Integrand size = 12, antiderivative size = 247 \[ \int \sqrt {1+\tan (e+f x)} \, dx=-\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{f}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f} \]
-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))*(2+2*2^(1/2))^(1/2)/f+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))*(2+2*2^(1/2))^(1/2)/f+1/2*ln(1+2^(1/2)-(2+2 *2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))/f/(2+2*2^(1/2))^(1/2)-1/2 *ln(1+2^(1/2)+(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))/f/(2+2* 2^(1/2))^(1/2)
Result contains complex when optimal does not.
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.27 \[ \int \sqrt {1+\tan (e+f x)} \, dx=-\frac {i \left (\sqrt {1-i} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )-\sqrt {1+i} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )\right )}{f} \]
((-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
Time = 0.44 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.00, 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 |
| \(\displaystyle \frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}-\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{\sqrt {2 \left (\sqrt {2}-1\right )}}+\frac {\arctan \left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )+\frac {1}{2} \left (\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )}}-\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )}}\right )\right )}{f}\) |
(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
3.4.86.3.1 Defintions of rubi rules used
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])
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]
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]
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]
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]
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]))
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]
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]
Time = 0.99 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.17
| method | result | size |
| derivativedivides | \(\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}-\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {\sqrt {2+2 \sqrt {2}}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}\) | \(288\) |
| default | \(\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}-\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {\sqrt {2+2 \sqrt {2}}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}\) | \(288\) |
1/4/f*(2+2*2^(1/2))^(1/2)*2^(1/2)*ln(1+2^(1/2)-(2+2*2^(1/2))^(1/2)*(1+tan( f*x+e))^(1/2)+tan(f*x+e))+1/f/(-2+2*2^(1/2))^(1/2)*arctan((2*(1+tan(f*x+e) )^(1/2)-(2+2*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))-1/4/f*(2+2*2^(1/2))^(1/ 2)*ln(1+2^(1/2)-(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))-1/4/f *(2+2*2^(1/2))^(1/2)*2^(1/2)*ln(1+2^(1/2)+(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e ))^(1/2)+tan(f*x+e))+1/f/(-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))+1/4/f*(2+2*2^(1/2))^(1/2)*ln (1+2^(1/2)+(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))
Time = 0.25 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.05 \[ \int \sqrt {1+\tan (e+f x)} \, dx=-\frac {1}{2} \, \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (f^{3} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + \frac {1}{2} \, \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-f^{3} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + \frac {1}{2} \, \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (f^{3} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \frac {1}{2} \, \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-f^{3} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \sqrt {-\frac {1}{f^{4}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) \]
-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))
\[ \int \sqrt {1+\tan (e+f x)} \, dx=\int \sqrt {\tan {\left (e + f x \right )} + 1}\, dx \]
Exception generated. \[ \int \sqrt {1+\tan (e+f x)} \, dx=\text {Exception raised: RuntimeError} \]
Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is 1which is not of the expected type LIST
Time = 0.74 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.05 \[ \int \sqrt {1+\tan (e+f x)} \, dx=\frac {{\left (f^{2} \sqrt {\sqrt {2} + 1} + f \sqrt {\sqrt {2} - 1} {\left | f \right |}\right )} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f^{3}} + \frac {{\left (f^{2} \sqrt {\sqrt {2} + 1} + f \sqrt {\sqrt {2} - 1} {\left | f \right |}\right )} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f^{3}} + \frac {{\left (f^{2} \sqrt {\sqrt {2} - 1} - f \sqrt {\sqrt {2} + 1} {\left | f \right |}\right )} \log \left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f^{3}} - \frac {{\left (f^{2} \sqrt {\sqrt {2} - 1} - f \sqrt {\sqrt {2} + 1} {\left | f \right |}\right )} \log \left (-2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f^{3}} \]
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
Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.30 \[ \int \sqrt {1+\tan (e+f x)} \, dx=2\,\mathrm {atanh}\left (4\,f^3\,{\left (\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}\right )}^{3/2}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}+2\,\mathrm {atanh}\left (4\,f^3\,{\left (\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}\right )}^{3/2}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}} \]