\(\int \frac {(A+C \cos ^2(c+d x)) \sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx\) [1249]

Optimal result

Integrand size = 37, antiderivative size = 174 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(19 A+3 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {(9 A-7 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \]

[Out]

-1/4*(A+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)/sec(d*x+c)^(1/2)-1/16*(9*A-7*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))
^(3/2)/sec(d*x+c)^(1/2)+1/32*(19*A+3*C)*arctan(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c)
)^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(5/2)/d*2^(1/2)

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4306, 3121, 3057, 12, 2861, 211} \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {(19 A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(9 A-7 C) \sin (c+d x)}{16 a d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{3/2}}-\frac {(A+C) \sin (c+d x)}{4 d \sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^{5/2}} \]

[In]

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

[Out]

((19*A + 3*C)*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])]*Sqrt[Cos[c
+ d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A + C)*Sin[c + d*x])/(4*d*(a + a*Cos[c + d*x])^(5/2)*Sq
rt[Sec[c + d*x]]) - ((9*A - 7*C)*Sin[c + d*x])/(16*a*d*(a + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2861

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[-2*(a/f), Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c +
d*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 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)]

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 {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{5/2}} \, dx \\ & = -\frac {(A+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} a (7 A-C)-a (A-3 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = -\frac {(A+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {(9 A-7 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {a^2 (19 A+3 C)}{4 \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {(A+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {(9 A-7 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}+\frac {\left ((19 A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{32 a^2} \\ & = -\frac {(A+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {(9 A-7 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}}-\frac {\left ((19 A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{2 a^2+a x^2} \, dx,x,-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 a d} \\ & = \frac {(19 A+3 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {(9 A-7 C) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(412\) vs. \(2(174)=348\).

Time = 2.53 (sec) , antiderivative size = 412, normalized size of antiderivative = 2.37 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {\cos \left (\frac {1}{2} (c+d x)\right ) (1+\cos (c+d x))^{3/2} \sqrt {\sec (c+d x)} \left (-3 C \arcsin \left (\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \sqrt {-1+\cos (c+d x)} \sqrt {1+\cos (c+d x)}-38 A \text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )} \sin \left (\frac {1}{2} (c+d x)\right )+11 A \sqrt {1+\cos (c+d x)} \sec ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\sec (c+d x)} \tan \left (\frac {1}{2} (c+d x)\right )-31 A \sqrt {1+\cos (c+d x)} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\sec (c+d x)} \tan ^3\left (\frac {1}{2} (c+d x)\right )+18 A \sqrt {1+\cos (c+d x)} \sqrt {1-\sec (c+d x)} \sin \left (\frac {1}{2} (c+d x)\right ) \tan ^4\left (\frac {1}{2} (c+d x)\right )-10 C \sqrt {\cos (c+d x)} \sin \left (\frac {1}{2} (c+d x)\right ) \sqrt {-\tan ^2\left (\frac {1}{2} (c+d x)\right )}-4 C \sqrt {\cos (c+d x)} \sin \left (\frac {1}{2} (c+d x)\right ) \left (-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2}\right )}{16 d (a (1+\cos (c+d x)))^{5/2} \sqrt {1-\sec (c+d x)}} \]

[In]

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

[Out]

-1/16*(Cos[(c + d*x)/2]*(1 + Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d*x]]*(-3*C*ArcSin[Sin[(c + d*x)/2]/Sqrt[Cos[(c
+ d*x)/2]^2]]*Sqrt[-1 + Cos[c + d*x]]*Sqrt[1 + Cos[c + d*x]] - 38*A*ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/
2]^2)]]*Sqrt[Cos[(c + d*x)/2]^2]*Sin[(c + d*x)/2] + 11*A*Sqrt[1 + Cos[c + d*x]]*Sec[(c + d*x)/2]^3*Sqrt[1 - Se
c[c + d*x]]*Tan[(c + d*x)/2] - 31*A*Sqrt[1 + Cos[c + d*x]]*Sec[(c + d*x)/2]*Sqrt[1 - Sec[c + d*x]]*Tan[(c + d*
x)/2]^3 + 18*A*Sqrt[1 + Cos[c + d*x]]*Sqrt[1 - Sec[c + d*x]]*Sin[(c + d*x)/2]*Tan[(c + d*x)/2]^4 - 10*C*Sqrt[C
os[c + d*x]]*Sin[(c + d*x)/2]*Sqrt[-Tan[(c + d*x)/2]^2] - 4*C*Sqrt[Cos[c + d*x]]*Sin[(c + d*x)/2]*(-Tan[(c + d
*x)/2]^2)^(3/2)))/(d*(a*(1 + Cos[c + d*x]))^(5/2)*Sqrt[1 - Sec[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(366\) vs. \(2(145)=290\).

Time = 1.38 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.11

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}{\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )-1}}\, \left (\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )-1\right ) \sqrt {\frac {a}{\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}}\, \left (-2 A \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \left (\csc ^{3}\left (d x +c \right )\right )-2 C \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \left (\csc ^{3}\left (d x +c \right )\right )-11 A \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+5 C \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-19 A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-3 C \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right )}{32 a^{3} d \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \left (\csc ^{2}\left (d x +c \right )\right )+1}}\)	\(367\)
parts	\(-\frac {A \sqrt {2}\, \left (9 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+13 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+19 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )+38 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+19 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sqrt {\sec }\left (d x +c \right )\right ) \cos \left (d x +c \right )}{32 d \,a^{3} \left (1+\cos \left (d x +c \right )\right )^{3} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}-\frac {C \sqrt {2}\, \left (-7 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \left (\cos ^{2}\left (d x +c \right )\right )-3 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+6 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+3 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sqrt {\sec }\left (d x +c \right )\right ) \cos \left (d x +c \right )}{32 d \,a^{3} \left (1+\cos \left (d x +c \right )\right )^{3} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(406\)

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/32/a^3/d*2^(1/2)*(-((1-cos(d*x+c))^2*csc(d*x+c)^2+1)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*((1-cos(d*x+c
))^2*csc(d*x+c)^2-1)*(a/((1-cos(d*x+c))^2*csc(d*x+c)^2+1))^(1/2)*(-2*A*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2
)*(1-cos(d*x+c))^3*csc(d*x+c)^3-2*C*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(1-cos(d*x+c))^3*csc(d*x+c)^3-11*
A*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))+5*C*(-(1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/
2)*(-cot(d*x+c)+csc(d*x+c))-19*A*arcsin(cot(d*x+c)-csc(d*x+c))-3*C*arcsin(cot(d*x+c)-csc(d*x+c)))/(-(1-cos(d*x
+c))^2*csc(d*x+c)^2+1)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.19 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 19 \, A + 3 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left ({\left (9 \, A - 7 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (13 \, A - 3 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/32*(sqrt(2)*((19*A + 3*C)*cos(d*x + c)^3 + 3*(19*A + 3*C)*cos(d*x + c)^2 + 3*(19*A + 3*C)*cos(d*x + c) + 19
*A + 3*C)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*((9*A
- 7*C)*cos(d*x + c)^2 + (13*A - 3*C)*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(cos(d*x + c)))/
(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

Sympy [F(-1)]

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

[In]

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**(1/2)/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F(-1)]

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

[In]

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(1/2))/(a + a*cos(c + d*x))^(5/2),x)

[Out]

int(((A + C*cos(c + d*x)^2)*(1/cos(c + d*x))^(1/2))/(a + a*cos(c + d*x))^(5/2), x)