\(\int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, dx\) [240]

Optimal result

Integrand size = 25, antiderivative size = 97 \[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}} \]

[Out]

1/4*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*sin(d
*x+c)*cos(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2843, 12, 2861, 211} \[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}} \]

[In]

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

[Out]

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) + (
Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2))

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 2843

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] - Dist[1/(a*b*(2*m + 1)),
Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[a*d*n - b*c*(m + 1) - b*d*(m + n + 1)*Sin[
e + f*x], x], 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] && LtQ[m, -1] && LtQ[0, n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))

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]

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {1+\cos (c+d x)} \left (\arcsin \left (\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \sqrt {1+\cos (c+d x)}+2 \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d (a (1+\cos (c+d x)))^{3/2}} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sqrt[1 + Cos[c + d*x]]*(ArcSin[Sin[(c + d*x)/2]/Sqrt[Cos[(c + d*x)/2]^2]]*Sqrt[1 + Cos[c + d
*x]] + 2*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sin[(c + d*x)/2]))/(2*d*(a*(1 + Cos[c + d*x]))^(3/2))

Maple [A] (verified)

Time = 3.48 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.35

method	result	size
default	\(-\frac {\left (-\sin \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \cos \left (d x +c \right )+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {2}}{4 d \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, a^{2}}\)	\(131\)

[In]

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

[Out]

-1/4/d*(-sin(d*x+c)*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+arcsin(cot(d*x+c)-csc(d*x+c))*cos(d*x+c)+arcsin(
cot(d*x+c)-csc(d*x+c)))*(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)*2^(1/2)/a^2

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.49 \[ \int \frac {\sqrt {\cos (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) + 2 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

[In]

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

[Out]

1/4*(sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*arctan(1/2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)
*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))) + 2*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*
x + c))*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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[In]

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

[Out]

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