\(\int \frac {\csc ^2(e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\) [77]

Optimal result

Integrand size = 23, antiderivative size = 82 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {3 \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{5/2} f}-\frac {3 \cot (e+f x)}{2 a^2 f}+\frac {\cot (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)\right )} \]

[Out]

-3/2*cot(f*x+e)/a^2/f-3/2*arctan(b^(1/2)*tan(f*x+e)/a^(1/2))*b^(1/2)/a^(5/2)/f+1/2*cot(f*x+e)/a/f/(a+b*tan(f*x
+e)^2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3744, 296, 331, 211} \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {3 \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{5/2} f}-\frac {3 \cot (e+f x)}{2 a^2 f}+\frac {\cot (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)\right )} \]

[In]

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

[Out]

(-3*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(2*a^(5/2)*f) - (3*Cot[e + f*x])/(2*a^2*f) + Cot[e + f*x]/
(2*a*f*(a + b*Tan[e + f*x]^2))

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 296

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

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 )^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\cot (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)\right )}+\frac {3 \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 a f} \\ & = -\frac {3 \cot (e+f x)}{2 a^2 f}+\frac {\cot (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)\right )}-\frac {(3 b) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^2 f} \\ & = -\frac {3 \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{5/2} f}-\frac {3 \cot (e+f x)}{2 a^2 f}+\frac {\cot (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.01 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {-3 \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )+\sqrt {a} \left (-2 \cot (e+f x)-\frac {b \sin (2 (e+f x))}{a+b+(a-b) \cos (2 (e+f x))}\right )}{2 a^{5/2} f} \]

[In]

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

[Out]

(-3*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] + Sqrt[a]*(-2*Cot[e + f*x] - (b*Sin[2*(e + f*x)])/(a + b +
(a - b)*Cos[2*(e + f*x)])))/(2*a^(5/2)*f)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {-\frac {1}{a^{2} \tan \left (f x +e \right )}-\frac {b \left (\frac {\tan \left (f x +e \right )}{2 a +2 b \tan \left (f x +e \right )^{2}}+\frac {3 \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{2}}}{f}\)	\(69\)
default	\(\frac {-\frac {1}{a^{2} \tan \left (f x +e \right )}-\frac {b \left (\frac {\tan \left (f x +e \right )}{2 a +2 b \tan \left (f x +e \right )^{2}}+\frac {3 \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{2}}}{f}\)	\(69\)
risch	\(-\frac {i \left (2 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-3 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+3 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+4 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-6 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+2 a^{2}-5 a b +3 b^{2}\right )}{f \left (a -b \right ) a^{2} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}-b \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a -b \right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{4 a^{3} f}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{4 a^{3} f}\)	\(264\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (68) = 136\).

Time = 0.32 (sec) , antiderivative size = 373, normalized size of antiderivative = 4.55 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\left [-\frac {4 \, {\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + b\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - a b \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) + 12 \, b \cos \left (f x + e\right )}{8 \, {\left (a^{2} b f + {\left (a^{3} - a^{2} b\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + b\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 6 \, b \cos \left (f x + e\right )}{4 \, {\left (a^{2} b f + {\left (a^{3} - a^{2} b\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ] \]

[In]

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

[Out]

[-1/8*(4*(2*a - 3*b)*cos(f*x + e)^3 - 3*((a - b)*cos(f*x + e)^2 + b)*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) + 12*b
*cos(f*x + e))/((a^2*b*f + (a^3 - a^2*b)*f*cos(f*x + e)^2)*sin(f*x + e)), -1/4*(2*(2*a - 3*b)*cos(f*x + e)^3 -
3*((a - b)*cos(f*x + e)^2 + b)*sqrt(b/a)*arctan(1/2*((a + b)*cos(f*x + e)^2 - b)*sqrt(b/a)/(b*cos(f*x + e)*si
n(f*x + e)))*sin(f*x + e) + 6*b*cos(f*x + e))/((a^2*b*f + (a^3 - a^2*b)*f*cos(f*x + e)^2)*sin(f*x + e))]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\frac {3 \, b \tan \left (f x + e\right )^{2} + 2 \, a}{a^{2} b \tan \left (f x + e\right )^{3} + a^{3} \tan \left (f x + e\right )} + \frac {3 \, b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}}}{2 \, f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.56 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.07 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} b}{\sqrt {a b} a^{2}} + \frac {3 \, b \tan \left (f x + e\right )^{2} + 2 \, a}{{\left (b \tan \left (f x + e\right )^{3} + a \tan \left (f x + e\right )\right )} a^{2}}}{2 \, f} \]

[In]

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

[Out]

-1/2*(3*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*b/(sqrt(a*b)*a^2) + (3*b*tan(
f*x + e)^2 + 2*a)/((b*tan(f*x + e)^3 + a*tan(f*x + e))*a^2))/f

Mupad [B] (verification not implemented)

Time = 10.64 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \frac {\csc ^2(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\frac {1}{a}+\frac {3\,b\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,a^2}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^3+a\,\mathrm {tan}\left (e+f\,x\right )\right )}-\frac {3\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a}}\right )}{2\,a^{5/2}\,f} \]

[In]

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

[Out]

- (1/a + (3*b*tan(e + f*x)^2)/(2*a^2))/(f*(a*tan(e + f*x) + b*tan(e + f*x)^3)) - (3*b^(1/2)*atan((b^(1/2)*tan(
e + f*x))/a^(1/2)))/(2*a^(5/2)*f)