3.7.37 \(\int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx\) [637]

Optimal. Leaf size=687 \[ \frac {5 a b^{2/3} \text {ArcTan}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right ) \sqrt [3]{d \sec (e+f x)}}{2 \sqrt {3} \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \text {ArcTan}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right ) \sqrt [3]{d \sec (e+f x)}}{2 \sqrt {3} \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \tanh ^{-1}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) \sqrt [3]{d \sec (e+f x)}}{3 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}+\frac {5 a b^{2/3} \log \left (\sqrt [3]{a^2+b^2}-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) \sqrt [3]{d \sec (e+f x)}}{12 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \log \left (\sqrt [3]{a^2+b^2}+\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) \sqrt [3]{d \sec (e+f x)}}{12 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}+\frac {F_1\left (\frac {1}{2};2,\frac {5}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan (e+f x)}{a^2 f \sqrt [6]{\sec ^2(e+f x)}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {5}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan ^3(e+f x)}{3 a^4 f \sqrt [6]{\sec ^2(e+f x)}}-\frac {a b \sqrt [3]{d \sec (e+f x)}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )} \]

[Out]

-5/3*a*b^(2/3)*arctanh(b^(1/3)*(sec(f*x+e)^2)^(1/6)/(a^2+b^2)^(1/6))*(d*sec(f*x+e))^(1/3)/(a^2+b^2)^(11/6)/f/(
sec(f*x+e)^2)^(1/6)+5/12*a*b^(2/3)*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)*(se
c(f*x+e)^2)^(1/3))*(d*sec(f*x+e))^(1/3)/(a^2+b^2)^(11/6)/f/(sec(f*x+e)^2)^(1/6)-5/12*a*b^(2/3)*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))*(d*sec(f*x+e))^(1/3)/(a^2+b^2)^
(11/6)/f/(sec(f*x+e)^2)^(1/6)-5/6*a*b^(2/3)*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))*(d*sec(f*x+e))^(1/3)/(a^2+b^2)^(11/6)/f/(sec(f*x+e)^2)^(1/6)*3^(1/2)-5/6*a*b^(2/3)*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))*(d*sec(f*x+e))^(1/3)/(a^2+b^2)^(11/6)/f/(sec(f*x
+e)^2)^(1/6)*3^(1/2)+AppellF1(1/2,2,5/6,3/2,b^2*tan(f*x+e)^2/a^2,-tan(f*x+e)^2)*(d*sec(f*x+e))^(1/3)*tan(f*x+e
)/a^2/f/(sec(f*x+e)^2)^(1/6)+1/3*b^2*AppellF1(3/2,2,5/6,5/2,b^2*tan(f*x+e)^2/a^2,-tan(f*x+e)^2)*(d*sec(f*x+e))
^(1/3)*tan(f*x+e)^3/a^4/f/(sec(f*x+e)^2)^(1/6)-a*b*(d*sec(f*x+e))^(1/3)/(a^2+b^2)/f/(a^2-b^2*tan(f*x+e)^2)

________________________________________________________________________________________

Rubi [A]
time = 0.62, 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 = {3593, 771, 440, 455, 44, 65, 216, 648, 632, 210, 642, 214, 524} \begin {gather*} \frac {\tan (e+f x) \sqrt [3]{d \sec (e+f x)} F_1\left (\frac {1}{2};2,\frac {5}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a^2 f \sqrt [6]{\sec ^2(e+f x)}}+\frac {5 a b^{2/3} \sqrt [3]{d \sec (e+f x)} \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 \sqrt {3} f \left (a^2+b^2\right )^{11/6} \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \sqrt [3]{d \sec (e+f x)} \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 \sqrt {3} f \left (a^2+b^2\right )^{11/6} \sqrt [6]{\sec ^2(e+f x)}}-\frac {a b \sqrt [3]{d \sec (e+f x)}}{f \left (a^2+b^2\right ) \left (a^2-b^2 \tan ^2(e+f x)\right )}+\frac {5 a b^{2/3} \sqrt [3]{d \sec (e+f x)} \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 f \left (a^2+b^2\right )^{11/6} \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \sqrt [3]{d \sec (e+f x)} \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 f \left (a^2+b^2\right )^{11/6} \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \sqrt [3]{d \sec (e+f x)} \tanh ^{-1}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right )}{3 f \left (a^2+b^2\right )^{11/6} \sqrt [6]{\sec ^2(e+f x)}}+\frac {b^2 \tan ^3(e+f x) \sqrt [3]{d \sec (e+f x)} F_1\left (\frac {3}{2};2,\frac {5}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{3 a^4 f \sqrt [6]{\sec ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(1/3)/(a + b*Tan[e + f*x])^2,x]

[Out]

(5*a*b^(2/3)*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
])^(1/3))/(2*Sqrt[3]*(a^2 + b^2)^(11/6)*f*(Sec[e + f*x]^2)^(1/6)) - (5*a*b^(2/3)*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])^(1/3))/(2*Sqrt[3]*(a^2 + b^2)^(11/6)*f*
(Sec[e + f*x]^2)^(1/6)) - (5*a*b^(2/3)*ArcTanh[(b^(1/3)*(Sec[e + f*x]^2)^(1/6))/(a^2 + b^2)^(1/6)]*(d*Sec[e +
f*x])^(1/3))/(3*(a^2 + b^2)^(11/6)*f*(Sec[e + f*x]^2)^(1/6)) + (5*a*b^(2/3)*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)]*(d*Sec[e + f*x])^(1/3))/(12*(a^2 + b^
2)^(11/6)*f*(Sec[e + f*x]^2)^(1/6)) - (5*a*b^(2/3)*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)]*(d*Sec[e + f*x])^(1/3))/(12*(a^2 + b^2)^(11/6)*f*(Sec[e + f*x]
^2)^(1/6)) + (AppellF1[1/2, 2, 5/6, 3/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*(d*Sec[e + f*x])^(1/3)*Tan
[e + f*x])/(a^2*f*(Sec[e + f*x]^2)^(1/6)) + (b^2*AppellF1[3/2, 2, 5/6, 5/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e +
f*x]^2]*(d*Sec[e + f*x])^(1/3)*Tan[e + f*x]^3)/(3*a^4*f*(Sec[e + f*x]^2)^(1/6)) - (a*b*(d*Sec[e + f*x])^(1/3)
)/((a^2 + b^2)*f*(a^2 - b^2*Tan[e + f*x]^2))

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 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 {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx &=\frac {\sqrt [3]{d \sec (e+f x)} \text {Subst}\left (\int \frac {1}{(a+x)^2 \left (1+\frac {x^2}{b^2}\right )^{5/6}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [6]{\sec ^2(e+f x)}}\\ &=\frac {\sqrt [3]{d \sec (e+f x)} \text {Subst}\left (\int \left (\frac {a^2}{\left (a^2-x^2\right )^2 \left (1+\frac {x^2}{b^2}\right )^{5/6}}-\frac {2 a x}{\left (a^2-x^2\right )^2 \left (1+\frac {x^2}{b^2}\right )^{5/6}}+\frac {x^2}{\left (-a^2+x^2\right )^2 \left (1+\frac {x^2}{b^2}\right )^{5/6}}\right ) \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [6]{\sec ^2(e+f x)}}\\ &=\frac {\sqrt [3]{d \sec (e+f x)} \text {Subst}\left (\int \frac {x^2}{\left (-a^2+x^2\right )^2 \left (1+\frac {x^2}{b^2}\right )^{5/6}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [6]{\sec ^2(e+f x)}}-\frac {\left (2 a \sqrt [3]{d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {x}{\left (a^2-x^2\right )^2 \left (1+\frac {x^2}{b^2}\right )^{5/6}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [6]{\sec ^2(e+f x)}}+\frac {\left (a^2 \sqrt [3]{d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-x^2\right )^2 \left (1+\frac {x^2}{b^2}\right )^{5/6}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [6]{\sec ^2(e+f x)}}\\ &=\frac {F_1\left (\frac {1}{2};2,\frac {5}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan (e+f x)}{a^2 f \sqrt [6]{\sec ^2(e+f x)}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {5}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan ^3(e+f x)}{3 a^4 f \sqrt [6]{\sec ^2(e+f x)}}-\frac {\left (a \sqrt [3]{d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-x\right )^2 \left (1+\frac {x}{b^2}\right )^{5/6}} \, dx,x,b^2 \tan ^2(e+f x)\right )}{b f \sqrt [6]{\sec ^2(e+f x)}}\\ &=\frac {F_1\left (\frac {1}{2};2,\frac {5}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan (e+f x)}{a^2 f \sqrt [6]{\sec ^2(e+f x)}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {5}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan ^3(e+f x)}{3 a^4 f \sqrt [6]{\sec ^2(e+f x)}}-\frac {a b \sqrt [3]{d \sec (e+f x)}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )}-\frac {\left (5 a \sqrt [3]{d \sec (e+f x)}\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 )}{6 b \left (a^2+b^2\right ) f \sqrt [6]{\sec ^2(e+f x)}}\\ &=\frac {F_1\left (\frac {1}{2};2,\frac {5}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan (e+f x)}{a^2 f \sqrt [6]{\sec ^2(e+f x)}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {5}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan ^3(e+f x)}{3 a^4 f \sqrt [6]{\sec ^2(e+f x)}}-\frac {a b \sqrt [3]{d \sec (e+f x)}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )}-\frac {\left (5 a b \sqrt [3]{d \sec (e+f x)}\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 \sqrt [6]{\sec ^2(e+f x)}}\\ &=\frac {F_1\left (\frac {1}{2};2,\frac {5}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan (e+f x)}{a^2 f \sqrt [6]{\sec ^2(e+f x)}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {5}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan ^3(e+f x)}{3 a^4 f \sqrt [6]{\sec ^2(e+f x)}}-\frac {a b \sqrt [3]{d \sec (e+f x)}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )}-\frac {\left (5 a b \sqrt [3]{d \sec (e+f x)}\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 )}{3 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {\left (5 a b \sqrt [3]{d \sec (e+f x)}\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 )}{3 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {\left (5 a b \sqrt [3]{d \sec (e+f x)}\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 )}{3 \left (a^2+b^2\right )^{5/3} f \sqrt [6]{\sec ^2(e+f x)}}\\ &=-\frac {5 a b^{2/3} \tanh ^{-1}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) \sqrt [3]{d \sec (e+f x)}}{3 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}+\frac {F_1\left (\frac {1}{2};2,\frac {5}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan (e+f x)}{a^2 f \sqrt [6]{\sec ^2(e+f x)}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {5}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan ^3(e+f x)}{3 a^4 f \sqrt [6]{\sec ^2(e+f x)}}-\frac {a b \sqrt [3]{d \sec (e+f x)}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )}+\frac {\left (5 a b^{2/3} \sqrt [3]{d \sec (e+f x)}\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 )}{12 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {\left (5 a b^{2/3} \sqrt [3]{d \sec (e+f x)}\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 )}{12 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {\left (5 a b \sqrt [3]{d \sec (e+f x)}\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 \sqrt [6]{\sec ^2(e+f x)}}-\frac {\left (5 a b \sqrt [3]{d \sec (e+f x)}\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 \sqrt [6]{\sec ^2(e+f x)}}\\ &=-\frac {5 a b^{2/3} \tanh ^{-1}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) \sqrt [3]{d \sec (e+f x)}}{3 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}+\frac {5 a b^{2/3} \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 ) \sqrt [3]{d \sec (e+f x)}}{12 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \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 ) \sqrt [3]{d \sec (e+f x)}}{12 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}+\frac {F_1\left (\frac {1}{2};2,\frac {5}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan (e+f x)}{a^2 f \sqrt [6]{\sec ^2(e+f x)}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {5}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan ^3(e+f x)}{3 a^4 f \sqrt [6]{\sec ^2(e+f x)}}-\frac {a b \sqrt [3]{d \sec (e+f x)}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )}-\frac {\left (5 a b^{2/3} \sqrt [3]{d \sec (e+f x)}\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 \sqrt [6]{\sec ^2(e+f x)}}+\frac {\left (5 a b^{2/3} \sqrt [3]{d \sec (e+f x)}\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 \sqrt [6]{\sec ^2(e+f x)}}\\ &=\frac {5 a b^{2/3} \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 ) \sqrt [3]{d \sec (e+f x)}}{2 \sqrt {3} \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \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 ) \sqrt [3]{d \sec (e+f x)}}{2 \sqrt {3} \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \tanh ^{-1}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) \sqrt [3]{d \sec (e+f x)}}{3 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}+\frac {5 a b^{2/3} \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 ) \sqrt [3]{d \sec (e+f x)}}{12 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \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 ) \sqrt [3]{d \sec (e+f x)}}{12 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}+\frac {F_1\left (\frac {1}{2};2,\frac {5}{6};\frac {3}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan (e+f x)}{a^2 f \sqrt [6]{\sec ^2(e+f x)}}+\frac {b^2 F_1\left (\frac {3}{2};2,\frac {5}{6};\frac {5}{2};\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan ^3(e+f x)}{3 a^4 f \sqrt [6]{\sec ^2(e+f x)}}-\frac {a b \sqrt [3]{d \sec (e+f x)}}{\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] Result contains complex when optimal does not.
time = 73.63, size = 4485, normalized size = 6.53 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Sec[e + f*x])^(1/3)/(a + b*Tan[e + f*x])^2,x]

[Out]

((d*Sec[e + f*x])^(1/3)*((5*(-1)^(5/6)*a*b^(2/3)*(-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*Arc
Tan[((-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)*Sq
rt[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)]))/(12*(a - I*b)*(a + I*b)*(a^2 + b^2)^(5/6)) + 3*((-2*b^2*AppellF1[7/6, 1/2, 1, 13/6, 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]^(10/3))/(21*(a^2 + b^2)^2*Sqrt[1 -
Sec[e + f*x]^2]) + (Sec[e + f*x]^(1/3)*((-(a*b) + b^2*Sqrt[1 - Cos[e + f*x]^2]*Sec[e + f*x])/(a^2 + b^2) + (7*
(3*a^2 - 2*b^2)*AppellF1[1/6, 1/2, 1, 7/6, 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])/((-1 + Sec[e + f*x]^2)*(7*(a^2 + b^2)*AppellF1[1/6, 1/2, 1, 7/6, Sec[e + f*x]^2, (b^2*Se
c[e + f*x]^2)/(a^2 + b^2)] + 3*(2*b^2*AppellF1[7/6, 1/2, 2, 13/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 +
b^2)] + (a^2 + b^2)*AppellF1[7/6, 3/2, 1, 13/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)])*Sec[e + f*x
]^2))))/(3*(a^2 - b^2*(-1 + Sec[e + f*x]^2))))))/(f*(a + b*Tan[e + f*x])^2*((5*(-1)^(5/6)*a*b^(2/3)*((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)*S
ec[e + f*x]^(2/3))))/(12*(a - I*b)*(a + I*b)*(a^2 + b^2)^(5/6)) + 3*((-2*b^2*AppellF1[7/6, 1/2, 1, 13/6, 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]^(19/3)*Sin[e + f*x])/(21*(a
^2 + b^2)^2*(1 - Sec[e + f*x]^2)^(3/2)) - (2*b^2*AppellF1[7/6, 1/2, 1, 13/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]
^2)/(a^2 + b^2)]*Sec[e + f*x]^(7/3)*Sin[e + f*x])/(21*(a^2 + b^2)^2*Sqrt[1 - Cos[e + f*x]^2]*Sqrt[1 - Sec[e +
f*x]^2]) - (20*b^2*AppellF1[7/6, 1/2, 1, 13/6, 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]^(13/3)*Sin[e + f*x])/(63*(a^2 + b^2)^2*Sqrt[1 - Sec[e + f*x]^2]) + (2*b^2*Sec[e + f*x
]^(10/3)*((-(a*b) + b^2*Sqrt[1 - Cos[e + f*x]^2]*Sec[e + f*x])/(a^2 + b^2) + (7*(3*a^2 - 2*b^2)*AppellF1[1/6,
1/2, 1, 7/6, 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])/((-1 + S
ec[e + f*x]^2)*(7*(a^2 + b^2)*AppellF1[1/6, 1/2, 1, 7/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)] + 3
*(2*b^2*AppellF1[7/6, 1/2, 2, 13/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)] + (a^2 + b^2)*AppellF1[7
/6, 3/2, 1, 13/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)])*Sec[e + f*x]^2)))*Sin[e + f*x])/(3*(a^2 -
b^2*(-1 + Sec[e + f*x]^2))^2) + (Sec[e + f*x]^(4/3)*((-(a*b) + b^2*Sqrt[1 - Cos[e + f*x]^2]*Sec[e + f*x])/(a^
2 + b^2) + (7*(3*a^2 - 2*b^2)*AppellF1[1/6, 1/2, 1, 7/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)]*Sqr
t[1 - Cos[e + f*x]^2]*Sec[e + f*x])/((-1 + Sec[e + f*x]^2)*(7*(a^2 + b^2)*AppellF1[1/6, 1/2, 1, 7/6, Sec[e + f
*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)] + 3*(2*b^2*AppellF1[7/6, 1/2, 2, 13/6, Sec[e + f*x]^2, (b^2*Sec[e + f
*x]^2)/(a^2 + b^2)] + (a^2 + b^2)*AppellF1[7/6, 3/2, 1, 13/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)
])*Sec[e + f*x]^2)))*Sin[e + f*x])/(9*(a^2 - b^2*(-1 + Sec[e + f*x]^2))) - (2*b^2*Sqrt[1 - Cos[e + f*x]^2]*Sec
[e + f*x]^(10/3)*((14*b^2*AppellF1[13/6, 1/2, 2, 19/6, 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])/(13*(a^2 + b^2)) + (7*AppellF1[13/6, 3/2, 1, 19/6, 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])/13))/(21*(a^2 + b^2)^2*Sqrt[1 - Sec[e + f*x]^2]) + (Sec[e + f*x]^(
1/3)*((7*(3*a^2 - 2*b^2)*AppellF1[1/6, 1/2, 1, 7/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)]*Sin[e +
f*x])/(Sqrt[1 - Cos[e + f*x]^2]*(-1 + Sec[e + f*x]^2)*(7*(a^2 + b^2)*AppellF1[1/6, 1/2, 1, 7/6, Sec[e + f*x]^2
, (b^2*Sec[e + f*x]^2)/(a^2 + b^2)] + 3*(2*b^2*AppellF1[7/6, 1/2, 2, 13/6, Sec[e + f*x]^2, (b^2*Sec[e + f*x]^2
)/(a^2 + b^2)] + (a^2 + b^2)*AppellF1[7/6, 3/2,...

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Maple [F]
time = 1.06, size = 0, normalized size = 0.00 \[\int \frac {\left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}}}{\left (a +b \tan \left (f x +e \right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(1/3)/(a+b*tan(f*x+e))^2,x)

[Out]

int((d*sec(f*x+e))^(1/3)/(a+b*tan(f*x+e))^2,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((d*sec(f*x+e))^(1/3)/(a+b*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(1/3)/(b*tan(f*x + e) + a)^2, 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((d*sec(f*x+e))^(1/3)/(a+b*tan(f*x+e))^2,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 {\sqrt [3]{d \sec {\left (e + f x \right )}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(1/3)/(a+b*tan(f*x+e))**2,x)

[Out]

Integral((d*sec(e + f*x))**(1/3)/(a + b*tan(e + f*x))**2, 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((d*sec(f*x+e))^(1/3)/(a+b*tan(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(1/3)/(b*tan(f*x + e) + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(1/3)/(a + b*tan(e + f*x))^2,x)

[Out]

int((d/cos(e + f*x))^(1/3)/(a + b*tan(e + f*x))^2, x)

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