3.146 \(\int \frac{\csc (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}} \, dx\)

Optimal. Leaf size=41 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{\sqrt{a} f} \]

[Out]

-(ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]]/(Sqrt[a]*f))

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Rubi [A]  time = 0.0759755, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3186, 377, 206} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{\sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]/Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

-(ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]]/(Sqrt[a]*f))

Rule 3186

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.149938, size = 48, normalized size = 1.17 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \cos (e+f x)}{\sqrt{2 a-b \cos (2 (e+f x))+b}}\right )}{\sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]/Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

-(ArcTanh[(Sqrt[2]*Sqrt[a]*Cos[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x)]]]/(Sqrt[a]*f))

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Maple [B]  time = 0.875, size = 112, normalized size = 2.7 \begin{align*} -{\frac{1}{2\,f\cos \left ( fx+e \right ) }\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) }\ln \left ({\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( \left ( a-b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\sqrt{a}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+a+b \right ) } \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2),x)

[Out]

-1/2*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)/a^(1/2)*ln(((a-b)*cos(f*x+e)^2+2*a^(1/2)*(-b*cos(f*x+e)^4+(a+b)*c
os(f*x+e)^2)^(1/2)+a+b)/sin(f*x+e)^2)/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.77761, size = 558, normalized size = 13.61 \begin{align*} \left [\frac{\log \left (\frac{2 \,{\left ({\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \,{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} +{\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{a} + a^{2} + 2 \, a b + b^{2}\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1}\right )}{4 \, \sqrt{a} f}, \frac{\sqrt{-a} \arctan \left (-\frac{{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a}}{2 \,{\left (a b \cos \left (f x + e\right )^{3} -{\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )}}\right )}{2 \, a f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*log(2*((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 2*(3*a^2 + 2*a*b - b^2)*cos(f*x + e)^2 - 4*((a - b)*cos(f*x +
e)^3 + (a + b)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) + a^2 + 2*a*b + b^2)/(cos(f*x + e)^4 - 2
*cos(f*x + e)^2 + 1))/(sqrt(a)*f), 1/2*sqrt(-a)*arctan(-1/2*((a - b)*cos(f*x + e)^2 + a + b)*sqrt(-b*cos(f*x +
e)^2 + a + b)*sqrt(-a)/(a*b*cos(f*x + e)^3 - (a^2 + a*b)*cos(f*x + e)))/(a*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc{\left (e + f x \right )}}{\sqrt{a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Integral(csc(e + f*x)/sqrt(a + b*sin(e + f*x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )}{\sqrt{b \sin \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(csc(f*x + e)/sqrt(b*sin(f*x + e)^2 + a), x)