3.635 \(\int \frac{1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))} \, dx\)

Optimal. Leaf size=581 \[ \frac{\tan (e+f x) \sec ^2(e+f x)^{5/6} F_1\left (\frac{1}{2};1,\frac{11}{6};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a f (d \sec (e+f x))^{5/3}}+\frac{3 b}{5 f \left (a^2+b^2\right ) (d \sec (e+f x))^{5/3}}+\frac{b^{8/3} \sec ^2(e+f x)^{5/6} \log \left (-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right )}{4 f \left (a^2+b^2\right )^{11/6} (d \sec (e+f x))^{5/3}}-\frac{b^{8/3} \sec ^2(e+f x)^{5/6} \log \left (\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right )}{4 f \left (a^2+b^2\right )^{11/6} (d \sec (e+f x))^{5/3}}+\frac{\sqrt{3} b^{8/3} \sec ^2(e+f x)^{5/6} \tan ^{-1}\left (\frac{1}{\sqrt{3}}-\frac{2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt{3} \sqrt [6]{a^2+b^2}}\right )}{2 f \left (a^2+b^2\right )^{11/6} (d \sec (e+f x))^{5/3}}-\frac{\sqrt{3} b^{8/3} \sec ^2(e+f x)^{5/6} \tan ^{-1}\left (\frac{2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt{3} \sqrt [6]{a^2+b^2}}+\frac{1}{\sqrt{3}}\right )}{2 f \left (a^2+b^2\right )^{11/6} (d \sec (e+f x))^{5/3}}-\frac{b^{8/3} \sec ^2(e+f x)^{5/6} \tanh ^{-1}\left (\frac{\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right )}{f \left (a^2+b^2\right )^{11/6} (d \sec (e+f x))^{5/3}} \]

[Out]

(3*b)/(5*(a^2 + b^2)*f*(d*Sec[e + f*x])^(5/3)) + (Sqrt[3]*b^(8/3)*ArcTan[1/Sqrt[3] - (2*b^(1/3)*(Sec[e + f*x]^
2)^(1/6))/(Sqrt[3]*(a^2 + b^2)^(1/6))]*(Sec[e + f*x]^2)^(5/6))/(2*(a^2 + b^2)^(11/6)*f*(d*Sec[e + f*x])^(5/3))
- (Sqrt[3]*b^(8/3)*ArcTan[1/Sqrt[3] + (2*b^(1/3)*(Sec[e + f*x]^2)^(1/6))/(Sqrt[3]*(a^2 + b^2)^(1/6))]*(Sec[e
+ f*x]^2)^(5/6))/(2*(a^2 + b^2)^(11/6)*f*(d*Sec[e + f*x])^(5/3)) - (b^(8/3)*ArcTanh[(b^(1/3)*(Sec[e + f*x]^2)^
(1/6))/(a^2 + b^2)^(1/6)]*(Sec[e + f*x]^2)^(5/6))/((a^2 + b^2)^(11/6)*f*(d*Sec[e + f*x])^(5/3)) + (b^(8/3)*Log
[(a^2 + b^2)^(1/3) - b^(1/3)*(a^2 + b^2)^(1/6)*(Sec[e + f*x]^2)^(1/6) + b^(2/3)*(Sec[e + f*x]^2)^(1/3)]*(Sec[e
+ f*x]^2)^(5/6))/(4*(a^2 + b^2)^(11/6)*f*(d*Sec[e + f*x])^(5/3)) - (b^(8/3)*Log[(a^2 + b^2)^(1/3) + b^(1/3)*(
a^2 + b^2)^(1/6)*(Sec[e + f*x]^2)^(1/6) + b^(2/3)*(Sec[e + f*x]^2)^(1/3)]*(Sec[e + f*x]^2)^(5/6))/(4*(a^2 + b^
2)^(11/6)*f*(d*Sec[e + f*x])^(5/3)) + (AppellF1[1/2, 1, 11/6, 3/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*
(Sec[e + f*x]^2)^(5/6)*Tan[e + f*x])/(a*f*(d*Sec[e + f*x])^(5/3))

________________________________________________________________________________________

Rubi [A]  time = 0.819225, antiderivative size = 581, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {3512, 757, 429, 444, 51, 63, 210, 634, 618, 204, 628, 208} \[ \frac{\tan (e+f x) \sec ^2(e+f x)^{5/6} F_1\left (\frac{1}{2};1,\frac{11}{6};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a f (d \sec (e+f x))^{5/3}}+\frac{3 b}{5 f \left (a^2+b^2\right ) (d \sec (e+f x))^{5/3}}+\frac{b^{8/3} \sec ^2(e+f x)^{5/6} \log \left (-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right )}{4 f \left (a^2+b^2\right )^{11/6} (d \sec (e+f x))^{5/3}}-\frac{b^{8/3} \sec ^2(e+f x)^{5/6} \log \left (\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right )}{4 f \left (a^2+b^2\right )^{11/6} (d \sec (e+f x))^{5/3}}+\frac{\sqrt{3} b^{8/3} \sec ^2(e+f x)^{5/6} \tan ^{-1}\left (\frac{1}{\sqrt{3}}-\frac{2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt{3} \sqrt [6]{a^2+b^2}}\right )}{2 f \left (a^2+b^2\right )^{11/6} (d \sec (e+f x))^{5/3}}-\frac{\sqrt{3} b^{8/3} \sec ^2(e+f x)^{5/6} \tan ^{-1}\left (\frac{2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt{3} \sqrt [6]{a^2+b^2}}+\frac{1}{\sqrt{3}}\right )}{2 f \left (a^2+b^2\right )^{11/6} (d \sec (e+f x))^{5/3}}-\frac{b^{8/3} \sec ^2(e+f x)^{5/6} \tanh ^{-1}\left (\frac{\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right )}{f \left (a^2+b^2\right )^{11/6} (d \sec (e+f x))^{5/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*Sec[e + f*x])^(5/3)*(a + b*Tan[e + f*x])),x]

[Out]

(3*b)/(5*(a^2 + b^2)*f*(d*Sec[e + f*x])^(5/3)) + (Sqrt[3]*b^(8/3)*ArcTan[1/Sqrt[3] - (2*b^(1/3)*(Sec[e + f*x]^
2)^(1/6))/(Sqrt[3]*(a^2 + b^2)^(1/6))]*(Sec[e + f*x]^2)^(5/6))/(2*(a^2 + b^2)^(11/6)*f*(d*Sec[e + f*x])^(5/3))
- (Sqrt[3]*b^(8/3)*ArcTan[1/Sqrt[3] + (2*b^(1/3)*(Sec[e + f*x]^2)^(1/6))/(Sqrt[3]*(a^2 + b^2)^(1/6))]*(Sec[e
+ f*x]^2)^(5/6))/(2*(a^2 + b^2)^(11/6)*f*(d*Sec[e + f*x])^(5/3)) - (b^(8/3)*ArcTanh[(b^(1/3)*(Sec[e + f*x]^2)^
(1/6))/(a^2 + b^2)^(1/6)]*(Sec[e + f*x]^2)^(5/6))/((a^2 + b^2)^(11/6)*f*(d*Sec[e + f*x])^(5/3)) + (b^(8/3)*Log
[(a^2 + b^2)^(1/3) - b^(1/3)*(a^2 + b^2)^(1/6)*(Sec[e + f*x]^2)^(1/6) + b^(2/3)*(Sec[e + f*x]^2)^(1/3)]*(Sec[e
+ f*x]^2)^(5/6))/(4*(a^2 + b^2)^(11/6)*f*(d*Sec[e + f*x])^(5/3)) - (b^(8/3)*Log[(a^2 + b^2)^(1/3) + b^(1/3)*(
a^2 + b^2)^(1/6)*(Sec[e + f*x]^2)^(1/6) + b^(2/3)*(Sec[e + f*x]^2)^(1/3)]*(Sec[e + f*x]^2)^(5/6))/(4*(a^2 + b^
2)^(11/6)*f*(d*Sec[e + f*x])^(5/3)) + (AppellF1[1/2, 1, 11/6, 3/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*
(Sec[e + f*x]^2)^(5/6)*Tan[e + f*x])/(a*f*(d*Sec[e + f*x])^(5/3))

Rule 3512

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(d^(2
*IntPart[m/2])*(d*Sec[e + f*x])^(2*FracPart[m/2]))/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]), Subst[Int[(a + x)^n*(
1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] &&
!IntegerQ[m/2]

Rule 757

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p, (d/(d
^2 - e^2*x^2) - (e*x)/(d^2 - e^2*x^2))^(-m), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&
!IntegerQ[p] && ILtQ[m, 0]

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 210

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[-(a/b), n]], s = Denominator[Rt[-(a/b
), n]], k, u}, Simp[u = Int[(r - s*Cos[(2*k*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x] + Int[(r +
s*Cos[(2*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; (2*r^2*Int[1/(r^2 - s^2*x^2), x])/(a*n) +
Dist[(2*r)/(a*n), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[a/b]

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]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))} \, dx &=\frac{\sec ^2(e+f x)^{5/6} \operatorname{Subst}\left (\int \frac{1}{(a+x) \left (1+\frac{x^2}{b^2}\right )^{11/6}} \, dx,x,b \tan (e+f x)\right )}{b f (d \sec (e+f x))^{5/3}}\\ &=\frac{\sec ^2(e+f x)^{5/6} \operatorname{Subst}\left (\int \left (\frac{a}{\left (a^2-x^2\right ) \left (1+\frac{x^2}{b^2}\right )^{11/6}}+\frac{x}{\left (-a^2+x^2\right ) \left (1+\frac{x^2}{b^2}\right )^{11/6}}\right ) \, dx,x,b \tan (e+f x)\right )}{b f (d \sec (e+f x))^{5/3}}\\ &=\frac{\sec ^2(e+f x)^{5/6} \operatorname{Subst}\left (\int \frac{x}{\left (-a^2+x^2\right ) \left (1+\frac{x^2}{b^2}\right )^{11/6}} \, dx,x,b \tan (e+f x)\right )}{b f (d \sec (e+f x))^{5/3}}+\frac{\left (a \sec ^2(e+f x)^{5/6}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a^2-x^2\right ) \left (1+\frac{x^2}{b^2}\right )^{11/6}} \, dx,x,b \tan (e+f x)\right )}{b f (d \sec (e+f x))^{5/3}}\\ &=\frac{F_1\left (\frac{1}{2};1,\frac{11}{6};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/6} \tan (e+f x)}{a f (d \sec (e+f x))^{5/3}}+\frac{\sec ^2(e+f x)^{5/6} \operatorname{Subst}\left (\int \frac{1}{\left (-a^2+x\right ) \left (1+\frac{x}{b^2}\right )^{11/6}} \, dx,x,b^2 \tan ^2(e+f x)\right )}{2 b f (d \sec (e+f x))^{5/3}}\\ &=\frac{3 b}{5 \left (a^2+b^2\right ) f (d \sec (e+f x))^{5/3}}+\frac{F_1\left (\frac{1}{2};1,\frac{11}{6};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/6} \tan (e+f x)}{a f (d \sec (e+f x))^{5/3}}+\frac{\left (b \sec ^2(e+f x)^{5/6}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-a^2+x\right ) \left (1+\frac{x}{b^2}\right )^{5/6}} \, dx,x,b^2 \tan ^2(e+f x)\right )}{2 \left (a^2+b^2\right ) f (d \sec (e+f x))^{5/3}}\\ &=\frac{3 b}{5 \left (a^2+b^2\right ) f (d \sec (e+f x))^{5/3}}+\frac{F_1\left (\frac{1}{2};1,\frac{11}{6};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/6} \tan (e+f x)}{a f (d \sec (e+f x))^{5/3}}+\frac{\left (3 b^3 \sec ^2(e+f x)^{5/6}\right ) \operatorname{Subst}\left (\int \frac{1}{-a^2-b^2+b^2 x^6} \, dx,x,\sqrt [6]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right ) f (d \sec (e+f x))^{5/3}}\\ &=\frac{3 b}{5 \left (a^2+b^2\right ) f (d \sec (e+f x))^{5/3}}+\frac{F_1\left (\frac{1}{2};1,\frac{11}{6};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/6} \tan (e+f x)}{a f (d \sec (e+f x))^{5/3}}-\frac{\left (b^3 \sec ^2(e+f x)^{5/6}\right ) \operatorname{Subst}\left (\int \frac{\sqrt [6]{a^2+b^2}-\frac{\sqrt [3]{b} x}{2}}{\sqrt [3]{a^2+b^2}-\sqrt [3]{b} \sqrt [6]{a^2+b^2} x+b^{2/3} x^2} \, dx,x,\sqrt [6]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}-\frac{\left (b^3 \sec ^2(e+f x)^{5/6}\right ) \operatorname{Subst}\left (\int \frac{\sqrt [6]{a^2+b^2}+\frac{\sqrt [3]{b} x}{2}}{\sqrt [3]{a^2+b^2}+\sqrt [3]{b} \sqrt [6]{a^2+b^2} x+b^{2/3} x^2} \, dx,x,\sqrt [6]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}-\frac{\left (b^3 \sec ^2(e+f x)^{5/6}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a^2+b^2}-b^{2/3} x^2} \, dx,x,\sqrt [6]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right )^{5/3} f (d \sec (e+f x))^{5/3}}\\ &=\frac{3 b}{5 \left (a^2+b^2\right ) f (d \sec (e+f x))^{5/3}}-\frac{b^{8/3} \tanh ^{-1}\left (\frac{\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) \sec ^2(e+f x)^{5/6}}{\left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}+\frac{F_1\left (\frac{1}{2};1,\frac{11}{6};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/6} \tan (e+f x)}{a f (d \sec (e+f x))^{5/3}}+\frac{\left (b^{8/3} \sec ^2(e+f x)^{5/6}\right ) \operatorname{Subst}\left (\int \frac{-\sqrt [3]{b} \sqrt [6]{a^2+b^2}+2 b^{2/3} x}{\sqrt [3]{a^2+b^2}-\sqrt [3]{b} \sqrt [6]{a^2+b^2} x+b^{2/3} x^2} \, dx,x,\sqrt [6]{\sec ^2(e+f x)}\right )}{4 \left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}-\frac{\left (b^{8/3} \sec ^2(e+f x)^{5/6}\right ) \operatorname{Subst}\left (\int \frac{\sqrt [3]{b} \sqrt [6]{a^2+b^2}+2 b^{2/3} x}{\sqrt [3]{a^2+b^2}+\sqrt [3]{b} \sqrt [6]{a^2+b^2} x+b^{2/3} x^2} \, dx,x,\sqrt [6]{\sec ^2(e+f x)}\right )}{4 \left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}-\frac{\left (3 b^3 \sec ^2(e+f x)^{5/6}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a^2+b^2}-\sqrt [3]{b} \sqrt [6]{a^2+b^2} x+b^{2/3} x^2} \, dx,x,\sqrt [6]{\sec ^2(e+f x)}\right )}{4 \left (a^2+b^2\right )^{5/3} f (d \sec (e+f x))^{5/3}}-\frac{\left (3 b^3 \sec ^2(e+f x)^{5/6}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a^2+b^2}+\sqrt [3]{b} \sqrt [6]{a^2+b^2} x+b^{2/3} x^2} \, dx,x,\sqrt [6]{\sec ^2(e+f x)}\right )}{4 \left (a^2+b^2\right )^{5/3} f (d \sec (e+f x))^{5/3}}\\ &=\frac{3 b}{5 \left (a^2+b^2\right ) f (d \sec (e+f x))^{5/3}}-\frac{b^{8/3} \tanh ^{-1}\left (\frac{\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) \sec ^2(e+f x)^{5/6}}{\left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}+\frac{b^{8/3} \log \left (\sqrt [3]{a^2+b^2}-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) \sec ^2(e+f x)^{5/6}}{4 \left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}-\frac{b^{8/3} \log \left (\sqrt [3]{a^2+b^2}+\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) \sec ^2(e+f x)^{5/6}}{4 \left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}+\frac{F_1\left (\frac{1}{2};1,\frac{11}{6};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/6} \tan (e+f x)}{a f (d \sec (e+f x))^{5/3}}-\frac{\left (3 b^{8/3} \sec ^2(e+f x)^{5/6}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right )}{2 \left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}+\frac{\left (3 b^{8/3} \sec ^2(e+f x)^{5/6}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\frac{2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right )}{2 \left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}\\ &=\frac{3 b}{5 \left (a^2+b^2\right ) f (d \sec (e+f x))^{5/3}}+\frac{\sqrt{3} b^{8/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}}{\sqrt{3}}\right ) \sec ^2(e+f x)^{5/6}}{2 \left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}-\frac{\sqrt{3} b^{8/3} \tan ^{-1}\left (\frac{1+\frac{2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}}{\sqrt{3}}\right ) \sec ^2(e+f x)^{5/6}}{2 \left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}-\frac{b^{8/3} \tanh ^{-1}\left (\frac{\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) \sec ^2(e+f x)^{5/6}}{\left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}+\frac{b^{8/3} \log \left (\sqrt [3]{a^2+b^2}-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) \sec ^2(e+f x)^{5/6}}{4 \left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}-\frac{b^{8/3} \log \left (\sqrt [3]{a^2+b^2}+\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) \sec ^2(e+f x)^{5/6}}{4 \left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}+\frac{F_1\left (\frac{1}{2};1,\frac{11}{6};\frac{3}{2};\frac{b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/6} \tan (e+f x)}{a f (d \sec (e+f x))^{5/3}}\\ \end{align*}

Mathematica [B]  time = 31.7372, size = 6862, normalized size = 11.81 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((d*Sec[e + f*x])^(5/3)*(a + b*Tan[e + f*x])),x]

[Out]

Result too large to show

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Maple [F]  time = 0.169, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{a+b\tan \left ( fx+e \right ) } \left ( d\sec \left ( fx+e \right ) \right ) ^{-{\frac{5}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x)

[Out]

int(1/(d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x, algorithm="maxima")

[Out]

integrate(1/((d*sec(f*x + e))^(5/3)*(b*tan(f*x + e) + a)), x)

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Fricas [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(1/(d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))**(5/3)/(a+b*tan(f*x+e)),x)

[Out]

Integral(1/((d*sec(e + f*x))**(5/3)*(a + b*tan(e + f*x))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate(1/((d*sec(f*x + e))^(5/3)*(b*tan(f*x + e) + a)), x)