∫ ∘ ________ − csc(e + fx) ∘ __________

3.20 \(\int \sqrt {-\csc (e+f x)} \sqrt {a-a \csc (e+f x)} \, dx\)

Optimal. Leaf size=38 \[ -\frac {2 \sqrt {a} \sinh ^{-1}\left (\frac {\sqrt {a} \cot (e+f x)}{\sqrt {a-a \csc (e+f x)}}\right )}{f} \]

[Out]

-2*arcsinh(cot(f*x+e)*a^(1/2)/(a-a*csc(f*x+e))^(1/2))*a^(1/2)/f

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Rubi [A] time = 0.07, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3801, 215} \[ -\frac {2 \sqrt {a} \sinh ^{-1}\left (\frac {\sqrt {a} \cot (e+f x)}{\sqrt {a-a \csc (e+f x)}}\right )}{f} \]

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[-Csc[e + f*x]]*Sqrt[a - a*Csc[e + f*x]],x]

[Out]

(-2*Sqrt[a]*ArcSinh[(Sqrt[a]*Cot[e + f*x])/Sqrt[a - a*Csc[e + f*x]]])/f

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},

x] && GtQ[a, 0] && PosQ[b]

Rule 3801

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq

rt[(a*d)/b])/(b*f), Subst[Int[1/Sqrt[1 + x^2/a], x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; Free

Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[(a*d)/b, 0]

Rubi steps

\begin {align*} \int \sqrt {-\csc (e+f x)} \sqrt {a-a \csc (e+f x)} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a}}} \, dx,x,-\frac {a \cot (e+f x)}{\sqrt {a-a \csc (e+f x)}}\right )}{f}\\ &=-\frac {2 \sqrt {a} \sinh ^{-1}\left (\frac {\sqrt {a} \cot (e+f x)}{\sqrt {a-a \csc (e+f x)}}\right )}{f}\\ \end {align*}

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Mathematica [B] time = 0.79, size = 101, normalized size = 2.66 \[ \frac {2 \tan \left (\frac {1}{2} (e+f x)\right ) \sqrt {-\csc (e+f x)} \sqrt {a-a \csc (e+f x)} \left (\tanh ^{-1}\left (\sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )}\right )+\sinh ^{-1}\left (\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{f \left (\tan \left (\frac {1}{2} (e+f x)\right )-1\right ) \sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )}} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[-Csc[e + f*x]]*Sqrt[a - a*Csc[e + f*x]],x]

[Out]

(2*(ArcSinh[Tan[(e + f*x)/2]] + ArcTanh[Sqrt[Sec[(e + f*x)/2]^2]])*Sqrt[-Csc[e + f*x]]*Sqrt[a - a*Csc[e + f*x]

]*Tan[(e + f*x)/2])/(f*Sqrt[Sec[(e + f*x)/2]^2]*(-1 + Tan[(e + f*x)/2]))

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fricas [B] time = 1.08, size = 296, normalized size = 7.79 \[ \left [\frac {\sqrt {a} \log \left (\frac {a \cos \left (f x + e\right )^{3} - 7 \, a \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) - 3\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 3\right )} \sqrt {a} \sqrt {\frac {a \sin \left (f x + e\right ) - a}{\sin \left (f x + e\right )}} \sqrt {-\frac {1}{\sin \left (f x + e\right )}} - 9 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right )^{2} + 8 \, a \cos \left (f x + e\right ) - a\right )} \sin \left (f x + e\right ) - a}{\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 1}\right )}{2 \, f}, \frac {\sqrt {-a} \arctan \left (-\frac {{\left (\cos \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) - 1\right )} \sqrt {-a} \sqrt {\frac {a \sin \left (f x + e\right ) - a}{\sin \left (f x + e\right )}} \sqrt {-\frac {1}{\sin \left (f x + e\right )}}}{2 \, a \cos \left (f x + e\right )}\right )}{f}\right ] \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*sqrt(a)*log((a*cos(f*x + e)^3 - 7*a*cos(f*x + e)^2 - 4*(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + (cos(f*x + e)

^2 - 2*cos(f*x + e) - 3)*sin(f*x + e) - cos(f*x + e) - 3)*sqrt(a)*sqrt((a*sin(f*x + e) - a)/sin(f*x + e))*sqrt

(-1/sin(f*x + e)) - 9*a*cos(f*x + e) - (a*cos(f*x + e)^2 + 8*a*cos(f*x + e) - a)*sin(f*x + e) - a)/(cos(f*x +

e)^3 + cos(f*x + e)^2 - (cos(f*x + e)^2 - 1)*sin(f*x + e) - cos(f*x + e) - 1))/f, sqrt(-a)*arctan(-1/2*(cos(f*

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

*x + e)))/f]

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/