3.406 \(\int \frac{\tan ^4(e+f x)}{\sqrt{1+\tan (e+f x)}} \, dx\)

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

[Out]

-(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/(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)

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Rubi [A]  time = 0.356478, antiderivative size = 311, normalized size of antiderivative = 1., 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 = {3566, 3647, 3631, 3485, 708, 1094, 634, 618, 204, 628} \[ \frac{2 \sqrt{\tan (e+f x)+1} \tan ^2(e+f x)}{5 f}-\frac{\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{\tan (e+f x)+1}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{2 f}+\frac{\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sqrt{\tan (e+f x)+1}+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{2 f}-\frac{8 \sqrt{\tan (e+f x)+1} \tan (e+f x)}{15 f}-\frac{14 \sqrt{\tan (e+f x)+1}}{15 f}-\frac{\log \left (\tan (e+f x)-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{\tan (e+f x)+1}+\sqrt{2}+1\right )}{4 \sqrt{1+\sqrt{2}} f}+\frac{\log \left (\tan (e+f x)+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{\tan (e+f x)+1}+\sqrt{2}+1\right )}{4 \sqrt{1+\sqrt{2}} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/(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 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 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 3631

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 3485

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 708

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 1094

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 634

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 618

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 204

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

Rule 628

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]

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{\operatorname{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 \operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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*}

Mathematica [C]  time = 0.778414, size = 112, normalized size = 0.36 \[ \frac{15 (1-i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1-i}}\right )+15 (1+i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1+i}}\right )-4 \sqrt{\tan (e+f x)+1} \sec ^2(e+f x) (2 \sin (2 (e+f x))+5 \cos (2 (e+f x))+2)}{30 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(1 - I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]] + 15*(1 + I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*x]
]/Sqrt[1 + I]] - 4*Sec[e + f*x]^2*(2 + 5*Cos[2*(e + f*x)] + 2*Sin[2*(e + f*x)])*Sqrt[1 + Tan[e + f*x]])/(30*f)

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Maple [A]  time = 0.054, size = 326, normalized size = 1.1 \begin{align*}{\frac{2}{5\,f} \left ( 1+\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{4}{3\,f} \left ( 1+\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{\sqrt{2\,\sqrt{2}+2}}{4\,f}\ln \left ( 1+\sqrt{2}-\sqrt{2\,\sqrt{2}+2}\sqrt{1+\tan \left ( fx+e \right ) }+\tan \left ( fx+e \right ) \right ) }+{\frac{\sqrt{2\,\sqrt{2}+2}\sqrt{2}}{8\,f}\ln \left ( 1+\sqrt{2}-\sqrt{2\,\sqrt{2}+2}\sqrt{1+\tan \left ( fx+e \right ) }+\tan \left ( fx+e \right ) \right ) }+{\frac{\sqrt{2}}{2\,f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( 2\,\sqrt{1+\tan \left ( fx+e \right ) }-\sqrt{2\,\sqrt{2}+2} \right ) } \right ) }+{\frac{\sqrt{2\,\sqrt{2}+2}}{4\,f}\ln \left ( 1+\sqrt{2}+\sqrt{2\,\sqrt{2}+2}\sqrt{1+\tan \left ( fx+e \right ) }+\tan \left ( fx+e \right ) \right ) }-{\frac{\sqrt{2\,\sqrt{2}+2}\sqrt{2}}{8\,f}\ln \left ( 1+\sqrt{2}+\sqrt{2\,\sqrt{2}+2}\sqrt{1+\tan \left ( fx+e \right ) }+\tan \left ( fx+e \right ) \right ) }+{\frac{\sqrt{2}}{2\,f\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{1}{\sqrt{-2+2\,\sqrt{2}}} \left ( \sqrt{2\,\sqrt{2}+2}+2\,\sqrt{1+\tan \left ( fx+e \right ) } \right ) } \right ) } \end{align*}

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)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )^{4}}{\sqrt{\tan \left (f x + e\right ) + 1}}\,{d x} \end{align*}

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]  time = 2.03117, size = 2607, normalized size = 8.38 \begin{align*} \text{result too large to display} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{4}{\left (e + f x \right )}}{\sqrt{\tan{\left (e + f x \right )} + 1}}\, dx \end{align*}

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

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]

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