\(\int \sqrt {a+a \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\) [1313]
Optimal result
Integrand size = 45, antiderivative size = 140 \[
\int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {2 \sqrt {a} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {2 a (A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}
\]
[Out]
2/3*A*sec(d*x+c)^(3/2)*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d+2*C*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2
))*a^(1/2)*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+2/3*a*(A+3*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(1
/2)
Rubi [A] (verified)
Time = 0.54 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of
steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4306, 3122, 3059, 2853, 222}
\[
\int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {2 a (A+3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}+\frac {2 \sqrt {a} C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}
\]
[In]
Int[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2),x]
[Out]
(2*Sqrt[a]*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/d
+ (2*a*(A + 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (2*A*Sqrt[a + a*Cos[c + d*x
]]*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d)
Rule 222
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]
Rule 2853
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]
Rule 3059
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n +
1)*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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] && LtQ[n, -1]
Rule 3122
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[(-(c^2*C - B*c*d + A*d^2))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[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
- B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(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, B, 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])
Rule 4306
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u,
x]
Rubi steps \begin{align*}
\text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 A \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)} \left (\frac {1}{2} a (A+3 B)+\frac {3}{2} a C \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{3 a} \\ & = \frac {2 a (A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\left (C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 a (A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}-\frac {\left (2 C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a}}} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{d} \\ & = \frac {2 \sqrt {a} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {2 a (A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.75
\[
\int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {\sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \left (3 \sqrt {2} C \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {3}{2}}(c+d x)+2 (A+(2 A+3 B) \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{3 d}
\]
[In]
Integrate[Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(5/2),x]
[Out]
(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sec[c + d*x]^(3/2)*(3*Sqrt[2]*C*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]*
Cos[c + d*x]^(3/2) + 2*(A + (2*A + 3*B)*Cos[c + d*x])*Sin[(c + d*x)/2]))/(3*d)
Maple [A] (verified)
Time = 1.85 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.29
| | |
| method | result | size |
| | |
| parts |
\(-\frac {2 A \left (\sec ^{\frac {5}{2}}\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (2 \left (\cos ^{2}\left (d x +c \right )\right )-\cos \left (d x +c \right )-1\right ) \cot \left (d x +c \right )}{3 d}-\frac {2 B \left (\sec ^{\frac {5}{2}}\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\cos \left (d x +c \right )-1\right ) \cos \left (d x +c \right ) \cot \left (d x +c \right )}{d}+\frac {2 C \left (\sec ^{\frac {5}{2}}\left (d x +c \right )\right ) \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )}{d}\) |
\(180\) |
| default |
\(\frac {2 \left (\sec ^{\frac {5}{2}}\left (d x +c \right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (3 C \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+3 C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+2 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 B \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+A \cos \left (d x +c \right ) \sin \left (d x +c \right )\right )}{3 d \left (1+\cos \left (d x +c \right )\right )}\) |
\(199\) |
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[In]
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)*(a+a*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
[Out]
-2/3*A/d*sec(d*x+c)^(5/2)*((1+cos(d*x+c))*a)^(1/2)*(2*cos(d*x+c)^2-cos(d*x+c)-1)*cot(d*x+c)-2*B/d*sec(d*x+c)^(
5/2)*((1+cos(d*x+c))*a)^(1/2)*(cos(d*x+c)-1)*cos(d*x+c)*cot(d*x+c)+2*C/d*sec(d*x+c)^(5/2)*arctan((cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)*tan(d*x+c))*((1+cos(d*x+c))*a)^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2
Fricas [A] (verification not implemented)
none
Time = 0.27 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.89
\[
\int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=-\frac {2 \, {\left (3 \, {\left (C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left ({\left (2 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{3 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}
\]
[In]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
-2/3*(3*(C*cos(d*x + c)^2 + C*cos(d*x + c))*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a
)*sin(d*x + c))) - ((2*A + 3*B)*cos(d*x + c) + A)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d
*cos(d*x + c)^2 + d*cos(d*x + c))
Sympy [F(-1)]
Timed out. \[
\int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\text {Timed out}
\]
[In]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(5/2)*(a+a*cos(d*x+c))**(1/2),x)
[Out]
Timed out
Maxima [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 1547 vs. \(2 (118) = 236\).
Time = 0.70 (sec) , antiderivative size = 1547, normalized size of antiderivative = 11.05
\[
\int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\text {Too large to display}
\]
[In]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
1/6*(3*(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*sqrt(a)*sin(1/2*arctan2(sin
(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(
1/4)*(cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(2*d*x + 2*c) - (cos(2*d*x + 2*c) + 1)*sin(3
/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) + ((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*c
os(2*d*x + 2*c) + 1)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2
*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - cos(1
/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))), (co
s(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos
(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) + 1) - (cos(2*d*x + 2*c)^2 + sin(2*d
*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) +
1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2
*c) + 1)) - cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d
*x + 2*c)))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*
d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(
2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) - 1) - (cos(2*d*x +
2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*co
s(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(
2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) +
(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x +
2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x +
2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
+ 1)) - 1))*sqrt(a))*C/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) + 4*A*(3*sqrt(2)*sq
rt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 4*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sqrt(2)*sqrt(a
)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((sin(d*x + c)/(cos(d*x + c
) + 1) + 1)^(5/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*
x + c)^4/(cos(d*x + c) + 1)^4 + 1)) + 12*B*(sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 2*sqrt(2)*sqrt(a
)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)*(sin(d*x + c)^2/(
cos(d*x + c) + 1)^2 + 1)^2/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)
^(5/2)*(2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)))/d
Giac [F(-1)]
Timed out. \[
\int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\text {Timed out}
\]
[In]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(5/2)*(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")
[Out]
Timed out
Mupad [F(-1)]
Timed out. \[
\int \sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x
\]
[In]
int((1/cos(c + d*x))^(5/2)*(a + a*cos(c + d*x))^(1/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)
[Out]
int((1/cos(c + d*x))^(5/2)*(a + a*cos(c + d*x))^(1/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2), x)