\(\int \frac {\csc ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [125]

Optimal result

Integrand size = 25, antiderivative size = 30 \[ \int \frac {\csc ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 30, 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 = {3744, 270} \[ \int \frac {\csc ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a f} \]

[In]

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

[Out]

-((Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(a*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 3744

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

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

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63 \[ \int \frac {\csc ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\cot (e+f x) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{\sqrt {2} a f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(-\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{f a \tan \left (f x +e \right )}\)	\(31\)
default	\(-\frac {\sqrt {a +b \tan \left (f x +e \right )^{2}}}{f a \tan \left (f x +e \right )}\)	\(31\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.63 \[ \int \frac {\csc ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a f \sin \left (f x + e\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {\csc ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a}}{a f \tan \left (f x + e\right )} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 11.83 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {\csc ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\mathrm {cot}\left (e+f\,x\right )\,\sqrt {a+\frac {b\,{\sin \left (e+f\,x\right )}^2}{{\cos \left (e+f\,x\right )}^2}}}{a\,f} \]

[In]

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

[Out]

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