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

Optimal result

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

[Out]

-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))*(2^(1/2)-1)^(1/2
)/f-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))*(1+2^(1/2))^(
1/2)/f+44/105*(1+tan(f*x+e))^(1/2)/f-22/105*(1+tan(f*x+e))^(1/2)*tan(f*x+e)/f-12/35*(1+tan(f*x+e))^(1/2)*tan(f
*x+e)^2/f+2/7*(1+tan(f*x+e))^(1/2)*tan(f*x+e)^3/f

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3647, 3728, 3729, 3711, 12, 3617, 3616, 209, 213} \[ \int \frac {\tan ^5(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {\sqrt {\sqrt {2}-1} \arctan \left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 f}-\frac {\sqrt {1+\sqrt {2}} \text {arctanh}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 f}+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}-\frac {12 \sqrt {\tan (e+f x)+1} \tan ^2(e+f x)}{35 f}-\frac {22 \sqrt {\tan (e+f x)+1} \tan (e+f x)}{105 f}+\frac {44 \sqrt {\tan (e+f x)+1}}{105 f} \]

[In]

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

[Out]

-1/2*(Sqrt[-1 + 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]])])/f - (Sqrt[1 + 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*f) + (44*Sqrt[1 + Tan[e + f*x]])/(105*f) - (22*Tan[e + f*x]*Sqrt[1 + Ta
n[e + f*x]])/(105*f) - (12*Tan[e + f*x]^2*Sqrt[1 + Tan[e + f*x]])/(35*f) + (2*Tan[e + f*x]^3*Sqrt[1 + Tan[e +
f*x]])/(7*f)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 213

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 3616

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-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 3617

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> With[{q =
Rt[a^2 + b^2, 2]}, Dist[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] - Dist[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] && NeQ[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] && (PerfectSquareQ[a^2 + b^2] || RationalQ[a, b, c, d])

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[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] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3711

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*Simp[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 3728

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] + Dist[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] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3729

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] + Dist[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])))

Rubi steps \begin{align*} \text {integral}& = \frac {2 \tan ^3(e+f x) \sqrt {1+\tan (e+f x)}}{7 f}+\frac {2}{7} \int \frac {\tan ^2(e+f x) \left (-3-\frac {7}{2} \tan (e+f x)-3 \tan ^2(e+f x)\right )}{\sqrt {1+\tan (e+f x)}} \, dx \\ & = -\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}+\frac {4}{35} \int \frac {\tan (e+f x) \left (6-\frac {11}{4} \tan ^2(e+f x)\right )}{\sqrt {1+\tan (e+f x)}} \, dx \\ & = -\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}+\frac {8}{105} \int \frac {\frac {11}{4}+\frac {105}{8} \tan (e+f x)+\frac {11}{4} \tan ^2(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx \\ & = \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}+\frac {8}{105} \int \frac {105 \tan (e+f x)}{8 \sqrt {1+\tan (e+f x)}} \, dx \\ & = \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}+\int \frac {\tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx \\ & = \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}-\frac {\int \frac {1+\left (-1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx}{2 \sqrt {2}}+\frac {\int \frac {1+\left (-1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx}{2 \sqrt {2}} \\ & = \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}+\frac {\left (4-3 \sqrt {2}\right ) \text {Subst}\left (\int \frac {1}{2 \left (-1+\sqrt {2}\right )-4 \left (-1+\sqrt {2}\right )^2+x^2} \, dx,x,\frac {1-2 \left (-1+\sqrt {2}\right )-\left (-1+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}}\right )}{2 f}+\frac {\left (4+3 \sqrt {2}\right ) \text {Subst}\left (\int \frac {1}{2 \left (-1-\sqrt {2}\right )-4 \left (-1-\sqrt {2}\right )^2+x^2} \, dx,x,\frac {1-2 \left (-1-\sqrt {2}\right )-\left (-1-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}}\right )}{2 f} \\ & = -\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} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.46 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=-\frac {\frac {\text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )}{\sqrt {1-i}}+\frac {\text {arctanh}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{\sqrt {1+i}}-\frac {2}{105} \sqrt {1+\tan (e+f x)} \left (40-26 \tan (e+f x)+3 \sec ^2(e+f x) (-6+5 \tan (e+f x))\right )}{f} \]

[In]

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

[Out]

-((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)

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.02

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

[In]

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

[Out]

1/f*(2/7*(1+tan(f*x+e))^(7/2)-6/5*(1+tan(f*x+e))^(5/2)+4/3*(1+tan(f*x+e))^(3/2)-1/4*2^(1/2)*(1/2*(2+2*2^(1/2))
^(1/2)*ln(1+2^(1/2)+(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))+2*(1-2^(1/2))/(-2+2*2^(1/2))^(1/2)*ar
ctan(((2+2*2^(1/2))^(1/2)+2*(1+tan(f*x+e))^(1/2))/(-2+2*2^(1/2))^(1/2)))-1/4*2^(1/2)*(-1/2*(2+2*2^(1/2))^(1/2)
*ln(1+2^(1/2)-(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))+2*(1-2^(1/2))/(-2+2*2^(1/2))^(1/2)*arctan((
2*(1+tan(f*x+e))^(1/2)-(2+2*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.45 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {105 \, \sqrt {\frac {1}{2}} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 105 \, \sqrt {\frac {1}{2}} f \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} - f\right )} \sqrt {\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} + 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) - 105 \, \sqrt {\frac {1}{2}} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 105 \, \sqrt {\frac {1}{2}} f \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} \log \left (-\sqrt {\frac {1}{2}} {\left (f^{3} \sqrt {-\frac {1}{f^{4}}} + f\right )} \sqrt {-\frac {f^{2} \sqrt {-\frac {1}{f^{4}}} - 1}{f^{2}}} + \sqrt {\tan \left (f x + e\right ) + 1}\right ) + 4 \, {\left (15 \, \tan \left (f x + e\right )^{3} - 18 \, \tan \left (f x + e\right )^{2} - 11 \, \tan \left (f x + e\right ) + 22\right )} \sqrt {\tan \left (f x + e\right ) + 1}}{210 \, f} \]

[In]

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

[Out]

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*s
qrt(-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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^5(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 5.11 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.49 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx=\frac {4\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{3/2}}{3\,f}-\frac {6\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{5/2}}{5\,f}+\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{7/2}}{7\,f}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,2{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,2{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \]

[In]

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

[Out]

(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