\(\int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx\) [604]

Optimal result

Integrand size = 25, antiderivative size = 396 \[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {\sqrt [4]{a^2+b^2} d^2 \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\sqrt [4]{a^2+b^2} d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {2 a d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a d^2 \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a d^2 \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}} \]

[Out]

2*d^2*(d*sec(f*x+e))^(1/2)/b/f-(a^2+b^2)^(1/4)*d^2*arctan((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*(d*sec
(f*x+e))^(1/2)/b^(3/2)/f/(sec(f*x+e)^2)^(1/4)-(a^2+b^2)^(1/4)*d^2*arctanh((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^
2)^(1/4))*(d*sec(f*x+e))^(1/2)/b^(3/2)/f/(sec(f*x+e)^2)^(1/4)-2*a*d^2*(cos(1/2*arctan(tan(f*x+e)))^2)^(1/2)/co
s(1/2*arctan(tan(f*x+e)))*EllipticF(sin(1/2*arctan(tan(f*x+e))),2^(1/2))*(d*sec(f*x+e))^(1/2)/b^2/f/(sec(f*x+e
)^2)^(1/4)+a*d^2*cot(f*x+e)*EllipticPi((sec(f*x+e)^2)^(1/4),-b/(a^2+b^2)^(1/2),I)*(d*sec(f*x+e))^(1/2)*(-tan(f
*x+e)^2)^(1/2)/b^2/f/(sec(f*x+e)^2)^(1/4)+a*d^2*cot(f*x+e)*EllipticPi((sec(f*x+e)^2)^(1/4),b/(a^2+b^2)^(1/2),I
)*(d*sec(f*x+e))^(1/2)*(-tan(f*x+e)^2)^(1/2)/b^2/f/(sec(f*x+e)^2)^(1/4)

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3593, 749, 858, 237, 761, 410, 109, 418, 1227, 551, 455, 65, 218, 214, 211} \[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\frac {a d^2 \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \sqrt {d \sec (e+f x)} \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a d^2 \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \sqrt {d \sec (e+f x)} \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {d^2 \sqrt [4]{a^2+b^2} \sqrt {d \sec (e+f x)} \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {d^2 \sqrt [4]{a^2+b^2} \sqrt {d \sec (e+f x)} \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {2 a d^2 \sqrt {d \sec (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f} \]

[In]

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

[Out]

(2*d^2*Sqrt[d*Sec[e + f*x]])/(b*f) - ((a^2 + b^2)^(1/4)*d^2*ArcTan[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2
)^(1/4)]*Sqrt[d*Sec[e + f*x]])/(b^(3/2)*f*(Sec[e + f*x]^2)^(1/4)) - ((a^2 + b^2)^(1/4)*d^2*ArcTanh[(Sqrt[b]*(S
ec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*Sqrt[d*Sec[e + f*x]])/(b^(3/2)*f*(Sec[e + f*x]^2)^(1/4)) - (2*a*d^2*E
llipticF[ArcTan[Tan[e + f*x]]/2, 2]*Sqrt[d*Sec[e + f*x]])/(b^2*f*(Sec[e + f*x]^2)^(1/4)) + (a*d^2*Cot[e + f*x]
*EllipticPi[-(b/Sqrt[a^2 + b^2]), ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*Sqrt[d*Sec[e + f*x]]*Sqrt[-Tan[e + f*x]^
2])/(b^2*f*(Sec[e + f*x]^2)^(1/4)) + (a*d^2*Cot[e + f*x]*EllipticPi[b/Sqrt[a^2 + b^2], ArcSin[(Sec[e + f*x]^2)
^(1/4)], -1]*Sqrt[d*Sec[e + f*x]]*Sqrt[-Tan[e + f*x]^2])/(b^2*f*(Sec[e + f*x]^2)^(1/4))

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 109

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(3/4)), x_Symbol] :> Dist[-4, Subst[
Int[1/((b*e - a*f - b*x^4)*Sqrt[c - d*(e/f) + d*(x^4/f)]), x], x, (e + f*x)^(1/4)], x] /; FreeQ[{a, b, c, d, e
, f}, x] && GtQ[-f/(d*e - c*f), 0]

Rule 211

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

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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 410

Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[Sqrt[(-b)*(x^2/a)]/(2*x), Subst[I
nt[1/(Sqrt[(-b)*(x/a)]*(a + b*x)^(3/4)*(c + d*x)), x], x, x^2], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1
- Rt[-d/c, 2]*x^2)), x], x] + Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-d/c, 2]*x^2)), x], x] /; FreeQ[{a,
b, c, d}, x] && NeQ[b*c - a*d, 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 551

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1/(a*Sqr
t[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b*(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-f/e, -d/c])

Rule 749

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + c*x^2)^p/(e
*(m + 2*p + 1))), x] + Dist[2*(p/(e*(m + 2*p + 1))), Int[(d + e*x)^m*Simp[a*e - c*d*x, x]*(a + c*x^2)^(p - 1),
x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 761

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(3/4)), x_Symbol] :> Dist[d, Int[1/((d^2 - e^2*x^2)*(a + c*x^
2)^(3/4)), x], x] - Dist[e, Int[x/((d^2 - e^2*x^2)*(a + c*x^2)^(3/4)), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ
[c*d^2 + a*e^2, 0]

Rule 858

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1227

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[Sqrt[-c],
Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c,
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*} \text {integral}& = \frac {\left (d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {\sqrt [4]{1+\frac {x^2}{b^2}}}{a+x} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}+\frac {\left (d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1-\frac {a x}{b^2}}{(a+x) \left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {\left (a d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{b^3 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (\left (1+\frac {a^2}{b^2}\right ) d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{(a+x) \left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {2 a d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\left (\left (1+\frac {a^2}{b^2}\right ) d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {x}{\left (a^2-x^2\right ) \left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (a \left (1+\frac {a^2}{b^2}\right ) d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-x^2\right ) \left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {2 a d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\left (\left (1+\frac {a^2}{b^2}\right ) d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-x\right ) \left (1+\frac {x}{b^2}\right )^{3/4}} \, dx,x,b^2 \tan ^2(e+f x)\right )}{2 b f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (a \left (1+\frac {a^2}{b^2}\right ) d^2 \cot (e+f x) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-x\right ) \sqrt {-\frac {x}{b^2}} \left (1+\frac {x}{b^2}\right )^{3/4}} \, dx,x,b^2 \tan ^2(e+f x)\right )}{2 b^2 f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {2 a d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\left (2 \left (1+\frac {a^2}{b^2}\right ) b d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{a^2+b^2-b^2 x^4} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\left (2 a \left (1+\frac {a^2}{b^2}\right ) d^2 \cot (e+f x) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^4} \left (-1-\frac {a^2}{b^2}+x^4\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {2 a d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\left (\left (1+\frac {a^2}{b^2}\right ) b d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a^2+b^2}-b x^2} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\sqrt {a^2+b^2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\left (\left (1+\frac {a^2}{b^2}\right ) b d^2 \sqrt {d \sec (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a^2+b^2}+b x^2} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\sqrt {a^2+b^2} f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (a \left (1+\frac {a^2}{b^2}\right ) d^2 \cot (e+f x) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (1-\frac {b x^2}{\sqrt {a^2+b^2}}\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (a \left (1+\frac {a^2}{b^2}\right ) d^2 \cot (e+f x) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {b x^2}{\sqrt {a^2+b^2}}\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {\sqrt [4]{a^2+b^2} d^2 \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\sqrt [4]{a^2+b^2} d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {2 a d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (a \left (1+\frac {a^2}{b^2}\right ) d^2 \cot (e+f x) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (1-\frac {b x^2}{\sqrt {a^2+b^2}}\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (a \left (1+\frac {a^2}{b^2}\right ) d^2 \cot (e+f x) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (1+\frac {b x^2}{\sqrt {a^2+b^2}}\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}} \\ & = \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {\sqrt [4]{a^2+b^2} d^2 \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\sqrt [4]{a^2+b^2} d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {2 a d^2 \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a d^2 \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a d^2 \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 27.35 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.73 \[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\frac {d^2 \cot (e+f x) \sqrt {d \sec (e+f x)} \left (-\sqrt {b} \sqrt [4]{a^2+b^2} \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \tan (e+f x)-\sqrt {b} \sqrt [4]{a^2+b^2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \tan (e+f x)+2 b \sqrt [4]{\sec ^2(e+f x)} \tan (e+f x)-a \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\tan ^2(e+f x)\right ) \tan ^2(e+f x)+a \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {-\tan ^2(e+f x)}+a \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {-\tan ^2(e+f x)}\right )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}} \]

[In]

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

[Out]

(d^2*Cot[e + f*x]*Sqrt[d*Sec[e + f*x]]*(-(Sqrt[b]*(a^2 + b^2)^(1/4)*ArcTan[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a
^2 + b^2)^(1/4)]*Tan[e + f*x]) - Sqrt[b]*(a^2 + b^2)^(1/4)*ArcTanh[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2
)^(1/4)]*Tan[e + f*x] + 2*b*(Sec[e + f*x]^2)^(1/4)*Tan[e + f*x] - a*Hypergeometric2F1[1/2, 3/4, 3/2, -Tan[e +
f*x]^2]*Tan[e + f*x]^2 + a*EllipticPi[-(b/Sqrt[a^2 + b^2]), ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*Sqrt[-Tan[e +
f*x]^2] + a*EllipticPi[b/Sqrt[a^2 + b^2], ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*Sqrt[-Tan[e + f*x]^2]))/(b^2*f*(
Sec[e + f*x]^2)^(1/4))

Maple [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 7346 vs. \(2 (369 ) = 738\).

Time = 30.43 (sec) , antiderivative size = 7347, normalized size of antiderivative = 18.55

method	result	size
default	\(\text {Expression too large to display}\)	\(7347\)

[In]

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

[Out]

result too large to display

Fricas [F]

\[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}}{b \tan \left (f x + e\right ) + a} \,d x } \]

[In]

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

[Out]

integral(sqrt(d*sec(f*x + e))*d^2*sec(f*x + e)^2/(b*tan(f*x + e) + a), x)

Sympy [F]

\[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\int \frac {\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}{a + b \tan {\left (e + f x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}}{b \tan \left (f x + e\right ) + a} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}}{b \tan \left (f x + e\right ) + a} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{a+b\,\mathrm {tan}\left (e+f\,x\right )} \,d x \]

[In]

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

[Out]

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