3.4.78 \(\int \tan ^5(e+f x) \sqrt {1+\tan (e+f x)} \, dx\) [378]

Optimal. Leaf size=264 \[ -\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {2}\right )} \text {ArcTan}\left (\frac {4-3 \sqrt {2}+\left (2-\sqrt {2}\right ) \tan (e+f x)}{2 \sqrt {-7+5 \sqrt {2}} \sqrt {1+\tan (e+f x)}}\right )}{f}-\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tanh ^{-1}\left (\frac {4+3 \sqrt {2}+\left (2+\sqrt {2}\right ) \tan (e+f x)}{2 \sqrt {7+5 \sqrt {2}} \sqrt {1+\tan (e+f x)}}\right )}{f}+\frac {2 \sqrt {1+\tan (e+f x)}}{f}+\frac {52 (1+\tan (e+f x))^{3/2}}{315 f}-\frac {26 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{105 f}-\frac {4 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{21 f}+\frac {2 \tan ^3(e+f x) (1+\tan (e+f x))^{3/2}}{9 f} \]

[Out]

-1/2*arctan(1/2*(4-3*2^(1/2)+(2-2^(1/2))*tan(f*x+e))/(-7+5*2^(1/2))^(1/2)/(1+tan(f*x+e))^(1/2))*(-2+2*2^(1/2))
^(1/2)/f-1/2*arctanh(1/2*(4+3*2^(1/2)+(2+2^(1/2))*tan(f*x+e))/(7+5*2^(1/2))^(1/2)/(1+tan(f*x+e))^(1/2))*(2+2*2
^(1/2))^(1/2)/f+2*(1+tan(f*x+e))^(1/2)/f+52/315*(1+tan(f*x+e))^(3/2)/f-26/105*tan(f*x+e)*(1+tan(f*x+e))^(3/2)/
f-4/21*tan(f*x+e)^2*(1+tan(f*x+e))^(3/2)/f+2/9*tan(f*x+e)^3*(1+tan(f*x+e))^(3/2)/f

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Rubi [A]
time = 0.38, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3647, 3728, 3729, 3711, 12, 3609, 3617, 3616, 209, 213} \begin {gather*} -\frac {\sqrt {\frac {1}{2} \left (\sqrt {2}-1\right )} \text {ArcTan}\left (\frac {\left (2-\sqrt {2}\right ) \tan (e+f x)-3 \sqrt {2}+4}{2 \sqrt {5 \sqrt {2}-7} \sqrt {\tan (e+f x)+1}}\right )}{f}+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^3(e+f x)}{9 f}-\frac {4 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{21 f}-\frac {26 (\tan (e+f x)+1)^{3/2} \tan (e+f x)}{105 f}+\frac {52 (\tan (e+f x)+1)^{3/2}}{315 f}+\frac {2 \sqrt {\tan (e+f x)+1}}{f}-\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tanh ^{-1}\left (\frac {\left (2+\sqrt {2}\right ) \tan (e+f x)+3 \sqrt {2}+4}{2 \sqrt {7+5 \sqrt {2}} \sqrt {\tan (e+f x)+1}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[(-1 + Sqrt[2])/2]*ArcTan[(4 - 3*Sqrt[2] + (2 - Sqrt[2])*Tan[e + f*x])/(2*Sqrt[-7 + 5*Sqrt[2]]*Sqrt[1 +
Tan[e + f*x]])])/f) - (Sqrt[(1 + Sqrt[2])/2]*ArcTanh[(4 + 3*Sqrt[2] + (2 + Sqrt[2])*Tan[e + f*x])/(2*Sqrt[7 +
5*Sqrt[2]]*Sqrt[1 + Tan[e + f*x]])])/f + (2*Sqrt[1 + Tan[e + f*x]])/f + (52*(1 + Tan[e + f*x])^(3/2))/(315*f)
- (26*Tan[e + f*x]*(1 + Tan[e + f*x])^(3/2))/(105*f) - (4*Tan[e + f*x]^2*(1 + Tan[e + f*x])^(3/2))/(21*f) + (
2*Tan[e + f*x]^3*(1 + Tan[e + f*x])^(3/2))/(9*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 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*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] && GtQ[m, 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*} \int \tan ^5(e+f x) \sqrt {1+\tan (e+f x)} \, dx &=\frac {2 \tan ^3(e+f x) (1+\tan (e+f x))^{3/2}}{9 f}+\frac {2}{9} \int \tan ^2(e+f x) \sqrt {1+\tan (e+f x)} \left (-3-\frac {9}{2} \tan (e+f x)-3 \tan ^2(e+f x)\right ) \, dx\\ &=-\frac {4 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{21 f}+\frac {2 \tan ^3(e+f x) (1+\tan (e+f x))^{3/2}}{9 f}+\frac {4}{63} \int \tan (e+f x) \sqrt {1+\tan (e+f x)} \left (6-\frac {39}{4} \tan ^2(e+f x)\right ) \, dx\\ &=-\frac {26 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{105 f}-\frac {4 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{21 f}+\frac {2 \tan ^3(e+f x) (1+\tan (e+f x))^{3/2}}{9 f}+\frac {8}{315} \int \sqrt {1+\tan (e+f x)} \left (\frac {39}{4}+\frac {315}{8} \tan (e+f x)+\frac {39}{4} \tan ^2(e+f x)\right ) \, dx\\ &=\frac {52 (1+\tan (e+f x))^{3/2}}{315 f}-\frac {26 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{105 f}-\frac {4 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{21 f}+\frac {2 \tan ^3(e+f x) (1+\tan (e+f x))^{3/2}}{9 f}+\frac {8}{315} \int \frac {315}{8} \tan (e+f x) \sqrt {1+\tan (e+f x)} \, dx\\ &=\frac {52 (1+\tan (e+f x))^{3/2}}{315 f}-\frac {26 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{105 f}-\frac {4 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{21 f}+\frac {2 \tan ^3(e+f x) (1+\tan (e+f x))^{3/2}}{9 f}+\int \tan (e+f x) \sqrt {1+\tan (e+f x)} \, dx\\ &=\frac {2 \sqrt {1+\tan (e+f x)}}{f}+\frac {52 (1+\tan (e+f x))^{3/2}}{315 f}-\frac {26 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{105 f}-\frac {4 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{21 f}+\frac {2 \tan ^3(e+f x) (1+\tan (e+f x))^{3/2}}{9 f}+\int \frac {-1+\tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=\frac {2 \sqrt {1+\tan (e+f x)}}{f}+\frac {52 (1+\tan (e+f x))^{3/2}}{315 f}-\frac {26 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{105 f}-\frac {4 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{21 f}+\frac {2 \tan ^3(e+f x) (1+\tan (e+f x))^{3/2}}{9 f}-\frac {\int \frac {\sqrt {2}+\left (-2-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx}{2 \sqrt {2}}+\frac {\int \frac {-\sqrt {2}+\left (-2+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx}{2 \sqrt {2}}\\ &=\frac {2 \sqrt {1+\tan (e+f x)}}{f}+\frac {52 (1+\tan (e+f x))^{3/2}}{315 f}-\frac {26 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{105 f}-\frac {4 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{21 f}+\frac {2 \tan ^3(e+f x) (1+\tan (e+f x))^{3/2}}{9 f}+\frac {\left (4-3 \sqrt {2}\right ) \text {Subst}\left (\int \frac {1}{-2 \sqrt {2} \left (-2+\sqrt {2}\right )-4 \left (-2+\sqrt {2}\right )^2+x^2} \, dx,x,\frac {-\sqrt {2}-2 \left (-2+\sqrt {2}\right )-\left (-2+\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}}\right )}{f}+\frac {\left (4+3 \sqrt {2}\right ) \text {Subst}\left (\int \frac {1}{2 \sqrt {2} \left (-2-\sqrt {2}\right )-4 \left (-2-\sqrt {2}\right )^2+x^2} \, dx,x,\frac {\sqrt {2}-2 \left (-2-\sqrt {2}\right )-\left (-2-\sqrt {2}\right ) \tan (e+f x)}{\sqrt {1+\tan (e+f x)}}\right )}{f}\\ &=-\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {4-3 \sqrt {2}+\left (2-\sqrt {2}\right ) \tan (e+f x)}{2 \sqrt {-7+5 \sqrt {2}} \sqrt {1+\tan (e+f x)}}\right )}{f}-\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tanh ^{-1}\left (\frac {4+3 \sqrt {2}+\left (2+\sqrt {2}\right ) \tan (e+f x)}{2 \sqrt {7+5 \sqrt {2}} \sqrt {1+\tan (e+f x)}}\right )}{f}+\frac {2 \sqrt {1+\tan (e+f x)}}{f}+\frac {52 (1+\tan (e+f x))^{3/2}}{315 f}-\frac {26 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{105 f}-\frac {4 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{21 f}+\frac {2 \tan ^3(e+f x) (1+\tan (e+f x))^{3/2}}{9 f}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.86, size = 150, normalized size = 0.57 \begin {gather*} \frac {\cos (e+f x) (1+\tan (e+f x)) \left (-315 \sqrt {1-i} \tanh ^{-1}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )-315 \sqrt {1+i} \tanh ^{-1}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )+2 \sqrt {1+\tan (e+f x)} \left (445+35 \sec ^4(e+f x)-18 \tan (e+f x)+\sec ^2(e+f x) (-139+5 \tan (e+f x))\right )\right )}{315 f (\cos (e+f x)+\sin (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(1 + Tan[e + f*x])*(-315*Sqrt[1 - I]*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]] - 315*Sqrt[1 +
I]*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 + I]] + 2*Sqrt[1 + Tan[e + f*x]]*(445 + 35*Sec[e + f*x]^4 - 18*Tan[e
+ f*x] + Sec[e + f*x]^2*(-139 + 5*Tan[e + f*x]))))/(315*f*(Cos[e + f*x] + Sin[e + f*x]))

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Maple [A]
time = 0.42, size = 242, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {\frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}-\frac {6 \left (1+\tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {4 \left (1+\tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+2 \sqrt {1+\tan \left (f x +e \right )}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4}-\frac {\left (\sqrt {2}-1\right ) \arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}}{f}\)	\(242\)
default	\(\frac {\frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}-\frac {6 \left (1+\tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {4 \left (1+\tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+2 \sqrt {1+\tan \left (f x +e \right )}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4}-\frac {\left (\sqrt {2}-1\right ) \arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4}+\frac {\left (1-\sqrt {2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{\sqrt {-2+2 \sqrt {2}}}}{f}\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(2/9*(1+tan(f*x+e))^(9/2)-6/7*(1+tan(f*x+e))^(7/2)+4/5*(1+tan(f*x+e))^(5/2)+2*(1+tan(f*x+e))^(1/2)+1/4*(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))-(2^(1/2)-1)/(-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))-1/4*(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^(1/2))/(-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1099 vs. \(2 (220) = 440\).
time = 0.99, size = 1099, normalized size = 4.16 \begin {gather*} \frac {1260 \cdot 2^{\frac {3}{4}} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} f \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} \sqrt {\frac {1}{2}} {\left (f^{5} \sqrt {\frac {1}{f^{4}}} + \sqrt {2} f^{3}\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 2^{\frac {1}{4}} {\left (\sqrt {2} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 2 \, f \cos \left (f x + e\right )\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} + 2 \, \cos \left (f x + e\right ) + 2 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} - \frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (f^{5} \sqrt {\frac {1}{f^{4}}} + \sqrt {2} f^{3}\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} - f^{2} \sqrt {\frac {1}{f^{4}}} - \sqrt {2}\right ) \cos \left (f x + e\right )^{4} + 1260 \cdot 2^{\frac {3}{4}} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} f \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {3}{4}} \sqrt {\frac {1}{2}} {\left (f^{5} \sqrt {\frac {1}{f^{4}}} + \sqrt {2} f^{3}\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - 2^{\frac {1}{4}} {\left (\sqrt {2} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 2 \, f \cos \left (f x + e\right )\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} + 2 \, \cos \left (f x + e\right ) + 2 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} - \frac {1}{2} \cdot 2^{\frac {3}{4}} {\left (f^{5} \sqrt {\frac {1}{f^{4}}} + \sqrt {2} f^{3}\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {3}{4}} + f^{2} \sqrt {\frac {1}{f^{4}}} + \sqrt {2}\right ) \cos \left (f x + e\right )^{4} - 315 \cdot 2^{\frac {1}{4}} {\left (\sqrt {2} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right )^{4} + 2 \, f \cos \left (f x + e\right )^{4}\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \frac {1}{f^{4}}^{\frac {1}{4}} \log \left (\frac {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 2^{\frac {1}{4}} {\left (\sqrt {2} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 2 \, f \cos \left (f x + e\right )\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} + 2 \, \cos \left (f x + e\right ) + 2 \, \sin \left (f x + e\right )}{2 \, \cos \left (f x + e\right )}\right ) + 315 \cdot 2^{\frac {1}{4}} {\left (\sqrt {2} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right )^{4} + 2 \, f \cos \left (f x + e\right )^{4}\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \frac {1}{f^{4}}^{\frac {1}{4}} \log \left (\frac {2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - 2^{\frac {1}{4}} {\left (\sqrt {2} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 2 \, f \cos \left (f x + e\right )\right )} \sqrt {-2 \, \sqrt {2} f^{2} \sqrt {\frac {1}{f^{4}}} + 4} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \frac {1}{f^{4}}^{\frac {1}{4}} + 2 \, \cos \left (f x + e\right ) + 2 \, \sin \left (f x + e\right )}{2 \, \cos \left (f x + e\right )}\right ) + 16 \, {\left (445 \, \cos \left (f x + e\right )^{4} - 139 \, \cos \left (f x + e\right )^{2} - {\left (18 \, \cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) + 35\right )} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{2520 \, f \cos \left (f x + e\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2520*(1260*2^(3/4)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(1/2*2^(3/4)*sqrt(1/2)*(f^5*
sqrt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e
) + 2^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 2*f*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*s
qrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e)
)*(f^(-4))^(3/4) - 1/2*2^(3/4)*(f^5*sqrt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((co
s(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) - f^2*sqrt(f^(-4)) - sqrt(2))*cos(f*x + e)^4 + 1260*2^
(3/4)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(1/2*2^(3/4)*sqrt(1/2)*(f^5*sqrt(f^(-4)) +
sqrt(2)*f^3)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 2^(1/4)*(sq
rt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 2*f*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x +
e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4
) - 1/2*2^(3/4)*(f^5*sqrt(f^(-4)) + sqrt(2)*f^3)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x + e) + si
n(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) + f^2*sqrt(f^(-4)) + sqrt(2))*cos(f*x + e)^4 - 315*2^(1/4)*(sqrt(2)*f
^3*sqrt(f^(-4))*cos(f*x + e)^4 + 2*f*cos(f*x + e)^4)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*log(
1/2*(2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 2^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 2*f*cos(f*x +
e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + 2*
cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e)) + 315*2^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e)^4 + 2*f*cos(
f*x + e)^4)*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*log(1/2*(2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x +
e) - 2^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e) + 2*f*cos(f*x + e))*sqrt(-2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)
*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x +
e)) + 16*(445*cos(f*x + e)^4 - 139*cos(f*x + e)^2 - (18*cos(f*x + e)^3 - 5*cos(f*x + e))*sin(f*x + e) + 35)*sq
rt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e)))/(f*cos(f*x + e)^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.68, size = 278, normalized size = 1.05 \begin {gather*} -\frac {\sqrt {2 \, \sqrt {2} - 2} \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 {2 \, \sqrt {2} - 2} \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 {2 \, \sqrt {2} + 2} \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 {2 \, \sqrt {2} + 2} \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 (35 \, f^{8} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {9}{2}} - 135 \, f^{8} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {7}{2}} + 126 \, f^{8} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {5}{2}} + 315 \, f^{8} \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{315 \, f^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2*sqrt(2) - 2)*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(2*sqrt(2) - 2)*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 - 1/4*sqrt(2*sqrt(2) + 2)*log(2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + sqrt(
2) + tan(f*x + e) + 1)/f + 1/4*sqrt(2*sqrt(2) + 2)*log(-2^(1/4)*sqrt(sqrt(2) + 2)*sqrt(tan(f*x + e) + 1) + sqr
t(2) + tan(f*x + e) + 1)/f + 2/315*(35*f^8*(tan(f*x + e) + 1)^(9/2) - 135*f^8*(tan(f*x + e) + 1)^(7/2) + 126*f
^8*(tan(f*x + e) + 1)^(5/2) + 315*f^8*sqrt(tan(f*x + e) + 1))/f^9

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Mupad [B]
time = 6.03, size = 135, normalized size = 0.51 \begin {gather*} \frac {2\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}}{f}+\frac {4\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{5/2}}{5\,f}-\frac {6\,{\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}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,\left (1-\mathrm {i}\right )\right )\,\sqrt {\frac {\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (f\,\sqrt {\frac {\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,\left (1+1{}\mathrm {i}\right )\right )\,\sqrt {\frac {\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \end {gather*}

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]

(2*(tan(e + f*x) + 1)^(1/2))/f + (4*(tan(e + f*x) + 1)^(5/2))/(5*f) - (6*(tan(e + f*x) + 1)^(7/2))/(7*f) + (2*
(tan(e + f*x) + 1)^(9/2))/(9*f) - atan(f*((1/4 - 1i/4)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2)*(1 - 1i))*((1/4 - 1
i/4)/f^2)^(1/2)*2i + atan(f*((1/4 + 1i/4)/f^2)^(1/2)*(tan(e + f*x) + 1)^(1/2)*(1 + 1i))*((1/4 + 1i/4)/f^2)^(1/
2)*2i

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