\(\int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\) [403]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [B] (warning: unable to verify)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 19, antiderivative size = 161 \[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {\sqrt {-1+\sqrt {2}} \arctan \left (\frac {3-2 \sqrt {2}+\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (-7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f}-\frac {2 \text {arctanh}\left (\sqrt {1+\tan (e+f x)}\right )}{f}+\frac {\sqrt {1+\sqrt {2}} \text {arctanh}\left (\frac {3+2 \sqrt {2}+\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f} \] Output:

1/2*(2^(1/2)-1)^(1/2)*arctan((3-2*2^(1/2)+(1-2^(1/2))*tan(f*x+e))/(-14+10* 
2^(1/2))^(1/2)/(1+tan(f*x+e))^(1/2))/f-2*arctanh((1+tan(f*x+e))^(1/2))/f+1 
/2*(1+2^(1/2))^(1/2)*arctanh((3+2*2^(1/2)+(1+2^(1/2))*tan(f*x+e))/(14+10*2 
^(1/2))^(1/2)/(1+tan(f*x+e))^(1/2))/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.52 \[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {2 \text {arctanh}\left (\sqrt {1+\tan (e+f x)}\right )}{f}+\frac {\text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )}{\sqrt {1-i} f}+\frac {\text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{\sqrt {1+i} f} \] Input:

Integrate[Cot[e + f*x]/Sqrt[1 + Tan[e + f*x]],x]
 

Output:

(-2*ArcTanh[Sqrt[1 + Tan[e + f*x]]])/f + ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sq 
rt[1 - I]]/(Sqrt[1 - I]*f) + ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]]/( 
Sqrt[1 + I]*f)
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.14, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {3042, 4057, 25, 3042, 4019, 3042, 4018, 216, 220, 4117, 73, 220}

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.

\(\displaystyle \int \frac {\cot (e+f x)}{\sqrt {\tan (e+f x)+1}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4057

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4019

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4018

\(\displaystyle \int \frac {\tan (e+f x)^2+1}{\tan (e+f x) \sqrt {\tan (e+f x)+1}}dx+\frac {\left (3-2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7-5 \sqrt {2}\right )}d\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}-\frac {\left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7+5 \sqrt {2}\right )}d\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}\)

\(\Big \downarrow \) 216

\(\displaystyle \int \frac {\tan (e+f x)^2+1}{\tan (e+f x) \sqrt {\tan (e+f x)+1}}dx-\frac {\left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7+5 \sqrt {2}\right )}d\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}+\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}\)

\(\Big \downarrow \) 220

\(\displaystyle \int \frac {\tan (e+f x)^2+1}{\tan (e+f x) \sqrt {\tan (e+f x)+1}}dx+\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}+\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {7+5 \sqrt {2}} f}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {\int \frac {\cot (e+f x)}{\sqrt {\tan (e+f x)+1}}d\tan (e+f x)}{f}+\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}+\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {7+5 \sqrt {2}} f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 \int \cot (e+f x)d\sqrt {\tan (e+f x)+1}}{f}+\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}+\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {7+5 \sqrt {2}} f}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}-\frac {2 \text {arctanh}\left (\sqrt {\tan (e+f x)+1}\right )}{f}+\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {7+5 \sqrt {2}} f}\)

Input:

Int[Cot[e + f*x]/Sqrt[1 + Tan[e + f*x]],x]
 

Output:

((3 - 2*Sqrt[2])*ArcTan[(3 - 2*Sqrt[2] + (1 - Sqrt[2])*Tan[e + f*x])/(Sqrt 
[2*(-7 + 5*Sqrt[2])]*Sqrt[1 + Tan[e + f*x]])])/(2*Sqrt[-7 + 5*Sqrt[2]]*f) 
- (2*ArcTanh[Sqrt[1 + Tan[e + f*x]]])/f + ((3 + 2*Sqrt[2])*ArcTanh[(3 + 2* 
Sqrt[2] + (1 + Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(7 + 5*Sqrt[2])]*Sqrt[1 + Ta 
n[e + f*x]])])/(2*Sqrt[7 + 5*Sqrt[2]]*f)
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, 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 220
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 
1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && 
 (LtQ[a, 0] || GtQ[b, 0])
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4018
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[-2*(d^2/f)   Subst[Int[1/(2*b*c*d - 4*a*d^2 
+ x^2), x], x, (b*c - 2*a*d - b*d*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]]], 
x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0 
] && NeQ[c^2 + d^2, 0] && EqQ[2*a*c*d - b*(c^2 - d^2), 0]
 

rule 4019
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> With[{q = Rt[a^2 + b^2, 2]}, Simp[1/(2*q)   Int[( 
a*c + b*d + c*q + (b*c - a*d + d*q)*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]], 
 x], x] - Simp[1/(2*q)   Int[(a*c + b*d - c*q + (b*c - a*d - d*q)*Tan[e + f 
*x])/Sqrt[a + b*Tan[e + f*x]], x], x]] /; FreeQ[{a, b, c, d, e, f}, x] && N 
eQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && NeQ[2*a*c*d - 
 b*(c^2 - d^2), 0] && NiceSqrtQ[a^2 + b^2]
 

rule 4057
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)/((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[1/(c^2 + d^2)   Int[(a + b*Tan[e + f*x])^m 
*(c - d*Tan[e + f*x]), x], x] + Simp[d^2/(c^2 + d^2)   Int[(a + b*Tan[e + f 
*x])^m*((1 + Tan[e + f*x]^2)/(c + d*Tan[e + f*x])), x], x] /; FreeQ[{a, b, 
c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d 
^2, 0] &&  !IntegerQ[m]
 

rule 4117
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[A/f   Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; 
FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
 
Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(2155\) vs. \(2(123)=246\).

Time = 86.68 (sec) , antiderivative size = 2156, normalized size of antiderivative = 13.39

method result size
default \(\text {Expression too large to display}\) \(2156\)

Input:

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

Output:

1/4/f*cot(f*x+e)*(7*(-(cos(f*x+e)+sin(f*x+e))*cos(f*x+e)/(2*sin(f*x+e)^2*2 
^(1/2)-2*2^(1/2)*sin(f*x+e)*cos(f*x+e)-2*sin(f*x+e)^2+2*sin(f*x+e)*cos(f*x 
+e)-1))^(1/2)*arctan(1/4*(-(4+3*2^(1/2))*(cos(f*x+e)+sin(f*x+e))*cos(f*x+e 
)*(3*2^(1/2)-4)/(2*sin(f*x+e)^2*2^(1/2)-2*2^(1/2)*sin(f*x+e)*cos(f*x+e)-2* 
sin(f*x+e)^2+2*sin(f*x+e)*cos(f*x+e)-1))^(1/2)/(2*cos(f*x+e)^2-1)*(4*sin(f 
*x+e)*cos(f*x+e)-1-tan(f*x+e))*(-2+2*2^(1/2))^(1/2)*(2*2^(1/2)+3)*(3*2^(1/ 
2)-4))*2^(1/2)*(-2+2*2^(1/2))^(1/2)*(1+2^(1/2))^(1/2)*sin(f*x+e)-3*(-(cos( 
f*x+e)+sin(f*x+e))*cos(f*x+e)/(2*sin(f*x+e)^2*2^(1/2)-2*2^(1/2)*sin(f*x+e) 
*cos(f*x+e)-2*sin(f*x+e)^2+2*sin(f*x+e)*cos(f*x+e)-1))^(1/2)*arctan(1/4*(- 
(4+3*2^(1/2))*(cos(f*x+e)+sin(f*x+e))*cos(f*x+e)*(3*2^(1/2)-4)/(2*sin(f*x+ 
e)^2*2^(1/2)-2*2^(1/2)*sin(f*x+e)*cos(f*x+e)-2*sin(f*x+e)^2+2*sin(f*x+e)*c 
os(f*x+e)-1))^(1/2)/(2*cos(f*x+e)^2-1)*(4*sin(f*x+e)*cos(f*x+e)-1-tan(f*x+ 
e))*(-2+2*2^(1/2))^(1/2)*(2*2^(1/2)+3)*(3*2^(1/2)-4))*cos(f*x+e)*2^(1/2)*( 
-2+2*2^(1/2))^(1/2)*(1+2^(1/2))^(1/2)-10*(-(cos(f*x+e)+sin(f*x+e))*cos(f*x 
+e)/(2*sin(f*x+e)^2*2^(1/2)-2*2^(1/2)*sin(f*x+e)*cos(f*x+e)-2*sin(f*x+e)^2 
+2*sin(f*x+e)*cos(f*x+e)-1))^(1/2)*arctan(1/4*(-(4+3*2^(1/2))*(cos(f*x+e)+ 
sin(f*x+e))*cos(f*x+e)*(3*2^(1/2)-4)/(2*sin(f*x+e)^2*2^(1/2)-2*2^(1/2)*sin 
(f*x+e)*cos(f*x+e)-2*sin(f*x+e)^2+2*sin(f*x+e)*cos(f*x+e)-1))^(1/2)/(2*cos 
(f*x+e)^2-1)*(4*sin(f*x+e)*cos(f*x+e)-1-tan(f*x+e))*(-2+2*2^(1/2))^(1/2)*( 
2*2^(1/2)+3)*(3*2^(1/2)-4))*(-2+2*2^(1/2))^(1/2)*(1+2^(1/2))^(1/2)*sin(...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (122) = 244\).

Time = 0.09 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.09 \[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {\sqrt {\frac {1}{2}} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \sqrt {\frac {1}{2}} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \sqrt {\frac {1}{2}} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + \sqrt {\frac {1}{2}} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 2 \, \log \left (\sqrt {\tan \left (f x + e\right ) + 1} + 1\right ) - 2 \, \log \left (\sqrt {\tan \left (f x + e\right ) + 1} - 1\right )}{2 \, f} \] Input:

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

Output:

-1/2*(sqrt(1/2)*f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2)*log(sqrt(1/2)*(f^3*sqrt 
(-1/f^4) - f)*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2) + sqrt(tan(f*x + e) + 1)) - 
 sqrt(1/2)*f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2)*log(-sqrt(1/2)*(f^3*sqrt(-1/ 
f^4) - f)*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2) + sqrt(tan(f*x + e) + 1)) - sqr 
t(1/2)*f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2)*log(sqrt(1/2)*(f^3*sqrt(-1/f^4) 
 + f)*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2) + sqrt(tan(f*x + e) + 1)) + sqrt(1 
/2)*f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2)*log(-sqrt(1/2)*(f^3*sqrt(-1/f^4) + 
 f)*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2) + sqrt(tan(f*x + e) + 1)) + 2*log(sq 
rt(tan(f*x + e) + 1) + 1) - 2*log(sqrt(tan(f*x + e) + 1) - 1))/f
 

Sympy [F]

\[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\int \frac {\cot {\left (e + f x \right )}}{\sqrt {\tan {\left (e + f x \right )} + 1}}\, dx \] Input:

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

Output:

Integral(cot(e + f*x)/sqrt(tan(e + f*x) + 1), x)
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\int { \frac {\cot \left (f x + e\right )}{\sqrt {\tan \left (f x + e\right ) + 1}} \,d x } \] Input:

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

Output:

integrate(cot(f*x + e)/sqrt(tan(f*x + e) + 1), x)
 

Giac [F]

\[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\int { \frac {\cot \left (f x + e\right )}{\sqrt {\tan \left (f x + e\right ) + 1}} \,d x } \] Input:

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

Output:

integrate(cot(f*x + e)/sqrt(tan(f*x + e) + 1), x)
 

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.53 \[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {2\,\mathrm {atanh}\left (\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )}{f}+2\,\mathrm {atanh}\left (2\,f\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}+2\,\mathrm {atanh}\left (2\,f\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}} \] Input:

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

Output:

2*atanh(2*f*((1/8 - 1i/8)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2))*((1/8 - 1i/ 
8)/f^2)^(1/2) - (2*atanh((tan(e + f*x) + 1)^(1/2)))/f + 2*atanh(2*f*((1/8 
+ 1i/8)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2))*((1/8 + 1i/8)/f^2)^(1/2)
 

Reduce [F]

\[ \int \frac {\cot (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\int \frac {\sqrt {\tan \left (f x +e \right )+1}\, \cot \left (f x +e \right )}{\tan \left (f x +e \right )+1}d x \] Input:

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

Output:

int((sqrt(tan(e + f*x) + 1)*cot(e + f*x))/(tan(e + f*x) + 1),x)