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*
sin(d*x + c)) + 3*(sqrt(2)*(-I*A - 9*I*B + 49*I*C)*cos(d*x + c)^4 + 3*sqrt(2)*(-I*A - 9*I*B + 49*I*C)*cos(d*x
+ c)^3 + 3*sqrt(2)*(-I*A - 9*I*B + 49*I*C)*cos(d*x + c)^2 + sqrt(2)*(-I*A - 9*I*B + 49*I*C)*cos(d*x + c))*weie
rstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(I*A + 9*I*B - 49*I
*C)*cos(d*x + c)^4 + 3*sqrt(2)*(I*A + 9*I*B - 49*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A + 9*I*B - 49*I*C)*cos(d*
x + c)^2 + sqrt(2)*(I*A + 9*I*B - 49*I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(
d*x + c) - I*sin(d*x + c))))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*c
os(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]

[In]

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

[Out]

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