\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx\) [117]

Optimal result

Integrand size = 35, antiderivative size = 217 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {(19 A+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^{3/2} d}-\frac {(13 A+5 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(7 A+2 C) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A+C) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}} \]

[Out]

1/4*(19*A+8*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d-1/4*(13*A+5*C)*arctanh(1/2*sin(d*x
+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/2*(A+C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c
))^(3/2)-1/4*(7*A+2*C)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)+1/2*(2*A+C)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(d*
x+c))^(1/2)

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3121, 3063, 3064, 2728, 212, 2852} \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {(19 A+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac {(13 A+5 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(7 A+2 C) \tan (c+d x)}{4 a d \sqrt {a \cos (c+d x)+a}}+\frac {(2 A+C) \tan (c+d x) \sec (c+d x)}{2 a d \sqrt {a \cos (c+d x)+a}}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}} \]

[In]

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

[Out]

((19*A + 8*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(4*a^(3/2)*d) - ((13*A + 5*C)*ArcTanh[
(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((7*A + 2*C)*Tan[c + d*x])
/(4*a*d*Sqrt[a + a*Cos[c + d*x]]) - ((A + C)*Sec[c + d*x]*Tan[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((2
*A + C)*Sec[c + d*x]*Tan[c + d*x])/(2*a*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 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 3121

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(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)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(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, 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+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\left (2 a (2 A+C)-\frac {1}{2} a (5 A+C) \cos (c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {(A+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A+C) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (-a^2 (7 A+2 C)+3 a^2 (2 A+C) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a^3} \\ & = -\frac {(7 A+2 C) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A+C) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (\frac {1}{2} a^3 (19 A+8 C)-\frac {1}{2} a^3 (7 A+2 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a^4} \\ & = -\frac {(7 A+2 C) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A+C) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}-\frac {(13 A+5 C) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a}+\frac {(19 A+8 C) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx}{8 a^2} \\ & = -\frac {(7 A+2 C) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A+C) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {(13 A+5 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 )}{2 a d}-\frac {(19 A+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 d} \\ & = \frac {(19 A+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{4 a^{3/2} d}-\frac {(13 A+5 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(7 A+2 C) \tan (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)}}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(2 A+C) \sec (c+d x) \tan (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.22 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.92 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\cos ^3\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (4 (13 A+5 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )\right )^2-2 \sqrt {2} (19 A+8 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )\right )^2+4 (3 A+2 C+6 A \cos (c+d x)+(7 A+2 C) \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{16 d (a (1+\cos (c+d x)))^{3/2} \left (-1+\sin ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]^3*Sec[c + d*x]^2*(4*(13*A + 5*C)*ArcTanh[Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Cos[(3*(c + d
*x))/2])^2 - 2*Sqrt[2]*(19*A + 8*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2]
)^2 + 4*(3*A + 2*C + 6*A*Cos[c + d*x] + (7*A + 2*C)*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(16*d*(a*(1 + Cos[c +
d*x]))^(3/2)*(-1 + Sin[(c + d*x)/2]^2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1099\) vs. \(2(186)=372\).

Time = 10.78 (sec) , antiderivative size = 1100, normalized size of antiderivative = 5.07

method	result	size
parts	\(\text {Expression too large to display}\)	\(1100\)
default	\(\text {Expression too large to display}\)	\(1540\)

[In]

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

[Out]

-1/2*A*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(104*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-76*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-76*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-104
*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+28*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+76*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+76*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+26*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)^2-22*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)-19*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)^2*a-19*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)^2*a+2*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(5/2
)/cos(1/2*d*x+1/2*c)/(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*c
os(1/2*d*x+1/2*c)^2)^(1/2)/d+1/4*C/a^(5/2)/cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*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)^2*a+2*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)^2*a-5*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)^2*a-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 = 0.33 (sec) , antiderivative size = 356, normalized size of antiderivative = 1.64 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {2 \, \sqrt {2} {\left ({\left (13 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (13 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + {\left ({\left (19 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (19 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (19 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \, {\left ({\left (7 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) - 2 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{16 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

1/16*(2*sqrt(2)*((13*A + 5*C)*cos(d*x + c)^4 + 2*(13*A + 5*C)*cos(d*x + c)^3 + (13*A + 5*C)*cos(d*x + c)^2)*sq
rt(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)) + ((19*A + 8*C)*cos(d*x + c)^4 + 2*(19*A + 8*C)*cos(d*x + c)^3 + (19
*A + 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*((7*A + 2*C)*cos(d*x + c)^2
+ 3*A*cos(d*x + c) - 2*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^
3 + a^2*d*cos(d*x + c)^2)

Sympy [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)**3/(a*(cos(c + d*x) + 1))**(3/2), x)

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Memory limit reached. Please jump to an outer pointer, quit progra
m and enlarge thememory limits before executing the program again.

Giac [A] (verification not implemented)

none

Time = 0.63 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.41 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {\frac {\sqrt {2} {\left (13 \, A \sqrt {a} + 5 \, C \sqrt {a}\right )} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (13 \, A \sqrt {a} + 5 \, C \sqrt {a}\right )} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {{\left (19 \, A + 8 \, C\right )} \log \left ({\left | \frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {{\left (19 \, A + 8 \, C\right )} \log \left ({\left | -\frac {1}{2} \, \sqrt {2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} {\left (A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{2} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \sqrt {2} {\left (10 \, A \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, A \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{8 \, d} \]

[In]

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

[Out]

-1/8*(sqrt(2)*(13*A*sqrt(a) + 5*C*sqrt(a))*log(sin(1/2*d*x + 1/2*c) + 1)/(a^2*sgn(cos(1/2*d*x + 1/2*c))) - sqr
t(2)*(13*A*sqrt(a) + 5*C*sqrt(a))*log(-sin(1/2*d*x + 1/2*c) + 1)/(a^2*sgn(cos(1/2*d*x + 1/2*c))) - (19*A + 8*C
)*log(abs(1/2*sqrt(2) + sin(1/2*d*x + 1/2*c)))/(a^(3/2)*sgn(cos(1/2*d*x + 1/2*c))) + (19*A + 8*C)*log(abs(-1/2
*sqrt(2) + sin(1/2*d*x + 1/2*c)))/(a^(3/2)*sgn(cos(1/2*d*x + 1/2*c))) - 2*sqrt(2)*(A*sqrt(a)*sin(1/2*d*x + 1/2
*c) + C*sqrt(a)*sin(1/2*d*x + 1/2*c))/((sin(1/2*d*x + 1/2*c)^2 - 1)*a^2*sgn(cos(1/2*d*x + 1/2*c))) - 2*sqrt(2)
*(10*A*sin(1/2*d*x + 1/2*c)^3 - 3*A*sin(1/2*d*x + 1/2*c))/((2*sin(1/2*d*x + 1/2*c)^2 - 1)^2*a^(3/2)*sgn(cos(1/
2*d*x + 1/2*c))))/d

Mupad [F(-1)]

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

[In]

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

[Out]

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