3.636 \(\int \frac {(d \sec (e+f x))^{5/3}}{(a+b \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=687 \[ \frac {\tan (e+f x) (d \sec (e+f x))^{5/3} F_1\left (\frac {1}{2};2,\frac {1}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a^2 f \sec ^2(e+f x)^{5/6}}-\frac {a b (d \sec (e+f x))^{5/3}}{f \left (a^2+b^2\right ) \left (a^2-b^2 \tan ^2(e+f x)\right )}+\frac {a (d \sec (e+f x))^{5/3} \log \left (-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right )}{12 b^{2/3} f \left (a^2+b^2\right )^{7/6} \sec ^2(e+f x)^{5/6}}-\frac {a (d \sec (e+f x))^{5/3} \log \left (\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right )}{12 b^{2/3} f \left (a^2+b^2\right )^{7/6} \sec ^2(e+f x)^{5/6}}-\frac {a (d \sec (e+f x))^{5/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right )}{2 \sqrt {3} b^{2/3} f \left (a^2+b^2\right )^{7/6} \sec ^2(e+f x)^{5/6}}+\frac {a (d \sec (e+f x))^{5/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}+\frac {1}{\sqrt {3}}\right )}{2 \sqrt {3} b^{2/3} f \left (a^2+b^2\right )^{7/6} \sec ^2(e+f x)^{5/6}}-\frac {a (d \sec (e+f x))^{5/3} \tanh ^{-1}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right )}{3 b^{2/3} f \left (a^2+b^2\right )^{7/6} \sec ^2(e+f x)^{5/6}}+\frac {b^2 \tan ^3(e+f x) (d \sec (e+f x))^{5/3} F_1\left (\frac {3}{2};2,\frac {1}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{3 a^4 f \sec ^2(e+f x)^{5/6}} \]

[Out]

-1/3*a*arctanh(b^(1/3)*(sec(f*x+e)^2)^(1/6)/(a^2+b^2)^(1/6))*(d*sec(f*x+e))^(5/3)/b^(2/3)/(a^2+b^2)^(7/6)/f/(s
ec(f*x+e)^2)^(5/6)+1/12*a*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))*(d*sec(f*x+e))^(5/3)/b^(2/3)/(a^2+b^2)^(7/6)/f/(sec(f*x+e)^2)^(5/6)-1/12*a*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))*(d*sec(f*x+e))^(5/3)/b^(2/3)/(a^2+b^2)^(7
/6)/f/(sec(f*x+e)^2)^(5/6)+1/6*a*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))
*(d*sec(f*x+e))^(5/3)/b^(2/3)/(a^2+b^2)^(7/6)/f/(sec(f*x+e)^2)^(5/6)*3^(1/2)+1/6*a*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))*(d*sec(f*x+e))^(5/3)/b^(2/3)/(a^2+b^2)^(7/6)/f/(sec(f*x+e)^2
)^(5/6)*3^(1/2)+AppellF1(1/2,2,1/6,3/2,b^2*tan(f*x+e)^2/a^2,-tan(f*x+e)^2)*(d*sec(f*x+e))^(5/3)*tan(f*x+e)/a^2
/f/(sec(f*x+e)^2)^(5/6)+1/3*b^2*AppellF1(3/2,2,1/6,5/2,b^2*tan(f*x+e)^2/a^2,-tan(f*x+e)^2)*(d*sec(f*x+e))^(5/3
)*tan(f*x+e)^3/a^4/f/(sec(f*x+e)^2)^(5/6)-a*b*(d*sec(f*x+e))^(5/3)/(a^2+b^2)/f/(a^2-b^2*tan(f*x+e)^2)

________________________________________________________________________________________

Rubi [A]  time = 0.94, antiderivative size = 687, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 13, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3512, 757, 429, 444, 51, 63, 296, 634, 618, 204, 628, 208, 510} \[ \frac {b^2 \tan ^3(e+f x) (d \sec (e+f x))^{5/3} F_1\left (\frac {3}{2};2,\frac {1}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{3 a^4 f \sec ^2(e+f x)^{5/6}}+\frac {\tan (e+f x) (d \sec (e+f x))^{5/3} F_1\left (\frac {1}{2};2,\frac {1}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a^2 f \sec ^2(e+f x)^{5/6}}+\frac {a (d \sec (e+f x))^{5/3} \log \left (-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right )}{12 b^{2/3} f \left (a^2+b^2\right )^{7/6} \sec ^2(e+f x)^{5/6}}-\frac {a (d \sec (e+f x))^{5/3} \log \left (\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right )}{12 b^{2/3} f \left (a^2+b^2\right )^{7/6} \sec ^2(e+f x)^{5/6}}-\frac {a (d \sec (e+f x))^{5/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right )}{2 \sqrt {3} b^{2/3} f \left (a^2+b^2\right )^{7/6} \sec ^2(e+f x)^{5/6}}+\frac {a (d \sec (e+f x))^{5/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}+\frac {1}{\sqrt {3}}\right )}{2 \sqrt {3} b^{2/3} f \left (a^2+b^2\right )^{7/6} \sec ^2(e+f x)^{5/6}}-\frac {a (d \sec (e+f x))^{5/3} \tanh ^{-1}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right )}{3 b^{2/3} f \left (a^2+b^2\right )^{7/6} \sec ^2(e+f x)^{5/6}}-\frac {a b (d \sec (e+f x))^{5/3}}{f \left (a^2+b^2\right ) \left (a^2-b^2 \tan ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(5/3)/(a + b*Tan[e + f*x])^2,x]

[Out]

-(a*ArcTan[1/Sqrt[3] - (2*b^(1/3)*(Sec[e + f*x]^2)^(1/6))/(Sqrt[3]*(a^2 + b^2)^(1/6))]*(d*Sec[e + f*x])^(5/3))
/(2*Sqrt[3]*b^(2/3)*(a^2 + b^2)^(7/6)*f*(Sec[e + f*x]^2)^(5/6)) + (a*ArcTan[1/Sqrt[3] + (2*b^(1/3)*(Sec[e + f*
x]^2)^(1/6))/(Sqrt[3]*(a^2 + b^2)^(1/6))]*(d*Sec[e + f*x])^(5/3))/(2*Sqrt[3]*b^(2/3)*(a^2 + b^2)^(7/6)*f*(Sec[
e + f*x]^2)^(5/6)) - (a*ArcTanh[(b^(1/3)*(Sec[e + f*x]^2)^(1/6))/(a^2 + b^2)^(1/6)]*(d*Sec[e + f*x])^(5/3))/(3
*b^(2/3)*(a^2 + b^2)^(7/6)*f*(Sec[e + f*x]^2)^(5/6)) + (a*Log[(a^2 + b^2)^(1/3) - b^(1/3)*(a^2 + b^2)^(1/6)*(S
ec[e + f*x]^2)^(1/6) + b^(2/3)*(Sec[e + f*x]^2)^(1/3)]*(d*Sec[e + f*x])^(5/3))/(12*b^(2/3)*(a^2 + b^2)^(7/6)*f
*(Sec[e + f*x]^2)^(5/6)) - (a*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)]*(d*Sec[e + f*x])^(5/3))/(12*b^(2/3)*(a^2 + b^2)^(7/6)*f*(Sec[e + f*x]^2)^(5/6)) + (
AppellF1[1/2, 2, 1/6, 3/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*(d*Sec[e + f*x])^(5/3)*Tan[e + f*x])/(a^
2*f*(Sec[e + f*x]^2)^(5/6)) + (b^2*AppellF1[3/2, 2, 1/6, 5/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*(d*Se
c[e + f*x])^(5/3)*Tan[e + f*x]^3)/(3*a^4*f*(Sec[e + f*x]^2)^(5/6)) - (a*b*(d*Sec[e + f*x])^(5/3))/((a^2 + b^2)
*f*(a^2 - b^2*Tan[e + f*x]^2))

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 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 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]

Rule 296

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

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 510

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

Rubi steps

\begin {align*} \int \frac {(d \sec (e+f x))^{5/3}}{(a+b \tan (e+f x))^2} \, dx &=\frac {(d \sec (e+f x))^{5/3} \operatorname {Subst}\left (\int \frac {1}{(a+x)^2 \sqrt [6]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{b f \sec ^2(e+f x)^{5/6}}\\ &=\frac {(d \sec (e+f x))^{5/3} \operatorname {Subst}\left (\int \left (\frac {a^2}{\left (a^2-x^2\right )^2 \sqrt [6]{1+\frac {x^2}{b^2}}}-\frac {2 a x}{\left (a^2-x^2\right )^2 \sqrt [6]{1+\frac {x^2}{b^2}}}+\frac {x^2}{\left (-a^2+x^2\right )^2 \sqrt [6]{1+\frac {x^2}{b^2}}}\right ) \, dx,x,b \tan (e+f x)\right )}{b f \sec ^2(e+f x)^{5/6}}\\ &=\frac {(d \sec (e+f x))^{5/3} \operatorname {Subst}\left (\int \frac {x^2}{\left (-a^2+x^2\right )^2 \sqrt [6]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{b f \sec ^2(e+f x)^{5/6}}-\frac {\left (2 a (d \sec (e+f x))^{5/3}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (a^2-x^2\right )^2 \sqrt [6]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{b f \sec ^2(e+f x)^{5/6}}+\frac {\left (a^2 (d \sec (e+f x))^{5/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a^2-x^2\right )^2 \sqrt [6]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{b f \sec ^2(e+f x)^{5/6}}\\ &=\frac {F_1\left (\frac {1}{2};2,\frac {1}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan (e+f x)}{a^2 f \sec ^2(e+f x)^{5/6}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {1}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan ^3(e+f x)}{3 a^4 f \sec ^2(e+f x)^{5/6}}-\frac {\left (a (d \sec (e+f x))^{5/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a^2-x\right )^2 \sqrt [6]{1+\frac {x}{b^2}}} \, dx,x,b^2 \tan ^2(e+f x)\right )}{b f \sec ^2(e+f x)^{5/6}}\\ &=\frac {F_1\left (\frac {1}{2};2,\frac {1}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan (e+f x)}{a^2 f \sec ^2(e+f x)^{5/6}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {1}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan ^3(e+f x)}{3 a^4 f \sec ^2(e+f x)^{5/6}}-\frac {a b (d \sec (e+f x))^{5/3}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )}-\frac {\left (a (d \sec (e+f x))^{5/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a^2-x\right ) \sqrt [6]{1+\frac {x}{b^2}}} \, dx,x,b^2 \tan ^2(e+f x)\right )}{6 b \left (a^2+b^2\right ) f \sec ^2(e+f x)^{5/6}}\\ &=\frac {F_1\left (\frac {1}{2};2,\frac {1}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan (e+f x)}{a^2 f \sec ^2(e+f x)^{5/6}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {1}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan ^3(e+f x)}{3 a^4 f \sec ^2(e+f x)^{5/6}}-\frac {a b (d \sec (e+f x))^{5/3}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )}-\frac {\left (a b (d \sec (e+f x))^{5/3}\right ) \operatorname {Subst}\left (\int \frac {x^4}{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 \sec ^2(e+f x)^{5/6}}\\ &=\frac {F_1\left (\frac {1}{2};2,\frac {1}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan (e+f x)}{a^2 f \sec ^2(e+f x)^{5/6}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {1}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan ^3(e+f x)}{3 a^4 f \sec ^2(e+f x)^{5/6}}-\frac {a b (d \sec (e+f x))^{5/3}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )}-\frac {\left (a (d \sec (e+f x))^{5/3}\right ) \operatorname {Subst}\left (\int \frac {-\frac {1}{2} \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 )}{3 \sqrt [3]{b} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}-\frac {\left (a (d \sec (e+f x))^{5/3}\right ) \operatorname {Subst}\left (\int \frac {-\frac {1}{2} \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 )}{3 \sqrt [3]{b} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}-\frac {\left (a (d \sec (e+f x))^{5/3}\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 )}{3 \sqrt [3]{b} \left (a^2+b^2\right ) f \sec ^2(e+f x)^{5/6}}\\ &=-\frac {a \tanh ^{-1}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) (d \sec (e+f x))^{5/3}}{3 b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}+\frac {F_1\left (\frac {1}{2};2,\frac {1}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan (e+f x)}{a^2 f \sec ^2(e+f x)^{5/6}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {1}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan ^3(e+f x)}{3 a^4 f \sec ^2(e+f x)^{5/6}}-\frac {a b (d \sec (e+f x))^{5/3}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )}+\frac {\left (a (d \sec (e+f x))^{5/3}\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 )}{12 b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}-\frac {\left (a (d \sec (e+f x))^{5/3}\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 )}{12 b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}+\frac {\left (a (d \sec (e+f x))^{5/3}\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 \sqrt [3]{b} \left (a^2+b^2\right ) f \sec ^2(e+f x)^{5/6}}+\frac {\left (a (d \sec (e+f x))^{5/3}\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 \sqrt [3]{b} \left (a^2+b^2\right ) f \sec ^2(e+f x)^{5/6}}\\ &=-\frac {a \tanh ^{-1}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) (d \sec (e+f x))^{5/3}}{3 b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}+\frac {a \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 ) (d \sec (e+f x))^{5/3}}{12 b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}-\frac {a \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 ) (d \sec (e+f x))^{5/3}}{12 b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}+\frac {F_1\left (\frac {1}{2};2,\frac {1}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan (e+f x)}{a^2 f \sec ^2(e+f x)^{5/6}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {1}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan ^3(e+f x)}{3 a^4 f \sec ^2(e+f x)^{5/6}}-\frac {a b (d \sec (e+f x))^{5/3}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )}+\frac {\left (a (d \sec (e+f x))^{5/3}\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 b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}-\frac {\left (a (d \sec (e+f x))^{5/3}\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 b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}\\ &=-\frac {a \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 ) (d \sec (e+f x))^{5/3}}{2 \sqrt {3} b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}+\frac {a \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 ) (d \sec (e+f x))^{5/3}}{2 \sqrt {3} b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}-\frac {a \tanh ^{-1}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) (d \sec (e+f x))^{5/3}}{3 b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}+\frac {a \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 ) (d \sec (e+f x))^{5/3}}{12 b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}-\frac {a \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 ) (d \sec (e+f x))^{5/3}}{12 b^{2/3} \left (a^2+b^2\right )^{7/6} f \sec ^2(e+f x)^{5/6}}+\frac {F_1\left (\frac {1}{2};2,\frac {1}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan (e+f x)}{a^2 f \sec ^2(e+f x)^{5/6}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {1}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan ^3(e+f x)}{3 a^4 f \sec ^2(e+f x)^{5/6}}-\frac {a b (d \sec (e+f x))^{5/3}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [C]  time = 39.52, size = 3398, normalized size = 4.95 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Sec[e + f*x])^(5/3)/(a + b*Tan[e + f*x])^2,x]

[Out]

(Sec[e + f*x]*(d*Sec[e + f*x])^(5/3)*(a*Cos[e + f*x] + b*Sin[e + f*x])^2*((b*Cos[e + f*x])/(a*(a - I*b)*(a + I
*b)) + Sin[e + f*x]/((a - I*b)*(a + I*b)) - b/((a - I*b)*(a + I*b)*(a*Cos[e + f*x] + b*Sin[e + f*x]))))/(f*(a
+ b*Tan[e + f*x])^2) - (4*AppellF1[1/3, 1/6, 1/6, 4/3, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e
+ f*x])]*(d*Sec[e + f*x])^(5/3)*(a*Cos[e + f*x] + b*Sin[e + f*x])^2)/(a*b*f*(a + b*Tan[e + f*x])*((a + I*b)*Ap
pellF1[4/3, 1/6, 7/6, 7/3, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])] + (a - I*b)*AppellF
1[4/3, 7/6, 1/6, 7/3, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])] + 8*AppellF1[1/3, 1/6, 1
/6, 4/3, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])]*(a + b*Tan[e + f*x]))) - (Sec[e + f*x
]*(d*Sec[e + f*x])^(5/3)*((6*(b + a*Sqrt[1 - Cos[e + f*x]^2]*Sec[e + f*x]))/((a^2 + b^2)*Sec[e + f*x]^(1/3)) +
(132*a*b^(2/3)*(5*a^2 + 3*b^2)*AppellF1[5/6, 1/2, 1, 11/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)]*
Sqrt[1 - Cos[e + f*x]^2]*Sec[e + f*x]^(8/3) - 240*a*b^(8/3)*AppellF1[11/6, 1/2, 1, 17/6, Sec[e + f*x]^2, (b^2*
Sec[e + f*x]^2)/(a^2 + b^2)]*Sqrt[1 - Cos[e + f*x]^2]*Sec[e + f*x]^(14/3) - 55*(-1)^(1/6)*(a^2 - b^2)*(a^2 + b
^2)^(5/6)*(-2*ArcTan[Sqrt[3] - (2*(-1)^(1/6)*b^(1/3)*Sec[e + f*x]^(1/3))/(a^2 + b^2)^(1/6)] + 2*ArcTan[Sqrt[3]
+ (2*(-1)^(1/6)*b^(1/3)*Sec[e + f*x]^(1/3))/(a^2 + b^2)^(1/6)] + 4*ArcTan[((-1)^(1/6)*b^(1/3)*Sec[e + f*x]^(1
/3))/(a^2 + b^2)^(1/6)] + Sqrt[3]*Log[(a^2 + b^2)^(1/3) - (-1)^(1/6)*Sqrt[3]*b^(1/3)*(a^2 + b^2)^(1/6)*Sec[e +
f*x]^(1/3) + (-1)^(1/3)*b^(2/3)*Sec[e + f*x]^(2/3)] - Sqrt[3]*Log[(a^2 + b^2)^(1/3) + (-1)^(1/6)*Sqrt[3]*b^(1
/3)*(a^2 + b^2)^(1/6)*Sec[e + f*x]^(1/3) + (-1)^(1/3)*b^(2/3)*Sec[e + f*x]^(2/3)])*Sqrt[1 - Sec[e + f*x]^2])/(
220*b^(2/3)*(a^2 + b^2)^2*Sqrt[1 - Sec[e + f*x]^2]))*(Cos[e + f*x] - Sin[e + f*x])*(Cos[e + f*x] + Sin[e + f*x
])*(a*Cos[e + f*x] + b*Sin[e + f*x]))/(6*a*f*(a + b*Tan[e + f*x])^2*((-2*Sec[e + f*x]^(2/3)*(b + a*Sqrt[1 - Co
s[e + f*x]^2]*Sec[e + f*x])*Sin[e + f*x])/(a^2 + b^2) + (Sec[e + f*x]^2*(132*a*b^(2/3)*(5*a^2 + 3*b^2)*AppellF
1[5/6, 1/2, 1, 11/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)]*Sqrt[1 - Cos[e + f*x]^2]*Sec[e + f*x]^(
8/3) - 240*a*b^(8/3)*AppellF1[11/6, 1/2, 1, 17/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)]*Sqrt[1 - C
os[e + f*x]^2]*Sec[e + f*x]^(14/3) - 55*(-1)^(1/6)*(a^2 - b^2)*(a^2 + b^2)^(5/6)*(-2*ArcTan[Sqrt[3] - (2*(-1)^
(1/6)*b^(1/3)*Sec[e + f*x]^(1/3))/(a^2 + b^2)^(1/6)] + 2*ArcTan[Sqrt[3] + (2*(-1)^(1/6)*b^(1/3)*Sec[e + f*x]^(
1/3))/(a^2 + b^2)^(1/6)] + 4*ArcTan[((-1)^(1/6)*b^(1/3)*Sec[e + f*x]^(1/3))/(a^2 + b^2)^(1/6)] + Sqrt[3]*Log[(
a^2 + b^2)^(1/3) - (-1)^(1/6)*Sqrt[3]*b^(1/3)*(a^2 + b^2)^(1/6)*Sec[e + f*x]^(1/3) + (-1)^(1/3)*b^(2/3)*Sec[e
+ f*x]^(2/3)] - Sqrt[3]*Log[(a^2 + b^2)^(1/3) + (-1)^(1/6)*Sqrt[3]*b^(1/3)*(a^2 + b^2)^(1/6)*Sec[e + f*x]^(1/3
) + (-1)^(1/3)*b^(2/3)*Sec[e + f*x]^(2/3)])*Sqrt[1 - Sec[e + f*x]^2])*Tan[e + f*x])/(220*b^(2/3)*(a^2 + b^2)^2
*(1 - Sec[e + f*x]^2)^(3/2)) + (6*((a*Sin[e + f*x])/Sqrt[1 - Cos[e + f*x]^2] + a*Sqrt[1 - Cos[e + f*x]^2]*Sec[
e + f*x]*Tan[e + f*x]))/((a^2 + b^2)*Sec[e + f*x]^(1/3)) + ((132*a*b^(2/3)*(5*a^2 + 3*b^2)*AppellF1[5/6, 1/2,
1, 11/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)]*Sec[e + f*x]^(5/3)*Sin[e + f*x])/Sqrt[1 - Cos[e + f
*x]^2] - (240*a*b^(8/3)*AppellF1[11/6, 1/2, 1, 17/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)]*Sec[e +
f*x]^(11/3)*Sin[e + f*x])/Sqrt[1 - Cos[e + f*x]^2] + 352*a*b^(2/3)*(5*a^2 + 3*b^2)*AppellF1[5/6, 1/2, 1, 11/6
, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)]*Sqrt[1 - Cos[e + f*x]^2]*Sec[e + f*x]^(11/3)*Sin[e + f*x]
- 1120*a*b^(8/3)*AppellF1[11/6, 1/2, 1, 17/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)]*Sqrt[1 - Cos[e
+ f*x]^2]*Sec[e + f*x]^(17/3)*Sin[e + f*x] - 55*(-1)^(1/6)*(a^2 - b^2)*(a^2 + b^2)^(5/6)*Sqrt[1 - Sec[e + f*x
]^2]*((4*(-1)^(1/6)*b^(1/3)*Sec[e + f*x]^(4/3)*Sin[e + f*x])/(3*(a^2 + b^2)^(1/6)*(1 + (Sqrt[3] - (2*(-1)^(1/6
)*b^(1/3)*Sec[e + f*x]^(1/3))/(a^2 + b^2)^(1/6))^2)) + (4*(-1)^(1/6)*b^(1/3)*Sec[e + f*x]^(4/3)*Sin[e + f*x])/
(3*(a^2 + b^2)^(1/6)*(1 + (Sqrt[3] + (2*(-1)^(1/6)*b^(1/3)*Sec[e + f*x]^(1/3))/(a^2 + b^2)^(1/6))^2)) + (4*(-1
)^(1/6)*b^(1/3)*Sec[e + f*x]^(4/3)*Sin[e + f*x])/(3*(a^2 + b^2)^(1/6)*(1 + ((-1)^(1/3)*b^(2/3)*Sec[e + f*x]^(2
/3))/(a^2 + b^2)^(1/3))) + (Sqrt[3]*(-(((-1)^(1/6)*b^(1/3)*(a^2 + b^2)^(1/6)*Sec[e + f*x]^(4/3)*Sin[e + f*x])/
Sqrt[3]) + (2*(-1)^(1/3)*b^(2/3)*Sec[e + f*x]^(5/3)*Sin[e + f*x])/3))/((a^2 + b^2)^(1/3) - (-1)^(1/6)*Sqrt[3]*
b^(1/3)*(a^2 + b^2)^(1/6)*Sec[e + f*x]^(1/3) + (-1)^(1/3)*b^(2/3)*Sec[e + f*x]^(2/3)) - (Sqrt[3]*(((-1)^(1/6)*
b^(1/3)*(a^2 + b^2)^(1/6)*Sec[e + f*x]^(4/3)*Sin[e + f*x])/Sqrt[3] + (2*(-1)^(1/3)*b^(2/3)*Sec[e + f*x]^(5/3)*
Sin[e + f*x])/3))/((a^2 + b^2)^(1/3) + (-1)^(1/6)*Sqrt[3]*b^(1/3)*(a^2 + b^2)^(1/6)*Sec[e + f*x]^(1/3) + (-1)^
(1/3)*b^(2/3)*Sec[e + f*x]^(2/3))) + (55*(-1)^(1/6)*(a^2 - b^2)*(a^2 + b^2)^(5/6)*(-2*ArcTan[Sqrt[3] - (2*(-1)
^(1/6)*b^(1/3)*Sec[e + f*x]^(1/3))/(a^2 + b^2)^(1/6)] + 2*ArcTan[Sqrt[3] + (2*(-1)^(1/6)*b^(1/3)*Sec[e + f*x]^
(1/3))/(a^2 + b^2)^(1/6)] + 4*ArcTan[((-1)^(1/6)*b^(1/3)*Sec[e + f*x]^(1/3))/(a^2 + b^2)^(1/6)] + Sqrt[3]*Log[
(a^2 + b^2)^(1/3) - (-1)^(1/6)*Sqrt[3]*b^(1/3)*(a^2 + b^2)^(1/6)*Sec[e + f*x]^(1/3) + (-1)^(1/3)*b^(2/3)*Sec[e
+ f*x]^(2/3)] - Sqrt[3]*Log[(a^2 + b^2)^(1/3) + (-1)^(1/6)*Sqrt[3]*b^(1/3)*(a^2 + b^2)^(1/6)*Sec[e + f*x]^(1/
3) + (-1)^(1/3)*b^(2/3)*Sec[e + f*x]^(2/3)])*Sec[e + f*x]^2*Tan[e + f*x])/Sqrt[1 - Sec[e + f*x]^2] + 132*a*b^(
2/3)*(5*a^2 + 3*b^2)*Sqrt[1 - Cos[e + f*x]^2]*Sec[e + f*x]^(8/3)*((10*b^2*AppellF1[11/6, 1/2, 2, 17/6, Sec[e +
f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)]*Sec[e + f*x]^2*Tan[e + f*x])/(11*(a^2 + b^2)) + (5*AppellF1[11/6, 3
/2, 1, 17/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)]*Sec[e + f*x]^2*Tan[e + f*x])/11) - 240*a*b^(8/3
)*Sqrt[1 - Cos[e + f*x]^2]*Sec[e + f*x]^(14/3)*((22*b^2*AppellF1[17/6, 1/2, 2, 23/6, Sec[e + f*x]^2, (b^2*Sec[
e + f*x]^2)/(a^2 + b^2)]*Sec[e + f*x]^2*Tan[e + f*x])/(17*(a^2 + b^2)) + (11*AppellF1[17/6, 3/2, 1, 23/6, Sec[
e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)]*Sec[e + f*x]^2*Tan[e + f*x])/17))/(220*b^(2/3)*(a^2 + b^2)^2*Sqr
t[1 - Sec[e + f*x]^2])))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(5/3)/(b*tan(f*x + e) + a)^2, x)

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maple [F]  time = 1.38, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}}}{\left (a +b \tan \left (f x +e \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e))^2,x)

[Out]

int((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e))^2,x)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(5/3)/(a + b*tan(e + f*x))^2,x)

[Out]

int((d/cos(e + f*x))^(5/3)/(a + b*tan(e + f*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(5/3)/(a+b*tan(f*x+e))**2,x)

[Out]

Integral((d*sec(e + f*x))**(5/3)/(a + b*tan(e + f*x))**2, x)

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