\(\int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\) [1118]

Optimal result

Integrand size = 35, antiderivative size = 151 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {4 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(A-5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {4 C \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}+\frac {(A-5 C) \sin (c+d x)}{3 a^2 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \]

[Out]

-4*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+1/3*(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+4*C*sin(d*x+c)/a^2/
d/cos(d*x+c)^(1/2)+1/3*(A-5*C)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))/cos(d*x+c)^(1/2)-1/3*(A+C)*sin(d*x+c)/d/(a+a*co
s(d*x+c))^2/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4199, 3121, 3057, 2827, 2716, 2719, 2720} \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {(A-5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {(A-5 C) \sin (c+d x)}{3 a^2 d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}-\frac {4 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {4 C \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

(-4*C*EllipticE[(c + d*x)/2, 2])/(a^2*d) + ((A - 5*C)*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) + (4*C*Sin[c + d*x]
)/(a^2*d*Sqrt[Cos[c + d*x]]) + ((A - 5*C)*Sin[c + d*x])/(3*a^2*d*Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])) - ((A
+ C)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*(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 {3}{2}}(c+d x) (a+a \cos (c+d x))^2} \, dx \\ & = -\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a (A+7 C)+\frac {3}{2} a (A-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))} \, dx}{3 a^2} \\ & = \frac {(A-5 C) \sin (c+d x)}{3 a^2 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {\int \frac {6 a^2 C+\frac {1}{2} a^2 (A-5 C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{3 a^4} \\ & = \frac {(A-5 C) \sin (c+d x)}{3 a^2 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}+\frac {(A-5 C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2}+\frac {(2 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{a^2} \\ & = \frac {(A-5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {4 C \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}+\frac {(A-5 C) \sin (c+d x)}{3 a^2 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2}-\frac {(2 C) \int \sqrt {\cos (c+d x)} \, dx}{a^2} \\ & = -\frac {4 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(A-5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {4 C \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}+\frac {(A-5 C) \sin (c+d x)}{3 a^2 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (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 = 11.41 (sec) , antiderivative size = 836, normalized size of antiderivative = 5.54 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {4 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 {20 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 {16 C \cot \left (\frac {c}{2}\right ) \sec (c)}{d}+\frac {16 C \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{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 {16 C \sec (c) \sec (c+d x) \sin (d x)}{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 {8 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]^(3/2)*(a + a*Sec[c + d*x])^2),x]

[Out]

(-4*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) + (20*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)*((16*C*Cot[c/2]*Sec[c])/d + (16*C*Sec[c/2]*Sec[c/2 + (d*x)/2]*Sin[
(d*x)/2])/d + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(3*d) + (16*C*Sec[c]*Sec[c +
d*x]*Sin[d*x])/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*S
ec[c + d*x])^2) + (8*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 + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + T
an[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]]]*Sqr
t[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)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(449\) vs. \(2(191)=382\).

Time = 1.88 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.98

method	result	size
default	\(-\frac {-2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-48 C \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A +43 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A +37 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(450\)

[In]

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

[Out]

-1/6*(-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(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)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+12*C*Elliptic
E(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)^2+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(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)*(A*EllipticF(cos(1/2*d*x+1/2*c),
2^(1/2))-5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+12*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2
*c)-48*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)^6+2*(-2*sin(1/2*d*x+1/2*c)^4+
sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+43*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+37*C)*sin(1/2*d*x+1/2*c)^2)/a^2/cos(1/2*d*x+1/2*c)^3/(-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 = 0.11 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.43 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {2 \, {\left (12 \, C \cos \left (d x + c\right )^{2} + {\left (A + 19 \, C\right )} \cos \left (d x + c\right ) + 6 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (-i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} {\left (i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A + 5 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} {\left (-i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A - 5 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 12 \, {\left (i \, \sqrt {2} C \cos \left (d x + c\right )^{3} + 2 i \, \sqrt {2} C \cos \left (d x + c\right )^{2} + i \, \sqrt {2} C \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 \, {\left (-i \, \sqrt {2} C \cos \left (d x + c\right )^{3} - 2 i \, \sqrt {2} C \cos \left (d x + c\right )^{2} - i \, \sqrt {2} C \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \]

[In]

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

[Out]

1/6*(2*(12*C*cos(d*x + c)^2 + (A + 19*C)*cos(d*x + c) + 6*C)*sqrt(cos(d*x + c))*sin(d*x + c) + (sqrt(2)*(-I*A
+ 5*I*C)*cos(d*x + c)^3 - 2*sqrt(2)*(I*A - 5*I*C)*cos(d*x + c)^2 + sqrt(2)*(-I*A + 5*I*C)*cos(d*x + c))*weiers
trassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(I*A - 5*I*C)*cos(d*x + c)^3 - 2*sqrt(2)*(-I*A
+ 5*I*C)*cos(d*x + c)^2 + sqrt(2)*(I*A - 5*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(
d*x + c)) - 12*(I*sqrt(2)*C*cos(d*x + c)^3 + 2*I*sqrt(2)*C*cos(d*x + c)^2 + I*sqrt(2)*C*cos(d*x + c))*weierstr
assZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 12*(-I*sqrt(2)*C*cos(d*x + c)^3 -
2*I*sqrt(2)*C*cos(d*x + c)^2 - I*sqrt(2)*C*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos
(d*x + c) - I*sin(d*x + c))))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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