\(\int \frac {\sqrt {\cos (c+d x)} (A+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^{3/2}} \, dx\) [206]

Optimal result

Integrand size = 37, antiderivative size = 188 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {3 C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{3/2} d}+\frac {(A+9 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(A+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}} \]

[Out]

-3*C*arcsin(sin(d*x+c)*a^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d-1/2*(A+C)*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*
cos(d*x+c))^(3/2)+1/4*(A+9*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))/a
^(3/2)/d*2^(1/2)+1/2*(A+3*C)*sin(d*x+c)*cos(d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {3121, 3062, 3061, 2861, 211, 2853, 222} \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {(A+9 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {3 C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{a^{3/2} d}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}+\frac {(A+3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 a d \sqrt {a \cos (c+d x)+a}} \]

[In]

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

[Out]

(-3*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(a^(3/2)*d) + ((A + 9*C)*ArcTan[(Sqrt[a]*Sin[c
+ d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A + C)*Cos[c + d*x]^
(3/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + ((A + 3*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*a*d*Sqrt
[a + a*Cos[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)]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {\int \frac {\sqrt {\cos (c+d x)} \left (\frac {1}{2} a (A-3 C)+a (A+3 C) \cos (c+d x)\right )}{\sqrt {a+a \cos (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {(A+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(A+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {\int \frac {\frac {1}{2} a^2 (A+3 C)-3 a^2 C \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{2 a^3} \\ & = -\frac {(A+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(A+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}-\frac {(3 C) \int \frac {\sqrt {a+a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx}{2 a^2}+\frac {(A+9 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \, dx}{4 a} \\ & = -\frac {(A+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(A+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}}+\frac {(3 C) \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 )}{a^2 d}-\frac {(A+9 C) \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 {3 C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{3/2} d}+\frac {(A+9 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 )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(A+3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \cos (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.70 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\left (-\sqrt {2} A \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right )-9 \sqrt {2} C \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right )-\sqrt {2} A \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \cos (c+d x)-9 \sqrt {2} C \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \cos (c+d x)+4 C \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)+2 A \sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}+6 C \sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}+6 C \arcsin \left (\sqrt {1-\cos (c+d x)}\right ) (1+\cos (c+d x))+18 C \arcsin \left (\sqrt {\cos (c+d x)}\right ) (1+\cos (c+d x))\right ) \sin (c+d x)}{4 d \sqrt {1-\cos (c+d x)} (a (1+\cos (c+d x)))^{3/2}} \]

[In]

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

[Out]

((-(Sqrt[2]*A*ArcTan[Sqrt[Cos[c + d*x]]/Sqrt[Sin[(c + d*x)/2]^2]]) - 9*Sqrt[2]*C*ArcTan[Sqrt[Cos[c + d*x]]/Sqr
t[Sin[(c + d*x)/2]^2]] - Sqrt[2]*A*ArcTan[Sqrt[Cos[c + d*x]]/Sqrt[Sin[(c + d*x)/2]^2]]*Cos[c + d*x] - 9*Sqrt[2
]*C*ArcTan[Sqrt[Cos[c + d*x]]/Sqrt[Sin[(c + d*x)/2]^2]]*Cos[c + d*x] + 4*C*Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]
^(3/2) + 2*A*Sqrt[-((-1 + Cos[c + d*x])*Cos[c + d*x])] + 6*C*Sqrt[-((-1 + Cos[c + d*x])*Cos[c + d*x])] + 6*C*A
rcSin[Sqrt[1 - Cos[c + d*x]]]*(1 + Cos[c + d*x]) + 18*C*ArcSin[Sqrt[Cos[c + d*x]]]*(1 + Cos[c + d*x]))*Sin[c +
d*x])/(4*d*Sqrt[1 - Cos[c + d*x]]*(a*(1 + Cos[c + d*x]))^(3/2))

Maple [A] (verified)

Time = 25.00 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.66

method	result	size
default	\(-\frac {\left (A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+9 C \cos \left (d x +c \right ) \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-4 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+A \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )-2 A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+9 C \sqrt {2}\, \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )+12 C \cos \left (d x +c \right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-6 C \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+12 C \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right )}{4 a^{2} d \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(313\)
parts	\(\frac {A \left (\sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{4 d \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{2}}+\frac {C \left (2 \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3 \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-6 \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right ) \cos \left (d x +c \right )-6 \sqrt {2}\, \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )-9 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )-9 \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {2}}{4 d \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{2}}\)	\(375\)

[In]

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

[Out]

-1/4/a^2/d*(A*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)+9*C*cos(d*x+c)*2^(1/2)*arcsin(cot(d*x+c)-csc(d*
x+c))-4*C*cos(d*x+c)*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+A*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))-2*A*
(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+9*C*2^(1/2)*arcsin(cot(d*x+c)-csc(d*x+c))+12*C*cos(d*x+c)*arctan(
tan(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-6*C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+12*C*arctan(tan
(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)))*(a*(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^(1/2)/(1+cos(d*x+c))^2/(cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 2.96 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (A + 9 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (A + 9 \, C\right )} \cos \left (d x + c\right ) + A + 9 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, {\left (2 \, C \cos \left (d x + c\right ) + A + 3 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 12 \, {\left (C \cos \left (d x + c\right )^{2} + 2 \, C \cos \left (d x + c\right ) + C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

integrate((A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/4*(sqrt(2)*((A + 9*C)*cos(d*x + c)^2 + 2*(A + 9*C)*cos(d*x + c) + A + 9*C)*sqrt(a)*arctan(sqrt(2)*sqrt(a*co
s(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - 2*(2*C*cos(d*x + c) + A + 3*C)*sqrt(a*cos(d*x + c
) + a)*sqrt(cos(d*x + c))*sin(d*x + c) - 12*(C*cos(d*x + c)^2 + 2*C*cos(d*x + c) + C)*sqrt(a)*arctan(sqrt(a*co
s(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2
*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate((A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F(-1)]

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

[In]

integrate((A+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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