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

Optimal. Leaf size=440 \[ \frac{a d^2 \sqrt{d \sec (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} f \left (a^2+b^2\right )^{3/4} \sqrt [4]{\sec ^2(e+f x)}}+\frac{a d^2 \sqrt{d \sec (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} f \left (a^2+b^2\right )^{3/4} \sqrt [4]{\sec ^2(e+f x)}}-\frac{a^2 d^2 \sqrt{-\tan ^2(e+f x)} \cot (e+f x) \sqrt{d \sec (e+f x)} \Pi \left (-\frac{b}{\sqrt{a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{2 b^2 f \left (a^2+b^2\right ) \sqrt [4]{\sec ^2(e+f x)}}-\frac{a^2 d^2 \sqrt{-\tan ^2(e+f x)} \cot (e+f x) \sqrt{d \sec (e+f x)} \Pi \left (\frac{b}{\sqrt{a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{2 b^2 f \left (a^2+b^2\right ) \sqrt [4]{\sec ^2(e+f x)}}-\frac{d^2 \sqrt{d \sec (e+f x)}}{b f (a+b \tan (e+f x))}+\frac{d^2 \sqrt{d \sec (e+f x)} F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.394753, antiderivative size = 440, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 15, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {3512, 733, 844, 231, 747, 401, 108, 409, 1213, 537, 444, 63, 212, 208, 205} \[ \frac{a d^2 \sqrt{d \sec (e+f x)} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} f \left (a^2+b^2\right )^{3/4} \sqrt [4]{\sec ^2(e+f x)}}+\frac{a d^2 \sqrt{d \sec (e+f x)} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{3/2} f \left (a^2+b^2\right )^{3/4} \sqrt [4]{\sec ^2(e+f x)}}-\frac{a^2 d^2 \sqrt{-\tan ^2(e+f x)} \cot (e+f x) \sqrt{d \sec (e+f x)} \Pi \left (-\frac{b}{\sqrt{a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{2 b^2 f \left (a^2+b^2\right ) \sqrt [4]{\sec ^2(e+f x)}}-\frac{a^2 d^2 \sqrt{-\tan ^2(e+f x)} \cot (e+f x) \sqrt{d \sec (e+f x)} \Pi \left (\frac{b}{\sqrt{a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{2 b^2 f \left (a^2+b^2\right ) \sqrt [4]{\sec ^2(e+f x)}}-\frac{d^2 \sqrt{d \sec (e+f x)}}{b f (a+b \tan (e+f x))}+\frac{d^2 \sqrt{d \sec (e+f x)} F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 733

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 + 1)), x] - Dist[(2*c*p)/(e*(m + 1)), Int[x*(d + e*x)^(m + 1)*(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] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m +
2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 844

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 231

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

Rule 747

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 401

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 108

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 409

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 1213

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 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), 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 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 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 212

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] &&
!GtQ[a/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 205

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

Rubi steps

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

Mathematica [C]  time = 21.8024, size = 3091, normalized size = 7.02 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((d*Sec[e + f*x])^(5/2)*(a*Cos[e + f*x] + b*Sin[e + f*x])^2*(-(1/(a*b)) + Sin[e + f*x]/(a*(a*Cos[e + f*x] + b*
Sin[e + f*x]))))/(f*(a + b*Tan[e + f*x])^2) - (((-2*I)*b*Sqrt[a^2 + b^2]*EllipticF[ArcSin[Sqrt[1 - I*Cos[e + f
*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(a - I*b + Sqrt[a^2 + b^2])*EllipticPi[((1 + I)*(a + I*(-b + Sqrt[a^2 + b
^2])))/(a + b - Sqrt[a^2 + b^2]), ArcSin[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(-a + I*b +
Sqrt[a^2 + b^2])*EllipticPi[((1 + I)*(a - I*(b + Sqrt[a^2 + b^2])))/(a + b + Sqrt[a^2 + b^2]), ArcSin[Sqrt[1 -
I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2])*(d*Sec[e + f*x])^(5/2)*Sqrt[Cos[(e + f*x)/2]^2*Sec[e + f*x]]*Sqr
t[I*Cos[e + f*x] - Sin[e + f*x]]*Sqrt[Cos[e + f*x]*(Cos[e + f*x] + I*Sin[e + f*x])]*Sin[e + f*x]*(a*Cos[e + f*
x] + b*Sin[e + f*x])*(I + Tan[(e + f*x)/2])^2)/(4*(a - I*b)*b^3*Sqrt[a^2 + b^2]*f*Sqrt[(1 + Cos[e + f*x])^(-1)
]*(a + b*Tan[e + f*x])^2*(-(((-2*I)*b*Sqrt[a^2 + b^2]*EllipticF[ArcSin[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]
/Sqrt[2]], 2] + a*(a - I*b + Sqrt[a^2 + b^2])*EllipticPi[((1 + I)*(a + I*(-b + Sqrt[a^2 + b^2])))/(a + b - Sqr
t[a^2 + b^2]), ArcSin[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(-a + I*b + Sqrt[a^2 + b^2])*El
lipticPi[((1 + I)*(a - I*(b + Sqrt[a^2 + b^2])))/(a + b + Sqrt[a^2 + b^2]), ArcSin[Sqrt[1 - I*Cos[e + f*x] + S
in[e + f*x]]/Sqrt[2]], 2])*Sec[(e + f*x)/2]^2*Sqrt[Cos[(e + f*x)/2]^2*Sec[e + f*x]]*Sqrt[I*Cos[e + f*x] - Sin[
e + f*x]]*Sqrt[Cos[e + f*x]*(Cos[e + f*x] + I*Sin[e + f*x])]*(I + Tan[(e + f*x)/2]))/(2*(a - I*b)*b^2*Sqrt[a^2
+ b^2]*Sqrt[(1 + Cos[e + f*x])^(-1)]) - (((-2*I)*b*Sqrt[a^2 + b^2]*EllipticF[ArcSin[Sqrt[1 - I*Cos[e + f*x] +
Sin[e + f*x]]/Sqrt[2]], 2] + a*(a - I*b + Sqrt[a^2 + b^2])*EllipticPi[((1 + I)*(a + I*(-b + Sqrt[a^2 + b^2]))
)/(a + b - Sqrt[a^2 + b^2]), ArcSin[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(-a + I*b + Sqrt[
a^2 + b^2])*EllipticPi[((1 + I)*(a - I*(b + Sqrt[a^2 + b^2])))/(a + b + Sqrt[a^2 + b^2]), ArcSin[Sqrt[1 - I*Co
s[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2])*Sqrt[Cos[(e + f*x)/2]^2*Sec[e + f*x]]*(-Cos[e + f*x] - I*Sin[e + f*x]
)*Sqrt[Cos[e + f*x]*(Cos[e + f*x] + I*Sin[e + f*x])]*(I + Tan[(e + f*x)/2])^2)/(4*(a - I*b)*b^2*Sqrt[a^2 + b^2
]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[I*Cos[e + f*x] - Sin[e + f*x]]) + (Sqrt[(1 + Cos[e + f*x])^(-1)]*((-2*I)*
b*Sqrt[a^2 + b^2]*EllipticF[ArcSin[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(a - I*b + Sqrt[a^
2 + b^2])*EllipticPi[((1 + I)*(a + I*(-b + Sqrt[a^2 + b^2])))/(a + b - Sqrt[a^2 + b^2]), ArcSin[Sqrt[1 - I*Cos
[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(-a + I*b + Sqrt[a^2 + b^2])*EllipticPi[((1 + I)*(a - I*(b + Sqrt[a
^2 + b^2])))/(a + b + Sqrt[a^2 + b^2]), ArcSin[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2])*Sqrt[Cos[
(e + f*x)/2]^2*Sec[e + f*x]]*Sqrt[I*Cos[e + f*x] - Sin[e + f*x]]*Sqrt[Cos[e + f*x]*(Cos[e + f*x] + I*Sin[e + f
*x])]*Sin[e + f*x]*(I + Tan[(e + f*x)/2])^2)/(4*(a - I*b)*b^2*Sqrt[a^2 + b^2]) - (((-2*I)*b*Sqrt[a^2 + b^2]*El
lipticF[ArcSin[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(a - I*b + Sqrt[a^2 + b^2])*EllipticPi
[((1 + I)*(a + I*(-b + Sqrt[a^2 + b^2])))/(a + b - Sqrt[a^2 + b^2]), ArcSin[Sqrt[1 - I*Cos[e + f*x] + Sin[e +
f*x]]/Sqrt[2]], 2] + a*(-a + I*b + Sqrt[a^2 + b^2])*EllipticPi[((1 + I)*(a - I*(b + Sqrt[a^2 + b^2])))/(a + b
+ Sqrt[a^2 + b^2]), ArcSin[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2])*Sqrt[Cos[(e + f*x)/2]^2*Sec[e
+ f*x]]*Sqrt[I*Cos[e + f*x] - Sin[e + f*x]]*(Cos[e + f*x]*(I*Cos[e + f*x] - Sin[e + f*x]) - (Cos[e + f*x] + I
*Sin[e + f*x])*Sin[e + f*x])*(I + Tan[(e + f*x)/2])^2)/(4*(a - I*b)*b^2*Sqrt[a^2 + b^2]*Sqrt[(1 + Cos[e + f*x]
)^(-1)]*Sqrt[Cos[e + f*x]*(Cos[e + f*x] + I*Sin[e + f*x])]) - (Sqrt[Cos[(e + f*x)/2]^2*Sec[e + f*x]]*Sqrt[I*Co
s[e + f*x] - Sin[e + f*x]]*Sqrt[Cos[e + f*x]*(Cos[e + f*x] + I*Sin[e + f*x])]*(((-I)*b*Sqrt[a^2 + b^2]*(Cos[e
+ f*x] + I*Sin[e + f*x]))/(Sqrt[2]*Sqrt[1 + (-1 + I*Cos[e + f*x] - Sin[e + f*x])/2]*Sqrt[I*Cos[e + f*x] - Sin[
e + f*x]]*Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]) + (a*(a - I*b + Sqrt[a^2 + b^2])*(Cos[e + f*x] + I*Sin[e +
f*x]))/(2*Sqrt[2]*Sqrt[1 + (-1 + I*Cos[e + f*x] - Sin[e + f*x])/2]*Sqrt[I*Cos[e + f*x] - Sin[e + f*x]]*Sqrt[1
- I*Cos[e + f*x] + Sin[e + f*x]]*(1 - ((1/2 + I/2)*(a + I*(-b + Sqrt[a^2 + b^2]))*(1 - I*Cos[e + f*x] + Sin[e
+ f*x]))/(a + b - Sqrt[a^2 + b^2]))) + (a*(-a + I*b + Sqrt[a^2 + b^2])*(Cos[e + f*x] + I*Sin[e + f*x]))/(2*Sqr
t[2]*Sqrt[1 + (-1 + I*Cos[e + f*x] - Sin[e + f*x])/2]*Sqrt[I*Cos[e + f*x] - Sin[e + f*x]]*Sqrt[1 - I*Cos[e + f
*x] + Sin[e + f*x]]*(1 - ((1/2 + I/2)*(a - I*(b + Sqrt[a^2 + b^2]))*(1 - I*Cos[e + f*x] + Sin[e + f*x]))/(a +
b + Sqrt[a^2 + b^2]))))*(I + Tan[(e + f*x)/2])^2)/(2*(a - I*b)*b^2*Sqrt[a^2 + b^2]*Sqrt[(1 + Cos[e + f*x])^(-1
)]) - (((-2*I)*b*Sqrt[a^2 + b^2]*EllipticF[ArcSin[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(a
- I*b + Sqrt[a^2 + b^2])*EllipticPi[((1 + I)*(a + I*(-b + Sqrt[a^2 + b^2])))/(a + b - Sqrt[a^2 + b^2]), ArcSin
[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(-a + I*b + Sqrt[a^2 + b^2])*EllipticPi[((1 + I)*(a
- I*(b + Sqrt[a^2 + b^2])))/(a + b + Sqrt[a^2 + b^2]), ArcSin[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]]
, 2])*Sqrt[I*Cos[e + f*x] - Sin[e + f*x]]*Sqrt[Cos[e + f*x]*(Cos[e + f*x] + I*Sin[e + f*x])]*(I + Tan[(e + f*x
)/2])^2*(-(Cos[(e + f*x)/2]*Sec[e + f*x]*Sin[(e + f*x)/2]) + Cos[(e + f*x)/2]^2*Sec[e + f*x]*Tan[e + f*x]))/(4
*(a - I*b)*b^2*Sqrt[a^2 + b^2]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[Cos[(e + f*x)/2]^2*Sec[e + f*x]])))

________________________________________________________________________________________

Maple [B]  time = 0.607, size = 5329, normalized size = 12.1 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \sec \left (f x + e\right )\right )^{\frac{5}{2}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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