\(\int \frac {\cos ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [533]

Optimal result

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

[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] (verified)

Time = 1.06 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4169, 4105, 4107, 4005, 3859, 209, 3880} \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(39 A-20 B+8 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B+43 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(63 A-35 B+11 C) \sin (c+d x)}{16 a^2 d \sqrt {a \sec (c+d x)+a}}+\frac {(31 A-15 B+7 C) \sin (c+d x) \cos (c+d x)}{16 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(19 A-11 B+3 C) \sin (c+d x) \cos (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[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 209

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

Rule 3859

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 3880

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 4005

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 4105

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 4107

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 4169

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*C
sc[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*} \text {integral}& = -\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) \text {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) \text {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) \arctan \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) \arctan \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 [A] (warning: unable to verify)

Time = 10.33 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \left (\frac {\left ((-219 A+115 B-43 C) \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )+4 \sqrt {2} (39 A-20 B+8 C) \arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}}}\right )\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {1+\sec (c+d x)}}{\sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+\frac {1}{4} (73 A-43 B+11 C+(89 A-55 B+15 C) \cos (c+d x)+2 (5 A-4 B) \cos (2 (c+d x))-2 A \cos (3 (c+d x))) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (\sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{4 d (a (1+\sec (c+d x)))^{5/2}} \]

[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]

(Cos[(c + d*x)/2]^4*Sec[c + d*x]^(5/2)*((((-219*A + 115*B - 43*C)*ArcSin[Tan[(c + d*x)/2]] + 4*Sqrt[2]*(39*A -
20*B + 8*C)*ArcTan[Tan[(c + d*x)/2]/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]])*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*
x])]*Sqrt[1 + Sec[c + d*x]])/Sqrt[Sec[(c + d*x)/2]^2] + ((73*A - 43*B + 11*C + (89*A - 55*B + 15*C)*Cos[c + d*
x] + 2*(5*A - 4*B)*Cos[2*(c + d*x)] - 2*A*Cos[3*(c + d*x)])*Sec[(c + d*x)/2]^3*Sqrt[Sec[c + d*x]]*(Sin[(c + d*
x)/2] - Sin[(3*(c + d*x))/2]))/4))/(4*d*(a*(1 + Sec[c + d*x]))^(5/2))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(2033\) vs. \(2(245)=490\).

Time = 3.04 (sec) , antiderivative size = 2034, normalized size of antiderivative = 7.26

method	result	size
default	\(\text {Expression too large to display}\)	\(2034\)

[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,method=_RETURNVERBOSE)

[Out]

1/32/a^3/d*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(156*A*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^
2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*2^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(1-cos(d*x+c))^4*csc(d*x+
c)^4+32*C*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*2^(1/2)*((1-cos(d*
x+c))^2*csc(d*x+c)^2-1)^(3/2)*(1-cos(d*x+c))^4*csc(d*x+c)^4+312*A*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)
^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*2^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(1-cos(d*x+c))^2*csc(d*x
+c)^2+64*C*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*2^(1/2)*((1-cos(d
*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(1-cos(d*x+c))^2*csc(d*x+c)^2-2*C*(1-cos(d*x+c))^3*csc(d*x+c)^3+85*A*(-cot(d*x+
c)+csc(d*x+c))-21*B*(1-cos(d*x+c))^9*csc(d*x+c)^9-36*B*(1-cos(d*x+c))^7*csc(d*x+c)^7+74*B*(1-cos(d*x+c))^5*csc
(d*x+c)^5+34*B*(1-cos(d*x+c))^3*csc(d*x+c)^3+156*A*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*arctanh(2^(1/2)/((1
-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*2^(1/2)+115*B*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)
^(3/2)*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))-160*B*2^(1/2)*((1-cos(d*x+c))^2*csc(d
*x+c)^2-1)^(3/2)*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*(1-cos(d*x+
c))^2*csc(d*x+c)^2-80*B*2^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(
d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*(1-cos(d*x+c))^4*csc(d*x+c)^4+2*B*(1-cos(d*x+c))^11*csc(d*x+c)^11+
13*C*(-cot(d*x+c)+csc(d*x+c))-2*A*(1-cos(d*x+c))^11*csc(d*x+c)^11-219*A*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c
))^2*csc(d*x+c)^2-1)^(1/2))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)-43*C*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+
c))^2*csc(d*x+c)^2-1)^(1/2))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)+29*A*(1-cos(d*x+c))^9*csc(d*x+c)^9+13*C*(
1-cos(d*x+c))^9*csc(d*x+c)^9+92*A*(1-cos(d*x+c))^7*csc(d*x+c)^7+4*C*(1-cos(d*x+c))^7*csc(d*x+c)^7-178*A*(1-cos
(d*x+c))^5*csc(d*x+c)^5-26*C*(1-cos(d*x+c))^5*csc(d*x+c)^5-26*A*(1-cos(d*x+c))^3*csc(d*x+c)^3+32*C*arctanh(2^(
1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))*2^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-
1)^(3/2)-2*C*(1-cos(d*x+c))^11*csc(d*x+c)^11-80*B*2^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*arctanh(2^(1
/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)))-219*A*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d
*x+c))^2*csc(d*x+c)^2-1)^(1/2))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(1-cos(d*x+c))^4*csc(d*x+c)^4-43*C*ln(
csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(1-cos(
d*x+c))^4*csc(d*x+c)^4-438*A*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*((1-cos(d*x+c))
^2*csc(d*x+c)^2-1)^(3/2)*(1-cos(d*x+c))^2*csc(d*x+c)^2-86*C*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x
+c)^2-1)^(1/2))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(1-cos(d*x+c))^2*csc(d*x+c)^2+115*B*((1-cos(d*x+c))^2*
csc(d*x+c)^2-1)^(3/2)*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*(1-cos(d*x+c))^4*csc(d
*x+c)^4+230*B*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-
1)^(1/2))*(1-cos(d*x+c))^2*csc(d*x+c)^2-53*B*(-cot(d*x+c)+csc(d*x+c)))/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)/((1-c
os(d*x+c))^2*csc(d*x+c)^2+1)^2

Fricas [A] (verification not implemented)

none

Time = 47.04 (sec) , antiderivative size = 835, normalized size of antiderivative = 2.98 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[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)]

Sympy [F]

\[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[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)

Maxima [F]

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

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 702 vs. \(2 (245) = 490\).

Time = 2.50 (sec) , antiderivative size = 702, normalized size of antiderivative = 2.51 \[ \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[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*sgn
(cos(d*x + c))) - sqrt(2)*(29*A*a^5 - 21*B*a^5 + 13*C*a^5)/(a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c) + sqr
t(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(cos(d*x + c))) + 8*(39*A - 20*B + 8*C)*log(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) - 9284550294640
35206174343168*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) - 928455029464035206174343168*a))/(sqrt(-a)*a*abs(a)*
sgn(cos(d*x + c))) - 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*t
an(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^2*sqrt(-a)*a*sgn(cos(d*x + c))))/d

Mupad [F(-1)]

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

[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)