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

Optimal result

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

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

Mathematica [C] (warning: unable to verify)

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

Time = 3.78 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.29 \[ \int \frac {\cot ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=-\frac {2 \cos ^2(e+f x) \cot (e+f x) \left (1+\frac {b \tan ^2(e+f x)}{a}\right ) \left (2 (a-b) \operatorname {Hypergeometric2F1}\left (2,2,\frac {5}{2},\frac {(a-b) \sin ^2(e+f x)}{a}\right ) \sin ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )+\frac {3 a \arcsin \left (\sqrt {\frac {(a-b) \sin ^2(e+f x)}{a}}\right ) \left (a+2 b \tan ^2(e+f x)\right )}{\sqrt {\frac {(a-b) \sin ^2(2 (e+f x)) \left (a+b \tan ^2(e+f x)\right )}{a^2}}}\right )}{3 a^2 f \sqrt {a+b \tan ^2(e+f x)}} \] Input:

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

Output:

(-2*Cos[e + f*x]^2*Cot[e + f*x]*(1 + (b*Tan[e + f*x]^2)/a)*(2*(a - b)*Hype 
rgeometric2F1[2, 2, 5/2, ((a - b)*Sin[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])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4153, 382, 25, 27, 291, 216}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (e+f x)^2 \sqrt {a+b \tan (e+f x)^2}}dx\)

\(\Big \downarrow \) 4153

\(\displaystyle \frac {\int \frac {\cot ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 382

\(\displaystyle \frac {\frac {\int -\frac {a}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {\int \frac {a}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)}{a}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \sqrt {b \tan ^2(e+f x)+a}}d\tan (e+f x)-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{f}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {-\int \frac {1}{1-\frac {(b-a) \tan ^2(e+f x)}{b \tan ^2(e+f x)+a}}d\frac {\tan (e+f x)}{\sqrt {b \tan ^2(e+f x)+a}}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {\arctan \left (\frac {\sqrt {a-b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{\sqrt {a-b}}-\frac {\cot (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{a}}{f}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

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

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 382

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) 
, x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/ 
(a*c*e*(m + 1))), x] - Simp[1/(a*c*e^2*(m + 1))   Int[(e*x)^(m + 2)*(a + b* 
x^2)^p*(c + d*x^2)^q*Simp[(b*c + a*d)*(m + 3) + 2*(b*c*p + a*d*q) + b*d*(m 
+ 2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[ 
b*c - a*d, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4153

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]}, Simp[c*(ff/f)   Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f 
f^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] && Ratio 
nalQ[n]))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(165\) vs. \(2(70)=140\).

Time = 23.52 (sec) , antiderivative size = 166, normalized size of antiderivative = 2.13

method	result	size
default	\(\frac {-a \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )}{\sqrt {a -b}\, \left (\cos \left (f x +e \right )-1\right )}\right ) \left (\sec \left (f x +e \right )+1\right )-\sqrt {a -b}\, a \cot \left (f x +e \right )-\sqrt {a -b}\, b \tan \left (f x +e \right )}{f a \sqrt {a -b}\, \sqrt {a +b \tan \left (f x +e \right )^{2}}}\)	\(166\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 289, normalized size of antiderivative = 3.71 \[ \int \frac {\cot ^2(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx=\left [-\frac {a \sqrt {-a + b} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (f x + e\right )^{2} + a^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{3} - a \tan \left (f x + e\right )\right )} \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right ) + 4 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a - b\right )}}{4 \, {\left (a^{2} - a b\right )} f \tan \left (f x + e\right )}, -\frac {\sqrt {a - b} a \arctan \left (-\frac {2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {a - b} \tan \left (f x + e\right )}{{\left (a - 2 \, b\right )} \tan \left (f x + e\right )^{2} - a}\right ) \tan \left (f x + e\right ) + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a - b\right )}}{2 \, {\left (a^{2} - a b\right )} f \tan \left (f x + e\right )}\right ] \] Input:

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

Output:

[-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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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