3.7.7 \(\int \frac {1}{\sqrt {d \sec (e+f x)} (a+b \tan (e+f x))} \, dx\) [607]

Optimal. Leaf size=451 \[ \frac {b^{3/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right )^{5/4} f \sqrt {d \sec (e+f x)}}-\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right )^{5/4} f \sqrt {d \sec (e+f x)}}+\frac {2 a E\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {2 a \tan (e+f x)}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {a b \cot (e+f x) \Pi \left (-\frac {b}{\sqrt {a^2+b^2}};\left .\text {ArcSin}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right ) \sqrt [4]{\sec ^2(e+f x)} \sqrt {-\tan ^2(e+f x)}}{\left (a^2+b^2\right )^{3/2} f \sqrt {d \sec (e+f x)}}+\frac {a b \cot (e+f x) \Pi \left (\frac {b}{\sqrt {a^2+b^2}};\left .\text {ArcSin}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right ) \sqrt [4]{\sec ^2(e+f x)} \sqrt {-\tan ^2(e+f x)}}{\left (a^2+b^2\right )^{3/2} f \sqrt {d \sec (e+f x)}}+\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}} \]

[Out]

b^(3/2)*arctan((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*(sec(f*x+e)^2)^(1/4)/(a^2+b^2)^(5/4)/f/(d*sec(f*x
+e))^(1/2)-b^(3/2)*arctanh((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*(sec(f*x+e)^2)^(1/4)/(a^2+b^2)^(5/4)/
f/(d*sec(f*x+e))^(1/2)+2*a*(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))*(sec(f*x+e)^2)^(1/4)/(a^2+b^2)/f/(d*sec(f*x+e))^(1/2)-a*b*cot(f*x+e)*EllipticPi(
(sec(f*x+e)^2)^(1/4),-b/(a^2+b^2)^(1/2),I)*(sec(f*x+e)^2)^(1/4)*(-tan(f*x+e)^2)^(1/2)/(a^2+b^2)^(3/2)/f/(d*sec
(f*x+e))^(1/2)+a*b*cot(f*x+e)*EllipticPi((sec(f*x+e)^2)^(1/4),b/(a^2+b^2)^(1/2),I)*(sec(f*x+e)^2)^(1/4)*(-tan(
f*x+e)^2)^(1/2)/(a^2+b^2)^(3/2)/f/(d*sec(f*x+e))^(1/2)-2*a*tan(f*x+e)/(a^2+b^2)/f/(d*sec(f*x+e))^(1/2)+2*(b+a*
tan(f*x+e))/(a^2+b^2)/f/(d*sec(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.32, antiderivative size = 451, 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, 755, 858, 233, 202, 760, 408, 504, 1227, 551, 455, 65, 304, 211, 214} \begin {gather*} -\frac {a b \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \sqrt [4]{\sec ^2(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 )}{f \left (a^2+b^2\right )^{3/2} \sqrt {d \sec (e+f x)}}+\frac {a b \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \sqrt [4]{\sec ^2(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 )}{f \left (a^2+b^2\right )^{3/2} \sqrt {d \sec (e+f x)}}+\frac {2 a \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right )}{f \left (a^2+b^2\right ) \sqrt {d \sec (e+f x)}}+\frac {b^{3/2} \sqrt [4]{\sec ^2(e+f x)} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{f \left (a^2+b^2\right )^{5/4} \sqrt {d \sec (e+f x)}}-\frac {2 a \tan (e+f x)}{f \left (a^2+b^2\right ) \sqrt {d \sec (e+f x)}}+\frac {2 (a \tan (e+f x)+b)}{f \left (a^2+b^2\right ) \sqrt {d \sec (e+f x)}}-\frac {b^{3/2} \sqrt [4]{\sec ^2(e+f x)} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{f \left (a^2+b^2\right )^{5/4} \sqrt {d \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])),x]

[Out]

(b^(3/2)*ArcTan[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*(Sec[e + f*x]^2)^(1/4))/((a^2 + b^2)^(5/4)
*f*Sqrt[d*Sec[e + f*x]]) - (b^(3/2)*ArcTanh[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*(Sec[e + f*x]^
2)^(1/4))/((a^2 + b^2)^(5/4)*f*Sqrt[d*Sec[e + f*x]]) + (2*a*EllipticE[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]
^2)^(1/4))/((a^2 + b^2)*f*Sqrt[d*Sec[e + f*x]]) - (2*a*Tan[e + f*x])/((a^2 + b^2)*f*Sqrt[d*Sec[e + f*x]]) - (a
*b*Cot[e + f*x]*EllipticPi[-(b/Sqrt[a^2 + b^2]), ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*(Sec[e + f*x]^2)^(1/4)*Sq
rt[-Tan[e + f*x]^2])/((a^2 + b^2)^(3/2)*f*Sqrt[d*Sec[e + f*x]]) + (a*b*Cot[e + f*x]*EllipticPi[b/Sqrt[a^2 + b^
2], ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*(Sec[e + f*x]^2)^(1/4)*Sqrt[-Tan[e + f*x]^2])/((a^2 + b^2)^(3/2)*f*Sqr
t[d*Sec[e + f*x]]) + (2*(b + a*Tan[e + f*x]))/((a^2 + b^2)*f*Sqrt[d*Sec[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 755

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(a*e + c*d*x)*
((a + c*x^2)^(p + 1)/(2*a*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^
m*Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*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] && LtQ[p, -1] && 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 {1}{\sqrt {d \sec (e+f x)} (a+b \tan (e+f x))} \, dx &=\frac {\sqrt [4]{\sec ^2(e+f x)} \text {Subst}\left (\int \frac {1}{(a+x) \left (1+\frac {x^2}{b^2}\right )^{5/4}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt {d \sec (e+f x)}}\\ &=\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {\left (2 b \sqrt [4]{\sec ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {\frac {1}{2} \left (-1+\frac {a^2}{b^2}\right )+\frac {a x}{2 b^2}}{(a+x) \sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}\\ &=\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {\left (a \sqrt [4]{\sec ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{b \left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}+\frac {\left (b \sqrt [4]{\sec ^2(e+f x)}\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 )}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}\\ &=-\frac {2 a \tan (e+f x)}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}+\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}+\frac {\left (a \sqrt [4]{\sec ^2(e+f x)}\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 )}{b \left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {\left (b \sqrt [4]{\sec ^2(e+f x)}\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 )}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}+\frac {\left (a b \sqrt [4]{\sec ^2(e+f x)}\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 )}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}\\ &=\frac {2 a E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {2 a \tan (e+f x)}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}+\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {\left (b \sqrt [4]{\sec ^2(e+f x)}\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 )}{2 \left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}+\frac {\left (2 a \cot (e+f x) \sqrt [4]{\sec ^2(e+f x)} \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 )}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}\\ &=\frac {2 a E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {2 a \tan (e+f x)}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}+\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {\left (2 b^3 \sqrt [4]{\sec ^2(e+f x)}\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 )}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}+\frac {\left (a b \cot (e+f x) \sqrt [4]{\sec ^2(e+f x)} \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 )}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {\left (a b \cot (e+f x) \sqrt [4]{\sec ^2(e+f x)} \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 )}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}\\ &=\frac {2 a E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {2 a \tan (e+f x)}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}+\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {\left (b^2 \sqrt [4]{\sec ^2(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 )}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}+\frac {\left (b^2 \sqrt [4]{\sec ^2(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 )}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}+\frac {\left (a b \cot (e+f x) \sqrt [4]{\sec ^2(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 (\sqrt {a^2+b^2}-b x^2\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {\left (a b \cot (e+f x) \sqrt [4]{\sec ^2(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 (\sqrt {a^2+b^2}+b x^2\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}\\ &=\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right )^{5/4} f \sqrt {d \sec (e+f x)}}-\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right )^{5/4} f \sqrt {d \sec (e+f x)}}+\frac {2 a E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {2 a \tan (e+f x)}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}-\frac {a b \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 [4]{\sec ^2(e+f x)} \sqrt {-\tan ^2(e+f x)}}{\left (a^2+b^2\right )^{3/2} f \sqrt {d \sec (e+f x)}}+\frac {a b \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 [4]{\sec ^2(e+f x)} \sqrt {-\tan ^2(e+f x)}}{\left (a^2+b^2\right )^{3/2} f \sqrt {d \sec (e+f x)}}+\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(4497\) vs. \(2(451)=902\).
time = 69.39, size = 4497, normalized size = 9.97 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])),x]

[Out]

-1/2*(Sec[e + f*x]^(3/2)*(a*Cos[e + f*x] + b*Sin[e + f*x])*(Sqrt[Sec[e + f*x]]/(2*(a*Cos[e + f*x] + b*Sin[e +
f*x])) + (Cos[2*(e + f*x)]*Sqrt[Sec[e + f*x]])/(2*(a*Cos[e + f*x] + b*Sin[e + f*x])))*Sqrt[(1 - Tan[(e + f*x)/
2]^2)^(-1)]*(-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 - (b^(3/2)*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*Sqr
t[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[a^2 + b*(b - Sqrt[a^2 + b^2])] - (b^(3/2)*Sqrt[b + Sqrt[a^2 + b^2]]*ArcTan[(2*b*(b + Sq
rt[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[a^2 + b*(b + S
qrt[a^2 + b^2])] - 4*a*EllipticE[ArcSin[Tan[(e + f*x)/2]], -1]*Sqrt[1 - Tan[(e + f*x)/2]^4] + (4*(a^2 + b^2)*E
llipticF[ArcSin[Tan[(e + f*x)/2]], -1]*Sqrt[1 - Tan[(e + f*x)/2]^4])/a - (4*b^2*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])/a - (4*b^2*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])/a))/((a
^2 + b^2)*f*Sqrt[d*Sec[e + f*x]]*Sqrt[1 + Tan[(e + f*x)/2]^2]*((Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]*Sqrt[(1 -
Tan[(e + f*x)/2]^2)^(-1)]*(-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 - (b^
(3/2)*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[a^2 + b*(b - Sqrt[a^2 + b^2])] - (b^(3/2)*Sqrt[b + Sqrt[a^2 + b^2]]*ArcTa
n[(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[
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] + (4
*(a^2 + b^2)*EllipticF[ArcSin[Tan[(e + f*x)/2]], -1]*Sqrt[1 - Tan[(e + f*x)/2]^4])/a - (4*b^2*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])/a - (4*b^2
*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])/a))/(4*(a^2 + b^2)*(1 + Tan[(e + f*x)/2]^2)^(3/2)) - (Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]*((1 - Tan[(e
+ f*x)/2]^2)^(-1))^(3/2)*(-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 - (b^(
3/2)*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[a^2 + b*(b - Sqrt[a^2 + b^2])] - (b^(3/2)*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[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] + (4*
(a^2 + b^2)*EllipticF[ArcSin[Tan[(e + f*x)/2]], -1]*Sqrt[1 - Tan[(e + f*x)/2]^4])/a - (4*b^2*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])/a - (4*b^2*
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])/a))/(4*(a^2 + b^2)*Sqrt[1 + Tan[(e + f*x)/2]^2]) - (Sqrt[(1 - Tan[(e + f*x)/2]^2)^(-1)]*(-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 + (b^(3/2)*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[a^2 + b*(b - Sqrt[a^2 + b^2])]*Sqrt[1 - Tan[(e + f*x)/2]^4]) + (b^(3/2)*S
qrt[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[a^2 + b*(b + Sqrt[a^2 + b^2])]*Sqrt[1 - Tan[(e + f*x)/2]^4]) + (4*a*El
lipticE[ArcSin[Tan[(e + f*x)/2]], -1]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]^3)/Sqrt[1 - Tan[(e + f*x)/2]^4] - (4
*(a^2 + b^2)*EllipticF[ArcSin[Tan[(e + f*x)/2]], -1]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]^3)/(a*Sqrt[1 - Tan[(e
+ f*x)/2]^4]) + (4*b^2*EllipticPi[a^2/(a^2 + 2*b^2 - 2*Sqrt[b^2*(a^2 + b^2)]), ArcSin[Tan[(e + f*x)/2]], -1]*
Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]^3)/(a*Sqrt[...

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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. 8824 vs. \(2 (418 ) = 836\).
time = 1.37, size = 8825, normalized size = 19.57

method	result	size
default	\(\text {Expression too large to display}\)	\(8825\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*sec(f*x+e))^(1/2)/(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(1/(d*sec(f*x+e))^(1/2)/(a+b*tan(f*x+e)),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(d*sec(f*x + e))*(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(1/(d*sec(f*x+e))^(1/2)/(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 {1}{\sqrt {d \sec {\left (e + f x \right )}} \left (a + b \tan {\left (e + f x \right )}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))**(1/2)/(a+b*tan(f*x+e)),x)

[Out]

Integral(1/(sqrt(d*sec(e + f*x))*(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(1/(d*sec(f*x+e))^(1/2)/(a+b*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate(1/(sqrt(d*sec(f*x + e))*(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 {1}{\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}\,\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d/cos(e + f*x))^(1/2)*(a + b*tan(e + f*x))),x)

[Out]

int(1/((d/cos(e + f*x))^(1/2)*(a + b*tan(e + f*x))), x)

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