\(\int \frac {\csc ^2(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) [65]

Optimal result

Integrand size = 23, antiderivative size = 48 \[ \int \frac {\csc ^2(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{3/2} f}-\frac {\cot (e+f x)}{a f} \]

[Out]

-cot(f*x+e)/a/f-arctan(b^(1/2)*tan(f*x+e)/a^(1/2))*b^(1/2)/a^(3/2)/f

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3744, 331, 211} \[ \int \frac {\csc ^2(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{3/2} f}-\frac {\cot (e+f x)}{a f} \]

[In]

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

[Out]

-((Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(a^(3/2)*f)) - Cot[e + f*x]/(a*f)

Rule 211

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

Rule 331

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] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {\cot (e+f x)}{a f}-\frac {b \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{a f} \\ & = -\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{3/2} f}-\frac {\cot (e+f x)}{a f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\csc ^2(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{3/2} f}-\frac {\cot (e+f x)}{a f} \]

[In]

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

[Out]

-((Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(a^(3/2)*f)) - Cot[e + f*x]/(a*f)

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {-\frac {b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{a \sqrt {a b}}-\frac {1}{a \tan \left (f x +e \right )}}{f}\)	\(44\)
default	\(\frac {-\frac {b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{a \sqrt {a b}}-\frac {1}{a \tan \left (f x +e \right )}}{f}\)	\(44\)
risch	\(-\frac {2 i}{f a \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{2 a^{2} f}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{2 a^{2} f}\)	\(119\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (40) = 80\).

Time = 0.30 (sec) , antiderivative size = 257, normalized size of antiderivative = 5.35 \[ \int \frac {\csc ^2(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\left [\frac {\sqrt {-\frac {b}{a}} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - a b \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) - 4 \, \cos \left (f x + e\right )}{4 \, a f \sin \left (f x + e\right )}, \frac {\sqrt {\frac {b}{a}} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )}{2 \, a f \sin \left (f x + e\right )}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.88 \[ \int \frac {\csc ^2(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a} + \frac {1}{a \tan \left (f x + e\right )}}{f} \]

[In]

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

[Out]

-(b*arctan(b*tan(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a) + 1/(a*tan(f*x + e)))/f

Giac [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.23 \[ \int \frac {\csc ^2(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} b}{\sqrt {a b} a} + \frac {1}{a \tan \left (f x + e\right )}}{f} \]

[In]

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

[Out]

-((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*b/(sqrt(a*b)*a) + 1/(a*tan(f*x + e)
))/f

Mupad [B] (verification not implemented)

Time = 10.33 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.83 \[ \int \frac {\csc ^2(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\mathrm {cot}\left (e+f\,x\right )}{a\,f}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a}}\right )}{a^{3/2}\,f} \]

[In]

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

[Out]

- cot(e + f*x)/(a*f) - (b^(1/2)*atan((b^(1/2)*tan(e + f*x))/a^(1/2)))/(a^(3/2)*f)