\(\int \tan ^4(e+f x) (1+\tan (e+f x))^{3/2} \, dx\) [395]

Optimal result

Integrand size = 21, antiderivative size = 227 \[ \int \tan ^4(e+f x) (1+\tan (e+f x))^{3/2} \, dx=-\frac {\sqrt {-1+\sqrt {2}} \arctan \left (\frac {3-2 \sqrt {2}+\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (-7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{f}-\frac {\sqrt {1+\sqrt {2}} \text {arctanh}\left (\frac {3+2 \sqrt {2}+\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{f}+\frac {2 \sqrt {1+\tan (e+f x)}}{f}-\frac {22 (1+\tan (e+f x))^{5/2}}{63 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{63 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{9 f} \] Output:

-(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)*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+2*( 
1+tan(f*x+e))^(1/2)/f-22/63*(1+tan(f*x+e))^(5/2)/f-8/63*tan(f*x+e)*(1+tan( 
f*x+e))^(5/2)/f+2/9*tan(f*x+e)^2*(1+tan(f*x+e))^(5/2)/f
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.29 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.68 \[ \int \tan ^4(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\frac {2 \cos ^2(e+f x) \left (-63 \left (\frac {\text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )}{\sqrt {1-i}}+\frac {\text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{\sqrt {1+i}}\right ) (1+\tan (e+f x))^2+(1+\tan (e+f x))^{5/2} \left (71+7 \sec ^4(e+f x)-36 \tan (e+f x)+2 \sec ^2(e+f x) (-13+5 \tan (e+f x))\right )\right )}{63 f (\cos (e+f x)+\sin (e+f x))^2} \] Input:

Integrate[Tan[e + f*x]^4*(1 + Tan[e + f*x])^(3/2),x]
 

Output:

(2*Cos[e + f*x]^2*(-63*(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])*(1 + Tan[ 
e + f*x])^2 + (1 + Tan[e + f*x])^(5/2)*(71 + 7*Sec[e + f*x]^4 - 36*Tan[e + 
 f*x] + 2*Sec[e + f*x]^2*(-13 + 5*Tan[e + f*x]))))/(63*f*(Cos[e + f*x] + S 
in[e + f*x])^2)
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.18, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.810, Rules used = {3042, 4049, 27, 3042, 4130, 27, 3042, 4114, 3042, 3963, 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.

\(\displaystyle \int \tan ^4(e+f x) (\tan (e+f x)+1)^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (e+f x)^4 (\tan (e+f x)+1)^{3/2}dx\)

\(\Big \downarrow \) 4049

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

\(\displaystyle \frac {1}{9} \left (-\frac {2}{7} \int -\frac {1}{2} (\tan (e+f x)+1)^{3/2} \left (8-55 \tan ^2(e+f x)\right )dx-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\right )+\frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \int (\tan (e+f x)+1)^{3/2} \left (8-55 \tan ^2(e+f x)\right )dx-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\right )+\frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \int (\tan (e+f x)+1)^{3/2} \left (8-55 \tan (e+f x)^2\right )dx-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\right )+\frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}\)

\(\Big \downarrow \) 4114

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (63 \int (\tan (e+f x)+1)^{3/2}dx-\frac {22 (\tan (e+f x)+1)^{5/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\right )+\frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (63 \int (\tan (e+f x)+1)^{3/2}dx-\frac {22 (\tan (e+f x)+1)^{5/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\right )+\frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}\)

\(\Big \downarrow \) 3963

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (63 \left (\int \frac {2 \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 (\tan (e+f x)+1)^{5/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\right )+\frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (63 \left (2 \int \frac {\tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 (\tan (e+f x)+1)^{5/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\right )+\frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (63 \left (2 \int \frac {\tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 (\tan (e+f x)+1)^{5/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\right )+\frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}\)

\(\Big \downarrow \) 4019

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (63 \left (2 \left (\frac {\int \frac {1-\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}-\frac {\int \frac {1-\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}\right )+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 (\tan (e+f x)+1)^{5/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\right )+\frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (63 \left (2 \left (\frac {\int \frac {1-\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}-\frac {\int \frac {1-\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {\tan (e+f x)+1}}dx}{2 \sqrt {2}}\right )+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 (\tan (e+f x)+1)^{5/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\right )+\frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}\)

\(\Big \downarrow \) 4018

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (63 \left (2 \left (\frac {\left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7+5 \sqrt {2}\right )}d\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}-\frac {\left (3-2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7-5 \sqrt {2}\right )}d\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}\right )+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 (\tan (e+f x)+1)^{5/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\right )+\frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (63 \left (2 \left (\frac {\left (3+2 \sqrt {2}\right ) \int \frac {1}{\frac {\left (\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3\right )^2}{\tan (e+f x)+1}-2 \left (7+5 \sqrt {2}\right )}d\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {\tan (e+f x)+1}}}{\sqrt {2} f}-\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}\right )+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 (\tan (e+f x)+1)^{5/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\right )+\frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (63 \left (2 \left (-\frac {\left (3-2 \sqrt {2}\right ) \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {5 \sqrt {2}-7} f}-\frac {\left (3+2 \sqrt {2}\right ) \text {arctanh}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 \sqrt {7+5 \sqrt {2}} f}\right )+\frac {2 \sqrt {\tan (e+f x)+1}}{f}\right )-\frac {22 (\tan (e+f x)+1)^{5/2}}{f}\right )-\frac {8 \tan (e+f x) (\tan (e+f x)+1)^{5/2}}{7 f}\right )+\frac {2 \tan ^2(e+f x) (\tan (e+f x)+1)^{5/2}}{9 f}\)

Input:

Int[Tan[e + f*x]^4*(1 + Tan[e + f*x])^(3/2),x]
 

Output:

(2*Tan[e + f*x]^2*(1 + Tan[e + f*x])^(5/2))/(9*f) + ((-8*Tan[e + f*x]*(1 + 
 Tan[e + f*x])^(5/2))/(7*f) + ((-22*(1 + Tan[e + f*x])^(5/2))/f + 63*(2*(- 
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)) + (2*Sqrt[1 + Tan[e + f*x]])/f))/7)/9
 

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 3963

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

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 4114

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

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {9}{2}}}{9}-\frac {4 \left (\tan \left (f x +e \right )+1\right )^{\frac {7}{2}}}{7}+2 \sqrt {\tan \left (f x +e \right )+1}-\frac {\sqrt {2}\, \left (-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}-\frac {\sqrt {2}\, \left (\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}}{f}\)	\(245\)
default	\(\frac {\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {9}{2}}}{9}-\frac {4 \left (\tan \left (f x +e \right )+1\right )^{\frac {7}{2}}}{7}+2 \sqrt {\tan \left (f x +e \right )+1}-\frac {\sqrt {2}\, \left (-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}-\frac {\sqrt {2}\, \left (\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{2}}{f}\)	\(245\)

Input:

int(tan(f*x+e)^4*(tan(f*x+e)+1)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/f*(2/9*(tan(f*x+e)+1)^(9/2)-4/7*(tan(f*x+e)+1)^(7/2)+2*(tan(f*x+e)+1)^(1 
/2)-1/2*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)*arcta 
n((2*(tan(f*x+e)+1)^(1/2)-(2*2^(1/2)+2)^(1/2))/(-2+2*2^(1/2))^(1/2)))-1/2* 
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. 368 vs. \(2 (180) = 360\).

Time = 0.08 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.62 \[ \int \tan ^4(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\frac {63 \, \sqrt {2} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 63 \, \sqrt {2} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 63 \, \sqrt {2} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 63 \, \sqrt {2} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-\sqrt {2} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 4 \, {\left (7 \, \tan \left (f x + e\right )^{4} + 10 \, \tan \left (f x + e\right )^{3} - 12 \, \tan \left (f x + e\right )^{2} - 26 \, \tan \left (f x + e\right ) + 52\right )} \sqrt {\tan \left (f x + e\right ) + 1}}{126 \, f} \] Input:

integrate(tan(f*x+e)^4*(1+tan(f*x+e))^(3/2),x, algorithm="fricas")
 

Output:

1/126*(63*sqrt(2)*f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2)*log(sqrt(2)*(f^3*sqrt 
(-1/f^4) - f)*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) 
 - 63*sqrt(2)*f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2)*log(-sqrt(2)*(f^3*sqrt(-1 
/f^4) - f)*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) - 
63*sqrt(2)*f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2)*log(sqrt(2)*(f^3*sqrt(-1/f^ 
4) + f)*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) + 63 
*sqrt(2)*f*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2)*log(-sqrt(2)*(f^3*sqrt(-1/f^4 
) + f)*sqrt(-(f^2*sqrt(-1/f^4) - 1)/f^2) + 2*sqrt(tan(f*x + e) + 1)) + 4*( 
7*tan(f*x + e)^4 + 10*tan(f*x + e)^3 - 12*tan(f*x + e)^2 - 26*tan(f*x + e) 
 + 52)*sqrt(tan(f*x + e) + 1))/f
 

Sympy [F]

\[ \int \tan ^4(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\int \left (\tan {\left (e + f x \right )} + 1\right )^{\frac {3}{2}} \tan ^{4}{\left (e + f x \right )}\, dx \] Input:

integrate(tan(f*x+e)**4*(1+tan(f*x+e))**(3/2),x)
 

Output:

Integral((tan(e + f*x) + 1)**(3/2)*tan(e + f*x)**4, x)
 

Maxima [F]

\[ \int \tan ^4(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\int { {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{4} \,d x } \] Input:

integrate(tan(f*x+e)^4*(1+tan(f*x+e))^(3/2),x, algorithm="maxima")
 

Output:

integrate((tan(f*x + e) + 1)^(3/2)*tan(f*x + e)^4, x)
 

Giac [F(-2)]

Exception generated. \[ \int \tan ^4(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(f*x+e)^4*(1+tan(f*x+e))^(3/2),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 = 3.23 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.52 \[ \int \tan ^4(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}-\frac {4\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{7/2}}{7\,f}+\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{9/2}}{9\,f}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,1{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{2}-\frac {1}{2}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,1{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{2}+\frac {1}{2}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \] Input:

int(tan(e + f*x)^4*(tan(e + f*x) + 1)^(3/2),x)
 

Output:

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

Reduce [F]

\[ \int \tan ^4(e+f x) (1+\tan (e+f x))^{3/2} \, dx=\int \sqrt {\tan \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{5}d x +\int \sqrt {\tan \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{4}d x \] Input:

int(tan(f*x+e)^4*(1+tan(f*x+e))^(3/2),x)
 

Output:

int(sqrt(tan(e + f*x) + 1)*tan(e + f*x)**5,x) + int(sqrt(tan(e + f*x) + 1) 
*tan(e + f*x)**4,x)