\(\int \frac {\tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\) [402]

Optimal result

Integrand size = 19, antiderivative size = 143 \[ \int \frac {\tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {\sqrt {-1+\sqrt {2}} \arctan \left (\frac {3-2 \sqrt {2}+\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (-7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f}-\frac {\sqrt {1+\sqrt {2}} \text {arctanh}\left (\frac {3+2 \sqrt {2}+\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f} \] 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-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 = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.47 \[ \int \frac {\tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )}{\sqrt {1-i} f}-\frac {\text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{\sqrt {1+i} f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 4019, 3042, 4018, 216, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4019

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4018

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 220

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

Input:

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

Output:

-1/2*((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]])])/(Sqrt[-7 + 5*Sqrt[2]]* 
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 + Tan[e + f*x]])])/(2*Sqrt[7 + 5*Sqrt[2]] 
*f)
 

Defintions of rubi rules used

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]
 

Maple [A] (verified)

Time = 1.16 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.46

method	result	size
derivativedivides	\(\frac {-\frac {\sqrt {2}\, \left (-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{4}-\frac {\sqrt {2}\, \left (\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{4}}{f}\)	\(209\)
default	\(\frac {-\frac {\sqrt {2}\, \left (-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{4}-\frac {\sqrt {2}\, \left (\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{4}}{f}\)	\(209\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (106) = 212\).

Time = 0.08 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.09 \[ \int \frac {\tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - \frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + \frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) \] Input:

integrate(tan(f*x+e)/(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*s 
qrt(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*sq 
rt(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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Warning, need to choose a branch fo 
r the root of a polynomial with parameters. This might be wrong.Non regula 
r value [
 

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.48 \[ \int \frac {\tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-2\,\mathrm {atanh}\left (2\,f\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}-2\,\mathrm {atanh}\left (2\,f\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}} \] Input:

int(tan(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 - 1 
i/8)/f^2)^(1/2) - 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]

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

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

Output:

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