\(\int \frac {\cot ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [330]

Optimal result

Integrand size = 25, antiderivative size = 78 \[ \int \frac {\cot ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b} f}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3751, 491, 12, 385, 209} \[ \int \frac {\cot ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{f \sqrt {a-b}}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a f} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 385

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 491

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*e*(m + 1))), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rule 3751

Int[((d_.)*tan[(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/f), Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2
+ ff^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 6.69 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.29 \[ \int \frac {\cot ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {2 \cos ^2(e+f x) \cot (e+f x) \left (1+\frac {b \tan ^2(e+f x)}{a}\right ) \left (2 (a-b) \operatorname {Hypergeometric2F1}\left (2,2,\frac {5}{2},\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )+\frac {3 a \arcsin \left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \left (a+2 b \tan ^2(e+f x)\right )}{\sqrt {\frac {(a-b) \sin ^2(2 (e+f x)) \left (a+b \tan ^2(e+f x)\right )}{a^2}}}\right )}{3 a^2 f \sqrt {a+b \tan ^2(e+f x)}} \]

[In]

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

[Out]

(-2*Cos[e + f*x]^2*Cot[e + f*x]*(1 + (b*Tan[e + f*x]^2)/a)*(2*(a - b)*Hypergeometric2F1[2, 2, 5/2, ((a - b)*Si
n[e + f*x]^2)/a]*Sin[e + f*x]^2*(a + b*Tan[e + f*x]^2) + (3*a*ArcSin[Sqrt[((a - b)*Sin[e + f*x]^2)/a]]*(a + 2*
b*Tan[e + f*x]^2))/Sqrt[((a - b)*Sin[2*(e + f*x)]^2*(a + b*Tan[e + f*x]^2))/a^2]))/(3*a^2*f*Sqrt[a + b*Tan[e +
f*x]^2])

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(251\) vs. \(2(70)=140\).

Time = 4.72 (sec) , antiderivative size = 252, normalized size of antiderivative = 3.23

method	result	size
default	\(\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {a -b}}\right ) a +\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {a -b}}\right ) a \sec \left (f x +e \right )-\sqrt {a -b}\, b \tan \left (f x +e \right )-\sqrt {a -b}\, a \cot \left (f x +e \right )}{f a \sqrt {a -b}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\)	\(252\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 289, normalized size of antiderivative = 3.71 \[ \int \frac {\cot ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [-\frac {a \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right ) + 4 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a - b\right )}}{4 \, {\left (a^{2} - a b\right )} f \tan \left (f x + e\right )}, -\frac {\sqrt {a - b} a \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) \tan \left (f x + e\right ) + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a - b\right )}}{2 \, {\left (a^{2} - a b\right )} f \tan \left (f x + e\right )}\right ] \]

[In]

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

[Out]

[-1/4*(a*sqrt(-a + b)*log(-((a^2 - 8*a*b + 8*b^2)*tan(f*x + e)^4 - 2*(3*a^2 - 4*a*b)*tan(f*x + e)^2 + a^2 + 4*
((a - 2*b)*tan(f*x + e)^3 - a*tan(f*x + e))*sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b))/(tan(f*x + e)^4 + 2*tan(f
*x + e)^2 + 1))*tan(f*x + e) + 4*sqrt(b*tan(f*x + e)^2 + a)*(a - b))/((a^2 - a*b)*f*tan(f*x + e)), -1/2*(sqrt(
a - b)*a*arctan(-2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b)*tan(f*x + e)/((a - 2*b)*tan(f*x + e)^2 - a))*tan(f*x
+ e) + 2*sqrt(b*tan(f*x + e)^2 + a)*(a - b))/((a^2 - a*b)*f*tan(f*x + e))]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate(cot(f*x + e)^2/sqrt(b*tan(f*x + e)^2 + a), x)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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