3.533 \(\int \frac {\cos ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=280 \[ \frac {(39 A-20 B+8 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B+43 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(63 A-35 B+11 C) \sin (c+d x)}{16 a^2 d \sqrt {a \sec (c+d x)+a}}+\frac {(31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{16 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[Out]

1/4*(39*A-20*B+8*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d-1/4*(A-B+C)*cos(d*x+c)*sin(d*x
+c)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(19*A-11*B+3*C)*cos(d*x+c)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)-1/32*(219*A
-115*B+43*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)-1/16*(63*A-35*B+1
1*C)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)+1/16*(31*A-15*B+7*C)*cos(d*x+c)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))
^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.87, antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4084, 4020, 4022, 3920, 3774, 203, 3795} \[ -\frac {(63 A-35 B+11 C) \sin (c+d x)}{16 a^2 d \sqrt {a \sec (c+d x)+a}}+\frac {(39 A-20 B+8 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B+43 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{16 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((39*A - 20*B + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*a^(5/2)*d) - ((219*A - 115*B
+ 43*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B +
C)*Cos[c + d*x]*Sin[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((19*A - 11*B + 3*C)*Cos[c + d*x]*Sin[c + d*x
])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) - ((63*A - 35*B + 11*C)*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Sec[c + d*x]
]) + ((31*A - 15*B + 7*C)*Cos[c + d*x]*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 3920

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 4022

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1
/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - A*b*(m + n + 1)*Csc[e + f*x
], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rule 4084

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[e + f*x])^m*(d*Csc[e + f*x])^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)
))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {\int \frac {\cos ^2(c+d x) \left (2 a (3 A-B+C)-\frac {1}{2} a (7 A-7 B-C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B+3 C) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\cos ^2(c+d x) \left (a^2 (31 A-15 B+7 C)-\frac {5}{4} a^2 (19 A-11 B+3 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B+3 C) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(31 A-15 B+7 C) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {\cos (c+d x) \left (-a^3 (63 A-35 B+11 C)+\frac {3}{2} a^3 (31 A-15 B+7 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{16 a^5}\\ &=-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B+3 C) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(63 A-35 B+11 C) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(31 A-15 B+7 C) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {2 a^4 (39 A-20 B+8 C)-\frac {1}{2} a^4 (63 A-35 B+11 C) \sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{16 a^6}\\ &=-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B+3 C) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(63 A-35 B+11 C) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(31 A-15 B+7 C) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(39 A-20 B+8 C) \int \sqrt {a+a \sec (c+d x)} \, dx}{8 a^3}-\frac {(219 A-115 B+43 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B+3 C) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(63 A-35 B+11 C) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(31 A-15 B+7 C) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(39 A-20 B+8 C) \operatorname {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 a^2 d}+\frac {(219 A-115 B+43 C) \operatorname {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{16 a^2 d}\\ &=\frac {(39 A-20 B+8 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B+43 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B+3 C) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(63 A-35 B+11 C) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(31 A-15 B+7 C) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 28.82, size = 17717, normalized size = 63.28 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

________________________________________________________________________________________

fricas [A]  time = 84.50, size = 835, normalized size = 2.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/64*(sqrt(2)*((219*A - 115*B + 43*C)*cos(d*x + c)^3 + 3*(219*A - 115*B + 43*C)*cos(d*x + c)^2 + 3*(219*A -
115*B + 43*C)*cos(d*x + c) + 219*A - 115*B + 43*C)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)
/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) - 3*a*cos(d*x + c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*cos(
d*x + c) + 1)) + 8*((39*A - 20*B + 8*C)*cos(d*x + c)^3 + 3*(39*A - 20*B + 8*C)*cos(d*x + c)^2 + 3*(39*A - 20*B
+ 8*C)*cos(d*x + c) + 39*A - 20*B + 8*C)*sqrt(-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) +
a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) - 4*(8*A*cos(d*x + c)^4
- 4*(5*A - 4*B)*cos(d*x + c)^3 - 5*(19*A - 11*B + 3*C)*cos(d*x + c)^2 - (63*A - 35*B + 11*C)*cos(d*x + c))*sqr
t((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*co
s(d*x + c) + a^3*d), 1/32*(sqrt(2)*((219*A - 115*B + 43*C)*cos(d*x + c)^3 + 3*(219*A - 115*B + 43*C)*cos(d*x +
c)^2 + 3*(219*A - 115*B + 43*C)*cos(d*x + c) + 219*A - 115*B + 43*C)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 8*((39*A - 20*B + 8*C)*cos(d*x + c)^3 + 3*(39*A
- 20*B + 8*C)*cos(d*x + c)^2 + 3*(39*A - 20*B + 8*C)*cos(d*x + c) + 39*A - 20*B + 8*C)*sqrt(a)*arctan(sqrt((a*
cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 2*(8*A*cos(d*x + c)^4 - 4*(5*A - 4*B)*c
os(d*x + c)^3 - 5*(19*A - 11*B + 3*C)*cos(d*x + c)^2 - (63*A - 35*B + 11*C)*cos(d*x + c))*sqrt((a*cos(d*x + c)
+ a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*
d)]

________________________________________________________________________________________

giac [B]  time = 5.08, size = 737, normalized size = 2.63 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(A*a^5 - B*a^5 + C*a^5)*tan(1/2*d*x + 1/2*c)^2/(a^8*sg
n(tan(1/2*d*x + 1/2*c)^2 - 1)) - sqrt(2)*(29*A*a^5 - 21*B*a^5 + 13*C*a^5)/(a^8*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)
))*tan(1/2*d*x + 1/2*c) + sqrt(2)*(219*A - 115*B + 43*C)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*
d*x + 1/2*c)^2 + a))^2)/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 8*(39*A - 20*B + 8*C)*log(abs(3094850
09821345068724781056*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 618970019642690
137449562112*sqrt(2)*abs(a) - 928455029464035206174343168*a)/abs(309485009821345068724781056*(sqrt(-a)*tan(1/2
*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 618970019642690137449562112*sqrt(2)*abs(a) - 92845502
9464035206174343168*a))/(sqrt(-a)*a*abs(a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - 32*sqrt(2)*(41*(sqrt(-a)*tan(1/2
*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A - 12*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2
*d*x + 1/2*c)^2 + a))^6*B - 209*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*a +
76*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*a + 91*(sqrt(-a)*tan(1/2*d*x + 1/
2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*a^2 - 36*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x
+ 1/2*c)^2 + a))^2*B*a^2 - 11*A*a^3 + 4*B*a^3)/(((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)
^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^2*sqrt(-a)*a*s
gn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d

________________________________________________________________________________________

maple [B]  time = 1.78, size = 2096, normalized size = 7.49 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64/d*(-1+cos(d*x+c))^2*(140*B*cos(d*x+c)^2-252*A*cos(d*x+c)^2+60*C*cos(d*x+c)^4-44*C*cos(d*x+c)^2+468*A*arct
anh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/
2)*cos(d*x+c)*2^(1/2)*sin(d*x+c)+300*A*cos(d*x+c)^4-156*B*cos(d*x+c)^4-240*B*2^(1/2)*cos(d*x+c)*sin(d*x+c)*(-2
*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/
2))-16*C*cos(d*x+c)^3-128*A*cos(d*x+c)^3+80*B*cos(d*x+c)^3+112*A*cos(d*x+c)^5-64*B*cos(d*x+c)^5+468*A*cos(d*x+
c)^2*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
sin(d*x+c)/cos(d*x+c)*2^(1/2))+156*A*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arct
anh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+219*A*(-2*cos(d*x+c)/(1+cos(d*x+c)
))^(3/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*sin(d*x+c)-115*B*(-2*co
s(d*x+c)/(1+cos(d*x+c)))^(3/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*s
in(d*x+c)+43*C*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*
x+c)+1)/sin(d*x+c))*sin(d*x+c)-80*B*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctanh(1/2*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)+32*C*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2
)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)+96*C*2^(1/2)*cos(
d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*
x+c)/cos(d*x+c)*2^(1/2))+657*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*s
in(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*cos(d*x+c)*sin(d*x+c)-345*B*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*
x+c)))^(3/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))+129*C*cos(d*x+c)*si
n(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1
)/sin(d*x+c))+219*A*cos(d*x+c)^3*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))-115*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*ln(((-2*cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3+43*C*(-2*cos(d*x+c)/(1+cos
(d*x+c)))^(3/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*sin(d*x+c)*cos(d
*x+c)^3-80*B*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctanh(1/2*(-2*cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+32*C*sin(d*x+c)*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(1+cos
(d*x+c)))^(3/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))-240*B*cos(d*x+
c)^2*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
sin(d*x+c)/cos(d*x+c)*2^(1/2))+96*C*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arcta
nh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+657*A*cos(d*x+c)^2*sin(d*x+c)*(-2*c
os(d*x+c)/(1+cos(d*x+c)))^(3/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))-
345*B*sin(d*x+c)*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*si
n(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))+129*C*sin(d*x+c)*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*ln(((-2*
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))+156*A*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*2^(1/2)*sin(d*x+c)-32*A*cos
(d*x+c)^6)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)^5/cos(d*x+c)/a^3

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*cos(c + d*x)**2/(a*(sec(c + d*x) + 1))**(5/2), x)

________________________________________________________________________________________