\(\int \frac {\cot ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [412]

Optimal result

Integrand size = 25, antiderivative size = 74 \[ \int \frac {\cot ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{\sqrt {a} f}-\frac {\cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{(a+b) f} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.72 \[ \int \frac {\cot ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {\sqrt {a+2 b+a \cos (2 (e+f x))} \sec (e+f x) \left ((a+b) \arctan \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )+\sqrt {a} \csc (e+f x) \sqrt {a+b-a \sin ^2(e+f x)}\right )}{\sqrt {2} \sqrt {a} (a+b) f \sqrt {a+b \sec ^2(e+f x)}} \] Input:

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

Output:

-((Sqrt[a + 2*b + a*Cos[2*(e + f*x)]]*Sec[e + f*x]*((a + b)*ArcTan[(Sqrt[a 
]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]] + Sqrt[a]*Csc[e + f*x]*Sqr 
t[a + b - a*Sin[e + f*x]^2]))/(Sqrt[2]*Sqrt[a]*(a + b)*f*Sqrt[a + b*Sec[e 
+ f*x]^2]))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4629, 2075, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 382

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

(-(ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/Sqrt[a]) 
- (Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(a + b))/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 2075

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa 
ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi 
alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  ! 
BinomialMatchQ[{u, v}, x]
 

rule 3042

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

rule 4629

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f 
_.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim 
p[ff/f   Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 
)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte 
gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(297\) vs. \(2(66)=132\).

Time = 2.02 (sec) , antiderivative size = 298, normalized size of antiderivative = 4.03

method	result	size
default	\(\frac {-\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) a \left (\sec \left (f x +e \right )+1\right )-\sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \ln \left (4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right )+4 \sqrt {-a}\, \sqrt {\frac {b +a \cos \left (f x +e \right )^{2}}{\left (1+\cos \left (f x +e \right )\right )^{2}}}-4 \sin \left (f x +e \right ) a \right ) b \left (\sec \left (f x +e \right )+1\right )-a \sqrt {-a}\, \cot \left (f x +e \right )-b \sqrt {-a}\, \sec \left (f x +e \right ) \csc \left (f x +e \right )}{f \left (a +b \right ) \sqrt {-a}\, \sqrt {a +b \sec \left (f x +e \right )^{2}}}\)	\(298\)

Input:

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

Output:

1/f*(-((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(-a)^(1/2)*((b+a*co 
s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+ 
e)^2)/(1+cos(f*x+e))^2)^(1/2)-4*sin(f*x+e)*a)*a*(sec(f*x+e)+1)-((b+a*cos(f 
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*ln(4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+co 
s(f*x+e))^2)^(1/2)*cos(f*x+e)+4*(-a)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+ 
e))^2)^(1/2)-4*sin(f*x+e)*a)*b*(sec(f*x+e)+1)-a*(-a)^(1/2)*cot(f*x+e)-b*(- 
a)^(1/2)*sec(f*x+e)*csc(f*x+e))/(a+b)/(-a)^(1/2)/(a+b*sec(f*x+e)^2)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (66) = 132\).

Time = 0.24 (sec) , antiderivative size = 525, normalized size of antiderivative = 7.09 \[ \int \frac {\cot ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [-\frac {\sqrt {-a} {\left (a + b\right )} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - a^{3} b\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 14 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 28 \, a^{3} b + 70 \, a^{2} b^{2} - 28 \, a b^{3} + b^{4} - 32 \, {\left (a^{4} - 7 \, a^{3} b + 7 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (f x + e\right )^{2} - 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 14 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) \sin \left (f x + e\right ) + 8 \, a \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{8 \, {\left (a^{2} + a b\right )} f \sin \left (f x + e\right )}, \frac {{\left (a + b\right )} \sqrt {a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} - a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 4 \, a \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{4 \, {\left (a^{2} + a b\right )} f \sin \left (f x + e\right )}\right ] \] Input:

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

Output:

[-1/8*(sqrt(-a)*(a + b)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos 
(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28* 
a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^ 
3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + 
e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7* 
a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + 
e)^2)*sin(f*x + e))*sin(f*x + e) + 8*a*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x 
 + e)^2)*cos(f*x + e))/((a^2 + a*b)*f*sin(f*x + e)), 1/4*((a + b)*sqrt(a)* 
arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6 
*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e) 
^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^ 
2)*sin(f*x + e)))*sin(f*x + e) - 4*a*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + 
 e)^2)*cos(f*x + e))/((a^2 + a*b)*f*sin(f*x + e))]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

Timed out. \[ \int \frac {\cot ^2(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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