\(\int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx\) [610]

Optimal result

Integrand size = 25, antiderivative size = 480 \[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=-\frac {3 a d^2 \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{2 b^{5/2} \sqrt [4]{a^2+b^2} f \sec ^2(e+f x)^{3/4}}+\frac {3 a d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{2 b^{5/2} \sqrt [4]{a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac {3 d^2 E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right ) (d \sec (e+f x))^{3/2}}{b^2 f \sec ^2(e+f x)^{3/4}}+\frac {3 d^2 \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{b^2 f}+\frac {3 a^2 d^2 \cot (e+f x) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}}{2 b^3 \sqrt {a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac {3 a^2 d^2 \cot (e+f x) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right ) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}}{2 b^3 \sqrt {a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac {d^2 (d \sec (e+f x))^{3/2}}{b f (a+b \tan (e+f x))} \]

[Out]

-3/2*a*d^2*arctan((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*(d*sec(f*x+e))^(3/2)/b^(5/2)/(a^2+b^2)^(1/4)/f
/(sec(f*x+e)^2)^(3/4)+3/2*a*d^2*arctanh((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*(d*sec(f*x+e))^(3/2)/b^(
5/2)/(a^2+b^2)^(1/4)/f/(sec(f*x+e)^2)^(3/4)-3*d^2*(cos(1/2*arctan(tan(f*x+e)))^2)^(1/2)/cos(1/2*arctan(tan(f*x
+e)))*EllipticE(sin(1/2*arctan(tan(f*x+e))),2^(1/2))*(d*sec(f*x+e))^(3/2)/b^2/f/(sec(f*x+e)^2)^(3/4)+3*d^2*cos
(f*x+e)*(d*sec(f*x+e))^(3/2)*sin(f*x+e)/b^2/f+3/2*a^2*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))^(3/2)*(-tan(f*x+e)^2)^(1/2)/b^3/f/(sec(f*x+e)^2)^(3/4)/(a^2+b^2)^(1/2)-3/2*a^2*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))^(3/2)*(-tan(f*x+e)^2)^(1/2)/b^
3/f/(sec(f*x+e)^2)^(3/4)/(a^2+b^2)^(1/2)-d^2*(d*sec(f*x+e))^(3/2)/b/f/(a+b*tan(f*x+e))

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 480, 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, 747, 858, 233, 202, 760, 408, 504, 1227, 551, 455, 65, 304, 211, 214} \[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=\frac {3 a^2 d^2 \sqrt {-\tan ^2(e+f x)} \cot (e+f x) (d \sec (e+f x))^{3/2} \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right )}{2 b^3 f \sqrt {a^2+b^2} \sec ^2(e+f x)^{3/4}}-\frac {3 a^2 d^2 \sqrt {-\tan ^2(e+f x)} \cot (e+f x) (d \sec (e+f x))^{3/2} \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\right ),-1\right )}{2 b^3 f \sqrt {a^2+b^2} \sec ^2(e+f x)^{3/4}}-\frac {3 a d^2 (d \sec (e+f x))^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{5/2} f \sqrt [4]{a^2+b^2} \sec ^2(e+f x)^{3/4}}+\frac {3 a d^2 (d \sec (e+f x))^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 b^{5/2} f \sqrt [4]{a^2+b^2} \sec ^2(e+f x)^{3/4}}-\frac {d^2 (d \sec (e+f x))^{3/2}}{b f (a+b \tan (e+f x))}-\frac {3 d^2 (d \sec (e+f x))^{3/2} E\left (\left .\frac {1}{2} \arctan (\tan (e+f x))\right |2\right )}{b^2 f \sec ^2(e+f x)^{3/4}}+\frac {3 d^2 \sin (e+f x) \cos (e+f x) (d \sec (e+f x))^{3/2}}{b^2 f} \]

[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 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 202

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

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 233

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 304

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 408

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 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 504

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), Int[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 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 747

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 760

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 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 (d \sec (e+f x))^{3/2}\right ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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} \arctan (\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 ) \text {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 ) \text {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} \arctan (\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 ) \text {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 ) \text {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 ) \text {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} \arctan (\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 ) \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 b^2 f \sec ^2(e+f x)^{3/4}}-\frac {\left (3 a d^2 (d \sec (e+f x))^{3/2}\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 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 ) \text {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 ) \text {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 \arctan \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 \text {arctanh}\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} \arctan (\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) \operatorname {EllipticPi}\left (-\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\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) \operatorname {EllipticPi}\left (\frac {b}{\sqrt {a^2+b^2}},\arcsin \left (\sqrt [4]{\sec ^2(e+f x)}\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] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 29.03 (sec) , antiderivative size = 1129, normalized size of antiderivative = 2.35 \[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=\frac {\cos (e+f x) (d \sec (e+f x))^{7/2} (a \cos (e+f x)+b \sin (e+f x))^2 \left (\frac {3 \cos (e+f x)}{a b}+\frac {3 \sin (e+f x)}{b^2}-\frac {1}{b (a \cos (e+f x)+b \sin (e+f x))}\right )}{f (a+b \tan (e+f x))^2}+\frac {3 (d \sec (e+f x))^{7/2} (a \cos (e+f x)+b \sin (e+f x))^2 \left (-\frac {a E\left (\left .\arcsin \left (\tan \left (\frac {1}{2} (e+f x)\right )\right )\right |-1\right ) \sqrt {1+\tan ^2\left (\frac {1}{2} (e+f x)\right )}}{\sqrt {1-\tan ^2\left (\frac {1}{2} (e+f x)\right )}}+\frac {2 a \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (e+f x)\right )\right ),-1\right ) \sqrt {1+\tan ^2\left (\frac {1}{2} (e+f x)\right )}}{\sqrt {1-\tan ^2\left (\frac {1}{2} (e+f x)\right )}}+\frac {-2 \sqrt {2} a b \sqrt {a^2+b^2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {(1+i) \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{i+\tan \left (\frac {1}{2} (e+f x)\right )}}}{\sqrt {2}}\right ),2\right ) \sqrt {-\frac {1+i \tan \left (\frac {1}{2} (e+f x)\right )}{i+\tan \left (\frac {1}{2} (e+f x)\right )}}+\sqrt {2} a^2 \sqrt {a^2+b^2} \operatorname {EllipticPi}\left (\frac {(1+i) \left (a-i \left (b+\sqrt {a^2+b^2}\right )\right )}{a+b+\sqrt {a^2+b^2}},\arcsin \left (\frac {\sqrt {\frac {(1+i) \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{i+\tan \left (\frac {1}{2} (e+f x)\right )}}}{\sqrt {2}}\right ),2\right ) \sqrt {-\frac {1+i \tan \left (\frac {1}{2} (e+f x)\right )}{i+\tan \left (\frac {1}{2} (e+f x)\right )}}+a^2 \left (a+i b+\sqrt {a^2+b^2}\right ) \operatorname {EllipticPi}\left (\frac {(1+i) \left (a+i \left (-b+\sqrt {a^2+b^2}\right )\right )}{a+b-\sqrt {a^2+b^2}},\arcsin \left (\frac {\sqrt {\frac {(1+i) \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{i+\tan \left (\frac {1}{2} (e+f x)\right )}}}{\sqrt {2}}\right ),2\right ) \sqrt {-\frac {2+2 i \tan \left (\frac {1}{2} (e+f x)\right )}{i+\tan \left (\frac {1}{2} (e+f x)\right )}}-a^3 \operatorname {EllipticPi}\left (\frac {(1+i) \left (a-i \left (b+\sqrt {a^2+b^2}\right )\right )}{a+b+\sqrt {a^2+b^2}},\arcsin \left (\frac {\sqrt {\frac {(1+i) \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{i+\tan \left (\frac {1}{2} (e+f x)\right )}}}{\sqrt {2}}\right ),2\right ) \sqrt {-\frac {2+2 i \tan \left (\frac {1}{2} (e+f x)\right )}{i+\tan \left (\frac {1}{2} (e+f x)\right )}}-i a^2 b \operatorname {EllipticPi}\left (\frac {(1+i) \left (a-i \left (b+\sqrt {a^2+b^2}\right )\right )}{a+b+\sqrt {a^2+b^2}},\arcsin \left (\frac {\sqrt {\frac {(1+i) \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{i+\tan \left (\frac {1}{2} (e+f x)\right )}}}{\sqrt {2}}\right ),2\right ) \sqrt {-\frac {2+2 i \tan \left (\frac {1}{2} (e+f x)\right )}{i+\tan \left (\frac {1}{2} (e+f x)\right )}}-2 b^2 \sqrt {a^2+b^2} \sqrt {\frac {-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )}{\left (i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^2}}-2 a b \sqrt {a^2+b^2} \tan \left (\frac {1}{2} (e+f x)\right ) \sqrt {\frac {-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )}{\left (i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^2}}}{2 b \sqrt {a^2+b^2} \sqrt {\frac {-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )}{\left (i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^2}}}\right )}{a b^2 f \sec ^{\frac {3}{2}}(e+f x) \sqrt {\frac {1+\tan ^2\left (\frac {1}{2} (e+f x)\right )}{1-\tan ^2\left (\frac {1}{2} (e+f x)\right )}} (a+b \tan (e+f x))^2} \]

[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] (warning: unable to verify)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 33210 vs. \(2 (439 ) = 878\).

Time = 132.90 (sec) , antiderivative size = 33211, normalized size of antiderivative = 69.19

method	result	size
default	\(\text {Expression too large to display}\)	\(33211\)

[In]

int((d*sec(f*x+e))^(7/2)/(a+b*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F]

\[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

\[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {7}{2}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^2} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{7/2}}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \]

[In]

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

[Out]

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