\(\int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [1245]

Optimal result

Integrand size = 37, antiderivative size = 285 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(8 A+19 C) \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 a^{3/2} d}-\frac {(5 A+13 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)}+\frac {(A+2 C) \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}-\frac {(2 A+7 C) \sin (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {4306, 3121, 3062, 3061, 2861, 211, 2853, 222} \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(8 A+19 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac {(5 A+13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(A+2 C) \sin (c+d x)}{2 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^{3/2}}-\frac {(2 A+7 C) \sin (c+d x)}{4 a d \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+a}} \]

[In]

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

[Out]

((8*A + 19*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]])/(
4*a^(3/2)*d) - ((5*A + 13*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]])/(2*Sqrt[2]*a^(3/2)*d) - ((A + C)*Sin[c + d*x])/(2*d*(a + a*Cos[c +
d*x])^(3/2)*Sec[c + d*x]^(5/2)) + ((A + 2*C)*Sin[c + d*x])/(2*a*d*Sqrt[a + a*Cos[c + d*x]]*Sec[c + d*x]^(3/2))
- ((2*A + 7*C)*Sin[c + d*x])/(4*a*d*Sqrt[a + a*Cos[c + d*x]]*Sqrt[Sec[c + d*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 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 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 3061

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[B/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[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 3062

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)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*
(m + n + 1))), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*b*c
*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*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] && GtQ[n, 0] &&
(IntegerQ[n] || EqQ[m + 1/2, 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)]

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 {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx \\ & = -\frac {(A+C) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (-\frac {1}{2} a (A+5 C)+2 a (A+2 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {(A+C) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)}+\frac {(A+2 C) \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\cos (c+d x)} \left (3 a^2 (A+2 C)-a^2 (2 A+7 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{4 a^3} \\ & = -\frac {(A+C) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)}+\frac {(A+2 C) \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}-\frac {(2 A+7 C) \sin (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {1}{2} a^3 (2 A+7 C)+\frac {1}{2} a^3 (8 A+19 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{4 a^4} \\ & = -\frac {(A+C) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)}+\frac {(A+2 C) \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}-\frac {(2 A+7 C) \sin (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}-\frac {\left ((5 A+13 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}{4 a}+\frac {\left ((8 A+19 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}{8 a^2} \\ & = -\frac {(A+C) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)}+\frac {(A+2 C) \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}-\frac {(2 A+7 C) \sin (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {\left ((5 A+13 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 )}{2 d}-\frac {\left ((8 A+19 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 )}{4 a^2 d} \\ & = \frac {(8 A+19 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)}}{4 a^{3/2} d}-\frac {(5 A+13 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)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)}+\frac {(A+2 C) \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}-\frac {(2 A+7 C) \sin (c+d x)}{4 a d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.10 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.35 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {i e^{\frac {1}{2} i (c+d x)} \left (\frac {\sqrt {2} e^{-2 i (c+d x)} \left (-1+e^{i (c+d x)}-e^{2 i (c+d x)}+e^{3 i (c+d x)}\right ) \left (-4 A e^{2 i (c+d x)}+C \left (1-3 e^{i (c+d x)}-12 e^{2 i (c+d x)}-3 e^{3 i (c+d x)}+e^{4 i (c+d x)}\right )\right )}{\sqrt {1+e^{2 i (c+d x)}}}+\sqrt {2} (8 A+19 C) \left (1+e^{i (c+d x)}\right )^2 \text {arcsinh}\left (e^{i (c+d x)}\right )+4 (5 A+13 C) \left (1+e^{i (c+d x)}\right )^2 \text {arctanh}\left (\frac {1-e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )-\sqrt {2} (8 A+19 C) \left (1+e^{i (c+d x)}\right )^2 \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \cos \left (\frac {1}{2} (c+d x)\right )}{16 d \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{3/2} \left (1+e^{2 i (c+d x)}\right )^{3/2} (a (1+\cos (c+d x)))^{3/2}} \]

[In]

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

[Out]

((-1/16*I)*E^((I/2)*(c + d*x))*((Sqrt[2]*(-1 + E^(I*(c + d*x)) - E^((2*I)*(c + d*x)) + E^((3*I)*(c + d*x)))*(-
4*A*E^((2*I)*(c + d*x)) + C*(1 - 3*E^(I*(c + d*x)) - 12*E^((2*I)*(c + d*x)) - 3*E^((3*I)*(c + d*x)) + E^((4*I)
*(c + d*x)))))/(E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]) + Sqrt[2]*(8*A + 19*C)*(1 + E^(I*(c + d*x))
)^2*ArcSinh[E^(I*(c + d*x))] + 4*(5*A + 13*C)*(1 + E^(I*(c + d*x)))^2*ArcTanh[(1 - E^(I*(c + d*x)))/(Sqrt[2]*S
qrt[1 + E^((2*I)*(c + d*x))])] - Sqrt[2]*(8*A + 19*C)*(1 + E^(I*(c + d*x)))^2*ArcTanh[Sqrt[1 + E^((2*I)*(c + d
*x))]])*Cos[(c + d*x)/2])/(d*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(3/2)*(1 + E^((2*I)*(c + d*x)))^(3/2)
*(a*(1 + Cos[c + d*x]))^(3/2))

Maple [A] (verified)

Time = 3.06 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.49

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (2 C \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )-3 C \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+8 A \sqrt {2}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )-2 A \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )+19 C \sqrt {2}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )-7 C \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )+10 A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+8 A \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \sqrt {2}\, \sec \left (d x +c \right )+26 C \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+19 C \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \sqrt {2}\, \sec \left (d x +c \right )+10 A \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sec \left (d x +c \right )+26 C \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sec \left (d x +c \right )\right )}{8 a^{2} d \left (1+\cos \left (d x +c \right )\right )^{2} \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(424\)
parts	\(\frac {A \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (4 \sqrt {2}\, \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 {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )+4 \sqrt {2}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \sec \left (d x +c \right )+5 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+5 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sec \left (d x +c \right )\right )}{4 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2} \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}+\frac {C \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (2 \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 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+19 \sqrt {2}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )-7 \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )+19 \sqrt {2}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) \sec \left (d x +c \right )+26 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+26 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sec \left (d x +c \right )\right )}{8 d \,a^{2} \left (1+\cos \left (d x +c \right )\right )^{2} \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(476\)

[In]

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

[Out]

1/8/a^2/d*2^(1/2)*((1+cos(d*x+c))*a)^(1/2)/(1+cos(d*x+c))^2/sec(d*x+c)^(3/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
*(2*C*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)-3*C*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*sin(d*x+c)+8*A*2^(1/2)*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))-2*A*2^(1/2)*(cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*tan(d*x+c)+19*C*2^(1/2)*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))-7*C*2^(1/2)*(co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c)+10*A*arcsin(cot(d*x+c)-csc(d*x+c))+8*A*arctan((cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)*tan(d*x+c))*2^(1/2)*sec(d*x+c)+26*C*arcsin(cot(d*x+c)-csc(d*x+c))+19*C*arctan((cos(d*x+c)/(1+cos(d
*x+c)))^(1/2)*tan(d*x+c))*2^(1/2)*sec(d*x+c)+10*A*arcsin(cot(d*x+c)-csc(d*x+c))*sec(d*x+c)+26*C*arcsin(cot(d*x
+c)-csc(d*x+c))*sec(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 5.74 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.91 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2} \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {2} {\left ({\left (5 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 13 \, C\right )} \cos \left (d x + c\right ) + 5 \, A + 13 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left ({\left (8 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, A + 19 \, C\right )} \cos \left (d x + c\right ) + 8 \, A + 19 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {{\left (2 \, C \cos \left (d x + c\right )^{3} - 3 \, C \cos \left (d x + c\right )^{2} - {\left (2 \, A + 7 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

1/4*(sqrt(2)*((5*A + 13*C)*cos(d*x + c)^2 + 2*(5*A + 13*C)*cos(d*x + c) + 5*A + 13*C)*sqrt(a)*arctan(sqrt(2)*s
qrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - ((8*A + 19*C)*cos(d*x + c)^2 + 2*(8*A + 1
9*C)*cos(d*x + c) + 8*A + 19*C)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x +
c))) + (2*C*cos(d*x + c)^3 - 3*C*cos(d*x + c)^2 - (2*A + 7*C)*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x +
c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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