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

Optimal. Leaf size=241 \[ -\frac {\sqrt {\sqrt {2}-1} \tan ^{-1}\left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 f}+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}-\frac {12 \sqrt {\tan (e+f x)+1} \tan ^2(e+f x)}{35 f}-\frac {22 \sqrt {\tan (e+f x)+1} \tan (e+f x)}{105 f}+\frac {44 \sqrt {\tan (e+f x)+1}}{105 f}-\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 f} \]

[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

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Rubi [A]  time = 0.37, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3566, 3647, 3648, 3630, 12, 3536, 3535, 203, 207} \[ \frac {2 \sqrt {\tan (e+f x)+1} \tan ^3(e+f x)}{7 f}-\frac {12 \sqrt {\tan (e+f x)+1} \tan ^2(e+f x)}{35 f}-\frac {\sqrt {\sqrt {2}-1} \tan ^{-1}\left (\frac {\left (1-\sqrt {2}\right ) \tan (e+f x)-2 \sqrt {2}+3}{\sqrt {2 \left (5 \sqrt {2}-7\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 f}-\frac {22 \sqrt {\tan (e+f x)+1} \tan (e+f x)}{105 f}+\frac {44 \sqrt {\tan (e+f x)+1}}{105 f}-\frac {\sqrt {1+\sqrt {2}} \tanh ^{-1}\left (\frac {\left (1+\sqrt {2}\right ) \tan (e+f x)+2 \sqrt {2}+3}{\sqrt {2 \left (7+5 \sqrt {2}\right )} \sqrt {\tan (e+f x)+1}}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[-1 + Sqrt[2]]*ArcTan[(3 - 2*Sqrt[2] + (1 - Sqrt[2])*Tan[e + f*x])/(Sqrt[2*(-7 + 5*Sqrt[2])]*Sqrt[1 + Ta
n[e + f*x]])])/(2*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 203

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

Rule 207

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

Rule 3535

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 3536

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 3566

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 3630

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 3647

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 3648

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*} \int \frac {\tan ^5(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx &=\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 ) \operatorname {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 ) \operatorname {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}} \tan ^{-1}\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}} \tanh ^{-1}\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*}

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Mathematica [C]  time = 0.86, size = 112, normalized size = 0.46 \[ \frac {4 \sqrt {\tan (e+f x)+1} \left (15 \tan ^3(e+f x)-18 \tan ^2(e+f x)-11 \tan (e+f x)+22\right )-\frac {210 \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1-i}}\right )}{\sqrt {1-i}}-\frac {210 \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1+i}}\right )}{\sqrt {1+i}}}{210 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-210*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]])/Sqrt[1 - I] - (210*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 +
I]])/Sqrt[1 + I] + 4*Sqrt[1 + Tan[e + f*x]]*(22 - 11*Tan[e + f*x] - 18*Tan[e + f*x]^2 + 15*Tan[e + f*x]^3))/(
210*f)

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fricas [B]  time = 0.50, size = 1008, normalized size = 4.18 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/420*(420*(1/2)^(3/4)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(2*(1/2)^(3/4)*(f^5*sqr
t(f^(-4)) + sqrt(1/2)*f^3)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*sqrt((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x
+ e) + (1/2)^(1/4)*(2*sqrt(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(
-4)) + 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + cos(f*x + e) + sin(f*x + e))/cos(f
*x + e))*(f^(-4))^(3/4) - 2*(1/2)^(3/4)*(f^5*sqrt(f^(-4)) + sqrt(1/2)*f^3)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4))
+ 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) - f^2*sqrt(f^(-4)) - 2*sqrt(1/2))*cos(f*x
+ e)^3 + 420*(1/2)^(3/4)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(2*(1/2)^(3/4)*(f^5*s
qrt(f^(-4)) + sqrt(1/2)*f^3)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*sqrt((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*
x + e) - (1/2)^(1/4)*(2*sqrt(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-4*sqrt(1/2)*f^2*sqrt(f
^(-4)) + 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + cos(f*x + e) + sin(f*x + e))/cos
(f*x + e))*(f^(-4))^(3/4) - 2*(1/2)^(3/4)*(f^5*sqrt(f^(-4)) + sqrt(1/2)*f^3)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)
) + 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) + f^2*sqrt(f^(-4)) + 2*sqrt(1/2))*cos(f
*x + e)^3 + 105*(1/2)^(1/4)*(sqrt(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e)^3 + f*cos(f*x + e)^3)*sqrt(-4*sqrt(1/2)*f
^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*log((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e) + (1/2)^(1/4)*(2*sqrt(1/2)*
f^3*sqrt(f^(-4))*cos(f*x + e) + f*cos(f*x + e))*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x + e) + s
in(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + cos(f*x + e) + sin(f*x + e))/cos(f*x + e)) - 105*(1/2)^(1/4)*(sqrt
(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e)^3 + f*cos(f*x + e)^3)*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/
4)*log((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e) - (1/2)^(1/4)*(2*sqrt(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e) + f
*cos(f*x + e))*sqrt(-4*sqrt(1/2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4
))^(1/4) + cos(f*x + e) + sin(f*x + e))/cos(f*x + e)) - 8*(40*cos(f*x + e)^3 - (26*cos(f*x + e)^2 - 15)*sin(f*
x + e) - 18*cos(f*x + e))*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e)))/(f*cos(f*x + e)^3)

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giac [A]  time = 1.15, size = 254, normalized size = 1.05 \[ \frac {\sqrt {\sqrt {2} - 1} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f} + \frac {\sqrt {\sqrt {2} - 1} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f} - \frac {\sqrt {\sqrt {2} + 1} \log \left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f} + \frac {\sqrt {\sqrt {2} + 1} \log \left (-2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f} + \frac {2 \, {\left (15 \, f^{6} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {7}{2}} - 63 \, f^{6} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {5}{2}} + 70 \, f^{6} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}\right )}}{105 \, f^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(sqrt(2) - 1)*arctan(1/2*2^(3/4)*(2^(1/4)*sqrt(sqrt(2) + 2) + 2*sqrt(tan(f*x + e) + 1))/sqrt(-sqrt(2)
+ 2))/f + 1/2*sqrt(sqrt(2) - 1)*arctan(-1/2*2^(3/4)*(2^(1/4)*sqrt(sqrt(2) + 2) - 2*sqrt(tan(f*x + e) + 1))/sqr
t(-sqrt(2) + 2))/f - 1/4*sqrt(sqrt(2) + 1)*log(2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + sqrt(2) + ta
n(f*x + e) + 1)/f + 1/4*sqrt(sqrt(2) + 1)*log(-2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + sqrt(2) + ta
n(f*x + e) + 1)/f + 2/105*(15*f^6*(tan(f*x + e) + 1)^(7/2) - 63*f^6*(tan(f*x + e) + 1)^(5/2) + 70*f^6*(tan(f*x
+ e) + 1)^(3/2))/f^7

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maple [A]  time = 0.22, size = 342, normalized size = 1.42 \[ \frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7 f}-\frac {6 \left (1+\tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}+\frac {4 \left (1+\tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f}+\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8 f}-\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{2 f \sqrt {-2+2 \sqrt {2}}}+\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8 f}-\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right ) \sqrt {2}}{2 f \sqrt {-2+2 \sqrt {2}}}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7/f*(1+tan(f*x+e))^(7/2)-6/5*(1+tan(f*x+e))^(5/2)/f+4/3*(1+tan(f*x+e))^(3/2)/f+1/8/f*(2*2^(1/2)+2)^(1/2)*2^(
1/2)*ln(1+2^(1/2)-(2*2^(1/2)+2)^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))-1/2/f/(-2+2*2^(1/2))^(1/2)*arctan((2*(1
+tan(f*x+e))^(1/2)-(2*2^(1/2)+2)^(1/2))/(-2+2*2^(1/2))^(1/2))*2^(1/2)+1/f/(-2+2*2^(1/2))^(1/2)*arctan((2*(1+ta
n(f*x+e))^(1/2)-(2*2^(1/2)+2)^(1/2))/(-2+2*2^(1/2))^(1/2))-1/8/f*(2*2^(1/2)+2)^(1/2)*2^(1/2)*ln(1+2^(1/2)+(2*2
^(1/2)+2)^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))-1/2/f/(-2+2*2^(1/2))^(1/2)*arctan(((2*2^(1/2)+2)^(1/2)+2*(1+t
an(f*x+e))^(1/2))/(-2+2*2^(1/2))^(1/2))*2^(1/2)+1/f/(-2+2*2^(1/2))^(1/2)*arctan(((2*2^(1/2)+2)^(1/2)+2*(1+tan(
f*x+e))^(1/2))/(-2+2*2^(1/2))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (f x + e\right )^{5}}{\sqrt {\tan \left (f x + e\right ) + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.42, size = 118, normalized size = 0.49 \[ \frac {4\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{3/2}}{3\,f}-\frac {6\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{5/2}}{5\,f}+\frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{7/2}}{7\,f}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,2{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,2{}\mathrm {i}\right )\,\sqrt {\frac {\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{5}{\left (e + f x \right )}}{\sqrt {\tan {\left (e + f x \right )} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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