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

Optimal. Leaf size=318 \[ -\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}+\frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}-\frac {8 (\tan (e+f x)+1)^{3/2} \tan (e+f x)}{35 f}-\frac {18 (\tan (e+f x)+1)^{3/2}}{35 f}+\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f} \]

[Out]

-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))*(2+2*2^(1/2))^(1/2)/f+1/2*arcta
n(((2+2*2^(1/2))^(1/2)+2*(1+tan(f*x+e))^(1/2))/(-2+2*2^(1/2))^(1/2))*(2+2*2^(1/2))^(1/2)/f+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))/f/(2+2*2^(1/2))^(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))/f/(2+2*2^(1/2))^(1/2)-18/35*(1+tan(f*x+e))^(3/2)/f-8/35*tan(f*x+e)*(1+tan(f*x
+e))^(3/2)/f+2/7*tan(f*x+e)^2*(1+tan(f*x+e))^(3/2)/f

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Rubi [A]  time = 0.43, antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3566, 3647, 3631, 3485, 700, 1127, 1161, 618, 204, 1164, 628} \[ \frac {2 (\tan (e+f x)+1)^{3/2} \tan ^2(e+f x)}{7 f}-\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )}-2 \sqrt {\tan (e+f x)+1}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}+\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {2 \sqrt {\tan (e+f x)+1}+\sqrt {2 \left (1+\sqrt {2}\right )}}{\sqrt {2 \left (\sqrt {2}-1\right )}}\right )}{f}-\frac {8 (\tan (e+f x)+1)^{3/2} \tan (e+f x)}{35 f}-\frac {18 (\tan (e+f x)+1)^{3/2}}{35 f}+\frac {\log \left (\tan (e+f x)-\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {\log \left (\tan (e+f x)+\sqrt {2 \left (1+\sqrt {2}\right )} \sqrt {\tan (e+f x)+1}+\sqrt {2}+1\right )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[(1 + Sqrt[2])/2]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/f)
+ (Sqrt[(1 + Sqrt[2])/2]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] + 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/f
+ Log[1 + Sqrt[2] + Tan[e + f*x] - Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/(2*Sqrt[2*(1 + Sqrt[2])]*f)
- Log[1 + Sqrt[2] + Tan[e + f*x] + Sqrt[2*(1 + Sqrt[2])]*Sqrt[1 + Tan[e + f*x]]]/(2*Sqrt[2*(1 + Sqrt[2])]*f) -
(18*(1 + Tan[e + f*x])^(3/2))/(35*f) - (8*Tan[e + f*x]*(1 + Tan[e + f*x])^(3/2))/(35*f) + (2*Tan[e + f*x]^2*(
1 + Tan[e + f*x])^(3/2))/(7*f)

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 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 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]

Rule 700

Int[Sqrt[(d_) + (e_.)*(x_)]/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[2*e, Subst[Int[x^2/(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 1127

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

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 1164

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

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 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 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 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])))

Rubi steps

\begin {align*} \int \tan ^4(e+f x) \sqrt {1+\tan (e+f x)} \, dx &=\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{7 f}+\frac {2}{7} \int \tan (e+f x) \sqrt {1+\tan (e+f x)} \left (-2-\frac {7}{2} \tan (e+f x)-2 \tan ^2(e+f x)\right ) \, dx\\ &=-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{35 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{7 f}+\frac {4}{35} \int \sqrt {1+\tan (e+f x)} \left (2-\frac {27}{4} \tan ^2(e+f x)\right ) \, dx\\ &=-\frac {18 (1+\tan (e+f x))^{3/2}}{35 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{35 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{7 f}+\int \sqrt {1+\tan (e+f x)} \, dx\\ &=-\frac {18 (1+\tan (e+f x))^{3/2}}{35 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{35 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{7 f}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {1+x}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {18 (1+\tan (e+f x))^{3/2}}{35 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{35 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{7 f}+\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{2-2 x^2+x^4} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f}\\ &=-\frac {18 (1+\tan (e+f x))^{3/2}}{35 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{35 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{7 f}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}-x^2}{2-2 x^2+x^4} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}+x^2}{2-2 x^2+x^4} \, dx,x,\sqrt {1+\tan (e+f x)}\right )}{f}\\ &=-\frac {18 (1+\tan (e+f x))^{3/2}}{35 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{35 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{7 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 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 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 )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} 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 )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} 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 )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} 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 )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {18 (1+\tan (e+f x))^{3/2}}{35 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{35 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{7 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 )}{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 )}{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 )}{\sqrt {2 \left (-1+\sqrt {2}\right )} 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 )}{\sqrt {2 \left (-1+\sqrt {2}\right )} 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 )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} 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 )}{2 \sqrt {2 \left (1+\sqrt {2}\right )} f}-\frac {18 (1+\tan (e+f x))^{3/2}}{35 f}-\frac {8 \tan (e+f x) (1+\tan (e+f x))^{3/2}}{35 f}+\frac {2 \tan ^2(e+f x) (1+\tan (e+f x))^{3/2}}{7 f}\\ \end {align*}

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Mathematica [C]  time = 0.71, size = 118, normalized size = 0.37 \[ \frac {-35 i \sqrt {1-i} \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1-i}}\right )+35 i \sqrt {1+i} \tanh ^{-1}\left (\frac {\sqrt {\tan (e+f x)+1}}{\sqrt {1+i}}\right )+2 \sqrt {\tan (e+f x)+1} \left ((5 \tan (e+f x)+1) \sec ^2(e+f x)-2 (9 \tan (e+f x)+5)\right )}{35 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-35*I)*Sqrt[1 - I]*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]] + (35*I)*Sqrt[1 + I]*ArcTanh[Sqrt[1 + Tan[e +
f*x]]/Sqrt[1 + I]] + 2*Sqrt[1 + Tan[e + f*x]]*(Sec[e + f*x]^2*(1 + 5*Tan[e + f*x]) - 2*(5 + 9*Tan[e + f*x])))
/(35*f)

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fricas [B]  time = 0.62, size = 898, normalized size = 2.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/280*(140*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)*sqrt(2*
sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^5*sqrt((2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((cos(f*x + e) +
sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4)*cos(f*x + e) + 2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 2*cos(f*x
+ e) + 2*sin(f*x + e))/cos(f*x + e))*(f^(-4))^(5/4) - 1/2*2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^5*sqr
t((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(5/4) - f^2*sqrt(f^(-4)) - sqrt(2))*cos(f*x + e)^3 + 14
0*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)*sqrt(2*sqrt(2)*f^
2*sqrt(f^(-4)) + 4)*f^5*sqrt(-(2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((cos(f*x + e) + sin(f*x +
e))/cos(f*x + e))*(f^(-4))^(3/4)*cos(f*x + e) - 2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 2*cos(f*x + e) - 2*
sin(f*x + e))/cos(f*x + e))*(f^(-4))^(5/4) - 1/2*2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^5*sqrt((cos(f*
x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(5/4) + f^2*sqrt(f^(-4)) + sqrt(2))*cos(f*x + e)^3 - 35*2^(1/4)*
(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e)^3 - 2*f*cos(f*x + e)^3)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*(f^(-4))^(
1/4)*log(1/2*(2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e)
)*(f^(-4))^(3/4)*cos(f*x + e) + 2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos
(f*x + e)) + 35*2^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e)^3 - 2*f*cos(f*x + e)^3)*sqrt(2*sqrt(2)*f^2*sqrt
(f^(-4)) + 4)*(f^(-4))^(1/4)*log(-1/2*(2^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((cos(f*x + e) + s
in(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4)*cos(f*x + e) - 2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 2*cos(f*x +
e) - 2*sin(f*x + e))/cos(f*x + e)) + 16*(10*cos(f*x + e)^3 + (18*cos(f*x + e)^2 - 5)*sin(f*x + e) - 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 = 1.06, size = 246, normalized size = 0.77 \[ \frac {\sqrt {2 \, \sqrt {2} + 2} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} + 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f} + \frac {\sqrt {2 \, \sqrt {2} + 2} \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} - 2 \, \sqrt {\tan \left (f x + e\right ) + 1}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right )}{2 \, f} - \frac {\sqrt {2 \, \sqrt {2} - 2} \log \left (2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f} + \frac {\sqrt {2 \, \sqrt {2} - 2} \log \left (-2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2} \sqrt {\tan \left (f x + e\right ) + 1} + \sqrt {2} + \tan \left (f x + e\right ) + 1\right )}{4 \, f} + \frac {2 \, {\left (5 \, f^{6} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {7}{2}} - 14 \, f^{6} {\left (\tan \left (f x + e\right ) + 1\right )}^{\frac {5}{2}}\right )}}{35 \, f^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+tan(f*x+e))^(1/2)*tan(f*x+e)^4,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) + sqrt
(2) + tan(f*x + e) + 1)/f + 2/35*(5*f^6*(tan(f*x + e) + 1)^(7/2) - 14*f^6*(tan(f*x + e) + 1)^(5/2))/f^7

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maple [A]  time = 0.23, size = 318, normalized size = 1.00 \[ \frac {2 \left (1+\tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7 f}-\frac {4 \left (1+\tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}+\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {2 \sqrt {1+\tan \left (f x +e \right )}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}-\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}-\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}-\frac {\sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {2}+2}+2 \sqrt {1+\tan \left (f x +e \right )}}{\sqrt {-2+2 \sqrt {2}}}\right )}{f \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2 \sqrt {2}+2}\, \ln \left (1+\sqrt {2}+\sqrt {2 \sqrt {2}+2}\, \sqrt {1+\tan \left (f x +e \right )}+\tan \left (f x +e \right )\right )}{4 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\tan \left (f x + e\right ) + 1} \tan \left (f x + e\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.11, size = 107, normalized size = 0.34 \[ \frac {2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{7/2}}{7\,f}-\frac {4\,{\left (\mathrm {tan}\left (e+f\,x\right )+1\right )}^{5/2}}{5\,f}-\mathrm {atan}\left (f^3\,{\left (\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}\right )}^{3/2}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,4{}\mathrm {i}\right )\,\sqrt {\frac {-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (f^3\,{\left (\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}\right )}^{3/2}\,\sqrt {\mathrm {tan}\left (e+f\,x\right )+1}\,4{}\mathrm {i}\right )\,\sqrt {\frac {-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}}{f^2}}\,2{}\mathrm {i} \]

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)^(7/2))/(7*f) - atan(f^3*((- 1/4 + 1i/4)/f^2)^(3/2)*(tan(e + f*x) + 1)^(1/2)*4i)*((- 1/4
+ 1i/4)/f^2)^(1/2)*2i - (4*(tan(e + f*x) + 1)^(5/2))/(5*f) - atan(f^3*((- 1/4 - 1i/4)/f^2)^(3/2)*(tan(e + f*x)
+ 1)^(1/2)*4i)*((- 1/4 - 1i/4)/f^2)^(1/2)*2i

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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