\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\) [423]

Optimal result

Integrand size = 43, antiderivative size = 217 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {(5 A-2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac {(115 A-43 B+3 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B+3 C) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \]

[Out]

-(5*A-2*B)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d+1/32*(115*A-43*B+3*C)*arctanh(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)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(5/
2)-1/16*(15*A-7*B-C)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(3/2)+1/16*(35*A-11*B+3*C)*tan(d*x+c)/a^2/d/(a+a*cos(d*x+
c))^(1/2)

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {3120, 3057, 3063, 3064, 2728, 212, 2852} \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(115 A-43 B+3 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(5 A-2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(35 A-11 B+3 C) \tan (c+d x)}{16 a^2 d \sqrt {a \cos (c+d x)+a}}-\frac {(15 A-7 B-C) \tan (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac {(A-B+C) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]

[In]

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

[Out]

-(((5*A - 2*B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(5/2)*d)) + ((115*A - 43*B + 3*C)*
ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Tan[
c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) - ((15*A - 7*B - C)*Tan[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2
)) + ((35*A - 11*B + 3*C)*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) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {\int \frac {\left (a (5 A-B+C)-\frac {1}{2} a (5 A-5 B-3 C) \cos (c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = -\frac {(A-B+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\left (\frac {1}{2} a^2 (35 A-11 B+3 C)-\frac {3}{4} a^2 (15 A-7 B-C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {(A-B+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B+3 C) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (-4 a^3 (5 A-2 B)+\frac {1}{4} a^3 (35 A-11 B+3 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{8 a^5} \\ & = -\frac {(A-B+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B+3 C) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}-\frac {(5 A-2 B) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx}{2 a^3}+\frac {(115 A-43 B+3 C) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2} \\ & = -\frac {(A-B+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B+3 C) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}}+\frac {(5 A-2 B) \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 )}{a^2 d}-\frac {(115 A-43 B+3 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 {(5 A-2 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac {(115 A-43 B+3 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 {(A-B+C) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {(15 A-7 B-C) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(35 A-11 B+3 C) \tan (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.00 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.87 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (4 (115 A-43 B+3 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \cos (c+d x)-64 \sqrt {2} (5 A-2 B) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \cos (c+d x)+(67 A-11 B+3 C+2 (55 A-15 B+7 C) \cos (c+d x)+(35 A-11 B+3 C) \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{32 a d (a (1+\cos (c+d x)))^{3/2}} \]

[In]

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

[Out]

(Sec[(c + d*x)/2]*Sec[c + d*x]*(4*(115*A - 43*B + 3*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^4*Cos[c + d*
x] - 64*Sqrt[2]*(5*A - 2*B)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^4*Cos[c + d*x] + (67*A - 11*B +
3*C + 2*(55*A - 15*B + 7*C)*Cos[c + d*x] + (35*A - 11*B + 3*C)*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(32*a*d*(
a*(1 + Cos[c + d*x]))^(3/2))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1089\) vs. \(2(188)=376\).

Time = 16.28 (sec) , antiderivative size = 1090, normalized size of antiderivative = 5.02

method	result	size
parts	\(\text {Expression too large to display}\)	\(1090\)
default	\(\text {Expression too large to display}\)	\(1292\)

[In]

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

[Out]

1/16*A*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(230*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-160*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-160*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-1
15*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+70*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+80*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+80*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-15*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*cos(1/2*
d*x+1/2*c)-2^(1/2))/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d+1/32*B/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*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+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+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)*(3*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+3*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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 421 vs. \(2 (188) = 376\).

Time = 0.34 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.94 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (115 \, A - 43 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (115 \, A - 43 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (115 \, A - 43 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (115 \, A - 43 \, B + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 16 \, {\left ({\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left ({\left (35 \, A - 11 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (55 \, A - 15 \, B + 7 \, C\right )} \cos \left (d x + c\right ) + 16 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]

[In]

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

[Out]

1/64*(sqrt(2)*((115*A - 43*B + 3*C)*cos(d*x + c)^4 + 3*(115*A - 43*B + 3*C)*cos(d*x + c)^3 + 3*(115*A - 43*B +
3*C)*cos(d*x + c)^2 + (115*A - 43*B + 3*C)*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*co
s(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)) - 16*((5
*A - 2*B)*cos(d*x + c)^4 + 3*(5*A - 2*B)*cos(d*x + c)^3 + 3*(5*A - 2*B)*cos(d*x + c)^2 + (5*A - 2*B)*cos(d*x +
c))*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*((35*A - 11*B + 3*C)*cos(d*x + c)^2 + (55*A - 15*
B + 7*C)*cos(d*x + c) + 16*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x +
c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2/(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (188) = 376\).

Time = 1.79 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.77 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\frac {\sqrt {2} {\left (115 \, A \sqrt {a} - 43 \, B \sqrt {a} + 3 \, 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 (115 \, A \sqrt {a} - 43 \, B \sqrt {a} + 3 \, 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 {32 \, {\left (5 \, A - 2 \, B\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 {32 \, {\left (5 \, A - 2 \, B\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 {64 \, \sqrt {2} A \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, {\left (19 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 11 \, \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, \sqrt {2} C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 13 \, \sqrt {2} B \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, \sqrt {2} C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\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)^2/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/64*(sqrt(2)*(115*A*sqrt(a) - 43*B*sqrt(a) + 3*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)*(115*A*sqrt(a) - 43*B*sqrt(a) + 3*C*sqrt(a))*log(-sin(1/2*d*x + 1/2*c) + 1)/(a^3*sgn(cos(
1/2*d*x + 1/2*c))) - 32*(5*A - 2*B)*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))) + 32*(5*A - 2*B)*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))) - 64
*sqrt(2)*A*sin(1/2*d*x + 1/2*c)/((2*sin(1/2*d*x + 1/2*c)^2 - 1)*a^(5/2)*sgn(cos(1/2*d*x + 1/2*c))) - 2*(19*sqr
t(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 - 11*sqrt(2)*B*sqrt(a)*sin(1/2*d*x + 1/2*c)^3 + 3*sqrt(2)*C*sqrt(a)*sin(
1/2*d*x + 1/2*c)^3 - 21*sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + 13*sqrt(2)*B*sqrt(a)*sin(1/2*d*x + 1/2*c) - 5
*sqrt(2)*C*sqrt(a)*sin(1/2*d*x + 1/2*c))/((sin(1/2*d*x + 1/2*c)^2 - 1)^2*a^3*sgn(cos(1/2*d*x + 1/2*c))))/d

Mupad [F(-1)]

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

[In]

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