\(\int \frac {\tan ^3(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\) [401]

Optimal result

Integrand size = 21, antiderivative size = 187 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {\sqrt {-1+\sqrt {2}} \arctan \left (\frac {3-2 \sqrt {2}+\left (1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (-7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f}+\frac {\sqrt {1+\sqrt {2}} \text {arctanh}\left (\frac {3+2 \sqrt {2}+\left (1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {1+\tan (e+f x)}}\right )}{2 f}-\frac {4 \sqrt {1+\tan (e+f x)}}{3 f}+\frac {2 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{3 f} \] 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-4/3*(1+tan(f*x+e))^(1/2)/f+2/3*tan(f*x+e)*(1+tan(f*x+e))^(1/2)/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.48 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {\frac {6 \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )}{\sqrt {1-i}}+\frac {6 \text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{\sqrt {1+i}}+4 (-2+\tan (e+f x)) \sqrt {1+\tan (e+f x)}}{6 f} \] Input:

Integrate[Tan[e + f*x]^3/Sqrt[1 + Tan[e + f*x]],x]
 

Output:

((6*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]])/Sqrt[1 - I] + (6*ArcTanh[ 
Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]])/Sqrt[1 + I] + 4*(-2 + Tan[e + f*x])*S 
qrt[1 + Tan[e + f*x]])/(6*f)
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.15, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4049, 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.

\(\displaystyle \int \frac {\tan ^3(e+f x)}{\sqrt {\tan (e+f x)+1}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (e+f x)^3}{\sqrt {\tan (e+f x)+1}}dx\)

\(\Big \downarrow \) 4049

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4113

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4019

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4018

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 220

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

Input:

Int[Tan[e + f*x]^3/Sqrt[1 + Tan[e + f*x]],x]
 

Output:

(2*Tan[e + f*x]*Sqrt[1 + Tan[e + f*x]])/(3*f) + (-3*(-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)) - (4*Sqrt[1 + Ta 
n[e + f*x]])/f)/3
 

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]
 

Maple [A] (verified)

Time = 1.13 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.25

method	result	size
derivativedivides	\(\frac {\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {3}{2}}}{3}-2 \sqrt {\tan \left (f x +e \right )+1}+\frac {\sqrt {2}\, \left (\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{4}+\frac {\sqrt {2}\, \left (-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{4}}{f}\)	\(233\)
default	\(\frac {\frac {2 \left (\tan \left (f x +e \right )+1\right )^{\frac {3}{2}}}{3}-2 \sqrt {\tan \left (f x +e \right )+1}+\frac {\sqrt {2}\, \left (\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1+\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {\tan \left (f x +e \right )+1}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{4}+\frac {\sqrt {2}\, \left (-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (\tan \left (f x +e \right )+1-\sqrt {\tan \left (f x +e \right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )}{2}+\frac {2 \left (1-\sqrt {2}\right ) \arctan \left (\frac {2 \sqrt {\tan \left (f x +e \right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}\right )}{4}}{f}\)	\(233\)

Input:

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

Output:

1/f*(2/3*(tan(f*x+e)+1)^(3/2)-2*(tan(f*x+e)+1)^(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)))+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*(tan(f*x+e)+1)^(1/2)-(2*2^(1/ 
2)+2)^(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. 328 vs. \(2 (142) = 284\).

Time = 0.08 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.75 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {3 \, \sqrt {\frac {1}{2}} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 3 \, \sqrt {\frac {1}{2}} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 3 \, \sqrt {\frac {1}{2}} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 3 \, \sqrt {\frac {1}{2}} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 4 \, \sqrt {\tan \left (f x + e\right ) + 1} {\left (\tan \left (f x + e\right ) - 2\right )}}{6 \, f} \] Input:

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

Output:

-1/6*(3*sqrt(1/2)*f*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2)*log(sqrt(1/2)*(f^3*sq 
rt(-1/f^4) - f)*sqrt((f^2*sqrt(-1/f^4) + 1)/f^2) + sqrt(tan(f*x + e) + 1)) 
 - 3*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)) - 
 3*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)) + 
3*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*sqrt(tan(f*x + e) + 1)*(tan(f*x + e) - 2))/f
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

integrate(tan(f*x+e)^3/(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,[14]%%%}+%%%{6,[12]%%%}+%%%{15,[10]%%%}+%%%{20,[8]%% 
%}+%%%{15
 

Mupad [B] (verification not implemented)

Time = 1.32 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.55 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{3/2}}{3\,f}-\frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}-\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,2{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,2{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \] Input:

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

Output:

(2*(tan(e + f*x) + 1)^(3/2))/(3*f) - (2*(tan(e + f*x) + 1)^(1/2))/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]

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

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

Output:

int((sqrt(tan(e + f*x) + 1)*tan(e + f*x)**3)/(tan(e + f*x) + 1),x)