\(\int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {-3-2 \cos (c+d x)}} \, dx\) [675]

Optimal result

Integrand size = 27, antiderivative size = 75 \[ \int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {-3-2 \cos (c+d x)}} \, dx=-\frac {3 \cot (c+d x) \operatorname {EllipticPi}\left (\frac {5}{2},\arcsin \left (\frac {\sqrt {-3-2 \cos (c+d x)}}{\sqrt {5} \sqrt {-\cos (c+d x)}}\right ),-5\right ) \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}{d} \]

[Out]

-3*cot(d*x+c)*EllipticPi(1/5*(-3-2*cos(d*x+c))^(1/2)*5^(1/2)/(-cos(d*x+c))^(1/2),5/2,I*5^(1/2))*(1-sec(d*x+c))
^(1/2)*(1+sec(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {2887} \[ \int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {-3-2 \cos (c+d x)}} \, dx=-\frac {3 \cot (c+d x) \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \operatorname {EllipticPi}\left (\frac {5}{2},\arcsin \left (\frac {\sqrt {-2 \cos (c+d x)-3}}{\sqrt {5} \sqrt {-\cos (c+d x)}}\right ),-5\right )}{d} \]

[In]

Int[Sqrt[-Cos[c + d*x]]/Sqrt[-3 - 2*Cos[c + d*x]],x]

[Out]

(-3*Cot[c + d*x]*EllipticPi[5/2, ArcSin[Sqrt[-3 - 2*Cos[c + d*x]]/(Sqrt[5]*Sqrt[-Cos[c + d*x]])], -5]*Sqrt[1 -
Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]])/d

Rule 2887

Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[2*c*Rt[b
*(c + d), 2]*Tan[e + f*x]*Sqrt[1 + Csc[e + f*x]]*(Sqrt[1 - Csc[e + f*x]]/(d*f*Sqrt[c^2 - d^2]))*EllipticPi[(c
+ d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ
[{b, c, d, e, f}, x] && GtQ[c^2 - d^2, 0] && PosQ[(c + d)/b] && GtQ[c^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 \cot (c+d x) \operatorname {EllipticPi}\left (\frac {5}{2},\arcsin \left (\frac {\sqrt {-3-2 \cos (c+d x)}}{\sqrt {5} \sqrt {-\cos (c+d x)}}\right ),-5\right ) \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}{d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.71 \[ \int \frac {\sqrt {-\cos (c+d x)}}{\sqrt {-3-2 \cos (c+d x)}} \, dx=-\frac {2 i \cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (\operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {5}}\right ),-5\right )-2 \operatorname {EllipticPi}\left (5,i \text {arcsinh}\left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {5}}\right ),-5\right )\right ) \sqrt {\cos (c+d x) (3+2 \cos (c+d x)) \sec ^4\left (\frac {1}{2} (c+d x)\right )}}{d \sqrt {-3-2 \cos (c+d x)} \sqrt {-\cos (c+d x)}} \]

[In]

Integrate[Sqrt[-Cos[c + d*x]]/Sqrt[-3 - 2*Cos[c + d*x]],x]

[Out]

((-2*I)*Cos[(c + d*x)/2]^2*(EllipticF[I*ArcSinh[Tan[(c + d*x)/2]/Sqrt[5]], -5] - 2*EllipticPi[5, I*ArcSinh[Tan
[(c + d*x)/2]/Sqrt[5]], -5])*Sqrt[Cos[c + d*x]*(3 + 2*Cos[c + d*x])*Sec[(c + d*x)/2]^4])/(d*Sqrt[-3 - 2*Cos[c
+ d*x]]*Sqrt[-Cos[c + d*x]])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (68 ) = 136\).

Time = 5.94 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.92

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {-\cos \left (d x +c \right )}\, \sqrt {-3-2 \cos \left (d x +c \right )}\, \left (F\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \frac {i \sqrt {5}}{5}\right )-2 \Pi \left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), -1, \frac {i \sqrt {5}}{5}\right )\right ) \sqrt {10}\, \sqrt {\frac {3+2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (1+\sec \left (d x +c \right )\right )}{5 d \left (3+2 \cos \left (d x +c \right )\right )}\)	\(144\)

[In]

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

[Out]

-1/5/d*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(-cos(d*x+c))^(1/2)*(-3-2*cos(d*x+c))^(1/2)*(EllipticF(cot(d*
x+c)-csc(d*x+c),1/5*I*5^(1/2))-2*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,1/5*I*5^(1/2)))*10^(1/2)*((3+2*cos(d*x+c)
)/(1+cos(d*x+c)))^(1/2)/(3+2*cos(d*x+c))*(1+sec(d*x+c))

Fricas [F]

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

[In]

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

[Out]

integral(-sqrt(-cos(d*x + c))*sqrt(-2*cos(d*x + c) - 3)/(2*cos(d*x + c) + 3), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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