\(\int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\) [1119]

Optimal result

Integrand size = 35, antiderivative size = 189 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {(A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {2 (A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {2 (A+5 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(A+7 C) \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}-\frac {(A+7 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x) (1+\cos (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \]

[Out]

(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))/a^2/d+2/3*(A+5*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^2/d+2/3*(A+5*C)*sin(
d*x+c)/a^2/d/cos(d*x+c)^(3/2)-1/3*(A+7*C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(3/2)/(1+cos(d*x+c))-1/3*(A+C)*sin(d*x+c
)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^2-(A+7*C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4199, 3121, 3057, 2827, 2716, 2720, 2719} \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {2 (A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {(A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {(A+7 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x)+1)}+\frac {2 (A+5 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(A+7 C) \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

((A + 7*C)*EllipticE[(c + d*x)/2, 2])/(a^2*d) + (2*(A + 5*C)*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) + (2*(A + 5*
C)*Sin[c + d*x])/(3*a^2*d*Cos[c + d*x]^(3/2)) - ((A + 7*C)*Sin[c + d*x])/(a^2*d*Sqrt[Cos[c + d*x]]) - ((A + 7*
C)*Sin[c + d*x])/(3*a^2*d*Cos[c + d*x]^(3/2)*(1 + Cos[c + d*x])) - ((A + C)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3
/2)*(a + a*Cos[c + d*x])^2)

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 3121

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(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)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(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, 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 4199

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {C+A \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx \\ & = -\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}+\frac {\int \frac {\frac {3}{2} a (A+3 C)+\frac {1}{2} a (A-5 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{3 a^2} \\ & = -\frac {(A+7 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x) (1+\cos (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}+\frac {\int \frac {3 a^2 (A+5 C)-\frac {3}{2} a^2 (A+7 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{3 a^4} \\ & = -\frac {(A+7 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x) (1+\cos (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}+\frac {(A+5 C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{a^2}-\frac {(A+7 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{2 a^2} \\ & = \frac {2 (A+5 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(A+7 C) \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}-\frac {(A+7 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x) (1+\cos (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2}+\frac {(A+5 C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 a^2}+\frac {(A+7 C) \int \sqrt {\cos (c+d x)} \, dx}{2 a^2} \\ & = \frac {(A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {2 (A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {2 (A+5 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(A+7 C) \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}-\frac {(A+7 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x) (1+\cos (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 10.30 (sec) , antiderivative size = 1155, normalized size of antiderivative = 6.11 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {8 A \cos ^4\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 ) \left (A+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+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))^2}-\frac {40 C \cos ^4\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 ) \left (A+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+A \cos (2 c+2 d x)) \sqrt {1+\cot ^2(c)} (a+a \sec (c+d x))^2}+\frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {4 (4 C+A \cos (c)+3 C \cos (c)) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec (c)}{d}-\frac {4 \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 )+C \sin \left (\frac {d x}{2}\right )\right )}{3 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 )+3 C \sin \left (\frac {d x}{2}\right )\right )}{d}+\frac {16 C \sec (c) \sec ^2(c+d x) \sin (d x)}{3 d}+\frac {16 \sec (c) \sec (c+d x) (C \sin (c)-6 C \sin (d x))}{3 d}-\frac {4 (A+C) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{3 d}\right )}{(A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2}-\frac {2 A \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (A+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 )}{d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2}-\frac {14 C \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (A+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 )}{d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2} \]

[In]

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

[Out]

(-8*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]
*(A + 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 + A*Cos[2*c + 2*d*x])*
Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^2) - (40*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2},
{5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + 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[Co
t[c]]]])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^2) + (Cos[c/2 + (d*x)/2]^
4*Sqrt[Cos[c + d*x]]*(A + C*Sec[c + d*x]^2)*((-4*(4*C + A*Cos[c] + 3*C*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/d - (
4*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]*(A*S
in[(d*x)/2] + 3*C*Sin[(d*x)/2]))/d + (16*C*Sec[c]*Sec[c + d*x]^2*Sin[d*x])/(3*d) + (16*Sec[c]*Sec[c + d*x]*(C*
Sin[c] - 6*C*Sin[d*x]))/(3*d) - (4*(A + C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/((A + 2*C + A*Cos[2*c + 2*d*
x])*(a + a*Sec[c + d*x])^2) - (2*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*((Hypergeomet
ricPFQ[{-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]]]*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]])
)/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) - (14*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*(A
+ 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 + A
rcTan[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 + Arc
Tan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(710\) vs. \(2(225)=450\).

Time = 2.30 (sec) , antiderivative size = 711, normalized size of antiderivative = 3.76

method	result	size
default	\(\text {Expression too large to display}\)	\(711\)

[In]

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

[Out]

-1/2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^2*(4*C*(cos(1/2*d*x+1/2*c)*(sin(1/2*d*x+1/2*c
)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(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)/cos(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d
*x+1/2*c)^2)^(1/2)+1/3*(A+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*EllipticF(cos
(1/2*d*x+1/2*c),2^(1/2))-3*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*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*Ellipti
cE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-12*sin(1/2*d*x+1/2*c)^6+20*sin(1/2*d*x+1/2*c)^4-7*sin(1/2*d
*x+1/2*c)^2)/cos(1/2*d*x+1/2*c)/(-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)^2-1)-
8*C/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(
sin(1/2*d*x+1/2*c)^2)^(1/2))+4*C*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)
/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d
*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(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)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 427, normalized size of antiderivative = 2.26 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {2 \, {\left (3 \, {\left (A + 7 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (A + 8 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, C \cos \left (d x + c\right ) - 2 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 2 \, {\left (\sqrt {2} {\left (i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, \sqrt {2} {\left (i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (\sqrt {2} {\left (-i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, \sqrt {2} {\left (-i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{2}\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 - 7 i \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, \sqrt {2} {\left (-i \, A - 7 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-i \, A - 7 i \, C\right )} \cos \left (d x + c\right )^{2}\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 + 7 i \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, \sqrt {2} {\left (i \, A + 7 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (i \, A + 7 i \, C\right )} \cos \left (d x + c\right )^{2}\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 )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

-1/6*(2*(3*(A + 7*C)*cos(d*x + c)^3 + 4*(A + 8*C)*cos(d*x + c)^2 + 8*C*cos(d*x + c) - 2*C)*sqrt(cos(d*x + c))*
sin(d*x + c) + 2*(sqrt(2)*(I*A + 5*I*C)*cos(d*x + c)^4 + 2*sqrt(2)*(I*A + 5*I*C)*cos(d*x + c)^3 + sqrt(2)*(I*A
+ 5*I*C)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 2*(sqrt(2)*(-I*A - 5*I*C
)*cos(d*x + c)^4 + 2*sqrt(2)*(-I*A - 5*I*C)*cos(d*x + c)^3 + sqrt(2)*(-I*A - 5*I*C)*cos(d*x + c)^2)*weierstras
sPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(-I*A - 7*I*C)*cos(d*x + c)^4 + 2*sqrt(2)*(-I*A -
7*I*C)*cos(d*x + c)^3 + sqrt(2)*(-I*A - 7*I*C)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4,
0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(I*A + 7*I*C)*cos(d*x + c)^4 + 2*sqrt(2)*(I*A + 7*I*C)*cos(d*
x + c)^3 + sqrt(2)*(I*A + 7*I*C)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c
) - I*sin(d*x + c))))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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