3.7.17 \(\int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^3} \, dx\) [617]

Optimal. Leaf size=583 \[ \frac {3 \left (a^2+2 b^2\right ) d^2 \text {ArcTan}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{8 b^{5/2} \left (a^2+b^2\right )^{5/4} f \sec ^2(e+f x)^{3/4}}-\frac {3 \left (a^2+2 b^2\right ) d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{8 b^{5/2} \left (a^2+b^2\right )^{5/4} f \sec ^2(e+f x)^{3/4}}+\frac {3 a d^2 E\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right ) (d \sec (e+f x))^{3/2}}{4 b^2 \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}-\frac {3 a d^2 \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{4 b^2 \left (a^2+b^2\right ) f}-\frac {3 a \left (a^2+2 b^2\right ) d^2 \cot (e+f x) \Pi \left (-\frac {b}{\sqrt {a^2+b^2}};\left .\text {ArcSin}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right ) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}}{8 b^3 \left (a^2+b^2\right )^{3/2} f \sec ^2(e+f x)^{3/4}}+\frac {3 a \left (a^2+2 b^2\right ) d^2 \cot (e+f x) \Pi \left (\frac {b}{\sqrt {a^2+b^2}};\left .\text {ArcSin}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right ) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}}{8 b^3 \left (a^2+b^2\right )^{3/2} f \sec ^2(e+f x)^{3/4}}-\frac {d^2 (d \sec (e+f x))^{3/2}}{2 b f (a+b \tan (e+f x))^2}+\frac {3 a d^2 (d \sec (e+f x))^{3/2}}{4 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))} \]

[Out]

3/8*(a^2+2*b^2)*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
)^(5/4)/f/(sec(f*x+e)^2)^(3/4)-3/8*(a^2+2*b^2)*d^2*arctanh((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*(d*se
c(f*x+e))^(3/2)/b^(5/2)/(a^2+b^2)^(5/4)/f/(sec(f*x+e)^2)^(3/4)+3/4*a*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/(a^2+b^2)
/f/(sec(f*x+e)^2)^(3/4)-3/4*a*d^2*cos(f*x+e)*(d*sec(f*x+e))^(3/2)*sin(f*x+e)/b^2/(a^2+b^2)/f-3/8*a*(a^2+2*b^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/(a^2+b^2)^(3/2)/f/(sec(f*x+e)^2)^(3/4)+3/8*a*(a^2+2*b^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/(a^2+b^2)^(3/2)/f/(sec(f*x+e)^2)^(3/4)-1/
2*d^2*(d*sec(f*x+e))^(3/2)/b/f/(a+b*tan(f*x+e))^2+3/4*a*d^2*(d*sec(f*x+e))^(3/2)/b/(a^2+b^2)/f/(a+b*tan(f*x+e)
)

________________________________________________________________________________________

Rubi [A]
time = 0.37, antiderivative size = 583, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 16, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3593, 747, 849, 858, 233, 202, 760, 408, 504, 1227, 551, 455, 65, 304, 211, 214} \begin {gather*} -\frac {3 a d^2 \left (a^2+2 b^2\right ) \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 .\text {ArcSin}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{8 b^3 f \left (a^2+b^2\right )^{3/2} \sec ^2(e+f x)^{3/4}}+\frac {3 a d^2 \left (a^2+2 b^2\right ) \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 .\text {ArcSin}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{8 b^3 f \left (a^2+b^2\right )^{3/2} \sec ^2(e+f x)^{3/4}}+\frac {3 a d^2 (d \sec (e+f x))^{3/2} E\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right )}{4 b^2 f \left (a^2+b^2\right ) \sec ^2(e+f x)^{3/4}}+\frac {3 d^2 \left (a^2+2 b^2\right ) (d \sec (e+f x))^{3/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{8 b^{5/2} f \left (a^2+b^2\right )^{5/4} \sec ^2(e+f x)^{3/4}}+\frac {3 a d^2 (d \sec (e+f x))^{3/2}}{4 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}-\frac {3 a d^2 \sin (e+f x) \cos (e+f x) (d \sec (e+f x))^{3/2}}{4 b^2 f \left (a^2+b^2\right )}-\frac {3 d^2 \left (a^2+2 b^2\right ) (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 )}{8 b^{5/2} f \left (a^2+b^2\right )^{5/4} \sec ^2(e+f x)^{3/4}}-\frac {d^2 (d \sec (e+f x))^{3/2}}{2 b f (a+b \tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(7/2)/(a + b*Tan[e + f*x])^3,x]

[Out]

(3*(a^2 + 2*b^2)*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))/(8*b^(
5/2)*(a^2 + b^2)^(5/4)*f*(Sec[e + f*x]^2)^(3/4)) - (3*(a^2 + 2*b^2)*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))/(8*b^(5/2)*(a^2 + b^2)^(5/4)*f*(Sec[e + f*x]^2)^(3/4)) + (3*a*d^
2*EllipticE[ArcTan[Tan[e + f*x]]/2, 2]*(d*Sec[e + f*x])^(3/2))/(4*b^2*(a^2 + b^2)*f*(Sec[e + f*x]^2)^(3/4)) -
(3*a*d^2*Cos[e + f*x]*(d*Sec[e + f*x])^(3/2)*Sin[e + f*x])/(4*b^2*(a^2 + b^2)*f) - (3*a*(a^2 + 2*b^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])/(8*b^3*(a^2 + b^2)^(3/2)*f*(Sec[e + f*x]^2)^(3/4)) + (3*a*(a^2 + 2*b^2)*d^2*Cot[e + f*x]*Ellipti
cPi[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])/(8*b^
3*(a^2 + b^2)^(3/2)*f*(Sec[e + f*x]^2)^(3/4)) - (d^2*(d*Sec[e + f*x])^(3/2))/(2*b*f*(a + b*Tan[e + f*x])^2) +
(3*a*d^2*(d*Sec[e + f*x])^(3/2))/(4*b*(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 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 849

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*
(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/((m + 1)*(c*d^2 + a*e^2))), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

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*} \int \frac {(d \sec (e+f x))^{7/2}}{(a+b \tan (e+f x))^3} \, dx &=\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)^3} \, 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}}{2 b f (a+b \tan (e+f x))^2}+\frac {\left (3 d^2 (d \sec (e+f x))^{3/2}\right ) \text {Subst}\left (\int \frac {x}{(a+x)^2 \sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{4 b^3 f \sec ^2(e+f x)^{3/4}}\\ &=-\frac {d^2 (d \sec (e+f x))^{3/2}}{2 b f (a+b \tan (e+f x))^2}+\frac {3 a d^2 (d \sec (e+f x))^{3/2}}{4 b \left (a^2+b^2\right ) 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+\frac {a x}{2 b^2}}{(a+x) \sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{4 b \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}\\ &=-\frac {d^2 (d \sec (e+f x))^{3/2}}{2 b f (a+b \tan (e+f x))^2}+\frac {3 a d^2 (d \sec (e+f x))^{3/2}}{4 b \left (a^2+b^2\right ) 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 [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{8 b^3 \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}-\frac {\left (3 \left (-2-\frac {a^2}{b^2}\right ) 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 )}{8 b \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}\\ &=-\frac {3 a d^2 \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{4 b^2 \left (a^2+b^2\right ) f}-\frac {d^2 (d \sec (e+f x))^{3/2}}{2 b f (a+b \tan (e+f x))^2}+\frac {3 a d^2 (d \sec (e+f x))^{3/2}}{4 b \left (a^2+b^2\right ) 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 (1+\frac {x^2}{b^2}\right )^{5/4}} \, dx,x,b \tan (e+f x)\right )}{8 b^3 \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}+\frac {\left (3 \left (-2-\frac {a^2}{b^2}\right ) 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 )}{8 b \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}-\frac {\left (3 a \left (-2-\frac {a^2}{b^2}\right ) 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 )}{8 b \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}\\ &=\frac {3 a d^2 E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) (d \sec (e+f x))^{3/2}}{4 b^2 \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}-\frac {3 a d^2 \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{4 b^2 \left (a^2+b^2\right ) f}-\frac {d^2 (d \sec (e+f x))^{3/2}}{2 b f (a+b \tan (e+f x))^2}+\frac {3 a d^2 (d \sec (e+f x))^{3/2}}{4 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {\left (3 \left (-2-\frac {a^2}{b^2}\right ) 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 )}{16 b \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}-\frac {\left (3 a \left (-2-\frac {a^2}{b^2}\right ) 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 )}{4 b^2 \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}\\ &=\frac {3 a d^2 E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) (d \sec (e+f x))^{3/2}}{4 b^2 \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}-\frac {3 a d^2 \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{4 b^2 \left (a^2+b^2\right ) f}-\frac {d^2 (d \sec (e+f x))^{3/2}}{2 b f (a+b \tan (e+f x))^2}+\frac {3 a d^2 (d \sec (e+f x))^{3/2}}{4 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {\left (3 \left (-2-\frac {a^2}{b^2}\right ) b 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 )}{4 \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}-\frac {\left (3 a \left (-2-\frac {a^2}{b^2}\right ) 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 )}{8 b \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}+\frac {\left (3 a \left (-2-\frac {a^2}{b^2}\right ) 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 )}{8 b \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}\\ &=\frac {3 a d^2 E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) (d \sec (e+f x))^{3/2}}{4 b^2 \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}-\frac {3 a d^2 \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{4 b^2 \left (a^2+b^2\right ) f}-\frac {d^2 (d \sec (e+f x))^{3/2}}{2 b f (a+b \tan (e+f x))^2}+\frac {3 a d^2 (d \sec (e+f x))^{3/2}}{4 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac {\left (3 \left (-2-\frac {a^2}{b^2}\right ) 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 )}{8 \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}-\frac {\left (3 \left (-2-\frac {a^2}{b^2}\right ) 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 )}{8 \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}-\frac {\left (3 a \left (-2-\frac {a^2}{b^2}\right ) 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 )}{8 b \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}+\frac {\left (3 a \left (-2-\frac {a^2}{b^2}\right ) 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 )}{8 b \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}\\ &=\frac {3 \left (a^2+2 b^2\right ) 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}}{8 b^{5/2} \left (a^2+b^2\right )^{5/4} f \sec ^2(e+f x)^{3/4}}-\frac {3 \left (a^2+2 b^2\right ) 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}}{8 b^{5/2} \left (a^2+b^2\right )^{5/4} f \sec ^2(e+f x)^{3/4}}+\frac {3 a d^2 E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) (d \sec (e+f x))^{3/2}}{4 b^2 \left (a^2+b^2\right ) f \sec ^2(e+f x)^{3/4}}-\frac {3 a d^2 \cos (e+f x) (d \sec (e+f x))^{3/2} \sin (e+f x)}{4 b^2 \left (a^2+b^2\right ) f}-\frac {3 a \left (a^2+2 b^2\right ) 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)}}{8 b^3 \left (a^2+b^2\right )^{3/2} f \sec ^2(e+f x)^{3/4}}+\frac {3 a \left (a^2+2 b^2\right ) 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)}}{8 b^3 \left (a^2+b^2\right )^{3/2} f \sec ^2(e+f x)^{3/4}}-\frac {d^2 (d \sec (e+f x))^{3/2}}{2 b f (a+b \tan (e+f x))^2}+\frac {3 a d^2 (d \sec (e+f x))^{3/2}}{4 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 69.83, size = 8652, normalized size = 14.84 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Sec[e + f*x])^(7/2)/(a + b*Tan[e + f*x])^3,x]

[Out]

Result too large to show

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 101371 vs. \(2 (532 ) = 1064\).
time = 2.52, size = 101372, normalized size = 173.88

method	result	size
default	\(\text {Expression too large to display}\)	\(101372\)

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))^3,x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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))^3,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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))**3,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3063 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^3,x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(7/2)/(b*tan(f*x + e) + a)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(7/2)/(a + b*tan(e + f*x))^3,x)

[Out]

int((d/cos(e + f*x))^(7/2)/(a + b*tan(e + f*x))^3, x)

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