\(\int \frac {\tan ^5(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\) [400]

Optimal result

Integrand size = 21, antiderivative size = 241 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {\sqrt {-1+\sqrt {2}} \arctan \left (\frac {3-2 \sqrt {2}+\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (-7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f}-\frac {\sqrt {1+\sqrt {2}} \text {arctanh}\left (\frac {3+2 \sqrt {2}+\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f}+\frac {44 \sqrt {1+\tan (e+f x)}}{105 f}-\frac {22 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{105 f}-\frac {12 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{35 f}+\frac {2 \tan ^3(e+f x) \sqrt {1+\tan (e+f x)}}{7 f} \] 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+44/105*(1+tan(f*x+e))^(1/2)/f-22/105*tan(f*x+e)*(1+tan(f*x+e))^(1/2)/ 
f-12/35*tan(f*x+e)^2*(1+tan(f*x+e))^(1/2)/f+2/7*tan(f*x+e)^3*(1+tan(f*x+e) 
)^(1/2)/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.98 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.46 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {\frac {\text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )}{\sqrt {1-i}}+\frac {\text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{\sqrt {1+i}}-\frac {2}{105} \sqrt {1+\tan (e+f x)} \left (40-26 \tan (e+f x)+3 \sec ^2(e+f x) (-6+5 \tan (e+f x))\right )}{f} \] Input:

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

Output:

-((ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]]/Sqrt[1 - I] + ArcTanh[Sqrt[ 
1 + Tan[e + f*x]]/Sqrt[1 + I]]/Sqrt[1 + I] - (2*Sqrt[1 + Tan[e + f*x]]*(40 
 - 26*Tan[e + f*x] + 3*Sec[e + f*x]^2*(-6 + 5*Tan[e + f*x])))/105)/f)
 

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.16, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {3042, 4049, 27, 3042, 4130, 27, 3042, 4131, 27, 3042, 4113, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

\(\displaystyle \frac {2}{7} \int -\frac {\tan ^2(e+f x) \left (6 \tan ^2(e+f x)+7 \tan (e+f x)+6\right )}{2 \sqrt {\tan (e+f x)+1}}dx+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan ^3(e+f x) \sqrt {\tan (e+f x)+1}}{7 f}-\frac {1}{7} \int \frac {\tan ^2(e+f x) \left (6 \tan ^2(e+f x)+7 \tan (e+f x)+6\right )}{\sqrt {\tan (e+f x)+1}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan ^3(e+f x) \sqrt {\tan (e+f x)+1}}{7 f}-\frac {1}{7} \int \frac {\tan (e+f x)^2 \left (6 \tan (e+f x)^2+7 \tan (e+f x)+6\right )}{\sqrt {\tan (e+f x)+1}}dx\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {1}{7} \left (-\frac {2}{5} \int -\frac {\tan (e+f x) \left (24-11 \tan ^2(e+f x)\right )}{2 \sqrt {\tan (e+f x)+1}}dx-\frac {12 \sqrt {\tan (e+f x)+1} \tan ^2(e+f x)}{5 f}\right )+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \frac {\tan (e+f x) \left (24-11 \tan ^2(e+f x)\right )}{\sqrt {\tan (e+f x)+1}}dx-\frac {12 \tan ^2(e+f x) \sqrt {\tan (e+f x)+1}}{5 f}\right )+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \frac {\tan (e+f x) \left (24-11 \tan (e+f x)^2\right )}{\sqrt {\tan (e+f x)+1}}dx-\frac {12 \tan ^2(e+f x) \sqrt {\tan (e+f x)+1}}{5 f}\right )+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

\(\Big \downarrow \) 4131

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {22 \tan ^2(e+f x)+105 \tan (e+f x)+22}{2 \sqrt {\tan (e+f x)+1}}dx-\frac {22 \tan (e+f x) \sqrt {\tan (e+f x)+1}}{3 f}\right )-\frac {12 \tan ^2(e+f x) \sqrt {\tan (e+f x)+1}}{5 f}\right )+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {22 \tan ^2(e+f x)+105 \tan (e+f x)+22}{\sqrt {\tan (e+f x)+1}}dx-\frac {22 \tan (e+f x) \sqrt {\tan (e+f x)+1}}{3 f}\right )-\frac {12 \tan ^2(e+f x) \sqrt {\tan (e+f x)+1}}{5 f}\right )+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {22 \tan (e+f x)^2+105 \tan (e+f x)+22}{\sqrt {\tan (e+f x)+1}}dx-\frac {22 \tan (e+f x) \sqrt {\tan (e+f x)+1}}{3 f}\right )-\frac {12 \tan ^2(e+f x) \sqrt {\tan (e+f x)+1}}{5 f}\right )+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (\int \frac {105 \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx+\frac {44 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 \tan (e+f x) \sqrt {\tan (e+f x)+1}}{3 f}\right )-\frac {12 \tan ^2(e+f x) \sqrt {\tan (e+f x)+1}}{5 f}\right )+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (105 \int \frac {\tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx+\frac {44 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 \tan (e+f x) \sqrt {\tan (e+f x)+1}}{3 f}\right )-\frac {12 \tan ^2(e+f x) \sqrt {\tan (e+f x)+1}}{5 f}\right )+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (105 \int \frac {\tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx+\frac {44 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 \tan (e+f x) \sqrt {\tan (e+f x)+1}}{3 f}\right )-\frac {12 \tan ^2(e+f x) \sqrt {\tan (e+f x)+1}}{5 f}\right )+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

\(\Big \downarrow \) 4019

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (105 \left (\frac {\int \frac {1-\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}-\frac {\int \frac {1-\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}\right )+\frac {44 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 \tan (e+f x) \sqrt {\tan (e+f x)+1}}{3 f}\right )-\frac {12 \tan ^2(e+f x) \sqrt {\tan (e+f x)+1}}{5 f}\right )+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (105 \left (\frac {\int \frac {1-\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}-\frac {\int \frac {1-\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}\right )+\frac {44 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 \tan (e+f x) \sqrt {\tan (e+f x)+1}}{3 f}\right )-\frac {12 \tan ^2(e+f x) \sqrt {\tan (e+f x)+1}}{5 f}\right )+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

\(\Big \downarrow \) 4018

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (105 \left (\frac {\left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7+5 \sqrt {2}\right )}d\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}-\frac {\left (3-2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7-5 \sqrt {2}\right )}d\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}\right )+\frac {44 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 \tan (e+f x) \sqrt {\tan (e+f x)+1}}{3 f}\right )-\frac {12 \tan ^2(e+f x) \sqrt {\tan (e+f x)+1}}{5 f}\right )+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (105 \left (\frac {\left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7+5 \sqrt {2}\right )}d\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}-\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}\right )+\frac {44 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 \tan (e+f x) \sqrt {\tan (e+f x)+1}}{3 f}\right )-\frac {12 \tan ^2(e+f x) \sqrt {\tan (e+f x)+1}}{5 f}\right )+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (\frac {1}{3} \left (105 \left (-\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}-\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {7+5 \sqrt {2}} f}\right )+\frac {44 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 \tan (e+f x) \sqrt {\tan (e+f x)+1}}{3 f}\right )-\frac {12 \tan ^2(e+f x) \sqrt {\tan (e+f x)+1}}{5 f}\right )+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}\)

Input:

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

Output:

(2*Tan[e + f*x]^3*Sqrt[1 + Tan[e + f*x]])/(7*f) + ((-12*Tan[e + f*x]^2*Sqr 
t[1 + Tan[e + f*x]])/(5*f) + ((-22*Tan[e + f*x]*Sqrt[1 + Tan[e + f*x]])/(3 
*f) + (105*(-1/2*((3 - 2*Sqrt[2])*ArcTan[(3 - 2*Sqrt[2] + (1 - Sqrt[2])*Ta 
n[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])*T 
an[e + f*x])/(Sqrt[2*(7 + 5*Sqrt[2])]*Sqrt[1 + Tan[e + f*x]])])/(2*Sqrt[7 
+ 5*Sqrt[2]]*f)) + (44*Sqrt[1 + Tan[e + f*x]])/f)/3)/5)/7
 

Defintions of rubi rules used

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 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 = 1.13 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {7}{2}}}{7}-\frac {6 \left (\tan \left (f x +e \right )+1\right )^{\frac {5}{2}}}{5}+\frac {4 \left (\tan \left (f x +e \right )+1\right )^{\frac {3}{2}}}{3}-\frac {\sqrt {2}\, \left (\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{4}-\frac {\sqrt {2}\, \left (-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{4}}{f}\)	\(245\)
default	\(\frac {\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {7}{2}}}{7}-\frac {6 \left (\tan \left (f x +e \right )+1\right )^{\frac {5}{2}}}{5}+\frac {4 \left (\tan \left (f x +e \right )+1\right )^{\frac {3}{2}}}{3}-\frac {\sqrt {2}\, \left (\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{4}-\frac {\sqrt {2}\, \left (-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{4}}{f}\)	\(245\)

Input:

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

Output:

1/f*(2/7*(tan(f*x+e)+1)^(7/2)-6/5*(tan(f*x+e)+1)^(5/2)+4/3*(tan(f*x+e)+1)^ 
(3/2)-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)*arct 
an(((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^(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*(t 
an(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 = 0.09 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.45 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {105 \, \sqrt {\frac {1}{2}} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 105 \, \sqrt {\frac {1}{2}} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 105 \, \sqrt {\frac {1}{2}} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 105 \, \sqrt {\frac {1}{2}} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 4 \, {\left (15 \, \tan \left (f x + e\right )^{3} - 18 \, \tan \left (f x + e\right )^{2} - 11 \, \tan \left (f x + e\right ) + 22\right )} \sqrt {\tan \left (f x + e\right ) + 1}}{210 \, f} \] Input:

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

Output:

1/210*(105*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)) - 105*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)) - 105*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)) + 105*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*(15*tan(f*x + e)^3 - 18*tan(f*x + e)^2 - 11*tan(f*x + e) + 22)* 
sqrt(tan(f*x + e) + 1))/f
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^5(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\text {Exception raised: TypeError} \] 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,[24]%%%}+%%%{10,[22]%%%}+%%%{45,[20]%%%}+%%%{120,[18 
]%%%}+%%%
 

Mupad [B] (verification not implemented)

Time = 1.60 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.49 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {4\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{3/2}}{3\,f}-\frac {6\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{5/2}}{5\,f}+\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{7/2}}{7\,f}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,2{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,2{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \] Input:

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

Output:

(4*(tan(e + f*x) + 1)^(3/2))/(3*f) - (6*(tan(e + f*x) + 1)^(5/2))/(5*f) + 
(2*(tan(e + f*x) + 1)^(7/2))/(7*f) + atan(f*((1/8 - 1i/8)/f^2)^(1/2)*(tan( 
e + f*x) + 1)^(1/2)*2i)*((1/8 - 1i/8)/f^2)^(1/2)*2i + atan(f*((1/8 + 1i/8) 
/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2)*2i)*((1/8 + 1i/8)/f^2)^(1/2)*2i
 

Reduce [F]

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