3.610 \(\int \frac{(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=480 \[ -\frac{3 a d^2 (d \sec (e+f x))^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{5/2} f \sqrt [4]{a^2+b^2} \sec ^2(e+f x)^{3/4}}+\frac{3 a d^2 (d \sec (e+f x))^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{5/2} f \sqrt [4]{a^2+b^2} \sec ^2(e+f x)^{3/4}}+\frac{3 a^2 d^2 \sqrt{-\tan ^2(e+f x)} \cot (e+f x) (d \sec (e+f x))^{3/2} \Pi \left (-\frac{b}{\sqrt{a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{2 b^3 f \sqrt{a^2+b^2} \sec ^2(e+f x)^{3/4}}-\frac{3 a^2 d^2 \sqrt{-\tan ^2(e+f x)} \cot (e+f x) (d \sec (e+f x))^{3/2} \Pi \left (\frac{b}{\sqrt{a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{2 b^3 f \sqrt{a^2+b^2} \sec ^2(e+f x)^{3/4}}-\frac{d^2 (d \sec (e+f x))^{3/2}}{b f (a+b \tan (e+f x))}+\frac{3 d^2 \sin (e+f x) \cos (e+f x) (d \sec (e+f x))^{3/2}}{b^2 f}-\frac{3 d^2 (d \sec (e+f x))^{3/2} E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{b^2 f \sec ^2(e+f x)^{3/4}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.382219, antiderivative size = 480, 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, 227, 196, 746, 399, 490, 1213, 537, 444, 63, 298, 205, 208} \[ -\frac{3 a d^2 (d \sec (e+f x))^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{5/2} f \sqrt [4]{a^2+b^2} \sec ^2(e+f x)^{3/4}}+\frac{3 a d^2 (d \sec (e+f x))^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{5/2} f \sqrt [4]{a^2+b^2} \sec ^2(e+f x)^{3/4}}+\frac{3 a^2 d^2 \sqrt{-\tan ^2(e+f x)} \cot (e+f x) (d \sec (e+f x))^{3/2} \Pi \left (-\frac{b}{\sqrt{a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{2 b^3 f \sqrt{a^2+b^2} \sec ^2(e+f x)^{3/4}}-\frac{3 a^2 d^2 \sqrt{-\tan ^2(e+f x)} \cot (e+f x) (d \sec (e+f x))^{3/2} \Pi \left (\frac{b}{\sqrt{a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{2 b^3 f \sqrt{a^2+b^2} \sec ^2(e+f x)^{3/4}}-\frac{d^2 (d \sec (e+f x))^{3/2}}{b f (a+b \tan (e+f x))}+\frac{3 d^2 \sin (e+f x) \cos (e+f x) (d \sec (e+f x))^{3/2}}{b^2 f}-\frac{3 d^2 (d \sec (e+f x))^{3/2} E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{b^2 f \sec ^2(e+f x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(7/2)/(a + b*Tan[e + f*x])^2,x]

[Out]

(-3*a*d^2*ArcTan[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*(d*Sec[e + f*x])^(3/2))/(2*b^(5/2)*(a^2 +
b^2)^(1/4)*f*(Sec[e + f*x]^2)^(3/4)) + (3*a*d^2*ArcTanh[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*(
d*Sec[e + f*x])^(3/2))/(2*b^(5/2)*(a^2 + b^2)^(1/4)*f*(Sec[e + f*x]^2)^(3/4)) - (3*d^2*EllipticE[ArcTan[Tan[e
+ f*x]]/2, 2]*(d*Sec[e + f*x])^(3/2))/(b^2*f*(Sec[e + f*x]^2)^(3/4)) + (3*d^2*Cos[e + f*x]*(d*Sec[e + f*x])^(3
/2)*Sin[e + f*x])/(b^2*f) + (3*a^2*d^2*Cot[e + f*x]*EllipticPi[-(b/Sqrt[a^2 + b^2]), ArcSin[(Sec[e + f*x]^2)^(
1/4)], -1]*(d*Sec[e + f*x])^(3/2)*Sqrt[-Tan[e + f*x]^2])/(2*b^3*Sqrt[a^2 + b^2]*f*(Sec[e + f*x]^2)^(3/4)) - (3
*a^2*d^2*Cot[e + f*x]*EllipticPi[b/Sqrt[a^2 + b^2], ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*(d*Sec[e + f*x])^(3/2)
*Sqrt[-Tan[e + f*x]^2])/(2*b^3*Sqrt[a^2 + b^2]*f*(Sec[e + f*x]^2)^(3/4)) - (d^2*(d*Sec[e + f*x])^(3/2))/(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 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 746

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

Rule 399

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

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]],
s = Denominator[Rt[-(a/b), 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), 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 298

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

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]

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]

Rubi steps

\begin{align*} \int \frac{(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx &=\frac{\left (d^2 (d \sec (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{x^2}{b^2}\right )^{3/4}}{(a+x)^2} \, dx,x,b \tan (e+f x)\right )}{b f \sec ^2(e+f x)^{3/4}}\\ &=-\frac{d^2 (d \sec (e+f x))^{3/2}}{b f (a+b \tan (e+f x))}+\frac{\left (3 d^2 (d \sec (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{x}{(a+x) \sqrt [4]{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{2 b^3 f \sec ^2(e+f x)^{3/4}}\\ &=-\frac{d^2 (d \sec (e+f x))^{3/2}}{b f (a+b \tan (e+f x))}+\frac{\left (3 d^2 (d \sec (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{2 b^3 f \sec ^2(e+f x)^{3/4}}-\frac{\left (3 a d^2 (d \sec (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{(a+x) \sqrt [4]{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{2 b^3 f \sec ^2(e+f x)^{3/4}}\\ &=\frac{3 d^2 \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{b^2 f}-\frac{d^2 (d \sec (e+f x))^{3/2}}{b f (a+b \tan (e+f x))}-\frac{\left (3 d^2 (d \sec (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+\frac{x^2}{b^2}\right )^{5/4}} \, dx,x,b \tan (e+f x)\right )}{2 b^3 f \sec ^2(e+f x)^{3/4}}+\frac{\left (3 a d^2 (d \sec (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{x}{\left (a^2-x^2\right ) \sqrt [4]{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{2 b^3 f \sec ^2(e+f x)^{3/4}}-\frac{\left (3 a^2 d^2 (d \sec (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a^2-x^2\right ) \sqrt [4]{1+\frac{x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{2 b^3 f \sec ^2(e+f x)^{3/4}}\\ &=-\frac{3 d^2 E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) (d \sec (e+f x))^{3/2}}{b^2 f \sec ^2(e+f x)^{3/4}}+\frac{3 d^2 \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{b^2 f}-\frac{d^2 (d \sec (e+f x))^{3/2}}{b f (a+b \tan (e+f x))}+\frac{\left (3 a d^2 (d \sec (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a^2-x\right ) \sqrt [4]{1+\frac{x}{b^2}}} \, dx,x,b^2 \tan ^2(e+f x)\right )}{4 b^3 f \sec ^2(e+f x)^{3/4}}-\frac{\left (3 a^2 d^2 \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt{-\tan ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\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 \sec ^2(e+f x)^{3/4}}\\ &=-\frac{3 d^2 E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) (d \sec (e+f x))^{3/2}}{b^2 f \sec ^2(e+f x)^{3/4}}+\frac{3 d^2 \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{b^2 f}-\frac{d^2 (d \sec (e+f x))^{3/2}}{b f (a+b \tan (e+f x))}+\frac{\left (3 a d^2 (d \sec (e+f x))^{3/2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{a^2+b^2-b^2 x^4} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}-\frac{\left (3 a^2 d^2 \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt{-\tan ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{a^2+b^2}-b x^2\right ) \sqrt{1-x^4}} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{2 b^3 f \sec ^2(e+f x)^{3/4}}+\frac{\left (3 a^2 d^2 \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt{-\tan ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\sqrt{a^2+b^2}+b x^2\right ) \sqrt{1-x^4}} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{2 b^3 f \sec ^2(e+f x)^{3/4}}\\ &=-\frac{3 d^2 E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) (d \sec (e+f x))^{3/2}}{b^2 f \sec ^2(e+f x)^{3/4}}+\frac{3 d^2 \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{b^2 f}-\frac{d^2 (d \sec (e+f x))^{3/2}}{b f (a+b \tan (e+f x))}+\frac{\left (3 a d^2 (d \sec (e+f x))^{3/2}\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^2 f \sec ^2(e+f x)^{3/4}}-\frac{\left (3 a d^2 (d \sec (e+f x))^{3/2}\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^2 f \sec ^2(e+f x)^{3/4}}-\frac{\left (3 a^2 d^2 \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt{-\tan ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+x^2} \left (\sqrt{a^2+b^2}-b x^2\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{2 b^3 f \sec ^2(e+f x)^{3/4}}+\frac{\left (3 a^2 d^2 \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt{-\tan ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+x^2} \left (\sqrt{a^2+b^2}+b x^2\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{2 b^3 f \sec ^2(e+f x)^{3/4}}\\ &=-\frac{3 a d^2 \tan ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{2 b^{5/2} \sqrt [4]{a^2+b^2} f \sec ^2(e+f x)^{3/4}}+\frac{3 a d^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{2 b^{5/2} \sqrt [4]{a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac{3 d^2 E\left (\left .\frac{1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) (d \sec (e+f x))^{3/2}}{b^2 f \sec ^2(e+f x)^{3/4}}+\frac{3 d^2 \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{b^2 f}+\frac{3 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 ) (d \sec (e+f x))^{3/2} \sqrt{-\tan ^2(e+f x)}}{2 b^3 \sqrt{a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac{3 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 ) (d \sec (e+f x))^{3/2} \sqrt{-\tan ^2(e+f x)}}{2 b^3 \sqrt{a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac{d^2 (d \sec (e+f x))^{3/2}}{b f (a+b \tan (e+f x))}\\ \end{align*}

Mathematica [C]  time = 22.1452, size = 1129, normalized size = 2.35 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Sec[e + f*x])^(7/2)/(a + b*Tan[e + f*x])^2,x]

[Out]

(Cos[e + f*x]*(d*Sec[e + f*x])^(7/2)*(a*Cos[e + f*x] + b*Sin[e + f*x])^2*((3*Cos[e + f*x])/(a*b) + (3*Sin[e +
f*x])/b^2 - 1/(b*(a*Cos[e + f*x] + b*Sin[e + f*x]))))/(f*(a + b*Tan[e + f*x])^2) + (3*(d*Sec[e + f*x])^(7/2)*(
a*Cos[e + f*x] + b*Sin[e + f*x])^2*(-((a*EllipticE[ArcSin[Tan[(e + f*x)/2]], -1]*Sqrt[1 + Tan[(e + f*x)/2]^2])
/Sqrt[1 - Tan[(e + f*x)/2]^2]) + (2*a*EllipticF[ArcSin[Tan[(e + f*x)/2]], -1]*Sqrt[1 + Tan[(e + f*x)/2]^2])/Sq
rt[1 - Tan[(e + f*x)/2]^2] + (-2*Sqrt[2]*a*b*Sqrt[a^2 + b^2]*EllipticF[ArcSin[Sqrt[((1 + I)*(1 + Tan[(e + f*x)
/2]))/(I + Tan[(e + f*x)/2])]/Sqrt[2]], 2]*Sqrt[-((1 + I*Tan[(e + f*x)/2])/(I + Tan[(e + f*x)/2]))] + Sqrt[2]*
a^2*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)*(1 + Tan[(e + f*x)/2]))/(I + Tan[(e + f*x)/2])]/Sqrt[2]], 2]*Sqrt[-((1 + I*Tan[(e + f*x)/2])/(I + Tan
[(e + f*x)/2]))] + a^2*(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)*(1 + Tan[(e + f*x)/2]))/(I + Tan[(e + f*x)/2])]/Sqrt[2]], 2]*Sqrt[-((
2 + (2*I)*Tan[(e + f*x)/2])/(I + Tan[(e + f*x)/2]))] - a^3*EllipticPi[((1 + I)*(a - I*(b + Sqrt[a^2 + b^2])))/
(a + b + Sqrt[a^2 + b^2]), ArcSin[Sqrt[((1 + I)*(1 + Tan[(e + f*x)/2]))/(I + Tan[(e + f*x)/2])]/Sqrt[2]], 2]*S
qrt[-((2 + (2*I)*Tan[(e + f*x)/2])/(I + Tan[(e + f*x)/2]))] - I*a^2*b*EllipticPi[((1 + I)*(a - I*(b + Sqrt[a^2
+ b^2])))/(a + b + Sqrt[a^2 + b^2]), ArcSin[Sqrt[((1 + I)*(1 + Tan[(e + f*x)/2]))/(I + Tan[(e + f*x)/2])]/Sqr
t[2]], 2]*Sqrt[-((2 + (2*I)*Tan[(e + f*x)/2])/(I + Tan[(e + f*x)/2]))] - 2*b^2*Sqrt[a^2 + b^2]*Sqrt[(-1 + Tan[
(e + f*x)/2]^2)/(I + Tan[(e + f*x)/2])^2] - 2*a*b*Sqrt[a^2 + b^2]*Tan[(e + f*x)/2]*Sqrt[(-1 + Tan[(e + f*x)/2]
^2)/(I + Tan[(e + f*x)/2])^2])/(2*b*Sqrt[a^2 + b^2]*Sqrt[(-1 + Tan[(e + f*x)/2]^2)/(I + Tan[(e + f*x)/2])^2]))
)/(a*b^2*f*Sec[e + f*x]^(3/2)*Sqrt[(1 + Tan[(e + f*x)/2]^2)/(1 - Tan[(e + f*x)/2]^2)]*(a + b*Tan[e + f*x])^2)

________________________________________________________________________________________

Maple [B]  time = 4.405, size = 44463, normalized size = 92.6 \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))^(7/2)/(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))^(7/2)/(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))^(7/2)/(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))**(7/2)/(a+b*tan(f*x+e))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \sec \left (f x + e\right )\right )^{\frac{7}{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))^(7/2)/(a+b*tan(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(7/2)/(b*tan(f*x + e) + a)^2, x)