\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\) [424]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 1.15 (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 = {3120, 3057, 3063, 3064, 2728, 212, 2852} \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(39 A-20 B+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 a^{5/2} d}-\frac {(219 A-115 B+43 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(63 A-35 B+11 C) \tan (c+d x)}{16 a^2 d \sqrt {a \cos (c+d x)+a}}+\frac {(31 A-15 B+7 C) \tan (c+d x) \sec (c+d x)}{16 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(19 A-11 B+3 C) \tan (c+d x) \sec (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B+C) \tan (c+d x) \sec (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[In]

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

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

Rule 212

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

Rule 2728

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

Rule 2852

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[-2*(
b/f), Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3057

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 3063

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 3064

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*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]

Rule 3120

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)]

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

Mathematica [A] (verified)

Time = 5.80 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.79 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \left ((-438 A+230 B-86 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {8 \sqrt {2} (39 A-20 B+8 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )-\frac {1}{4} (158 A-110 B+30 C+(269 A-169 B+33 C) \cos (c+d x)+10 (19 A-11 B+3 C) \cos (2 (c+d x))+63 A \cos (3 (c+d x))-35 B \cos (3 (c+d x))+11 C \cos (3 (c+d x))) \sec ^2(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )}{\left (-1+\sin ^2\left (\frac {1}{2} (c+d x)\right )\right )^2}\right )}{8 d (a (1+\cos (c+d x)))^{5/2}} \]

[In]

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(Cos[(c + d*x)/2]^5*((-438*A + 230*B - 86*C)*ArcTanh[Sin[(c + d*x)/2]] + (8*Sqrt[2]*(39*A - 20*B + 8*C)*ArcTan
h[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^4 - ((158*A - 110*B + 30*C + (269*A - 169*B + 33*C)*Cos[c + d*x]
+ 10*(19*A - 11*B + 3*C)*Cos[2*(c + d*x)] + 63*A*Cos[3*(c + d*x)] - 35*B*Cos[3*(c + d*x)] + 11*C*Cos[3*(c + d*
x)])*Sec[c + d*x]^2*Sin[(c + d*x)/2])/4)/(-1 + Sin[(c + d*x)/2]^2)^2))/(8*d*(a*(1 + Cos[c + d*x]))^(5/2))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1759\) vs. \(2(245)=490\).

Time = 17.52 (sec) , antiderivative size = 1760, normalized size of antiderivative = 6.29

method	result	size
parts	\(\text {Expression too large to display}\)	\(1760\)
default	\(\text {Expression too large to display}\)	\(2366\)

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+cos(d*x+c)*a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/8*A*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(876*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2
*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^8*a-624*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(
1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^8*a-624*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2
))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^8*a-8
76*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^6*a+252*
2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^6+624*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(
2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^6*a+624*l
n(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2
)+2*a))*cos(1/2*d*x+1/2*c)^6*a+219*2^(1/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2
*c))*a*cos(1/2*d*x+1/2*c)^4-188*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2*d*x+1/2*c)^4-156*ln(-4/
(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*
a))*cos(1/2*d*x+1/2*c)^4*a-156*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*si
n(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a+19*cos(1/2*d*x+1/2*c)^2*(a*sin(1/2*d*x+1/2*c)^2
)^(1/2)*2^(1/2)*a^(1/2)+2*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(7/2)/cos(1/2*d*x+1/2*c)^3/(2*cos(
1/2*d*x+1/2*c)+2^(1/2))^2/(2*cos(1/2*d*x+1/2*c)-2^(1/2))^2/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
-1/16*B*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(80*ln(2/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2
^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*2^(1/2)*cos(1/2*d*x+1/2*c)^6*a+80*ln(-2/(2*cos(1/2*d*x+1/2
*c)-2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*2^(1/2)*cos(1/
2*d*x+1/2*c)^6*a-230*ln(2*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^6*
a-40*2^(1/2)*ln(2/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2
)^(1/2)*a^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a-40*2^(1/2)*ln(-2/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(2^(1/2)*a*cos(1/
2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^4*a-70*cos(1/2*d*x+1/2*c)
^4*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+115*ln(2*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*
c))*cos(1/2*d*x+1/2*c)^4*a+15*cos(1/2*d*x+1/2*c)^2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a^(1/2)*(a*sin(1/2
*d*x+1/2*c)^2)^(1/2))/a^(7/2)/cos(1/2*d*x+1/2*c)^3/(2*cos(1/2*d*x+1/2*c)+2^(1/2))/(2*cos(1/2*d*x+1/2*c)-2^(1/2
))/sin(1/2*d*x+1/2*c)*2^(1/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+1/32*C/a^(7/2)/cos(1/2*d*x+1/2*c)^3*(a*sin(1/2*
d*x+1/2*c)^2)^(1/2)*(16*2^(1/2)*ln(2/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(2^(1/2)*a*cos(1/2*d*x+1/2*c)+2^(1/2)*(a*s
in(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*cos(1/2*d*x+1/2*c)^4*a+16*2^(1/2)*ln(-2/(2*cos(1/2*d*x+1/2*c)-2^(1/2)
)*(2^(1/2)*a*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))*cos(1/2*d*x+1/2*c)^4*a-43
*ln(2*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*cos(1/2*d*x+1/2*c)^4*a-11*cos(1/2*d*x+1/2
*c)^2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2))/sin(1/2*d*x+1/2*c)*2^(1
/2)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.63 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (219 \, A - 115 \, B + 43 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (219 \, A - 115 \, B + 43 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (219 \, A - 115 \, B + 43 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (219 \, A - 115 \, B + 43 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (39 \, A - 20 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (39 \, A - 20 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (39 \, A - 20 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (39 \, A - 20 \, B + 8 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \, {\left ({\left (63 \, A - 35 \, B + 11 \, C\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (19 \, A - 11 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (5 \, A - 4 \, B\right )} \cos \left (d x + c\right ) - 8 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/64*(sqrt(2)*((219*A - 115*B + 43*C)*cos(d*x + c)^5 + 3*(219*A - 115*B + 43*C)*cos(d*x + c)^4 + 3*(219*A - 11
5*B + 43*C)*cos(d*x + c)^3 + (219*A - 115*B + 43*C)*cos(d*x + c)^2)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)
*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)
) + 4*((39*A - 20*B + 8*C)*cos(d*x + c)^5 + 3*(39*A - 20*B + 8*C)*cos(d*x + c)^4 + 3*(39*A - 20*B + 8*C)*cos(d
*x + c)^3 + (39*A - 20*B + 8*C)*cos(d*x + c)^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*
cos(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) - 4*((63*A
- 35*B + 11*C)*cos(d*x + c)^3 + 5*(19*A - 11*B + 3*C)*cos(d*x + c)^2 + 4*(5*A - 4*B)*cos(d*x + c) - 8*A)*sqrt
(a*cos(d*x + c) + a)*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a
^3*d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [A] (verification not implemented)

none

Time = 1.76 (sec) , antiderivative size = 446, normalized size of antiderivative = 1.59 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {\frac {\sqrt {2} {\left (219 \, A \sqrt {a} - 115 \, B \sqrt {a} + 43 \, C \sqrt {a}\right )} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (219 \, A \sqrt {a} - 115 \, B \sqrt {a} + 43 \, C \sqrt {a}\right )} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {8 \, {\left (39 \, A - 20 \, B + 8 \, C\right )} \log \left ({\left | \frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {8 \, {\left (39 \, A - 20 \, B + 8 \, C\right )} \log \left ({\left | -\frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} {\left (252 \, A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 140 \, B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 44 \, C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 568 \, A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 320 \, B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 96 \, C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 399 \, A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 231 \, B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 63 \, C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 85 \, A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 53 \, B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 13 \, C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 3 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{64 \, d} \]

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/64*(sqrt(2)*(219*A*sqrt(a) - 115*B*sqrt(a) + 43*C*sqrt(a))*log(sin(1/2*d*x + 1/2*c) + 1)/(a^3*sgn(cos(1/2*d
*x + 1/2*c))) - sqrt(2)*(219*A*sqrt(a) - 115*B*sqrt(a) + 43*C*sqrt(a))*log(-sin(1/2*d*x + 1/2*c) + 1)/(a^3*sgn
(cos(1/2*d*x + 1/2*c))) - 8*(39*A - 20*B + 8*C)*log(abs(1/2*sqrt(2) + sin(1/2*d*x + 1/2*c)))/(a^(5/2)*sgn(cos(
1/2*d*x + 1/2*c))) + 8*(39*A - 20*B + 8*C)*log(abs(-1/2*sqrt(2) + sin(1/2*d*x + 1/2*c)))/(a^(5/2)*sgn(cos(1/2*
d*x + 1/2*c))) - 2*sqrt(2)*(252*A*sqrt(a)*sin(1/2*d*x + 1/2*c)^7 - 140*B*sqrt(a)*sin(1/2*d*x + 1/2*c)^7 + 44*C
*sqrt(a)*sin(1/2*d*x + 1/2*c)^7 - 568*A*sqrt(a)*sin(1/2*d*x + 1/2*c)^5 + 320*B*sqrt(a)*sin(1/2*d*x + 1/2*c)^5
- 96*C*sqrt(a)*sin(1/2*d*x + 1/2*c)^5 + 399*A*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 - 231*B*sqrt(a)*sin(1/2*d*x + 1/2
*c)^3 + 63*C*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 - 85*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + 53*B*sqrt(a)*sin(1/2*d*x + 1
/2*c) - 13*C*sqrt(a)*sin(1/2*d*x + 1/2*c))/((2*sin(1/2*d*x + 1/2*c)^4 - 3*sin(1/2*d*x + 1/2*c)^2 + 1)^2*a^3*sg
n(cos(1/2*d*x + 1/2*c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a*cos(c + d*x))^(5/2)),x)

[Out]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a*cos(c + d*x))^(5/2)), x)