\(\int \frac {\sqrt {d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx\) [613]

Optimal result

Integrand size = 25, antiderivative size = 430 \[ \int \frac {\sqrt {d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=-\frac {3 a \sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{2 \left (a^2+b^2\right )^{7/4} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {3 a \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{2 \left (a^2+b^2\right )^{7/4} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}+\frac {3 a^2 \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{2 \left (a^2+b^2\right )^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {3 a^2 \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{2 \left (a^2+b^2\right )^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {b \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))} \]

[Out]

-(cos(1/2*arctan(tan(f*x+e)))^2)^(1/2)/cos(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)/(a^2+b^2)/f/(sec(f*x+e)^2)^(1/4)-3/2*a*arctan((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^
(1/4))*b^(1/2)*(d*sec(f*x+e))^(1/2)/(a^2+b^2)^(7/4)/f/(sec(f*x+e)^2)^(1/4)-3/2*a*arctanh((sec(f*x+e)^2)^(1/4)*
b^(1/2)/(a^2+b^2)^(1/4))*b^(1/2)*(d*sec(f*x+e))^(1/2)/(a^2+b^2)^(7/4)/f/(sec(f*x+e)^2)^(1/4)+3/2*a^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)/(a^2+b^2)^2
/f/(sec(f*x+e)^2)^(1/4)+3/2*a^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)/(a^2+b^2)^2/f/(sec(f*x+e)^2)^(1/4)-b*(d*sec(f*x+e))^(1/2)/(a^2+b^2)/f/(a+b*tan(f*
x+e))

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 430, 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, 759, 858, 237, 761, 410, 109, 418, 1227, 551, 455, 65, 218, 214, 211} \[ \int \frac {\sqrt {d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\frac {3 a^2 \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \sqrt {d \sec (e+f x)} \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right )}{2 f \left (a^2+b^2\right )^2 \sqrt [4]{\sec ^2(e+f x)}}+\frac {3 a^2 \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \sqrt {d \sec (e+f x)} \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right )}{2 f \left (a^2+b^2\right )^2 \sqrt [4]{\sec ^2(e+f x)}}-\frac {3 a \sqrt {b} \sqrt {d \sec (e+f x)} \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 f \left (a^2+b^2\right )^{7/4} \sqrt [4]{\sec ^2(e+f x)}}-\frac {\sqrt {d \sec (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right )}{f \left (a^2+b^2\right ) \sqrt [4]{\sec ^2(e+f x)}}-\frac {3 a \sqrt {b} \sqrt {d \sec (e+f x)} \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 f \left (a^2+b^2\right )^{7/4} \sqrt [4]{\sec ^2(e+f x)}}-\frac {b \sqrt {d \sec (e+f x)}}{f \left (a^2+b^2\right ) (a+b \tan (e+f x))} \]

[In]

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

[Out]

(-3*a*Sqrt[b]*ArcTan[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*Sqrt[d*Sec[e + f*x]])/(2*(a^2 + b^2)^
(7/4)*f*(Sec[e + f*x]^2)^(1/4)) - (3*a*Sqrt[b]*ArcTanh[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*Sqr
t[d*Sec[e + f*x]])/(2*(a^2 + b^2)^(7/4)*f*(Sec[e + f*x]^2)^(1/4)) - (EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*Sqrt
[d*Sec[e + f*x]])/((a^2 + b^2)*f*(Sec[e + f*x]^2)^(1/4)) + (3*a^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])/(2*(a^2 + b^2)^2*f*(Sec[e +
f*x]^2)^(1/4)) + (3*a^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])/(2*(a^2 + b^2)^2*f*(Sec[e + f*x]^2)^(1/4)) - (b*Sqrt[d*Sec[e + f*x]])/((a
^2 + b^2)*f*(a + b*Tan[e + f*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 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 759

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m + 1)*((a + c*x^2)^(p
+ 1)/((m + 1)*(c*d^2 + a*e^2))), x] + Dist[c/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*Simp[d*(m + 1)
- e*(m + 2*p + 3)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[
m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ[p]) || ILtQ
[Simplify[m + 2*p + 3], 0])

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 {\sqrt {d \sec (e+f x)} \text {Subst}\left (\int \frac {1}{(a+x)^2 \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 {b \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac {\sqrt {d \sec (e+f x)} \text {Subst}\left (\int \frac {-a+\frac {x}{2}}{(a+x) \left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{b \left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}} \\ & = -\frac {b \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac {\sqrt {d \sec (e+f x)} \text {Subst}\left (\int \frac {1}{\left (1+\frac {x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{2 b \left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (3 a \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 )}{2 b \left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}} \\ & = -\frac {\operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}-\frac {b \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac {\left (3 a \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 )}{2 b \left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (3 a^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 )}{2 b \left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}} \\ & = -\frac {\operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}-\frac {b \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac {\left (3 a \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 )}{4 b \left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (3 a^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 )}{4 b^2 \left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}} \\ & = -\frac {\operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}-\frac {b \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac {\left (3 a b \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 )}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\left (3 a^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 \left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}} \\ & = -\frac {\operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}-\frac {b \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac {\left (3 a b \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 )}{2 \left (a^2+b^2\right )^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\left (3 a b \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 )}{2 \left (a^2+b^2\right )^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (3 a^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 )}{2 \left (a^2+b^2\right )^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (3 a^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 )}{2 \left (a^2+b^2\right )^2 f \sqrt [4]{\sec ^2(e+f x)}} \\ & = -\frac {3 a \sqrt {b} \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)}}{2 \left (a^2+b^2\right )^{7/4} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {3 a \sqrt {b} \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)}}{2 \left (a^2+b^2\right )^{7/4} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}-\frac {b \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {\left (3 a^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 )}{2 \left (a^2+b^2\right )^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (3 a^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 )}{2 \left (a^2+b^2\right )^2 f \sqrt [4]{\sec ^2(e+f x)}} \\ & = -\frac {3 a \sqrt {b} \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)}}{2 \left (a^2+b^2\right )^{7/4} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {3 a \sqrt {b} \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)}}{2 \left (a^2+b^2\right )^{7/4} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\operatorname {EllipticF}\left (\frac {1}{2} \arctan (\tan (e+f x)),2\right ) \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}+\frac {3 a^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)}}{2 \left (a^2+b^2\right )^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {3 a^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)}}{2 \left (a^2+b^2\right )^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {b \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right ) f (a+b \tan (e+f x))} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 10.76 (sec) , antiderivative size = 422, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\frac {\sec ^2(e+f x) \sqrt {d \sec (e+f x)} (a \cos (e+f x)+b \sin (e+f x))^2 \left (-\frac {b}{a (a-i b) (a+i b)}+\frac {b^2 \sin (e+f x)}{a (a-i b) (a+i b) (a \cos (e+f x)+b \sin (e+f x))}\right )}{f (a+b \tan (e+f x))^2}+\frac {\sqrt {d \sec (e+f x)} \sec ^2(e+f x)^{3/4} (a \cos (e+f x)+b \sin (e+f x))^2 \left (-\left (\left (a^2+b^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)\right )+3 a \left (-\sqrt {b} \sqrt [4]{a^2+b^2} \left (\arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )+\text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )\right )+a \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {-\tan ^2(e+f x)}+a \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) \sqrt {-\tan ^2(e+f x)}\right )\right )}{2 \left (a^2+b^2\right )^2 f (a+b \tan (e+f x))^2} \]

[In]

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

[Out]

(Sec[e + f*x]^2*Sqrt[d*Sec[e + f*x]]*(a*Cos[e + f*x] + b*Sin[e + f*x])^2*(-(b/(a*(a - I*b)*(a + I*b))) + (b^2*
Sin[e + f*x])/(a*(a - I*b)*(a + I*b)*(a*Cos[e + f*x] + b*Sin[e + f*x]))))/(f*(a + b*Tan[e + f*x])^2) + (Sqrt[d
*Sec[e + f*x]]*(Sec[e + f*x]^2)^(3/4)*(a*Cos[e + f*x] + b*Sin[e + f*x])^2*(-((a^2 + b^2)*Hypergeometric2F1[1/2
, 3/4, 3/2, -Tan[e + f*x]^2]*Tan[e + f*x]) + 3*a*(-(Sqrt[b]*(a^2 + b^2)^(1/4)*(ArcTan[(Sqrt[b]*(Sec[e + f*x]^2
)^(1/4))/(a^2 + b^2)^(1/4)] + ArcTanh[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)])) + a*Cot[e + f*x]*E
llipticPi[-(b/Sqrt[a^2 + b^2]), ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*Sqrt[-Tan[e + f*x]^2] + a*Cot[e + f*x]*Ell
ipticPi[b/Sqrt[a^2 + b^2], ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*Sqrt[-Tan[e + f*x]^2])))/(2*(a^2 + b^2)^2*f*(a
+ b*Tan[e + f*x])^2)

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. 12205 vs. \(2 (395 ) = 790\).

Time = 11.21 (sec) , antiderivative size = 12206, normalized size of antiderivative = 28.39

method	result	size
default	\(\text {Expression too large to display}\)	\(12206\)

[In]

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

[Out]

result too large to display

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt {d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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