3.1119 \(\int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\)
Optimal. Leaf size=189 \[ \frac {2 (A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac {(A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {(A+7 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x)+1)}+\frac {2 (A+5 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(A+7 C) \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]
[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)
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Rubi [A] time = 0.41, antiderivative size = 189, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used =
{4114, 3042, 2978, 2748, 2636, 2641, 2639} \[ \frac {2 (A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac {(A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {(A+7 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x)+1)}+\frac {2 (A+5 C) \sin (c+d x)}{3 a^2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {(A+7 C) \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}-\frac {(A+C) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
[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 2636
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 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 2748
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 2978
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 3042
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 4114
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*} \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+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) F\left (\left .\frac {1}{2} (c+d x)\right |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*}
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Mathematica [C] time = 7.51, size = 1391, normalized size = 7.36 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(A + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^2),x]
[Out]
(I*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/
4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x)
)*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))
*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c]) - (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*
Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*
I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[
c])))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + ((7*I)*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2
]*(A + C*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hypergeometric2F1[1/2, 3/4, 7/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c]
)^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x
)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c]])/((3*I)*d*(1 + E^((2*I)*d*x))*Cos[c] - 3*d*(-1 + E^((2*I)*d*x))*Sin[c])
- (2*Hypergeometric2F1[-1/4, 1/2, 3/4, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sqrt[(2*(1 + E^((2*I)*d*x))*Co
s[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x)]*Sqrt[1 + E^((2*I)*d*x)*Cos[2*c] + I*E^((2*I)*d*x)*Sin[2*c
]])/((-I)*d*(1 + E^((2*I)*d*x))*Cos[c] + d*(-1 + E^((2*I)*d*x))*Sin[c])))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a +
a*Sec[c + d*x])^2) - (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]]]]*Sqr
t[-(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]*Hypergeo
metricPFQ[{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) +
(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]*S
ec[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]*Se
c[c/2 + (d*x)/2]*(A*Sin[(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*Se
c[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)
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fricas [F] time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sqrt {\cos \left (d x + c\right )}}{a^{2} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right ) + a^{2} \cos \left (d x + c\right )^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[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]
integral((C*sec(d*x + c)^2 + A)*sqrt(cos(d*x + c))/(a^2*cos(d*x + c)^3*sec(d*x + c)^2 + 2*a^2*cos(d*x + c)^3*s
ec(d*x + c) + a^2*cos(d*x + c)^3), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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maple [B] time = 15.94, size = 738, normalized size = 3.90 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x)
[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*(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)))*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)^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)))*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)/(-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)/(-1+sin(1/2*d*x+1/2*c)^2)-8*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)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-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)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*s
in(1/2*d*x+1/2*c)^2-1)+4*C*(cos(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(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)+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)/(-1/2+cos(1/2*d*x+1/2*c)^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
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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+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
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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+C*sec(d*x+c)**2)/cos(d*x+c)**(5/2)/(a+a*sec(d*x+c))**2,x)
[Out]
Timed out
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