\(\int \tan ^5(e+f x) \sqrt {1+\tan (e+f x)} \, dx\) [378]

Optimal result

Integrand size = 21, antiderivative size = 264 \[ \int \tan ^5(e+f x) \sqrt {1+\tan (e+f x)} \, dx=-\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {2}\right )} \arctan \left (\frac {4-3 \sqrt {2}+\left (2-\sqrt {2}\right ) \tan (e+f x)}{2 \sqrt {-7+5 \sqrt {2}} \sqrt {1+\tan (e+f x)}}\right )}{f}-\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \text {arctanh}\left (\frac {4+3 \sqrt {2}+\left (2+\sqrt {2}\right ) \tan (e+f x)}{2 \sqrt {7+5 \sqrt {2}} \sqrt {1+\tan (e+f x)}}\right )}{f}+\frac {2 \sqrt {1+\tan (e+f x)}}{f}+\frac {52 (1+\tan (e+f x))^{3/2}}{315 f}-\frac {26 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{105 f}-\frac {4 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{21 f}+\frac {2 \tan ^3(e+f x) (1+\tan (e+f x))^{3/2}}{9 f} \] 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+52/315*(1+tan(f*x+e))^(3/2)/f-26/10 
5*tan(f*x+e)*(1+tan(f*x+e))^(3/2)/f-4/21*tan(f*x+e)^2*(1+tan(f*x+e))^(3/2) 
/f+2/9*tan(f*x+e)^3*(1+tan(f*x+e))^(3/2)/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.52 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.57 \[ \int \tan ^5(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\frac {\cos (e+f x) (1+\tan (e+f x)) \left (-315 \sqrt {1-i} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )-315 \sqrt {1+i} \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )+2 \sqrt {1+\tan (e+f x)} \left (445+35 \sec ^4(e+f x)-18 \tan (e+f x)+\sec ^2(e+f x) (-139+5 \tan (e+f x))\right )\right )}{315 f (\cos (e+f x)+\sin (e+f x))} \] Input:

Integrate[Tan[e + f*x]^5*Sqrt[1 + Tan[e + f*x]],x]
 

Output:

(Cos[e + f*x]*(1 + Tan[e + f*x])*(-315*Sqrt[1 - I]*ArcTanh[Sqrt[1 + Tan[e 
+ f*x]]/Sqrt[1 - I]] - 315*Sqrt[1 + I]*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt 
[1 + I]] + 2*Sqrt[1 + Tan[e + f*x]]*(445 + 35*Sec[e + f*x]^4 - 18*Tan[e + 
f*x] + Sec[e + f*x]^2*(-139 + 5*Tan[e + f*x]))))/(315*f*(Cos[e + f*x] + Si 
n[e + f*x]))
 

Rubi [A] (verified)

Time = 1.30 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.14, number of steps used = 22, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4049, 27, 3042, 4130, 27, 3042, 4131, 27, 3042, 4113, 27, 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.

\(\displaystyle \int \tan ^5(e+f x) \sqrt {\tan (e+f x)+1} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x)^5 \sqrt {\tan (e+f x)+1}dx\)

\(\Big \downarrow \) 4049

\(\displaystyle \frac {2}{9} \int -\frac {3}{2} \tan ^2(e+f x) \sqrt {\tan (e+f x)+1} \left (2 \tan ^2(e+f x)+3 \tan (e+f x)+2\right )dx+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan ^3(e+f x) (\tan (e+f x)+1)^{3/2}}{9 f}-\frac {1}{3} \int \tan ^2(e+f x) \sqrt {\tan (e+f x)+1} \left (2 \tan ^2(e+f x)+3 \tan (e+f x)+2\right )dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^3(e+f x) (\tan (e+f x)+1)^{3/2}}{9 f}-\frac {1}{3} \int \tan (e+f x)^2 \sqrt {\tan (e+f x)+1} \left (2 \tan (e+f x)^2+3 \tan (e+f x)+2\right )dx\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {1}{3} \left (-\frac {2}{7} \int -\frac {1}{2} \tan (e+f x) \sqrt {\tan (e+f x)+1} \left (8-13 \tan ^2(e+f x)\right )dx-\frac {4 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \int \tan (e+f x) \sqrt {\tan (e+f x)+1} \left (8-13 \tan ^2(e+f x)\right )dx-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \int \tan (e+f x) \sqrt {\tan (e+f x)+1} \left (8-13 \tan (e+f x)^2\right )dx-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 4131

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {2}{5} \int \frac {1}{2} \sqrt {\tan (e+f x)+1} \left (26 \tan ^2(e+f x)+105 \tan (e+f x)+26\right )dx-\frac {26 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {1}{5} \int \sqrt {\tan (e+f x)+1} \left (26 \tan ^2(e+f x)+105 \tan (e+f x)+26\right )dx-\frac {26 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {1}{5} \int \sqrt {\tan (e+f x)+1} \left (26 \tan (e+f x)^2+105 \tan (e+f x)+26\right )dx-\frac {26 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {1}{5} \left (\int 105 \tan (e+f x) \sqrt {\tan (e+f x)+1}dx+\frac {52 (\tan (e+f x)+1)^{3/2}}{3 f}\right )-\frac {26 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {1}{5} \left (105 \int \tan (e+f x) \sqrt {\tan (e+f x)+1}dx+\frac {52 (\tan (e+f x)+1)^{3/2}}{3 f}\right )-\frac {26 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {1}{5} \left (105 \int \tan (e+f x) \sqrt {\tan (e+f x)+1}dx+\frac {52 (\tan (e+f x)+1)^{3/2}}{3 f}\right )-\frac {26 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {1}{5} \left (105 \left (\int \frac {\tan (e+f x)-1}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )+\frac {52 (\tan (e+f x)+1)^{3/2}}{3 f}\right )-\frac {26 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {1}{5} \left (105 \left (\int \frac {\tan (e+f x)-1}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )+\frac {52 (\tan (e+f x)+1)^{3/2}}{3 f}\right )-\frac {26 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 4019

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {1}{5} \left (105 \left (\frac {\int -\frac {\left (2-\sqrt {2}\right ) \tan (e+f x)+\sqrt {2}}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2}-\left (2+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )+\frac {52 (\tan (e+f x)+1)^{3/2}}{3 f}\right )-\frac {26 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {1}{5} \left (105 \left (-\frac {\int \frac {\left (2-\sqrt {2}\right ) \tan (e+f x)+\sqrt {2}}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2}-\left (2+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )+\frac {52 (\tan (e+f x)+1)^{3/2}}{3 f}\right )-\frac {26 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {1}{5} \left (105 \left (-\frac {\int \frac {\left (2-\sqrt {2}\right ) \tan (e+f x)+\sqrt {2}}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2}-\left (2+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )+\frac {52 (\tan (e+f x)+1)^{3/2}}{3 f}\right )-\frac {26 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 4018

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {1}{5} \left (105 \left (\frac {\sqrt {2} \left (3-2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (2-\sqrt {2}\right ) \tan (e+f x)-3 \sqrt {2}+4\right )^2}{\tan (e+f x)+1}-4 \left (7-5 \sqrt {2}\right )}d\left (-\frac {\left (2-\sqrt {2}\right ) \tan (e+f x)-3 \sqrt {2}+4}{\sqrt {\tan (e+f x)+1}}\right )}{f}+\frac {\sqrt {2} \left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4\right )^2}{\tan (e+f x)+1}-4 \left (7+5 \sqrt {2}\right )}d\frac {\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4}{\sqrt {\tan (e+f x)+1}}}{f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )+\frac {52 (\tan (e+f x)+1)^{3/2}}{3 f}\right )-\frac {26 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {1}{5} \left (105 \left (\frac {\sqrt {2} \left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4\right )^2}{\tan (e+f x)+1}-4 \left (7+5 \sqrt {2}\right )}d\frac {\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4}{\sqrt {\tan (e+f x)+1}}}{f}-\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (2-\sqrt {2}\right ) \tan (e+f x)-3 \sqrt {2}+4}{2 \sqrt {5 \sqrt {2}-7} \sqrt {\tan (e+f x)+1}}\right )}{\sqrt {2 \left (5 \sqrt {2}-7\right )} f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )+\frac {52 (\tan (e+f x)+1)^{3/2}}{3 f}\right )-\frac {26 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {1}{3} \left (\frac {1}{7} \left (\frac {1}{5} \left (105 \left (-\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (2-\sqrt {2}\right ) \tan (e+f x)-3 \sqrt {2}+4}{2 \sqrt {5 \sqrt {2}-7} \sqrt {\tan (e+f x)+1}}\right )}{\sqrt {2 \left (5 \sqrt {2}-7\right )} f}-\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4}{2 \sqrt {7+5 \sqrt {2}} \sqrt {\tan (e+f x)+1}}\right )}{\sqrt {2 \left (7+5 \sqrt {2}\right )} f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )+\frac {52 (\tan (e+f x)+1)^{3/2}}{3 f}\right )-\frac {26 \tan (e+f x) (\tan (e+f x)+1)^{3/2}}{5 f}\right )-\frac {4 \tan ^2(e+f x) (\tan (e+f x)+1)^{3/2}}{7 f}\right )+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}\)

Input:

Int[Tan[e + f*x]^5*Sqrt[1 + Tan[e + f*x]],x]
 

Output:

(2*Tan[e + f*x]^3*(1 + Tan[e + f*x])^(3/2))/(9*f) + ((-4*Tan[e + f*x]^2*(1 
 + Tan[e + f*x])^(3/2))/(7*f) + ((-26*Tan[e + f*x]*(1 + Tan[e + f*x])^(3/2 
))/(5*f) + ((52*(1 + Tan[e + f*x])^(3/2))/(3*f) + 105*(-(((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 + T 
an[e + f*x]])/f))/5)/7)/3
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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]
 

rule 4049

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c 
+ d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) 
 Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 
, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) 
)
 

rule 4113

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + 
 b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si 
mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && 
NeQ[A*b^2 - a*b*B + a^2*C, 0] &&  !LeQ[m, -1]
 

rule 4130

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_. 
) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ 
e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a 
+ b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, 
b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && 
 NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

rule 4131

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

Maple [A] (verified)

Time = 1.86 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {9}{2}}}{9}-\frac {6 \left (\tan \left (f x +e \right )+1\right )^{\frac {7}{2}}}{7}+\frac {4 \left (\tan \left (f x +e \right )+1\right )^{\frac {5}{2}}}{5}+2 \sqrt {\tan \left (f x +e \right )+1}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4}-\frac {\left (\sqrt {2}-1\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}}{f}\)	\(242\)
default	\(\frac {\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {9}{2}}}{9}-\frac {6 \left (\tan \left (f x +e \right )+1\right )^{\frac {7}{2}}}{7}+\frac {4 \left (\tan \left (f x +e \right )+1\right )^{\frac {5}{2}}}{5}+2 \sqrt {\tan \left (f x +e \right )+1}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4}-\frac {\left (\sqrt {2}-1\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{4}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}}{f}\)	\(242\)

Input:

int(tan(f*x+e)^5*(tan(f*x+e)+1)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/f*(2/9*(tan(f*x+e)+1)^(9/2)-6/7*(tan(f*x+e)+1)^(7/2)+4/5*(tan(f*x+e)+1)^ 
(5/2)+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))-(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))-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)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.08 \[ \int \tan ^5(e+f x) \sqrt {1+\tan (e+f x)} \, dx=-\frac {315 \, f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 315 \, f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 315 \, f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 315 \, f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 4 \, {\left (35 \, \tan \left (f x + e\right )^{4} + 5 \, \tan \left (f x + e\right )^{3} - 69 \, \tan \left (f x + e\right )^{2} - 13 \, \tan \left (f x + e\right ) + 341\right )} \sqrt {\tan \left (f x + e\right ) + 1}}{630 \, f} \] Input:

integrate(tan(f*x+e)^5*(1+tan(f*x+e))^(1/2),x, algorithm="fricas")
 

Output:

-1/630*(315*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)) - 315*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)) + 3 
15*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)) - 315*f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2)*l 
og(-f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2) + sqrt(tan(f*x + e) + 1)) - 4*(35* 
tan(f*x + e)^4 + 5*tan(f*x + e)^3 - 69*tan(f*x + e)^2 - 13*tan(f*x + e) + 
341)*sqrt(tan(f*x + e) + 1))/f
 

Sympy [F]

\[ \int \tan ^5(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\int \sqrt {\tan {\left (e + f x \right )} + 1} \tan ^{5}{\left (e + f x \right )}\, dx \] Input:

integrate(tan(f*x+e)**5*(1+tan(f*x+e))**(1/2),x)
 

Output:

Integral(sqrt(tan(e + f*x) + 1)*tan(e + f*x)**5, x)
 

Maxima [F]

\[ \int \tan ^5(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\int { \sqrt {\tan \left (f x + e\right ) + 1} \tan \left (f x + e\right )^{5} \,d x } \] Input:

integrate(tan(f*x+e)^5*(1+tan(f*x+e))^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(tan(f*x + e) + 1)*tan(f*x + e)^5, x)
 

Giac [F(-2)]

Exception generated. \[ \int \tan ^5(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(f*x+e)^5*(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,[29]%%%}+%%%{12,[27]%%%}+%%%{66,[25]%%%}+%%%{220,[23 
]%%%}+%%%
 

Mupad [B] (verification not implemented)

Time = 3.26 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.51 \[ \int \tan ^5(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}+\frac {4\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{5/2}}{5\,f}-\frac {6\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{7/2}}{7\,f}+\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{9/2}}{9\,f}-\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,\left (1-\mathrm {i}\right )\right )\,\sqrt {\frac {\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,\left (1+1{}\mathrm {i}\right )\right )\,\sqrt {\frac {\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \] Input:

int(tan(e + f*x)^5*(tan(e + f*x) + 1)^(1/2),x)
 

Output:

(2*(tan(e + f*x) + 1)^(1/2))/f + (4*(tan(e + f*x) + 1)^(5/2))/(5*f) - (6*( 
tan(e + f*x) + 1)^(7/2))/(7*f) + (2*(tan(e + f*x) + 1)^(9/2))/(9*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]

\[ \int \tan ^5(e+f x) \sqrt {1+\tan (e+f x)} \, dx=\int \sqrt {\tan \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{5}d x \] Input:

int(tan(f*x+e)^5*(1+tan(f*x+e))^(1/2),x)
 

Output:

int(sqrt(tan(e + f*x) + 1)*tan(e + f*x)**5,x)