3.1226 \(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^2} \, dx\)
Optimal. Leaf size=180 \[ \frac {(10 A-5 B+2 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac {(7 A-4 B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {(7 A-4 B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {(10 A-5 B+2 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a^2 d}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]
[Out]
-(7*A-4*B+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*(
10*A-5*B+2*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-1/3*
(7*A-4*B+C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A-B+C)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*co
s(d*x+c))^2+1/3*(10*A-5*B+2*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/a^2/d
________________________________________________________________________________________
Rubi [A] time = 0.45, antiderivative size = 180, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used =
{4112, 3041, 2977, 2748, 2639, 2635, 2641} \[ \frac {(10 A-5 B+2 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac {(7 A-4 B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac {(7 A-4 B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac {(10 A-5 B+2 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 a^2 d}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^2,x]
[Out]
-(((7*A - 4*B + C)*EllipticE[(c + d*x)/2, 2])/(a^2*d)) + ((10*A - 5*B + 2*C)*EllipticF[(c + d*x)/2, 2])/(3*a^2
*d) + ((10*A - 5*B + 2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a^2*d) - ((7*A - 4*B + C)*Cos[c + d*x]^(3/2)*Sin
[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((A - B + C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*
x])^2)
Rule 2635
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[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 2977
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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 3041
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 4112
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*} \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (C+B \cos (c+d x)+A \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^2} \, dx\\ &=-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (-\frac {1}{2} a (5 A-5 B-C)+\frac {3}{2} a (3 A-B+C) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac {(7 A-4 B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \sqrt {\cos (c+d x)} \left (-\frac {3}{2} a^2 (7 A-4 B+C)+\frac {3}{2} a^2 (10 A-5 B+2 C) \cos (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {(7 A-4 B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {(7 A-4 B+C) \int \sqrt {\cos (c+d x)} \, dx}{2 a^2}+\frac {(10 A-5 B+2 C) \int \cos ^{\frac {3}{2}}(c+d x) \, dx}{2 a^2}\\ &=-\frac {(7 A-4 B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(10 A-5 B+2 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}-\frac {(7 A-4 B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {(10 A-5 B+2 C) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{6 a^2}\\ &=-\frac {(7 A-4 B+C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(10 A-5 B+2 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac {(10 A-5 B+2 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}-\frac {(7 A-4 B+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end {align*}
________________________________________________________________________________________
Mathematica [C] time = 6.91, size = 2064, normalized size = 11.47 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^2,x]
[Out]
((-7*I)*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hyp
ergeometric2F1[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 + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + ((4*I
)*B*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*E^((2*I)*d*x)*Hypergeom
etric2F1[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 + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) - (I*C*Cos[c/
2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*(A + B*Sec[c + d*x] + 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 + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) - (40*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 + 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]*Si
n[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])^2) + (20*B*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Hypergeometric
PFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(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])^2) - (8*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 + 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])^2) +
(Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((8*(3*A - 2*B + C + 4*A*Cos
[c] - 2*B*Cos[c])*Csc[c])/d + (16*A*Cos[d*x]*Sin[c])/(3*d) + (8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(3*A*Sin[(d*x)/2]
- 2*B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/d - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] +
C*Sin[(d*x)/2]))/(3*d) + (16*A*Cos[c]*Sin[d*x])/(3*d) - (4*(A - B + C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/
((A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2)
________________________________________________________________________________________
fricas [F] time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) \sec \left (d x + c\right ) + A \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}}{a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
[Out]
integral((C*cos(d*x + c)*sec(d*x + c)^2 + B*cos(d*x + c)*sec(d*x + c) + A*cos(d*x + c))*sqrt(cos(d*x + c))/(a^
2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a^2), x)
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*sec(d*x + c) + a)^2, x)
________________________________________________________________________________________
maple [B] time = 5.27, size = 472, normalized size = 2.62 \[ -\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 (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (10 A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+21 A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 B \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 B \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 C \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \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 ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (10 A \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+21 A \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 B \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 B \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 C \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \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 )+16 A \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-76 A +24 B -12 C \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (84 A -34 B +16 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-25 A +11 B -5 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \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} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x)
[Out]
-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-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)*(10*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+21*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*B*El
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*C*EllipticF(cos(1/2*d*x+1/2*c
),2^(1/2))+3*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)^2+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)*(10*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+21*A*EllipticE(cos
(1/2*d*x+1/2*c),2^(1/2))-5*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+
2*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)+16*A*s
in(1/2*d*x+1/2*c)^8+(-76*A+24*B-12*C)*sin(1/2*d*x+1/2*c)^6+(84*A-34*B+16*C)*sin(1/2*d*x+1/2*c)^4+(-25*A+11*B-5
*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
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(3/2)/(a*sec(d*x + c) + a)^2, x)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^2,x)
[Out]
int((cos(c + d*x)^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^2, x)
________________________________________________________________________________________
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(cos(d*x+c)**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2,x)
[Out]
Timed out
________________________________________________________________________________________