\(\int \frac {1}{\sqrt {1+\tan (e+f x)}} \, dx\) [408]

Optimal result

Integrand size = 12, antiderivative size = 183 \[ \int \frac {1}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{2 f}+\frac {\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{2 f}+\frac {\text {arctanh}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}{1+\sqrt {2}+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.36 \[ \int \frac {1}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {(1-i)^{3/2} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )+(1+i)^{3/2} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{2 f} \] Input:

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

Output:

((1 - I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]] + (1 + I)^(3/2) 
*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]])/(2*f)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.42, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 3966, 484, 1407, 1142, 25, 1083, 217, 1103}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3966

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

\(\Big \downarrow \) 484

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

\(\Big \downarrow \) 1407

\(\displaystyle \frac {2 \left (\frac {\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-\sqrt {\tan (e+f x)+1}}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{4 \sqrt {1+\sqrt {2}}}+\frac {\int \frac {\sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{4 \sqrt {1+\sqrt {2}}}\right )}{f}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {2 \left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \int \frac {1}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}-\frac {1}{2} \int -\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{4 \sqrt {1+\sqrt {2}}}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \int \frac {1}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}+\frac {1}{2} \int \frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{4 \sqrt {1+\sqrt {2}}}\right )}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \int \frac {1}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}+\frac {1}{2} \int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{4 \sqrt {1+\sqrt {2}}}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \int \frac {1}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}+\frac {1}{2} \int \frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}}{4 \sqrt {1+\sqrt {2}}}\right )}{f}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {2 \left (\frac {\frac {1}{2} \int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )} \int \frac {1}{-\tan (e+f x)+2 \left (1-\sqrt {2}\right )-1}d\left (2 \sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )}\right )}{4 \sqrt {1+\sqrt {2}}}+\frac {\frac {1}{2} \int \frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )} \int \frac {1}{-\tan (e+f x)+2 \left (1-\sqrt {2}\right )-1}d\left (2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}\right )}{4 \sqrt {1+\sqrt {2}}}\right )}{f}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {2 \left (\frac {\frac {1}{2} \int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}+\sqrt {\frac {1+\sqrt {2}}{\sqrt {2}-1}} \arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{4 \sqrt {1+\sqrt {2}}}+\frac {\frac {1}{2} \int \frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1}d\sqrt {\tan (e+f x)+1}+\sqrt {\frac {1+\sqrt {2}}{\sqrt {2}-1}} \arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{4 \sqrt {1+\sqrt {2}}}\right )}{f}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {2 \left (\frac {\sqrt {\frac {1+\sqrt {2}}{\sqrt {2}-1}} \arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )-\frac {1}{2} \log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}}}+\frac {\sqrt {\frac {1+\sqrt {2}}{\sqrt {2}-1}} \arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )+\frac {1}{2} \log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}}}\right )}{f}\)

Input:

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

Output:

(2*((Sqrt[(1 + Sqrt[2])/(-1 + Sqrt[2])]*ArcTan[(-Sqrt[2*(1 + Sqrt[2])] + 2 
*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]] - Log[1 + Sqrt[2] + Tan[e 
 + f*x] - Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/2)/(4*Sqrt[1 + Sqr 
t[2]]) + (Sqrt[(1 + Sqrt[2])/(-1 + Sqrt[2])]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] 
 + 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]] + Log[1 + Sqrt[2] + T 
an[e + f*x] + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/2)/(4*Sqrt[1 + 
 Sqrt[2]])))/f
 

Defintions of rubi rules used

rule 25

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

rule 217

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

rule 484

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*((a_) + (b_.)*(x_)^2)), x_Symbol] :> Simp[2* 
d   Subst[Int[1/(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4), x], x, Sqrt[c + d*x]], 
 x] /; FreeQ[{a, b, c, d}, x]
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1407

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[a/ 
c, 2]}, With[{r = Rt[2*q - b/c, 2]}, Simp[1/(2*c*q*r)   Int[(r - x)/(q - r* 
x + x^2), x], x] + Simp[1/(2*c*q*r)   Int[(r + x)/(q + r*x + x^2), x], x]]] 
 /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[b^2 - 4*a*c]
 

rule 3042

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

rule 3966

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Su 
bst[Int[(a + x)^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c 
, d, n}, x] && NeQ[a^2 + b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(295\) vs. \(2(137)=274\).

Time = 1.17 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.62

method	result	size
derivativedivides	\(-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}+\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{8 f}+\frac {\arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{2 f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}-\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{8 f}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{2 f \sqrt {-2+2 \sqrt {2}}}\)	\(296\)
default	\(-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}+\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{8 f}+\frac {\arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{2 f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}-\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{8 f}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{2 f \sqrt {-2+2 \sqrt {2}}}\)	\(296\)

Input:

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

Output:

-1/4/f*(2*2^(1/2)+2)^(1/2)*ln(tan(f*x+e)+1-(tan(f*x+e)+1)^(1/2)*(2*2^(1/2) 
+2)^(1/2)+2^(1/2))+1/8/f*(2*2^(1/2)+2)^(1/2)*2^(1/2)*ln(tan(f*x+e)+1-(tan( 
f*x+e)+1)^(1/2)*(2*2^(1/2)+2)^(1/2)+2^(1/2))+1/2/f/(-2+2*2^(1/2))^(1/2)*ar 
ctan((2*(tan(f*x+e)+1)^(1/2)-(2*2^(1/2)+2)^(1/2))/(-2+2*2^(1/2))^(1/2))*2^ 
(1/2)+1/4/f*(2*2^(1/2)+2)^(1/2)*ln(tan(f*x+e)+1+(tan(f*x+e)+1)^(1/2)*(2*2^ 
(1/2)+2)^(1/2)+2^(1/2))-1/8/f*(2*2^(1/2)+2)^(1/2)*2^(1/2)*ln(tan(f*x+e)+1+ 
(tan(f*x+e)+1)^(1/2)*(2*2^(1/2)+2)^(1/2)+2^(1/2))+1/2/f/(-2+2*2^(1/2))^(1/ 
2)*arctan(((2*2^(1/2)+2)^(1/2)+2*(tan(f*x+e)+1)^(1/2))/(-2+2*2^(1/2))^(1/2 
))*2^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (137) = 274\).

Time = 0.08 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.63 \[ \int \frac {1}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + \frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) \] Input:

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

Output:

1/2*sqrt(1/2)*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)) - 1 
/2*sqrt(1/2)*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)) - 1 
/2*sqrt(1/2)*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)) + 1/2* 
sqrt(1/2)*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))
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {1+\tan (e+f x)}} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: In function CAR, the value of 
the first argument is  1which is not of the expected type LIST
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (137) = 274\).

Time = 0.23 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.51 \[ \int \frac {1}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {{\left (f^{2} \sqrt {2 \, \sqrt {2} - 2} + f \sqrt {2 \, \sqrt {2} + 2} {\left | f \right |}\right )} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{4 \, f^{3}} + \frac {{\left (f^{2} \sqrt {2 \, \sqrt {2} - 2} + f \sqrt {2 \, \sqrt {2} + 2} {\left | f \right |}\right )} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{4 \, f^{3}} + \frac {{\left (f^{2} \sqrt {2 \, \sqrt {2} + 2} - f \sqrt {2 \, \sqrt {2} - 2} {\left | f \right |}\right )} \log \left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{8 \, f^{3}} - \frac {{\left (f^{2} \sqrt {2 \, \sqrt {2} + 2} - f \sqrt {2 \, \sqrt {2} - 2} {\left | f \right |}\right )} \log \left (-2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{8 \, f^{3}} \] Input:

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

Output:

1/4*(f^2*sqrt(2*sqrt(2) - 2) + f*sqrt(2*sqrt(2) + 2)*abs(f))*arctan(1/2*2^ 
(3/4)*(2^(1/4)*sqrt(sqrt(2) + 2) + 2*sqrt(tan(f*x + e) + 1))/sqrt(-sqrt(2) 
 + 2))/f^3 + 1/4*(f^2*sqrt(2*sqrt(2) - 2) + f*sqrt(2*sqrt(2) + 2)*abs(f))* 
arctan(-1/2*2^(3/4)*(2^(1/4)*sqrt(sqrt(2) + 2) - 2*sqrt(tan(f*x + e) + 1)) 
/sqrt(-sqrt(2) + 2))/f^3 + 1/8*(f^2*sqrt(2*sqrt(2) + 2) - f*sqrt(2*sqrt(2) 
 - 2)*abs(f))*log(2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + sqrt( 
2) + tan(f*x + e) + 1)/f^3 - 1/8*(f^2*sqrt(2*sqrt(2) + 2) - f*sqrt(2*sqrt( 
2) - 2)*abs(f))*log(-2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + sq 
rt(2) + tan(f*x + e) + 1)/f^3
 

Mupad [B] (verification not implemented)

Time = 1.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.39 \[ \int \frac {1}{\sqrt {1+\tan (e+f x)}} \, dx=\mathrm {atan}\left (2\,f\,\sqrt {\frac {-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (2\,f\,\sqrt {\frac {-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \] Input:

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

Output:

atan(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)*2i - atan(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)*2i
 

Reduce [F]

\[ \int \frac {1}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\left (\int \frac {\sqrt {\tan \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )+1}d x \right ) f}{f} \] Input:

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

Output:

(2*sqrt(tan(e + f*x) + 1) - int((sqrt(tan(e + f*x) + 1)*tan(e + f*x)**2)/( 
tan(e + f*x) + 1),x)*f)/f