\(\int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [166]

Optimal result

Integrand size = 33, antiderivative size = 264 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(39 A-20 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {7 (9 A-5 B) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(31 A-15 B) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}} \]

[Out]

1/4*(39*A-20*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d-1/4*(A-B)*cos(d*x+c)*sin(d*x+c)/d/
(a+a*sec(d*x+c))^(5/2)-1/16*(19*A-11*B)*cos(d*x+c)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)-1/32*(219*A-115*B)*ar
ctan(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)-7/16*(9*A-5*B)*sin(d*x+c)/a^2/d/
(a+a*sec(d*x+c))^(1/2)+1/16*(31*A-15*B)*cos(d*x+c)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 1.31 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4105, 4107, 4005, 3859, 209, 3880} \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(39 A-20 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {7 (9 A-5 B) \sin (c+d x)}{16 a^2 d \sqrt {a \sec (c+d x)+a}}+\frac {(31 A-15 B) \sin (c+d x) \cos (c+d x)}{16 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(19 A-11 B) \sin (c+d x) \cos (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[In]

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

[Out]

((39*A - 20*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*a^(5/2)*d) - ((219*A - 115*B)*ArcTa
n[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B)*Cos[c + d*x]*S
in[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((19*A - 11*B)*Cos[c + d*x]*Sin[c + d*x])/(16*a*d*(a + a*Sec[c
+ d*x])^(3/2)) - (7*(9*A - 5*B)*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Sec[c + d*x]]) + ((31*A - 15*B)*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]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \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)-\frac {7}{2} a (A-B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = -\frac {(A-B) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B) \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)-\frac {5}{4} a^2 (19 A-11 B) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {(A-B) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(31 A-15 B) \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 (-7 a^3 (9 A-5 B)+\frac {3}{2} a^3 (31 A-15 B) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{16 a^5} \\ & = -\frac {(A-B) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {7 (9 A-5 B) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(31 A-15 B) \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)-\frac {7}{2} a^4 (9 A-5 B) \sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{16 a^6} \\ & = -\frac {(A-B) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {7 (9 A-5 B) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(31 A-15 B) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(219 A-115 B) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}+\frac {(39 A-20 B) \int \sqrt {a+a \sec (c+d x)} \, dx}{8 a^3} \\ & = -\frac {(A-B) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {7 (9 A-5 B) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(31 A-15 B) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(219 A-115 B) \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) \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 {(39 A-20 B) \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) \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) \cos (c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(19 A-11 B) \cos (c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {7 (9 A-5 B) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(31 A-15 B) \cos (c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 4.00 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.32 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {32 B \sin (c+d x)+152 A (1+\cos (c+d x)) \sin (c+d x)+120 B (1+\sec (c+d x)) \sin (c+d x)-280 B (1+\sec (c+d x))^2 \sin (c+d x)+16 A \sin (2 (c+d x))+760 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\sec (c+d x)\right ) (1+\sec (c+d x))^2 \tan (c+d x)+\frac {640 B \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) (1+\sec (c+d x))^2 \tan (c+d x)}{\sqrt {1-\sec (c+d x)}}-\frac {460 \sqrt {2} B \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) (1+\sec (c+d x))^2 \tan (c+d x)}{\sqrt {1-\sec (c+d x)}}-\frac {219 A \left (7 \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )-4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+\cos (c+d x) (-1+2 \cos (c+d x)) \sqrt {1-\sec (c+d x)}\right ) (1+\sec (c+d x))^2 \tan (c+d x)}{\sqrt {1-\sec (c+d x)}}}{128 d (a (1+\sec (c+d x)))^{5/2}} \]

[In]

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

[Out]

-1/128*(32*B*Sin[c + d*x] + 152*A*(1 + Cos[c + d*x])*Sin[c + d*x] + 120*B*(1 + Sec[c + d*x])*Sin[c + d*x] - 28
0*B*(1 + Sec[c + d*x])^2*Sin[c + d*x] + 16*A*Sin[2*(c + d*x)] + 760*A*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c
+ d*x]]*(1 + Sec[c + d*x])^2*Tan[c + d*x] + (640*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*(1 + Sec[c + d*x])^2*Tan[c
+ d*x])/Sqrt[1 - Sec[c + d*x]] - (460*Sqrt[2]*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*(1 + Sec[c + d*x])^2*
Tan[c + d*x])/Sqrt[1 - Sec[c + d*x]] - (219*A*(7*ArcTanh[Sqrt[1 - Sec[c + d*x]]] - 4*Sqrt[2]*ArcTanh[Sqrt[1 -
Sec[c + d*x]]/Sqrt[2]] + Cos[c + d*x]*(-1 + 2*Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]])*(1 + Sec[c + d*x])^2*Tan[c
+ d*x])/Sqrt[1 - Sec[c + d*x]])/(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. \(1385\) vs. \(2(229)=458\).

Time = 31.77 (sec) , antiderivative size = 1386, normalized size of antiderivative = 5.25

method	result	size
default	\(\text {Expression too large to display}\)	\(1386\)

[In]

int(cos(d*x+c)^2*(A+B*sec(d*x+c))/(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)*(-2*A*(1-cos(d*x+c))^11*csc(d*x+c)^11+2*B*(1-cos(d*x
+c))^11*csc(d*x+c)^11+156*A*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+29*A*(1-cos(d*x+c))^9*csc(d*x+c)
^9-80*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)^4-21*B*(1-cos(d*x+c))^9*csc(d*x+c)^9-219*A*((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+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+312*A*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+92*A*(1
-cos(d*x+c))^7*csc(d*x+c)^7-160*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)^2-36*B*(1-cos(d*x+c))^7*csc(
d*x+c)^7-438*A*((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+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+156*A*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)))-178*A*(1-co
s(d*x+c))^5*csc(d*x+c)^5-80*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)))+74*B*(1-cos(d*x+c))^5*csc(d*x+c)^5-219*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)+115*B*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)-26*A*(1-cos(d*x+c))^3
*csc(d*x+c)^3+34*B*(1-cos(d*x+c))^3*csc(d*x+c)^3+85*A*(-cot(d*x+c)+csc(d*x+c))-53*B*(-cot(d*x+c)+csc(d*x+c)))/
((1-cos(d*x+c))^2*csc(d*x+c)^2-1)/((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^2

Fricas [A] (verification not implemented)

none

Time = 6.33 (sec) , antiderivative size = 776, normalized size of antiderivative = 2.94 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [\frac {\sqrt {2} {\left ({\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right ) + 219 \, A - 115 \, B\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 8 \, {\left ({\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right ) + 39 \, A - 20 \, B\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (8 \, A \cos \left (d x + c\right )^{4} - 4 \, {\left (5 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{3} - 5 \, {\left (19 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} - 7 \, {\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, \frac {\sqrt {2} {\left ({\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (219 \, A - 115 \, B\right )} \cos \left (d x + c\right ) + 219 \, A - 115 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 8 \, {\left ({\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (39 \, A - 20 \, B\right )} \cos \left (d x + c\right ) + 39 \, A - 20 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left (8 \, A \cos \left (d x + c\right )^{4} - 4 \, {\left (5 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{3} - 5 \, {\left (19 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} - 7 \, {\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]

[In]

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

[Out]

[1/64*(sqrt(2)*((219*A - 115*B)*cos(d*x + c)^3 + 3*(219*A - 115*B)*cos(d*x + c)^2 + 3*(219*A - 115*B)*cos(d*x
+ c) + 219*A - 115*B)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*si
n(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)*cos(d*x + c)^3 + 3*(39*A - 20*B)*cos(d*x + c)^2 + 3*(39*A - 20*B)*cos(d*x + c) + 39*A - 20*B)*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)*cos(d
*x + c)^2 - 7*(9*A - 5*B)*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), 1/32*(sqrt(2)*((219*A - 115*B)*cos(d*x + c)^3
+ 3*(219*A - 115*B)*cos(d*x + c)^2 + 3*(219*A - 115*B)*cos(d*x + c) + 219*A - 115*B)*sqrt(a)*arctan(sqrt(2)*sq
rt((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)*cos(d*x + c)^3 +
3*(39*A - 20*B)*cos(d*x + c)^2 + 3*(39*A - 20*B)*cos(d*x + c) + 39*A - 20*B)*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)*cos(d*x +
c)^3 - 5*(19*A - 11*B)*cos(d*x + c)^2 - 7*(9*A - 5*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*si
n(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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate((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. 685 vs. \(2 (229) = 458\).

Time = 2.28 (sec) , antiderivative size = 685, normalized size of antiderivative = 2.59 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^2*(A+B*sec(d*x+c))/(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)*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)/(a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c) + sqrt(2)*(219*A - 115*B
)*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)*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) - 928455029464035206174343168*a)/abs(3094850098
21345068724781056*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 618970019642690137
449562112*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*tan(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. \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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