3.389 \(\int \tan ^5(e+f x) (1+\tan (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=369 \[ \frac{2 (\tan (e+f x)+1)^{5/2} \tan ^3(e+f x)}{11 f}-\frac{4 (\tan (e+f x)+1)^{5/2} \tan ^2(e+f x)}{33 f}+\frac{\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{\tan (e+f x)+1}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{f}-\frac{\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sqrt{\tan (e+f x)+1}+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{f}-\frac{50 (\tan (e+f x)+1)^{5/2} \tan (e+f x)}{231 f}+\frac{20 (\tan (e+f x)+1)^{5/2}}{231 f}+\frac{2 (\tan (e+f x)+1)^{3/2}}{3 f}+\frac{2 \sqrt{\tan (e+f x)+1}}{f}+\frac{\log \left (\tan (e+f x)-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{\tan (e+f x)+1}+\sqrt{2}+1\right )}{2 \sqrt{1+\sqrt{2}} f}-\frac{\log \left (\tan (e+f x)+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{\tan (e+f x)+1}+\sqrt{2}+1\right )}{2 \sqrt{1+\sqrt{2}} f} \]

[Out]

(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/f - (Sqr
t[1 + Sqrt[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[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[1 + Sqrt[2]]*f) + (2*Sqrt[1 + Tan[e
+ f*x]])/f + (2*(1 + Tan[e + f*x])^(3/2))/(3*f) + (20*(1 + Tan[e + f*x])^(5/2))/(231*f) - (50*Tan[e + f*x]*(1
+ Tan[e + f*x])^(5/2))/(231*f) - (4*Tan[e + f*x]^2*(1 + Tan[e + f*x])^(5/2))/(33*f) + (2*Tan[e + f*x]^3*(1 + T
an[e + f*x])^(5/2))/(11*f)

________________________________________________________________________________________

Rubi [A]  time = 0.566518, antiderivative size = 369, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 13, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.619, Rules used = {3566, 3647, 3648, 3630, 12, 3528, 3485, 708, 1094, 634, 618, 204, 628} \[ \frac{2 (\tan (e+f x)+1)^{5/2} \tan ^3(e+f x)}{11 f}-\frac{4 (\tan (e+f x)+1)^{5/2} \tan ^2(e+f x)}{33 f}+\frac{\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 \sqrt{\tan (e+f x)+1}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{f}-\frac{\sqrt{1+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sqrt{\tan (e+f x)+1}+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )}{f}-\frac{50 (\tan (e+f x)+1)^{5/2} \tan (e+f x)}{231 f}+\frac{20 (\tan (e+f x)+1)^{5/2}}{231 f}+\frac{2 (\tan (e+f x)+1)^{3/2}}{3 f}+\frac{2 \sqrt{\tan (e+f x)+1}}{f}+\frac{\log \left (\tan (e+f x)-\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{\tan (e+f x)+1}+\sqrt{2}+1\right )}{2 \sqrt{1+\sqrt{2}} f}-\frac{\log \left (\tan (e+f x)+\sqrt{2 \left (1+\sqrt{2}\right )} \sqrt{\tan (e+f x)+1}+\sqrt{2}+1\right )}{2 \sqrt{1+\sqrt{2}} f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^5*(1 + Tan[e + f*x])^(3/2),x]

[Out]

(Sqrt[1 + Sqrt[2]]*ArcTan[(Sqrt[2*(1 + Sqrt[2])] - 2*Sqrt[1 + Tan[e + f*x]])/Sqrt[2*(-1 + Sqrt[2])]])/f - (Sqr
t[1 + Sqrt[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[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[1 + Sqrt[2]]*f) + (2*Sqrt[1 + Tan[e
+ f*x]])/f + (2*(1 + Tan[e + f*x])^(3/2))/(3*f) + (20*(1 + Tan[e + f*x])^(5/2))/(231*f) - (50*Tan[e + f*x]*(1
+ Tan[e + f*x])^(5/2))/(231*f) - (4*Tan[e + f*x]^2*(1 + Tan[e + f*x])^(5/2))/(33*f) + (2*Tan[e + f*x]^3*(1 + T
an[e + f*x])^(5/2))/(11*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 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])))

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3528

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 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 \tan ^5(e+f x) (1+\tan (e+f x))^{3/2} \, dx &=\frac{2 \tan ^3(e+f x) (1+\tan (e+f x))^{5/2}}{11 f}+\frac{2}{11} \int \tan ^2(e+f x) (1+\tan (e+f x))^{3/2} \left (-3-\frac{11}{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))^{5/2}}{33 f}+\frac{2 \tan ^3(e+f x) (1+\tan (e+f x))^{5/2}}{11 f}+\frac{4}{99} \int \tan (e+f x) (1+\tan (e+f x))^{3/2} \left (6-\frac{75}{4} \tan ^2(e+f x)\right ) \, dx\\ &=-\frac{50 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{231 f}-\frac{4 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{33 f}+\frac{2 \tan ^3(e+f x) (1+\tan (e+f x))^{5/2}}{11 f}+\frac{8}{693} \int (1+\tan (e+f x))^{3/2} \left (\frac{75}{4}+\frac{693}{8} \tan (e+f x)+\frac{75}{4} \tan ^2(e+f x)\right ) \, dx\\ &=\frac{20 (1+\tan (e+f x))^{5/2}}{231 f}-\frac{50 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{231 f}-\frac{4 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{33 f}+\frac{2 \tan ^3(e+f x) (1+\tan (e+f x))^{5/2}}{11 f}+\frac{8}{693} \int \frac{693}{8} \tan (e+f x) (1+\tan (e+f x))^{3/2} \, dx\\ &=\frac{20 (1+\tan (e+f x))^{5/2}}{231 f}-\frac{50 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{231 f}-\frac{4 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{33 f}+\frac{2 \tan ^3(e+f x) (1+\tan (e+f x))^{5/2}}{11 f}+\int \tan (e+f x) (1+\tan (e+f x))^{3/2} \, dx\\ &=\frac{2 (1+\tan (e+f x))^{3/2}}{3 f}+\frac{20 (1+\tan (e+f x))^{5/2}}{231 f}-\frac{50 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{231 f}-\frac{4 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{33 f}+\frac{2 \tan ^3(e+f x) (1+\tan (e+f x))^{5/2}}{11 f}+\int (-1+\tan (e+f x)) \sqrt{1+\tan (e+f x)} \, dx\\ &=\frac{2 \sqrt{1+\tan (e+f x)}}{f}+\frac{2 (1+\tan (e+f x))^{3/2}}{3 f}+\frac{20 (1+\tan (e+f x))^{5/2}}{231 f}-\frac{50 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{231 f}-\frac{4 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{33 f}+\frac{2 \tan ^3(e+f x) (1+\tan (e+f x))^{5/2}}{11 f}+\int -\frac{2}{\sqrt{1+\tan (e+f x)}} \, dx\\ &=\frac{2 \sqrt{1+\tan (e+f x)}}{f}+\frac{2 (1+\tan (e+f x))^{3/2}}{3 f}+\frac{20 (1+\tan (e+f x))^{5/2}}{231 f}-\frac{50 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{231 f}-\frac{4 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{33 f}+\frac{2 \tan ^3(e+f x) (1+\tan (e+f x))^{5/2}}{11 f}-2 \int \frac{1}{\sqrt{1+\tan (e+f x)}} \, dx\\ &=\frac{2 \sqrt{1+\tan (e+f x)}}{f}+\frac{2 (1+\tan (e+f x))^{3/2}}{3 f}+\frac{20 (1+\tan (e+f x))^{5/2}}{231 f}-\frac{50 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{231 f}-\frac{4 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{33 f}+\frac{2 \tan ^3(e+f x) (1+\tan (e+f x))^{5/2}}{11 f}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x} \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{2 \sqrt{1+\tan (e+f x)}}{f}+\frac{2 (1+\tan (e+f x))^{3/2}}{3 f}+\frac{20 (1+\tan (e+f x))^{5/2}}{231 f}-\frac{50 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{231 f}-\frac{4 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{33 f}+\frac{2 \tan ^3(e+f x) (1+\tan (e+f x))^{5/2}}{11 f}-\frac{4 \operatorname{Subst}\left (\int \frac{1}{2-2 x^2+x^4} \, dx,x,\sqrt{1+\tan (e+f x)}\right )}{f}\\ &=\frac{2 \sqrt{1+\tan (e+f x)}}{f}+\frac{2 (1+\tan (e+f x))^{3/2}}{3 f}+\frac{20 (1+\tan (e+f x))^{5/2}}{231 f}-\frac{50 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{231 f}-\frac{4 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{33 f}+\frac{2 \tan ^3(e+f x) (1+\tan (e+f x))^{5/2}}{11 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 )}{\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 )}{\sqrt{1+\sqrt{2}} f}\\ &=\frac{2 \sqrt{1+\tan (e+f x)}}{f}+\frac{2 (1+\tan (e+f x))^{3/2}}{3 f}+\frac{20 (1+\tan (e+f x))^{5/2}}{231 f}-\frac{50 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{231 f}-\frac{4 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{33 f}+\frac{2 \tan ^3(e+f x) (1+\tan (e+f x))^{5/2}}{11 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 )}{\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 )}{\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 )}{2 \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 )}{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 )}{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 )}{2 \sqrt{1+\sqrt{2}} f}+\frac{2 \sqrt{1+\tan (e+f x)}}{f}+\frac{2 (1+\tan (e+f x))^{3/2}}{3 f}+\frac{20 (1+\tan (e+f x))^{5/2}}{231 f}-\frac{50 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{231 f}-\frac{4 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{33 f}+\frac{2 \tan ^3(e+f x) (1+\tan (e+f x))^{5/2}}{11 f}+\frac{\sqrt{2} \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{\sqrt{2} \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{-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 )}{\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 )}{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 )}{2 \sqrt{1+\sqrt{2}} f}+\frac{2 \sqrt{1+\tan (e+f x)}}{f}+\frac{2 (1+\tan (e+f x))^{3/2}}{3 f}+\frac{20 (1+\tan (e+f x))^{5/2}}{231 f}-\frac{50 \tan (e+f x) (1+\tan (e+f x))^{5/2}}{231 f}-\frac{4 \tan ^2(e+f x) (1+\tan (e+f x))^{5/2}}{33 f}+\frac{2 \tan ^3(e+f x) (1+\tan (e+f x))^{5/2}}{11 f}\\ \end{align*}

Mathematica [C]  time = 1.10219, size = 132, normalized size = 0.36 \[ \frac{2 \sqrt{\tan (e+f x)+1} \left (21 \tan ^5(e+f x)+28 \tan ^4(e+f x)-32 \tan ^3(e+f x)-54 \tan ^2(e+f x)+72 \tan (e+f x)+318\right )-231 (1-i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1-i}}\right )-231 (1+i)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{\tan (e+f x)+1}}{\sqrt{1+i}}\right )}{231 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^5*(1 + Tan[e + f*x])^(3/2),x]

[Out]

(-231*(1 - I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f*x]]/Sqrt[1 - I]] - 231*(1 + I)^(3/2)*ArcTanh[Sqrt[1 + Tan[e + f
*x]]/Sqrt[1 + I]] + 2*Sqrt[1 + Tan[e + f*x]]*(318 + 72*Tan[e + f*x] - 54*Tan[e + f*x]^2 - 32*Tan[e + f*x]^3 +
28*Tan[e + f*x]^4 + 21*Tan[e + f*x]^5))/(231*f)

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Maple [A]  time = 0.032, size = 371, normalized size = 1. \begin{align*}{\frac{2}{11\,f} \left ( 1+\tan \left ( fx+e \right ) \right ) ^{{\frac{11}{2}}}}-{\frac{2}{3\,f} \left ( 1+\tan \left ( fx+e \right ) \right ) ^{{\frac{9}{2}}}}+{\frac{4}{7\,f} \left ( 1+\tan \left ( fx+e \right ) \right ) ^{{\frac{7}{2}}}}+{\frac{2}{3\,f} \left ( 1+\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+2\,{\frac{\sqrt{1+\tan \left ( fx+e \right ) }}{f}}-{\frac{\sqrt{2\,\sqrt{2}+2}\sqrt{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}}{2\,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}}{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}\sqrt{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}}{2\,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}}{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)^5*(1+tan(f*x+e))^(3/2),x)

[Out]

2/11/f*(1+tan(f*x+e))^(11/2)-2/3/f*(1+tan(f*x+e))^(9/2)+4/7/f*(1+tan(f*x+e))^(7/2)+2/3*(1+tan(f*x+e))^(3/2)/f+
2*(1+tan(f*x+e))^(1/2)/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))+1/2/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/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)*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^(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))*2^(1/2)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [B]  time = 2.13283, size = 2692, normalized size = 7.3 \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)^5*(1+tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

1/1848*(924*8^(1/4)*sqrt(2)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(-1/8*8^(3/4)*sqrt(2)*
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) + 1/8
*8^(3/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 8^(1/4)*sqrt
(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4)*cos(f*x + e
) + 2*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) - f^2*sqrt(f^(-4)) - sqrt(2))*cos(f*x + e)^5
+ 924*8^(1/4)*sqrt(2)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f*(f^(-4))^(1/4)*arctan(-1/8*8^(3/4)*sqrt(2)*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) + 1/8*8^(3
/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f^3*sqrt((2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 8^(1/4)*sqrt(2*sq
rt(2)*f^2*sqrt(f^(-4)) + 4)*f*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4)*cos(f*x + e) + 2
*cos(f*x + e) + 2*sin(f*x + e))/cos(f*x + e))*(f^(-4))^(3/4) + f^2*sqrt(f^(-4)) + sqrt(2))*cos(f*x + e)^5 + 23
1*8^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e)^5 - 2*f*cos(f*x + e)^5)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*
(f^(-4))^(1/4)*log(2*(2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) + 8^(1/4)*sqrt(2*sqrt(2)*f^2*sqrt(f^(-4)) + 4)*f
*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4)*cos(f*x + e) + 2*cos(f*x + e) + 2*sin(f*x + e
))/cos(f*x + e)) - 231*8^(1/4)*(sqrt(2)*f^3*sqrt(f^(-4))*cos(f*x + e)^5 - 2*f*cos(f*x + e)^5)*sqrt(2*sqrt(2)*f
^2*sqrt(f^(-4)) + 4)*(f^(-4))^(1/4)*log(2*(2*sqrt(2)*f^2*sqrt(f^(-4))*cos(f*x + e) - 8^(1/4)*sqrt(2*sqrt(2)*f^
2*sqrt(f^(-4)) + 4)*f*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e))*(f^(-4))^(1/4)*cos(f*x + e) + 2*cos(f*x
+ e) + 2*sin(f*x + e))/cos(f*x + e)) + 16*(400*cos(f*x + e)^5 - 110*cos(f*x + e)^3 + (125*cos(f*x + e)^4 - 74
*cos(f*x + e)^2 + 21)*sin(f*x + e) + 28*cos(f*x + e))*sqrt((cos(f*x + e) + sin(f*x + e))/cos(f*x + e)))/(f*cos
(f*x + e)^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\tan{\left (e + f x \right )} + 1\right )^{\frac{3}{2}} \tan ^{5}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**5*(1+tan(f*x+e))**(3/2),x)

[Out]

Integral((tan(e + f*x) + 1)**(3/2)*tan(e + f*x)**5, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\tan \left (f x + e\right ) + 1\right )}^{\frac{3}{2}} \tan \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5*(1+tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((tan(f*x + e) + 1)^(3/2)*tan(f*x + e)^5, x)