\(\int \frac {\csc ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [103]

Optimal result

Integrand size = 25, antiderivative size = 33 \[ \int \frac {\csc ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{(a+b) f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {4217, 270} \[ \int \frac {\csc ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f (a+b)} \]

[In]

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

[Out]

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

Rule 270

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

Rule 4217

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1
+ ff^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{(a+b) f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.67 \[ \int \frac {\csc ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \csc (e+f x) \sec (e+f x)}{2 (a+b) f \sqrt {a+b \sec ^2(e+f x)}} \]

[In]

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

[Out]

-1/2*((a + 2*b + a*Cos[2*(e + f*x)])*Csc[e + f*x]*Sec[e + f*x])/((a + b)*f*Sqrt[a + b*Sec[e + f*x]^2])

Maple [A] (verified)

Time = 2.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.45

method	result	size
default	\(-\frac {\cot \left (f x +e \right ) a +b \sec \left (f x +e \right ) \csc \left (f x +e \right )}{f \left (a +b \right ) \sqrt {a +b \sec \left (f x +e \right )^{2}}}\)	\(48\)

[In]

int(csc(f*x+e)^2/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {\csc ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a + b\right )} f \sin \left (f x + e\right )} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\csc ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\csc ^{2}{\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {\csc ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a + b}}{{\left (a + b\right )} f \tan \left (f x + e\right )} \]

[In]

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

[Out]

-sqrt(b*tan(f*x + e)^2 + a + b)/((a + b)*f*tan(f*x + e))

Giac [F]

\[ \int \frac {\csc ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\csc \left (f x + e\right )^{2}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 18.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.24 \[ \int \frac {\csc ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\left (2\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )\right )\,\sqrt {\frac {a+2\,b+a\,\cos \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}}{2\,f\,{\sin \left (2\,e+2\,f\,x\right )}^2\,\left (a+b\right )} \]

[In]

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

[Out]

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