\(\int \tan (e+f x) \sqrt {1+\tan (e+f x)} \, dx\) [380]

Optimal result

Integrand size = 19, antiderivative size = 166 \[ \int \tan (e+f x) \sqrt {1+\tan (e+f x)} \, dx=-\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {2}\right )} \arctan \left (\frac {4-3 \sqrt {2}+\left (2-\sqrt {2}\right ) \tan (e+f x)}{2 \sqrt {-7+5 \sqrt {2}} \sqrt {1+\tan (e+f x)}}\right )}{f}-\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \text {arctanh}\left (\frac {4+3 \sqrt {2}+\left (2+\sqrt {2}\right ) \tan (e+f x)}{2 \sqrt {7+5 \sqrt {2}} \sqrt {1+\tan (e+f x)}}\right )}{f}+\frac {2 \sqrt {1+\tan (e+f x)}}{f} \] Output:

-1/2*(-2+2*2^(1/2))^(1/2)*arctan(1/2*(4-3*2^(1/2)+(2-2^(1/2))*tan(f*x+e))/ 
(-7+5*2^(1/2))^(1/2)/(1+tan(f*x+e))^(1/2))/f-1/2*(2+2*2^(1/2))^(1/2)*arcta 
nh(1/2*(4+3*2^(1/2)+(2+2^(1/2))*tan(f*x+e))/(7+5*2^(1/2))^(1/2)/(1+tan(f*x 
+e))^(1/2))/f+2*(1+tan(f*x+e))^(1/2)/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.47 \[ \int \tan (e+f x) \sqrt {1+\tan (e+f x)} \, dx=-\frac {\sqrt {1-i} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )+\sqrt {1+i} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )-2 \sqrt {1+\tan (e+f x)}}{f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3042, 4011, 3042, 4019, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4019

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4018

\(\displaystyle \frac {\sqrt {2} \left (3-2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (2-\sqrt {2}\right ) \tan (e+f x)-3 \sqrt {2}+4\right )^2}{\tan (e+f x)+1}-4 \left (7-5 \sqrt {2}\right )}d\left (-\frac {\left (2-\sqrt {2}\right ) \tan (e+f x)-3 \sqrt {2}+4}{\sqrt {\tan (e+f x)+1}}\right )}{f}+\frac {\sqrt {2} \left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4\right )^2}{\tan (e+f x)+1}-4 \left (7+5 \sqrt {2}\right )}d\frac {\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4}{\sqrt {\tan (e+f x)+1}}}{f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\sqrt {2} \left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4\right )^2}{\tan (e+f x)+1}-4 \left (7+5 \sqrt {2}\right )}d\frac {\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4}{\sqrt {\tan (e+f x)+1}}}{f}-\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (2-\sqrt {2}\right ) \tan (e+f x)-3 \sqrt {2}+4}{2 \sqrt {5 \sqrt {2}-7} \sqrt {\tan (e+f x)+1}}\right )}{\sqrt {2 \left (5 \sqrt {2}-7\right )} f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\)

\(\Big \downarrow \) 220

\(\displaystyle -\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (2-\sqrt {2}\right ) \tan (e+f x)-3 \sqrt {2}+4}{2 \sqrt {5 \sqrt {2}-7} \sqrt {\tan (e+f x)+1}}\right )}{\sqrt {2 \left (5 \sqrt {2}-7\right )} f}-\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4}{2 \sqrt {7+5 \sqrt {2}} \sqrt {\tan (e+f x)+1}}\right )}{\sqrt {2 \left (7+5 \sqrt {2}\right )} f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\)

Input:

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

Output:

-(((3 - 2*Sqrt[2])*ArcTan[(4 - 3*Sqrt[2] + (2 - Sqrt[2])*Tan[e + f*x])/(2* 
Sqrt[-7 + 5*Sqrt[2]]*Sqrt[1 + Tan[e + f*x]])])/(Sqrt[2*(-7 + 5*Sqrt[2])]*f 
)) - ((3 + 2*Sqrt[2])*ArcTanh[(4 + 3*Sqrt[2] + (2 + Sqrt[2])*Tan[e + f*x]) 
/(2*Sqrt[7 + 5*Sqrt[2]]*Sqrt[1 + Tan[e + f*x]])])/(Sqrt[2*(7 + 5*Sqrt[2])] 
*f) + (2*Sqrt[1 + Tan[e + f*x]])/f
 

Defintions of rubi rules used

rule 25

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

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 4011

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

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 = 1.00 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.24

method	result	size
derivativedivides	\(\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4}-\frac {\left (\sqrt {2}-1\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}}{f}\)	\(206\)
default	\(\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4}-\frac {\left (\sqrt {2}-1\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}}{f}\)	\(206\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.46 \[ \int \tan (e+f x) \sqrt {1+\tan (e+f x)} \, dx=-\frac {f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 4 \, \sqrt {\tan \left (f x + e\right ) + 1}}{2 \, f} \] Input:

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

Output:

-1/2*(f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2)*log(f*sqrt((f^2*sqrt(-1/f^4) + 1) 
/f^2) + sqrt(tan(f*x + e) + 1)) - f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2)*log(- 
f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2) + sqrt(tan(f*x + e) + 1)) + f*sqrt(-(f^ 
2*sqrt(-1/f^4) - 1)/f^2)*log(f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2) + sqrt(ta 
n(f*x + e) + 1)) - f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2)*log(-f*sqrt(-(f^2*s 
qrt(-1/f^4) - 1)/f^2) + sqrt(tan(f*x + e) + 1)) - 4*sqrt(tan(f*x + e) + 1) 
)/f
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \tan (e+f x) \sqrt {1+\tan (e+f x)} \, dx=\text {Exception raised: TypeError} \] 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:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[9]%%%}+%%%{4,[7]%%%}+%%%{6,[5]%%%}+%%%{4,[3]%%%}+%% 
%{1,[1]%%
 

Mupad [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.54 \[ \int \tan (e+f x) \sqrt {1+\tan (e+f x)} \, dx=\frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}-\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,\left (1-\mathrm {i}\right )\right )\,\sqrt {\frac {\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,\left (1+1{}\mathrm {i}\right )\right )\,\sqrt {\frac {\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \] Input:

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

Output:

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

Reduce [F]

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