\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx\) [109]

Optimal result

Integrand size = 35, antiderivative size = 200 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {(9 A+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 \sqrt {a} d}+\frac {\sqrt {2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {(7 A+8 C) \tan (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}-\frac {A \sec (c+d x) \tan (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}} \]

[Out]

-1/8*(9*A+8*C)*arctanh(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/d/a^(1/2)+(A+C)*arctanh(1/2*sin(d*x+c)*a^(1/
2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+1/8*(7*A+8*C)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-1/12*A*
sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+1/3*A*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 200, 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 = {3123, 3063, 3064, 2728, 212, 2852} \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {(9 A+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{8 \sqrt {a} d}+\frac {\sqrt {2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {(7 A+8 C) \tan (c+d x)}{8 d \sqrt {a \cos (c+d x)+a}}+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}-\frac {A \tan (c+d x) \sec (c+d x)}{12 d \sqrt {a \cos (c+d x)+a}} \]

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[a + a*Cos[c + d*x]],x]

[Out]

-1/8*((9*A + 8*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(Sqrt[a]*d) + (Sqrt[2]*(A + C)*Arc
Tanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d) + ((7*A + 8*C)*Tan[c + d*x])/(8*d
*Sqrt[a + a*Cos[c + d*x]]) - (A*Sec[c + d*x]*Tan[c + d*x])/(12*d*Sqrt[a + a*Cos[c + d*x]]) + (A*Sec[c + d*x]^2
*Tan[c + d*x])/(3*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 3123

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

Rubi steps \begin{align*} \text {integral}& = \frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (-\frac {a A}{2}+\frac {1}{2} a (5 A+6 C) \cos (c+d x)\right ) \sec ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{3 a} \\ & = -\frac {A \sec (c+d x) \tan (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (\frac {3}{4} a^2 (7 A+8 C)-\frac {3}{4} a^2 A \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{6 a^2} \\ & = \frac {(7 A+8 C) \tan (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}-\frac {A \sec (c+d x) \tan (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\left (-\frac {3}{8} a^3 (9 A+8 C)+\frac {3}{8} a^3 (7 A+8 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx}{6 a^3} \\ & = \frac {(7 A+8 C) \tan (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}-\frac {A \sec (c+d x) \tan (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+(A+C) \int \frac {1}{\sqrt {a+a \cos (c+d x)}} \, dx-\frac {(9 A+8 C) \int \sqrt {a+a \cos (c+d x)} \sec (c+d x) \, dx}{16 a} \\ & = \frac {(7 A+8 C) \tan (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}-\frac {A \sec (c+d x) \tan (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}-\frac {(2 (A+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 )}{d}+\frac {(9 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 )}{8 d} \\ & = -\frac {(9 A+8 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{8 \sqrt {a} d}+\frac {\sqrt {2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {(7 A+8 C) \tan (c+d x)}{8 d \sqrt {a+a \cos (c+d x)}}-\frac {A \sec (c+d x) \tan (c+d x)}{12 d \sqrt {a+a \cos (c+d x)}}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.66 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (48 (A+C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-3 \sqrt {2} (9 A+8 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+(37 A+24 C-4 A \cos (c+d x)+3 (7 A+8 C) \cos (2 (c+d x))) \sec ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{24 d \sqrt {a (1+\cos (c+d x))}} \]

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/Sqrt[a + a*Cos[c + d*x]],x]

[Out]

(Cos[(c + d*x)/2]*(48*(A + C)*ArcTanh[Sin[(c + d*x)/2]] - 3*Sqrt[2]*(9*A + 8*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/
2]] + (37*A + 24*C - 4*A*Cos[c + d*x] + 3*(7*A + 8*C)*Cos[2*(c + d*x)])*Sec[c + d*x]^3*Sin[(c + d*x)/2]))/(24*
d*Sqrt[a*(1 + Cos[c + d*x])])

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1357\) vs. \(2(171)=342\).

Time = 10.64 (sec) , antiderivative size = 1358, normalized size of antiderivative = 6.79

method	result	size
parts	\(\text {Expression too large to display}\)	\(1358\)
default	\(\text {Expression too large to display}\)	\(1661\)

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+cos(d*x+c)*a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/6*A*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*a*(-16*2^(1/2)*ln(4*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)
^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))+9*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))+9*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)))*sin(1/2*d*x+1/2*c)^6+(576*2^(1/2)*ln(4*(a^(1/2)*(a*si
n(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*a-324*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))*a-324*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))*a+168*2^(1/2)*(a*sin(1/2*d*x+1
/2*c)^2)^(1/2)*a^(1/2))*sin(1/2*d*x+1/2*c)^4+(-288*2^(1/2)*ln(4*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos
(1/2*d*x+1/2*c))*a+162*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))*a+162*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))*a-160*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))*sin(1/2*
d*x+1/2*c)^2+48*2^(1/2)*ln(4*(a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*a-27*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))*a-27
*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))*a+54*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(3/2)/(2*cos(1/2*d*x+1/2*c)+2^(1/2))^3/(2*co
s(1/2*d*x+1/2*c)-2^(1/2))^3/sin(1/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d-1/2*C*cos(1/2*d*x+1/2*c)*(a*si
n(1/2*d*x+1/2*c)^2)^(1/2)*(-2*a*(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))+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))-4*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)))*sin(1/2*d*x+1/2*c)^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))*a+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))*a-4*a^(1/2)*(a*sin(1/2*
d*x+1/2*c)^2)^(1/2)-4*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))*a)/a^(3/2)/(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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.48 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {3 \, {\left ({\left (9 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (9 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \, {\left (3 \, {\left (7 \, A + 8 \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) + 8 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right ) + \frac {48 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{4} + {\left (A + C\right )} a \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{96 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \]

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/96*(3*((9*A + 8*C)*cos(d*x + c)^4 + (9*A + 8*C)*cos(d*x + c)^3)*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*(3*(7*A + 8*C)*cos(d*x + c)^2 - 2*A*cos(d*x + c) + 8*A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c) + 48
*sqrt(2)*((A + C)*a*cos(d*x + c)^4 + (A + C)*a*cos(d*x + c)^3)*log(-(cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x
+ c) + a)*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos
(d*x + c)^4 + a*d*cos(d*x + c)^3)

Sympy [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}\, dx \]

[In]

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+a*cos(d*x+c))**(1/2),x)

[Out]

Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)**4/sqrt(a*(cos(c + d*x) + 1)), x)

Maxima [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^4\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \]

[In]

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a*cos(c + d*x))^(1/2)),x)

[Out]

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a*cos(c + d*x))^(1/2)), x)