3.5.6 \(\int \frac {\tan ^4(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\) [406]

Optimal. Leaf size=311 \[ -\frac {\sqrt {1+\sqrt {2}} \text {ArcTan}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{2 f}+\frac {\sqrt {1+\sqrt {2}} \text {ArcTan}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{2 f}-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f} \]

[Out]

-1/4*ln(1+2^(1/2)-(2+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))/f/(1+2^(1/2))^(1/2)+1/4*ln(1+2^(1/2)+(2
+2*2^(1/2))^(1/2)*(1+tan(f*x+e))^(1/2)+tan(f*x+e))/f/(1+2^(1/2))^(1/2)-1/2*arctan(((2+2*2^(1/2))^(1/2)-2*(1+ta
n(f*x+e))^(1/2))/(-2+2*2^(1/2))^(1/2))*(1+2^(1/2))^(1/2)/f+1/2*arctan(((2+2*2^(1/2))^(1/2)+2*(1+tan(f*x+e))^(1
/2))/(-2+2*2^(1/2))^(1/2))*(1+2^(1/2))^(1/2)/f-14/15*(1+tan(f*x+e))^(1/2)/f-8/15*(1+tan(f*x+e))^(1/2)*tan(f*x+
e)/f+2/5*(1+tan(f*x+e))^(1/2)*tan(f*x+e)^2/f

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Rubi [A]
time = 0.24, antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3647, 3728, 3712, 3566, 722, 1108, 648, 632, 210, 642} \begin {gather*} -\frac {\sqrt {1+\sqrt {2}} \text {ArcTan}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{2 f}+\frac {\sqrt {1+\sqrt {2}} \text {ArcTan}\left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{2 f}+\frac {2 \sqrt {\tan (e+f x)+1} \tan ^2(e+f x)}{5 f}-\frac {8 \sqrt {\tan (e+f x)+1} \tan (e+f x)}{15 f}-\frac {14 \sqrt {\tan (e+f x)+1}}{15 f}-\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}} f}+\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{4 \sqrt {1+\sqrt {2}} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^4/Sqrt[1 + Tan[e + f*x]],x]

[Out]

-1/2*(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/f +
(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/(2*f) -
Log[1 + Sqrt[2] + Tan[e + f*x] - Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/(4*Sqrt[1 + Sqrt[2]]*f) + Log[
1 + Sqrt[2] + Tan[e + f*x] + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/(4*Sqrt[1 + Sqrt[2]]*f) - (14*Sqrt[
1 + Tan[e + f*x]])/(15*f) - (8*Tan[e + f*x]*Sqrt[1 + Tan[e + f*x]])/(15*f) + (2*Tan[e + f*x]^2*Sqrt[1 + Tan[e
+ f*x]])/(5*f)

Rule 210

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

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 722

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2*e, Subst[Int[1/(c*d^2 + a*e^2 - 2*c
*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 1108

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}
, Dist[1/(2*c*q*r), Int[(r - x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(r + x)/(q + r*x + x^2), x], x
]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[b^2 - 4*a*c]

Rule 3566

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

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 3712

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] + Dist[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 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])))

Rubi steps

\begin {align*} \int \frac {\tan ^4(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx &=\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}+\frac {2}{5} \int \frac {\tan (e+f x) \left (-2-\frac {5}{2} \tan (e+f x)-2 \tan ^2(e+f x)\right )}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}+\frac {4}{15} \int \frac {2-\frac {7}{4} \tan ^2(e+f x)}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}+\int \frac {1}{\sqrt {1+\tan (e+f x)}} \, dx\\ &=-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {1+x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}+\frac {2 \text {Subst}\left (\int \frac {1}{2-2 x^2+x^4} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f}\\ &=-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}+\frac {\text {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}-x}{\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f}+\frac {\text {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}+x}{\sqrt {2}+\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {1+\sqrt {2}} f}\\ &=-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2} f}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2}+\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{2 \sqrt {2} f}-\frac {\text {Subst}\left (\int \frac {-\sqrt {2 \left (1+\sqrt {2}\right )}+2 x}{\sqrt {2}-\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}+\frac {\text {Subst}\left (\int \frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 x}{\sqrt {2}+\sqrt {2 \left (1+\sqrt {2}\right )} x+x^2} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}\\ &=-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}-\frac {\text {Subst}\left (\int \frac {1}{2 \left (1-\sqrt {2}\right )-x^2} \, dx,x,-\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}\right )}{\sqrt {2} f}-\frac {\text {Subst}\left (\int \frac {1}{2 \left (1-\sqrt {2}\right )-x^2} \, dx,x,\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}\right )}{\sqrt {2} f}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{2 \sqrt {-1+\sqrt {2}} f}+\frac {\tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}+2 \sqrt {1+\tan (e+f x)}}{\sqrt {2 \left (-1+\sqrt {2}\right )}}\right )}{2 \sqrt {-1+\sqrt {2}} f}-\frac {\log \left (1+\sqrt {2}+\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}+\frac {\log \left (1+\sqrt {2}+\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {1+\tan (e+f x)}\right )}{4 \sqrt {1+\sqrt {2}} f}-\frac {14 \sqrt {1+\tan (e+f x)}}{15 f}-\frac {8 \tan (e+f x) \sqrt {1+\tan (e+f x)}}{15 f}+\frac {2 \tan ^2(e+f x) \sqrt {1+\tan (e+f x)}}{5 f}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.73, size = 100, normalized size = 0.32 \begin {gather*} \frac {(1-i)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1-i}}\right )}{2 f}+\frac {(1+i)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {1+\tan (e+f x)}}{\sqrt {1+i}}\right )}{2 f}+\frac {2 (1+\tan (e+f x))^{3/2} (-7+3 \tan (e+f x))}{15 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^4/Sqrt[1 + Tan[e + f*x]],x]

[Out]

((1 - I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]])/(2*f) + ((1 + I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*
x]]/Sqrt[1 + I]])/(2*f) + (2*(1 + Tan[e + f*x])^(3/2)*(-7 + 3*Tan[e + f*x]))/(15*f)

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Maple [A]
time = 0.13, size = 327, normalized size = 1.05

method	result	size
derivativedivides	\(\frac {\frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {4 \left (1+\tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8}-\frac {\left (-2 \sqrt {2}+\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \sqrt {2 \sqrt {2}+2}}{2}\right ) \arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}+\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8}+\frac {\left (2 \sqrt {2}-\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \sqrt {2 \sqrt {2}+2}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}}{f}\)	\(327\)
default	\(\frac {\frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {4 \left (1+\tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8}-\frac {\left (-2 \sqrt {2}+\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \sqrt {2 \sqrt {2}+2}}{2}\right ) \arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}+\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{8}+\frac {\left (2 \sqrt {2}-\frac {\left (-\sqrt {2 \sqrt {2}+2}\, \sqrt {2}+2 \sqrt {2 \sqrt {2}+2}\right ) \sqrt {2 \sqrt {2}+2}}{2}\right ) \arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{2 \sqrt {-2+2 \sqrt {2}}}}{f}\)	\(327\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(2/5*(1+tan(f*x+e))^(5/2)-4/3*(1+tan(f*x+e))^(3/2)-1/8*(-(2*2^(1/2)+2)^(1/2)*2^(1/2)+2*(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*(-2*2^(1/2)+1/2*(-(2*2^(1/2)+2)^(1/2)*
2^(1/2)+2*(2*2^(1/2)+2)^(1/2))*(2*2^(1/2)+2)^(1/2))/(-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))+1/8*(-(2*2^(1/2)+2)^(1/2)*2^(1/2)+2*(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*(2*2^(1/2)-1/2*(-(2*2^(1/2)+2)^(1/2)*2^(1/2)+2*(2*2^(1/2
)+2)^(1/2))*(2*2^(1/2)+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(tan(f*x+e)^4/(1+tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 898 vs. \(2 (250) = 500\).
time = 1.26, size = 898, normalized size = 2.89 \begin {gather*} -\frac {60 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f^{3} \sqrt {\frac {2 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f \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}} \cos \left (f x + e\right ) + \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}} - 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f^{3} \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}}} - 2 \, \sqrt {\frac {1}{2}}\right ) \cos \left (f x + e\right )^{2} + 60 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f \frac {1}{f^{4}}^{\frac {1}{4}} \arctan \left (2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f^{3} \sqrt {\frac {2 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f \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}} \cos \left (f x + e\right ) + \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}} - 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f^{3} \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}}} + 2 \, \sqrt {\frac {1}{2}}\right ) \cos \left (f x + e\right )^{2} + 15 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (\sqrt {\frac {1}{2}} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right )^{2} - f \cos \left (f x + e\right )^{2}\right )} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} \frac {1}{f^{4}}^{\frac {1}{4}} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) + 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f \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}} \cos \left (f x + e\right ) + \cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) - 15 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} {\left (\sqrt {\frac {1}{2}} f^{3} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right )^{2} - f \cos \left (f x + e\right )^{2}\right )} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} \frac {1}{f^{4}}^{\frac {1}{4}} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} \cos \left (f x + e\right ) - 2 \, \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {\sqrt {\frac {1}{2}} f^{2} \sqrt {\frac {1}{f^{4}}} + 1} f \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}} \cos \left (f x + e\right ) + \cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}\right ) + 4 \, {\left (10 \, \cos \left (f x + e\right )^{2} + 4 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3\right )} \sqrt {\frac {\cos \left (f x + e\right ) + \sin \left (f x + e\right )}{\cos \left (f x + e\right )}}}{30 \, f \cos \left (f x + e\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(60*(1/2)^(3/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*f*(f^(-4))^(1/4)*arctan(2*(1/2)^(1/4)*sqrt(sqrt(1/2
)*f^2*sqrt(f^(-4)) + 1)*f^3*sqrt((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2
*sqrt(f^(-4)) + 1)*f*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4)*cos(f*x + e) + cos(f*x +
e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) - 2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*f^3*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)^2 +
60*(1/2)^(3/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*f*(f^(-4))^(1/4)*arctan(2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*s
qrt(f^(-4)) + 1)*f^3*sqrt((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*sqrt(f
^(-4)) + 1)*f*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4)*cos(f*x + e) + cos(f*x + e) + si
n(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) - 2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*f^3*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)^2 + 15*(1/2
)^(1/4)*(sqrt(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e)^2 - f*cos(f*x + e)^2)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) + 1)*(f
^(-4))^(1/4)*log((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 2*(1/2)^(1/4)*sqrt(sqrt(1/2)*f^2*sqrt(f^(-4)) +
1)*f*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4)*cos(f*x + e) + cos(f*x + e) + sin(f*x + e
))/cos(f*x + e)) - 15*(1/2)^(1/4)*(sqrt(1/2)*f^3*sqrt(f^(-4))*cos(f*x + e)^2 - f*cos(f*x + e)^2)*sqrt(sqrt(1/2
)*f^2*sqrt(f^(-4)) + 1)*(f^(-4))^(1/4)*log((2*sqrt(1/2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 2*(1/2)^(1/4)*sqrt(sqr
t(1/2)*f^2*sqrt(f^(-4)) + 1)*f*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4)*cos(f*x + e) +
cos(f*x + e) + sin(f*x + e))/cos(f*x + e)) + 4*(10*cos(f*x + e)^2 + 4*cos(f*x + e)*sin(f*x + e) - 3)*sqrt((cos
(f*x + e) + sin(f*x + e))/cos(f*x + e)))/(f*cos(f*x + e)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{4}{\left (e + f x \right )}}{\sqrt {\tan {\left (e + f x \right )} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.72, size = 238, normalized size = 0.77 \begin {gather*} \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 (3 \, f^{4} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {5}{2}} - 10 \, f^{4} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {3}{2}}\right )}}{15 \, f^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4/(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 + 2/15*(3*f^4*(tan(f*x + e) + 1)^(5/2) - 10*f^4*(tan(f*x + e) + 1)^(3/2))/f^5

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Mupad [B]
time = 0.55, size = 101, normalized size = 0.32 \begin {gather*} \frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{5/2}}{5\,f}-\frac {4\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{3/2}}{3\,f}+\mathrm {atan}\left (2\,f\,\sqrt {\frac {-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (2\,f\,\sqrt {\frac {-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}}{f^2}}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\right )\,\sqrt {\frac {-\frac {1}{8}+\frac {1}{8}{}\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)^4/(tan(e + f*x) + 1)^(1/2),x)

[Out]

(2*(tan(e + f*x) + 1)^(5/2))/(5*f) - (4*(tan(e + f*x) + 1)^(3/2))/(3*f) + atan(2*f*((- 1/8 - 1i/8)/f^2)^(1/2)*
(tan(e + f*x) + 1)^(1/2))*((- 1/8 - 1i/8)/f^2)^(1/2)*2i - atan(2*f*((- 1/8 + 1i/8)/f^2)^(1/2)*(tan(e + f*x) +
1)^(1/2))*((- 1/8 + 1i/8)/f^2)^(1/2)*2i

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