3.5.7 \(\int \frac {\tan ^2(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\) [407]

3.5.7.1 Optimal result

Integrand size = 21, antiderivative size = 257 \[ \int \frac {\tan ^2(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{2 f}-\frac {\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{2 f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}+\frac {2 \sqrt {1+\tan (e+f x)}}{f} \]

output

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

3.5.7.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.31 \[ \int \frac {\tan ^2(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {(1-i)^{3/2} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )+(1+i)^{3/2} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )-4 \sqrt {1+\tan (e+f x)}}{2 f} \]

input

Integrate[Tan[e + f*x]^2/Sqrt[1 + Tan[e + f*x]],x]
 

output

-1/2*((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]] - 4*Sqrt[1 + Tan[e + f*x 
]])/f
 

3.5.7.3 Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4026, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4026

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3966

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

\(\Big \downarrow \) 484

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

\(\Big \downarrow \) 1407

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

input

Int[Tan[e + f*x]^2/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 + Sq 
rt[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 
+ Sqrt[2]])))/f + (2*Sqrt[1 + Tan[e + f*x]])/f
 

3.5.7.3.1 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]
 

rule 4026

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

3.5.7.4 Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(\frac {2 \sqrt {1+\tan \left (f x +e \right )}}{f}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}-\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8 f}-\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{2 f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8 f}-\frac {\arctan \left (\frac {\sqrt {2+2 \sqrt {2}}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{2 f \sqrt {-2+2 \sqrt {2}}}\)	\(311\)
default	\(\frac {2 \sqrt {1+\tan \left (f x +e \right )}}{f}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}-\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}-\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8 f}-\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{2 f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (1+\sqrt {2}+\sqrt {2+2 \sqrt {2}}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8 f}-\frac {\arctan \left (\frac {\sqrt {2+2 \sqrt {2}}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{2 f \sqrt {-2+2 \sqrt {2}}}\)	\(311\)

input

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

output

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

3.5.7.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.24 \[ \int \frac {\tan ^2(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {\sqrt {\frac {1}{2}} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \sqrt {\frac {1}{2}} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \sqrt {\frac {1}{2}} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + \sqrt {\frac {1}{2}} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 4 \, \sqrt {\tan \left (f x + e\right ) + 1}}{2 \, f} \]

input

integrate(tan(f*x+e)^2/(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*sqr 
t(-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)) - 
 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)) - 4*sqrt(t 
an(f*x + e) + 1))/f
 

3.5.7.6 Sympy [F]

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

input

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

output

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

3.5.7.7 Maxima [F]

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

input

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

output

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

3.5.7.8 Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^2(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.5.7.9 Mupad [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.33 \[ \int \frac {\tan ^2(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}-\mathrm {atan}\left (2\,f\,\sqrt {\frac {-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (2\,f\,\sqrt {\frac {-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \]

input

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

output

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