3.7.35 \(\int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))} \, dx\) [635]

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

[Out]

3/5*b/(a^2+b^2)/f/(d*sec(f*x+e))^(5/3)-b^(8/3)*arctanh(b^(1/3)*(sec(f*x+e)^2)^(1/6)/(a^2+b^2)^(1/6))*(sec(f*x+
e)^2)^(5/6)/(a^2+b^2)^(11/6)/f/(d*sec(f*x+e))^(5/3)+1/4*b^(8/3)*ln((a^2+b^2)^(1/3)-b^(1/3)*(a^2+b^2)^(1/6)*(se
c(f*x+e)^2)^(1/6)+b^(2/3)*(sec(f*x+e)^2)^(1/3))*(sec(f*x+e)^2)^(5/6)/(a^2+b^2)^(11/6)/f/(d*sec(f*x+e))^(5/3)-1
/4*b^(8/3)*ln((a^2+b^2)^(1/3)+b^(1/3)*(a^2+b^2)^(1/6)*(sec(f*x+e)^2)^(1/6)+b^(2/3)*(sec(f*x+e)^2)^(1/3))*(sec(
f*x+e)^2)^(5/6)/(a^2+b^2)^(11/6)/f/(d*sec(f*x+e))^(5/3)-1/2*b^(8/3)*arctan(-1/3*3^(1/2)+2/3*b^(1/3)*(sec(f*x+e
)^2)^(1/6)/(a^2+b^2)^(1/6)*3^(1/2))*(sec(f*x+e)^2)^(5/6)*3^(1/2)/(a^2+b^2)^(11/6)/f/(d*sec(f*x+e))^(5/3)-1/2*b
^(8/3)*arctan(1/3*3^(1/2)+2/3*b^(1/3)*(sec(f*x+e)^2)^(1/6)/(a^2+b^2)^(1/6)*3^(1/2))*(sec(f*x+e)^2)^(5/6)*3^(1/
2)/(a^2+b^2)^(11/6)/f/(d*sec(f*x+e))^(5/3)+AppellF1(1/2,1,11/6,3/2,b^2*tan(f*x+e)^2/a^2,-tan(f*x+e)^2)*(sec(f*
x+e)^2)^(5/6)*tan(f*x+e)/a/f/(d*sec(f*x+e))^(5/3)

________________________________________________________________________________________

Rubi [A]
time = 0.58, antiderivative size = 581, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3593, 771, 440, 455, 53, 65, 216, 648, 632, 210, 642, 214} \begin {gather*} \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 {\sqrt {3} b^{8/3} \sec ^2(e+f x)^{5/6} \text {ArcTan}\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} \text {ArcTan}\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 {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 {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}} \end {gather*}

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 53

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 65

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_)^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 214

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

Rule 216

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*C
os[(2*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; 2*(r^2/(a*n))*Int[1/(r^2 - s^2*x^2), x] + Dis
t[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 440

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 455

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 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 771

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 3593

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]

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} \text {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} \text {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} \text {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 ) \text {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} \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(6862\) vs. \(2(581)=1162\).
time = 129.53, size = 6862, normalized size = 11.81 \begin {gather*} \text {Result too large to show} \end {gather*}

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 = 0.90, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}} \left (a +b \tan \left (f x +e \right )\right )}\, dx\]

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 = 0.00, 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(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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}\,\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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