3.7.4 \(\int \frac {(d \sec (e+f x))^{5/2}}{a+b \tan (e+f x)} \, dx\) [604]

Optimal. Leaf size=396 \[ \frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f}-\frac {\sqrt [4]{a^2+b^2} d^2 \text {ArcTan}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\sqrt [4]{a^2+b^2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {2 a d^2 F\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right ) \sqrt {d \sec (e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a d^2 \cot (e+f x) \Pi \left (-\frac {b}{\sqrt {a^2+b^2}};\left .\text {ArcSin}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a d^2 \cot (e+f x) \Pi \left (\frac {b}{\sqrt {a^2+b^2}};\left .\text {ArcSin}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{b^2 f \sqrt [4]{\sec ^2(e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.28, antiderivative size = 396, 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, 749, 858, 237, 761, 410, 109, 418, 1227, 551, 455, 65, 218, 214, 211} \begin {gather*} \frac {a d^2 \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \sqrt {d \sec (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 )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a d^2 \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \sqrt {d \sec (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 )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}-\frac {d^2 \sqrt [4]{a^2+b^2} \sqrt {d \sec (e+f x)} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {d^2 \sqrt [4]{a^2+b^2} \sqrt {d \sec (e+f x)} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{b^{3/2} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {2 a d^2 \sqrt {d \sec (e+f x)} F\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right )}{b^2 f \sqrt [4]{\sec ^2(e+f x)}}+\frac {2 d^2 \sqrt {d \sec (e+f x)}}{b f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Sec[e + f*x])^(5/2)/(a + b*Tan[e + f*x]),x]

[Out]

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

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 109

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 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 218

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] &&  !Gt
Q[a/b, 0]

Rule 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 410

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 418

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 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 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 749

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 + 2*p + 1))), x] + Dist[2*(p/(e*(m + 2*p + 1))), Int[(d + e*x)^m*Simp[a*e - c*d*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] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 761

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(2853\) vs. \(2(396)=792\).
time = 67.63, size = 2853, normalized size = 7.20 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Sec[e + f*x])^(5/2)/(a + b*Tan[e + f*x]),x]

[Out]

(Cos[(e + f*x)/2]^2*Cos[e + f*x]*((Sqrt[a^2 + b^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 - Sq
rt[a^2 + b^2])]*Sqrt[1 - Tan[(e + f*x)/2]^4])])/Sqrt[a^2 + b*(b - Sqrt[a^2 + b^2])] - (Sqrt[a^2 + b^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*Sq
rt[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[a^2 +
b*(b + Sqrt[a^2 + b^2])] + (4*b^(3/2)*EllipticF[ArcSin[Tan[(e + f*x)/2]], -1])/a + (4*Sqrt[b^2*(a^2 + b^2)]*E
llipticPi[a^2/(a^2 + 2*b^2 - 2*Sqrt[b^2*(a^2 + b^2)]), ArcSin[Tan[(e + f*x)/2]], -1])/(a*Sqrt[b]) - (4*a*b^(3/
2)*EllipticPi[a^2/(a^2 + 2*(b^2 + Sqrt[b^2*(a^2 + b^2)])), ArcSin[Tan[(e + f*x)/2]], -1])/Sqrt[b^2*(a^2 + b^2)
] - (4*b^(7/2)*EllipticPi[a^2/(a^2 + 2*(b^2 + Sqrt[b^2*(a^2 + b^2)])), ArcSin[Tan[(e + f*x)/2]], -1])/(a*Sqrt[
b^2*(a^2 + b^2)]))*(d*Sec[e + f*x])^(5/2)*(a*Cos[e + f*x] + b*Sin[e + f*x])*Sqrt[(1 - Tan[(e + f*x)/2]^2)^(-1)
]*Sqrt[1 - Tan[(e + f*x)/2]^2]*(-a^2 - 2*b^2 - 2*Sqrt[b^2*(a^2 + b^2)] + a^2*Tan[(e + f*x)/2]^2)*(-a^2 - 2*b^2
+ 2*Sqrt[b^2*(a^2 + b^2)] + a^2*Tan[(e + f*x)/2]^2)*(1 - Tan[(e + f*x)/2]^4)*(a^4 + 8*a^2*b^2 + 8*b^4 + 4*a^2
*b*Sqrt[a^2 + b^2] + 8*b^3*Sqrt[a^2 + b^2] - 2*a^4*Tan[(e + f*x)/2]^2 - 4*a^2*b^2*Tan[(e + f*x)/2]^2 - 4*a^2*b
*Sqrt[a^2 + b^2]*Tan[(e + f*x)/2]^2 + a^4*Tan[(e + f*x)/2]^4)*(a^4 + 8*a^2*b^2 + 8*b^4 - 4*a^2*b*Sqrt[a^2 + b^
2] - 8*b^3*Sqrt[a^2 + b^2] - 2*a^4*Tan[(e + f*x)/2]^2 - 4*a^2*b^2*Tan[(e + f*x)/2]^2 + 4*a^2*b*Sqrt[a^2 + b^2]
*Tan[(e + f*x)/2]^2 + a^4*Tan[(e + f*x)/2]^4)*(b - a*Tan[e + f*x]))/(2*a^6*b^(3/2)*f*Sqrt[-(-1 + Tan[(e + f*x)
/2]^2)^(-1)]*(a^5*b*Sqrt[1 + Tan[(e + f*x)/2]^2] - 3*a^5*b*Tan[(e + f*x)/2]^2*Sqrt[1 + Tan[(e + f*x)/2]^2] - 8
*a^3*b^3*Tan[(e + f*x)/2]^2*Sqrt[1 + Tan[(e + f*x)/2]^2] + a^5*b*Tan[(e + f*x)/2]^4*Sqrt[1 + Tan[(e + f*x)/2]^
2] + 8*a^3*b^3*Tan[(e + f*x)/2]^4*Sqrt[1 + Tan[(e + f*x)/2]^2] + 16*a*b^5*Tan[(e + f*x)/2]^4*Sqrt[1 + Tan[(e +
f*x)/2]^2] + 5*a^5*b*Tan[(e + f*x)/2]^6*Sqrt[1 + Tan[(e + f*x)/2]^2] + 16*a^3*b^3*Tan[(e + f*x)/2]^6*Sqrt[1 +
Tan[(e + f*x)/2]^2] + 16*a*b^5*Tan[(e + f*x)/2]^6*Sqrt[1 + Tan[(e + f*x)/2]^2] - 5*a^5*b*Tan[(e + f*x)/2]^8*S
qrt[1 + Tan[(e + f*x)/2]^2] - 16*a^3*b^3*Tan[(e + f*x)/2]^8*Sqrt[1 + Tan[(e + f*x)/2]^2] - 16*a*b^5*Tan[(e + f
*x)/2]^8*Sqrt[1 + Tan[(e + f*x)/2]^2] - a^5*b*Tan[(e + f*x)/2]^10*Sqrt[1 + Tan[(e + f*x)/2]^2] - 8*a^3*b^3*Tan
[(e + f*x)/2]^10*Sqrt[1 + Tan[(e + f*x)/2]^2] - 16*a*b^5*Tan[(e + f*x)/2]^10*Sqrt[1 + Tan[(e + f*x)/2]^2] + 3*
a^5*b*Tan[(e + f*x)/2]^12*Sqrt[1 + Tan[(e + f*x)/2]^2] + 8*a^3*b^3*Tan[(e + f*x)/2]^12*Sqrt[1 + Tan[(e + f*x)/
2]^2] - a^5*b*Tan[(e + f*x)/2]^14*Sqrt[1 + Tan[(e + f*x)/2]^2] - 2*a^6*Tan[(e + f*x)/2]*Sqrt[1 - Tan[(e + f*x)
/2]^2]*Sqrt[1 - Tan[(e + f*x)/2]^4] - 2*a^4*b^2*Tan[(e + f*x)/2]*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[1 - Tan[(e
+ f*x)/2]^4] + 10*a^6*Tan[(e + f*x)/2]^3*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[1 - Tan[(e + f*x)/2]^4] + 26*a^4*b^
2*Tan[(e + f*x)/2]^3*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[1 - Tan[(e + f*x)/2]^4] + 16*a^2*b^4*Tan[(e + f*x)/2]^3
*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[1 - Tan[(e + f*x)/2]^4] - 20*a^6*Tan[(e + f*x)/2]^5*Sqrt[1 - Tan[(e + f*x)/
2]^2]*Sqrt[1 - Tan[(e + f*x)/2]^4] - 68*a^4*b^2*Tan[(e + f*x)/2]^5*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[1 - Tan[(
e + f*x)/2]^4] - 80*a^2*b^4*Tan[(e + f*x)/2]^5*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[1 - Tan[(e + f*x)/2]^4] - 32*
b^6*Tan[(e + f*x)/2]^5*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[1 - Tan[(e + f*x)/2]^4] + 20*a^6*Tan[(e + f*x)/2]^7*S
qrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[1 - Tan[(e + f*x)/2]^4] + 68*a^4*b^2*Tan[(e + f*x)/2]^7*Sqrt[1 - Tan[(e + f*x
)/2]^2]*Sqrt[1 - Tan[(e + f*x)/2]^4] + 80*a^2*b^4*Tan[(e + f*x)/2]^7*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[1 - Tan
[(e + f*x)/2]^4] + 32*b^6*Tan[(e + f*x)/2]^7*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[1 - Tan[(e + f*x)/2]^4] - 10*a^
6*Tan[(e + f*x)/2]^9*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[1 - Tan[(e + f*x)/2]^4] - 26*a^4*b^2*Tan[(e + f*x)/2]^9
*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[1 - Tan[(e + f*x)/2]^4] - 16*a^2*b^4*Tan[(e + f*x)/2]^9*Sqrt[1 - Tan[(e + f
*x)/2]^2]*Sqrt[1 - Tan[(e + f*x)/2]^4] + 2*a^6*Tan[(e + f*x)/2]^11*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[1 - Tan[(
e + f*x)/2]^4] + 2*a^4*b^2*Tan[(e + f*x)/2]^11*Sqrt[1 - Tan[(e + f*x)/2]^2]*Sqrt[1 - Tan[(e + f*x)/2]^4])*(a +
b*Tan[e + f*x])^2) + (2*Cos[e + f*x]*(d*Sec[e + f*x])^(5/2)*(a*Cos[e + f*x] + b*Sin[e + f*x]))/(b*f*(a + b*Ta
n[e + f*x]))

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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. 10703 vs. \(2 (369 ) = 738\).
time = 0.80, size = 10704, normalized size = 27.03

method	result	size
default	\(\text {Expression too large to display}\)	\(10704\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(5/2)/(a+b*tan(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))^(5/2)/(a+b*tan(f*x+e)),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(5/2)/(b*tan(f*x + e) + a), x)

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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))^(5/2)/(a+b*tan(f*x+e)),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}{a + b \tan {\left (e + f x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(5/2)/(a+b*tan(f*x+e)),x)

[Out]

Integral((d*sec(e + f*x))**(5/2)/(a + b*tan(e + f*x)), x)

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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))^(5/2)/(a+b*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(5/2)/(b*tan(f*x + e) + a), 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 )}^{5/2}}{a+b\,\mathrm {tan}\left (e+f\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/cos(e + f*x))^(5/2)/(a + b*tan(e + f*x)),x)

[Out]

int((d/cos(e + f*x))^(5/2)/(a + b*tan(e + f*x)), x)

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