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

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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])

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 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]

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 = 0.17, 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))

________________________________________________________________________________________

fricas [B]  time = 0.51, size = 219, normalized size = 5.34 \[ \left [\frac {\log \left (\frac {2 \, {\left ({\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (3 \, a^{2} + 2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} + {\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a} + a^{2} + 2 \, a b + b^{2}\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1}\right )}{4 \, \sqrt {a} f}, \frac {\sqrt {-a} \arctan \left (-\frac {{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + a + b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a}}{2 \, {\left (a b \cos \left (f x + e\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (f x + e\right )\right )}}\right )}{2 \, a f}\right ] \]

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)]

________________________________________________________________________________________

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

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was
done assuming [a,b]=[-81,22]Warning, need to choose a branch for the root of a polynomial with parameters. Thi
s might be wrong.The choice was done assuming [a,b]=[53,42]Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_no
step/2)Evaluation time: 0.41Error: Bad Argument Type

________________________________________________________________________________________

maple [B]  time = 1.39, size = 112, normalized size = 2.73 \[ -\frac {\sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \ln \left (\frac {\left (a -b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+a +b}{\sin \left (f x +e \right )^{2}}\right )}{2 \sqrt {a}\, \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]

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

________________________________________________________________________________________

maxima [B]  time = 0.44, size = 109, normalized size = 2.66 \[ -\frac {\frac {\log \left (b - \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a}}{\cos \left (f x + e\right ) - 1} - \frac {a}{\cos \left (f x + e\right ) - 1}\right )}{\sqrt {a}} - \frac {\log \left (-b + \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a}}{\cos \left (f x + e\right ) + 1} + \frac {a}{\cos \left (f x + e\right ) + 1}\right )}{\sqrt {a}}}{2 \, f} \]

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]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sin \left (e+f\,x\right )\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc {\left (e + f x \right )}}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \]

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)

________________________________________________________________________________________