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

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

[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]
time = 0.28, antiderivative size = 480, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 15, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3593, 747, 858, 233, 202, 760, 408, 504, 1227, 551, 455, 65, 304, 211, 214} \begin {gather*} \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 .\text {ArcSin}\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 .\text {ArcSin}\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 d^2 (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 )}{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 {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} \text {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} \end {gather*}

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 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*} \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 ) \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} \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 ) \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} \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 ) \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} \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 ) \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 \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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(4735\) vs. \(2(480)=960\).
time = 69.23, size = 4735, normalized size = 9.86 \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])^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*((3*Sqrt[Sec[e + f*x]])/(4*a*(a*Cos[e + f*x] + b*Sin[e + f*x])) - (9*a*Sqrt
[Sec[e + f*x]])/(4*b^2*(a*Cos[e + f*x] + b*Sin[e + f*x])) - (3*Cos[2*(e + f*x)]*Sqrt[Sec[e + f*x]])/(4*a*(a*Co
s[e + f*x] + b*Sin[e + f*x])) - (3*a*Cos[2*(e + f*x)]*Sqrt[Sec[e + f*x]])/(4*b^2*(a*Cos[e + f*x] + b*Sin[e + f
*x])))*Sqrt[(1 + Tan[(e + f*x)/2]^2)/(1 - Tan[(e + f*x)/2]^2)]*(4*b + 4*a*Tan[(e + f*x)/2] - 4*b*Tan[(e + f*x)
/2]^2 - 4*a*Tan[(e + f*x)/2]^3 + (a^2*Sqrt[b - Sqrt[a^2 + b^2]]*ArcTan[(2*b*(b - Sqrt[a^2 + b^2])*Tan[(e + f*x
)/2]^2 + a^2*(-1 + Tan[(e + f*x)/2]^2))/(2*Sqrt[b]*Sqrt[b - Sqrt[a^2 + b^2]]*Sqrt[a^2 + b*(b - Sqrt[a^2 + b^2]
)]*Sqrt[1 - Tan[(e + f*x)/2]^4])]*Sqrt[1 - Tan[(e + f*x)/2]^4])/(Sqrt[b]*Sqrt[a^2 + b*(b - Sqrt[a^2 + b^2])])
+ (a^2*Sqrt[b + Sqrt[a^2 + b^2]]*ArcTan[(2*b*(b + Sqrt[a^2 + b^2])*Tan[(e + f*x)/2]^2 + a^2*(-1 + Tan[(e + f*x
)/2]^2))/(2*Sqrt[b]*Sqrt[b + Sqrt[a^2 + b^2]]*Sqrt[a^2 + b*(b + Sqrt[a^2 + b^2])]*Sqrt[1 - Tan[(e + f*x)/2]^4]
)]*Sqrt[1 - Tan[(e + f*x)/2]^4])/(Sqrt[b]*Sqrt[a^2 + b*(b + Sqrt[a^2 + b^2])]) + 4*a*EllipticE[ArcSin[Tan[(e +
f*x)/2]], -1]*Sqrt[1 - Tan[(e + f*x)/2]^4] - 8*a*EllipticF[ArcSin[Tan[(e + f*x)/2]], -1]*Sqrt[1 - Tan[(e + f*
x)/2]^4] + 4*a*EllipticPi[a^2/(a^2 + 2*b^2 - 2*Sqrt[b^2*(a^2 + b^2)]), ArcSin[Tan[(e + f*x)/2]], -1]*Sqrt[1 -
Tan[(e + f*x)/2]^4] + 4*a*EllipticPi[a^2/(a^2 + 2*(b^2 + Sqrt[b^2*(a^2 + b^2)])), ArcSin[Tan[(e + f*x)/2]], -1
]*Sqrt[1 - Tan[(e + f*x)/2]^4]))/(4*a*b^2*f*Sec[e + f*x]^(3/2)*(1 + Tan[(e + f*x)/2]^2)*((3*Sec[(e + f*x)/2]^2
*Tan[(e + f*x)/2]*Sqrt[(1 + Tan[(e + f*x)/2]^2)/(1 - Tan[(e + f*x)/2]^2)]*(4*b + 4*a*Tan[(e + f*x)/2] - 4*b*Ta
n[(e + f*x)/2]^2 - 4*a*Tan[(e + f*x)/2]^3 + (a^2*Sqrt[b - Sqrt[a^2 + b^2]]*ArcTan[(2*b*(b - Sqrt[a^2 + b^2])*T
an[(e + f*x)/2]^2 + a^2*(-1 + Tan[(e + f*x)/2]^2))/(2*Sqrt[b]*Sqrt[b - Sqrt[a^2 + b^2]]*Sqrt[a^2 + b*(b - Sqrt
[a^2 + b^2])]*Sqrt[1 - Tan[(e + f*x)/2]^4])]*Sqrt[1 - Tan[(e + f*x)/2]^4])/(Sqrt[b]*Sqrt[a^2 + b*(b - Sqrt[a^2
+ b^2])]) + (a^2*Sqrt[b + Sqrt[a^2 + b^2]]*ArcTan[(2*b*(b + Sqrt[a^2 + b^2])*Tan[(e + f*x)/2]^2 + a^2*(-1 + T
an[(e + f*x)/2]^2))/(2*Sqrt[b]*Sqrt[b + Sqrt[a^2 + b^2]]*Sqrt[a^2 + b*(b + Sqrt[a^2 + b^2])]*Sqrt[1 - Tan[(e +
f*x)/2]^4])]*Sqrt[1 - Tan[(e + f*x)/2]^4])/(Sqrt[b]*Sqrt[a^2 + b*(b + Sqrt[a^2 + b^2])]) + 4*a*EllipticE[ArcS
in[Tan[(e + f*x)/2]], -1]*Sqrt[1 - Tan[(e + f*x)/2]^4] - 8*a*EllipticF[ArcSin[Tan[(e + f*x)/2]], -1]*Sqrt[1 -
Tan[(e + f*x)/2]^4] + 4*a*EllipticPi[a^2/(a^2 + 2*b^2 - 2*Sqrt[b^2*(a^2 + b^2)]), ArcSin[Tan[(e + f*x)/2]], -1
]*Sqrt[1 - Tan[(e + f*x)/2]^4] + 4*a*EllipticPi[a^2/(a^2 + 2*(b^2 + Sqrt[b^2*(a^2 + b^2)])), ArcSin[Tan[(e + f
*x)/2]], -1]*Sqrt[1 - Tan[(e + f*x)/2]^4]))/(4*a*b^2*(1 + Tan[(e + f*x)/2]^2)^2) - (3*((Sec[(e + f*x)/2]^2*Tan
[(e + f*x)/2])/(1 - Tan[(e + f*x)/2]^2) + (Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]*(1 + Tan[(e + f*x)/2]^2))/(1 -
Tan[(e + f*x)/2]^2)^2)*(4*b + 4*a*Tan[(e + f*x)/2] - 4*b*Tan[(e + f*x)/2]^2 - 4*a*Tan[(e + f*x)/2]^3 + (a^2*Sq
rt[b - Sqrt[a^2 + b^2]]*ArcTan[(2*b*(b - Sqrt[a^2 + b^2])*Tan[(e + f*x)/2]^2 + a^2*(-1 + Tan[(e + f*x)/2]^2))/
(2*Sqrt[b]*Sqrt[b - Sqrt[a^2 + b^2]]*Sqrt[a^2 + b*(b - Sqrt[a^2 + b^2])]*Sqrt[1 - Tan[(e + f*x)/2]^4])]*Sqrt[1
- Tan[(e + f*x)/2]^4])/(Sqrt[b]*Sqrt[a^2 + b*(b - Sqrt[a^2 + b^2])]) + (a^2*Sqrt[b + Sqrt[a^2 + b^2]]*ArcTan[
(2*b*(b + Sqrt[a^2 + b^2])*Tan[(e + f*x)/2]^2 + a^2*(-1 + Tan[(e + f*x)/2]^2))/(2*Sqrt[b]*Sqrt[b + Sqrt[a^2 +
b^2]]*Sqrt[a^2 + b*(b + Sqrt[a^2 + b^2])]*Sqrt[1 - Tan[(e + f*x)/2]^4])]*Sqrt[1 - Tan[(e + f*x)/2]^4])/(Sqrt[b
]*Sqrt[a^2 + b*(b + Sqrt[a^2 + b^2])]) + 4*a*EllipticE[ArcSin[Tan[(e + f*x)/2]], -1]*Sqrt[1 - Tan[(e + f*x)/2]
^4] - 8*a*EllipticF[ArcSin[Tan[(e + f*x)/2]], -1]*Sqrt[1 - Tan[(e + f*x)/2]^4] + 4*a*EllipticPi[a^2/(a^2 + 2*b
^2 - 2*Sqrt[b^2*(a^2 + b^2)]), ArcSin[Tan[(e + f*x)/2]], -1]*Sqrt[1 - Tan[(e + f*x)/2]^4] + 4*a*EllipticPi[a^2
/(a^2 + 2*(b^2 + Sqrt[b^2*(a^2 + b^2)])), ArcSin[Tan[(e + f*x)/2]], -1]*Sqrt[1 - Tan[(e + f*x)/2]^4]))/(8*a*b^
2*(1 + Tan[(e + f*x)/2]^2)*Sqrt[(1 + Tan[(e + f*x)/2]^2)/(1 - Tan[(e + f*x)/2]^2)]) - (3*Sqrt[(1 + Tan[(e + f*
x)/2]^2)/(1 - Tan[(e + f*x)/2]^2)]*(2*a*Sec[(e + f*x)/2]^2 - 4*b*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2] - 6*a*Sec
[(e + f*x)/2]^2*Tan[(e + f*x)/2]^2 - (a^2*Sqrt[b - Sqrt[a^2 + b^2]]*ArcTan[(2*b*(b - Sqrt[a^2 + b^2])*Tan[(e +
f*x)/2]^2 + a^2*(-1 + Tan[(e + f*x)/2]^2))/(2*Sqrt[b]*Sqrt[b - Sqrt[a^2 + b^2]]*Sqrt[a^2 + b*(b - Sqrt[a^2 +
b^2])]*Sqrt[1 - Tan[(e + f*x)/2]^4])]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]^3)/(Sqrt[b]*Sqrt[a^2 + b*(b - Sqrt[a
^2 + b^2])]*Sqrt[1 - Tan[(e + f*x)/2]^4]) - (a^2*Sqrt[b + Sqrt[a^2 + b^2]]*ArcTan[(2*b*(b + Sqrt[a^2 + b^2])*T
an[(e + f*x)/2]^2 + a^2*(-1 + Tan[(e + f*x)/2]^2))/(2*Sqrt[b]*Sqrt[b + Sqrt[a^2 + b^2]]*Sqrt[a^2 + b*(b + Sqrt
[a^2 + b^2])]*Sqrt[1 - Tan[(e + f*x)/2]^4])]*Se...

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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. 44462 vs. \(2 (439 ) = 878\).
time = 1.56, size = 44463, normalized size = 92.63

method	result	size
default	\(\text {Expression too large to display}\)	\(44463\)

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,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))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [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))^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)

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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))**2,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))^2,x, algorithm="giac")

[Out]

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

[Out]

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

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