\(\int \sqrt {1+\tan (e+f x)} \, dx\) [386]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F(-2)]
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 12, antiderivative size = 186 \[ \int \sqrt {1+\tan (e+f x)} \, dx=-\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{f}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{f}-\frac {\text {arctanh}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}{1+\sqrt {2}+\tan (e+f x)}\right )}{\sqrt {2 \left (1+\sqrt {2}\right )} f} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.36 \[ \int \sqrt {1+\tan (e+f x)} \, dx=-\frac {i \left (\sqrt {1-i} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )-\sqrt {1+i} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )\right )}{f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.33, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {3042, 3966, 483, 1447, 1475, 1083, 217, 1478, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3966

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

\(\Big \downarrow \) 483

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

\(\Big \downarrow \) 1447

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

\(\Big \downarrow \) 1475

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1478

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(2*((ArcTan[(-Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 
 + Sqrt[2])]]/Sqrt[2*(-1 + Sqrt[2])] + ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*S 
qrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]]/Sqrt[2*(-1 + Sqrt[2])])/2 + 
 (Log[1 + Sqrt[2] + Tan[e + f*x] - Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + 
f*x]]]/(2*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*Sqrt[2*(1 + Sqrt[2])]))/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 483
Int[Sqrt[(c_) + (d_.)*(x_)]/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[2*d 
Subst[Int[x^2/(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 1447
Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a/c, 2]}, Simp[1/2   Int[(q + x^2)/(a + b*x^2 + c*x^4), x], x] - Simp[1/2 
 Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && LtQ[b 
^2 - 4*a*c, 0] && PosQ[a*c]
 

rule 1475
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[2*(d/e) - b/c, 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^ 
2, x], x], x] + Simp[e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; F 
reeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && 
 (GtQ[2*(d/e) - b/c, 0] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2] 
, 0]))
 

rule 1478
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[-2*(d/e) - b/c, 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e 
 + q*x - x^2, x], x], x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - 
x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ 
[c*d^2 - a*e^2, 0] &&  !GtQ[b^2 - 4*a*c, 0]
 

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. \(287\) vs. \(2(143)=286\).

Time = 1.59 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.55

method result size
derivativedivides \(\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}+\frac {\arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}-\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}\) \(288\)
default \(\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}+\frac {\arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}-\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4 f}\) \(288\)

Input:

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

Output:

1/4/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/f/(-2+2*2^(1/2))^(1/2)*arctan((2*(tan(f*x+e)+1 
)^(1/2)-(2*2^(1/2)+2)^(1/2))/(-2+2*2^(1/2))^(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))-1/4/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/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))+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))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.39 \[ \int \sqrt {1+\tan (e+f x)} \, dx=-\frac {1}{2} \, \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (f^{3} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + \frac {1}{2} \, \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-f^{3} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + \frac {1}{2} \, \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (f^{3} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \frac {1}{2} \, \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-f^{3} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((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 [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.40 \[ \int \sqrt {1+\tan (e+f x)} \, dx=\frac {{\left (f^{2} \sqrt {\sqrt {2} + 1} + f \sqrt {\sqrt {2} - 1} {\left | f \right |}\right )} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f^{3}} + \frac {{\left (f^{2} \sqrt {\sqrt {2} + 1} + f \sqrt {\sqrt {2} - 1} {\left | f \right |}\right )} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f^{3}} + \frac {{\left (f^{2} \sqrt {\sqrt {2} - 1} - f \sqrt {\sqrt {2} + 1} {\left | f \right |}\right )} \log \left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f^{3}} - \frac {{\left (f^{2} \sqrt {\sqrt {2} - 1} - f \sqrt {\sqrt {2} + 1} {\left | f \right |}\right )} \log \left (-2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f^{3}} \] Input:

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

Output:

1/2*(f^2*sqrt(sqrt(2) + 1) + f*sqrt(sqrt(2) - 1)*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 + 1/2*(f^2*sqrt(sqrt(2) + 1) + f*sqrt(sqrt(2) - 1)*abs(f))*arctan(- 
1/2*2^(3/4)*(2^(1/4)*sqrt(sqrt(2) + 2) - 2*sqrt(tan(f*x + e) + 1))/sqrt(-s 
qrt(2) + 2))/f^3 + 1/4*(f^2*sqrt(sqrt(2) - 1) - f*sqrt(sqrt(2) + 1)*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 - 1/4*(f^2*sqrt(sqrt(2) - 1) - f*sqrt(sqrt(2) + 1)*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
 

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.39 \[ \int \sqrt {1+\tan (e+f x)} \, dx=2\,\mathrm {atanh}\left (4\,f^3\,{\left (\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}\right )}^{3/2}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}+2\,\mathrm {atanh}\left (4\,f^3\,{\left (\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}\right )}^{3/2}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}} \] Input:

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

Output:

2*atanh(4*f^3*((- 1/4 - 1i/4)/f^2)^(3/2)*(tan(e + f*x) + 1)^(1/2))*((- 1/4 
 - 1i/4)/f^2)^(1/2) + 2*atanh(4*f^3*((- 1/4 + 1i/4)/f^2)^(3/2)*(tan(e + f* 
x) + 1)^(1/2))*((- 1/4 + 1i/4)/f^2)^(1/2)
 

Reduce [F]

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

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

Output:

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