3.608 \(\int \frac{1}{(d \sec (e+f x))^{3/2} (a+b \tan (e+f x))} \, dx\)

Optimal. Leaf size=422 \[ -\frac{b^{5/2} \sec ^2(e+f x)^{3/4} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{f \left (a^2+b^2\right )^{7/4} (d \sec (e+f x))^{3/2}}-\frac{b^{5/2} \sec ^2(e+f x)^{3/4} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{f \left (a^2+b^2\right )^{7/4} (d \sec (e+f x))^{3/2}}+\frac{2 (a \tan (e+f x)+b)}{3 f \left (a^2+b^2\right ) (d \sec (e+f x))^{3/2}}+\frac{2 a \sec ^2(e+f x)^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{3 f \left (a^2+b^2\right ) (d \sec (e+f x))^{3/2}}+\frac{a b^2 \sqrt{-\tan ^2(e+f x)} \cot (e+f x) \sec ^2(e+f x)^{3/4} \Pi \left (-\frac{b}{\sqrt{a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{f \left (a^2+b^2\right )^2 (d \sec (e+f x))^{3/2}}+\frac{a b^2 \sqrt{-\tan ^2(e+f x)} \cot (e+f x) \sec ^2(e+f x)^{3/4} \Pi \left (\frac{b}{\sqrt{a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{f \left (a^2+b^2\right )^2 (d \sec (e+f x))^{3/2}} \]

[Out]

-((b^(5/2)*ArcTan[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*(Sec[e + f*x]^2)^(3/4))/((a^2 + b^2)^(7/
4)*f*(d*Sec[e + f*x])^(3/2))) - (b^(5/2)*ArcTanh[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*(Sec[e +
f*x]^2)^(3/4))/((a^2 + b^2)^(7/4)*f*(d*Sec[e + f*x])^(3/2)) + (2*a*EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e
+ f*x]^2)^(3/4))/(3*(a^2 + b^2)*f*(d*Sec[e + f*x])^(3/2)) + (a*b^2*Cot[e + f*x]*EllipticPi[-(b/Sqrt[a^2 + b^2
]), ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*(Sec[e + f*x]^2)^(3/4)*Sqrt[-Tan[e + f*x]^2])/((a^2 + b^2)^2*f*(d*Sec[
e + f*x])^(3/2)) + (a*b^2*Cot[e + f*x]*EllipticPi[b/Sqrt[a^2 + b^2], ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*(Sec[
e + f*x]^2)^(3/4)*Sqrt[-Tan[e + f*x]^2])/((a^2 + b^2)^2*f*(d*Sec[e + f*x])^(3/2)) + (2*(b + a*Tan[e + f*x]))/(
3*(a^2 + b^2)*f*(d*Sec[e + f*x])^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.431916, antiderivative size = 422, 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, 741, 844, 231, 747, 401, 108, 409, 1213, 537, 444, 63, 212, 208, 205} \[ -\frac{b^{5/2} \sec ^2(e+f x)^{3/4} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{f \left (a^2+b^2\right )^{7/4} (d \sec (e+f x))^{3/2}}-\frac{b^{5/2} \sec ^2(e+f x)^{3/4} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{f \left (a^2+b^2\right )^{7/4} (d \sec (e+f x))^{3/2}}+\frac{2 (a \tan (e+f x)+b)}{3 f \left (a^2+b^2\right ) (d \sec (e+f x))^{3/2}}+\frac{2 a \sec ^2(e+f x)^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{3 f \left (a^2+b^2\right ) (d \sec (e+f x))^{3/2}}+\frac{a b^2 \sqrt{-\tan ^2(e+f x)} \cot (e+f x) \sec ^2(e+f x)^{3/4} \Pi \left (-\frac{b}{\sqrt{a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{f \left (a^2+b^2\right )^2 (d \sec (e+f x))^{3/2}}+\frac{a b^2 \sqrt{-\tan ^2(e+f x)} \cot (e+f x) \sec ^2(e+f x)^{3/4} \Pi \left (\frac{b}{\sqrt{a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{f \left (a^2+b^2\right )^2 (d \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^(5/2)*ArcTan[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*(Sec[e + f*x]^2)^(3/4))/((a^2 + b^2)^(7/
4)*f*(d*Sec[e + f*x])^(3/2))) - (b^(5/2)*ArcTanh[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*(Sec[e +
f*x]^2)^(3/4))/((a^2 + b^2)^(7/4)*f*(d*Sec[e + f*x])^(3/2)) + (2*a*EllipticF[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e
+ f*x]^2)^(3/4))/(3*(a^2 + b^2)*f*(d*Sec[e + f*x])^(3/2)) + (a*b^2*Cot[e + f*x]*EllipticPi[-(b/Sqrt[a^2 + b^2
]), ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*(Sec[e + f*x]^2)^(3/4)*Sqrt[-Tan[e + f*x]^2])/((a^2 + b^2)^2*f*(d*Sec[
e + f*x])^(3/2)) + (a*b^2*Cot[e + f*x]*EllipticPi[b/Sqrt[a^2 + b^2], ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*(Sec[
e + f*x]^2)^(3/4)*Sqrt[-Tan[e + f*x]^2])/((a^2 + b^2)^2*f*(d*Sec[e + f*x])^(3/2)) + (2*(b + a*Tan[e + f*x]))/(
3*(a^2 + b^2)*f*(d*Sec[e + f*x])^(3/2))

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 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*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] && LtQ[p, -1] && 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{1}{(d \sec (e+f x))^{3/2} (a+b \tan (e+f x))} \, dx &=\frac{\sec ^2(e+f x)^{3/4} \operatorname{Subst}\left (\int \frac{1}{(a+x) \left (1+\frac{x^2}{b^2}\right )^{7/4}} \, dx,x,b \tan (e+f x)\right )}{b f (d \sec (e+f x))^{3/2}}\\ &=\frac{2 (b+a \tan (e+f x))}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}-\frac{\left (2 b \sec ^2(e+f x)^{3/4}\right ) \operatorname{Subst}\left (\int \frac{\frac{1}{2} \left (-3-\frac{a^2}{b^2}\right )-\frac{a x}{2 b^2}}{(a+x) \left (1+\frac{x^2}{b^2}\right )^{3/4}} \, dx,x,b \tan (e+f x)\right )}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}\\ &=\frac{2 (b+a \tan (e+f x))}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}+\frac{\left (a \sec ^2(e+f x)^{3/4}\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 )}{3 b \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}+\frac{\left (b \sec ^2(e+f x)^{3/4}\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 )}{\left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}\\ &=\frac{2 a F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sec ^2(e+f x)^{3/4}}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}+\frac{2 (b+a \tan (e+f x))}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}-\frac{\left (b \sec ^2(e+f x)^{3/4}\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 )}{\left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}+\frac{\left (a b \sec ^2(e+f x)^{3/4}\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 )}{\left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}\\ &=\frac{2 a F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sec ^2(e+f x)^{3/4}}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}+\frac{2 (b+a \tan (e+f x))}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}-\frac{\left (b \sec ^2(e+f x)^{3/4}\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 )}{2 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}+\frac{\left (a \cot (e+f x) \sec ^2(e+f x)^{3/4} \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 )}{2 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}\\ &=\frac{2 a F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sec ^2(e+f x)^{3/4}}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}+\frac{2 (b+a \tan (e+f x))}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}-\frac{\left (2 b^3 \sec ^2(e+f x)^{3/4}\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 )}{\left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}-\frac{\left (2 a \cot (e+f x) \sec ^2(e+f x)^{3/4} \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 )}{\left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}\\ &=\frac{2 a F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sec ^2(e+f x)^{3/4}}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}+\frac{2 (b+a \tan (e+f x))}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}-\frac{\left (b^3 \sec ^2(e+f x)^{3/4}\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 )}{\left (a^2+b^2\right )^{3/2} f (d \sec (e+f x))^{3/2}}-\frac{\left (b^3 \sec ^2(e+f x)^{3/4}\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 )}{\left (a^2+b^2\right )^{3/2} f (d \sec (e+f x))^{3/2}}+\frac{\left (a b^2 \cot (e+f x) \sec ^2(e+f x)^{3/4} \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 )}{\left (a^2+b^2\right )^2 f (d \sec (e+f x))^{3/2}}+\frac{\left (a b^2 \cot (e+f x) \sec ^2(e+f x)^{3/4} \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 )}{\left (a^2+b^2\right )^2 f (d \sec (e+f x))^{3/2}}\\ &=-\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sec ^2(e+f x)^{3/4}}{\left (a^2+b^2\right )^{7/4} f (d \sec (e+f x))^{3/2}}-\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sec ^2(e+f x)^{3/4}}{\left (a^2+b^2\right )^{7/4} f (d \sec (e+f x))^{3/2}}+\frac{2 a F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sec ^2(e+f x)^{3/4}}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}+\frac{2 (b+a \tan (e+f x))}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}+\frac{\left (a b^2 \cot (e+f x) \sec ^2(e+f x)^{3/4} \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 )}{\left (a^2+b^2\right )^2 f (d \sec (e+f x))^{3/2}}+\frac{\left (a b^2 \cot (e+f x) \sec ^2(e+f x)^{3/4} \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 )}{\left (a^2+b^2\right )^2 f (d \sec (e+f x))^{3/2}}\\ &=-\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sec ^2(e+f x)^{3/4}}{\left (a^2+b^2\right )^{7/4} f (d \sec (e+f x))^{3/2}}-\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sec ^2(e+f x)^{3/4}}{\left (a^2+b^2\right )^{7/4} f (d \sec (e+f x))^{3/2}}+\frac{2 a F\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sec ^2(e+f x)^{3/4}}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}+\frac{a b^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 ) \sec ^2(e+f x)^{3/4} \sqrt{-\tan ^2(e+f x)}}{\left (a^2+b^2\right )^2 f (d \sec (e+f x))^{3/2}}+\frac{a b^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 ) \sec ^2(e+f x)^{3/4} \sqrt{-\tan ^2(e+f x)}}{\left (a^2+b^2\right )^2 f (d \sec (e+f x))^{3/2}}+\frac{2 (b+a \tan (e+f x))}{3 \left (a^2+b^2\right ) f (d \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 26.5739, size = 9313, normalized size = 22.07 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B]  time = 0.313, size = 6252, normalized size = 14.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}{\left (b \tan \left (f x + e\right ) + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \sec{\left (e + f x \right )}\right )^{\frac{3}{2}} \left (a + b \tan{\left (e + f x \right )}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}{\left (b \tan \left (f x + e\right ) + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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