3.22 \(\int \frac{\csc (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}} \, dx\)

Optimal. Leaf size=46 \[ \frac{\cos (e+f x) \log (\tan (e+f x))}{f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}} \]

[Out]

(Cos[e + f*x]*Log[Tan[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.193503, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {2946, 2620, 29} \[ \frac{\cos (e+f x) \log (\tan (e+f x))}{f \sqrt{a \sin (e+f x)+a} \sqrt{c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]),x]

[Out]

(Cos[e + f*x]*Log[Tan[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])

Rule 2946

Int[1/(sin[(e_.) + (f_.)*(x_)]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)]]), x_Symbol] :> Dist[Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), Int[1/(Cos[e + f*
x]*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[c^2
- d^2, 0]

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps

\begin{align*} \int \frac{\csc (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}} \, dx &=\frac{\cos (e+f x) \int \csc (e+f x) \sec (e+f x) \, dx}{\sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{\cos (e+f x) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,\tan (e+f x)\right )}{f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ &=\frac{\cos (e+f x) \log (\tan (e+f x))}{f \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.193308, size = 63, normalized size = 1.37 \[ -\frac{\sec (e+f x) \sqrt{a (\sin (e+f x)+1)} \sqrt{c-c \sin (e+f x)} (\log (\cos (e+f x))-\log (\sin (e+f x)))}{a c f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]),x]

[Out]

-(((Log[Cos[e + f*x]] - Log[Sin[e + f*x]])*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]])/(
a*c*f))

________________________________________________________________________________________

Maple [B]  time = 0.328, size = 111, normalized size = 2.4 \begin{align*} -{\frac{\cos \left ( fx+e \right ) }{f} \left ( \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) -\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) -\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \right ){\frac{1}{\sqrt{a \left ( 1+\sin \left ( fx+e \right ) \right ) }}}{\frac{1}{\sqrt{-c \left ( -1+\sin \left ( fx+e \right ) \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(f*x+e)/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2),x)

[Out]

-1/f*(ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-ln(-(-1+cos(f*x+e)
)/sin(f*x+e)))*cos(f*x+e)/(a*(1+sin(f*x+e)))^(1/2)/(-c*(-1+sin(f*x+e)))^(1/2)

________________________________________________________________________________________

Maxima [C]  time = 1.97753, size = 257, normalized size = 5.59 \begin{align*} -\frac{\left (-1\right )^{4 \, \cos \left (2 \, f x + 2 \, e\right )} \cosh \left (4 \, \pi \sin \left (2 \, f x + 2 \, e\right )\right ) \log \left (\frac{16 \,{\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}}{a c{\left | e^{\left (2 i \, f x + 2 i \, e\right )} - 1 \right |}^{2}}\right ) - 2 i \, \left (-1\right )^{4 \, \cos \left (2 \, f x + 2 \, e\right )} \arctan \left (\frac{4 \, \sin \left (2 \, f x + 2 \, e\right )}{\sqrt{a} \sqrt{c}{\left | e^{\left (2 i \, f x + 2 i \, e\right )} - 1 \right |}}, \frac{4 \,{\left (\cos \left (2 \, f x + 2 \, e\right ) + 1\right )}}{\sqrt{a} \sqrt{c}{\left | e^{\left (2 i \, f x + 2 i \, e\right )} - 1 \right |}}\right ) \sinh \left (4 \, \pi \sin \left (2 \, f x + 2 \, e\right )\right )}{2 \, \sqrt{a} \sqrt{c} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(f*x+e)/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-1/2*((-1)^(4*cos(2*f*x + 2*e))*cosh(4*pi*sin(2*f*x + 2*e))*log(16*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 +
2*cos(2*f*x + 2*e) + 1)/(a*c*abs(e^(2*I*f*x + 2*I*e) - 1)^2)) - 2*I*(-1)^(4*cos(2*f*x + 2*e))*arctan2(4*sin(2*
f*x + 2*e)/(sqrt(a)*sqrt(c)*abs(e^(2*I*f*x + 2*I*e) - 1)), 4*(cos(2*f*x + 2*e) + 1)/(sqrt(a)*sqrt(c)*abs(e^(2*
I*f*x + 2*I*e) - 1)))*sinh(4*pi*sin(2*f*x + 2*e)))/(sqrt(a)*sqrt(c)*f)

________________________________________________________________________________________

Fricas [A]  time = 2.77296, size = 475, normalized size = 10.33 \begin{align*} \left [\frac{\sqrt{a c} \log \left (-\frac{4 \,{\left (2 \, a c \cos \left (f x + e\right )^{5} - 2 \, a c \cos \left (f x + e\right )^{3} + a c \cos \left (f x + e\right ) - \sqrt{a c}{\left (2 \, \cos \left (f x + e\right )^{2} - 1\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}\right )}}{\cos \left (f x + e\right )^{5} - \cos \left (f x + e\right )^{3}}\right )}{2 \, a c f}, \frac{\sqrt{-a c} \arctan \left (\frac{\sqrt{-a c} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{-c \sin \left (f x + e\right ) + c}}{2 \, a c \cos \left (f x + e\right )^{3} - a c \cos \left (f x + e\right )}\right )}{a c f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(f*x+e)/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

[1/2*sqrt(a*c)*log(-4*(2*a*c*cos(f*x + e)^5 - 2*a*c*cos(f*x + e)^3 + a*c*cos(f*x + e) - sqrt(a*c)*(2*cos(f*x +
e)^2 - 1)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(cos(f*x + e)^5 - cos(f*x + e)^3))/(a*c*f), sqr
t(-a*c)*arctan(sqrt(-a*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(2*a*c*cos(f*x + e)^3 - a*c*cos(f
*x + e)))/(a*c*f)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )} \sqrt{- c \left (\sin{\left (e + f x \right )} - 1\right )} \sin{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(f*x+e)/(a+a*sin(f*x+e))**(1/2)/(c-c*sin(f*x+e))**(1/2),x)

[Out]

Integral(1/(sqrt(a*(sin(e + f*x) + 1))*sqrt(-c*(sin(e + f*x) - 1))*sin(e + f*x)), x)

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(f*x+e)/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError