3.7.6 \(\int \frac {\sqrt {d \sec (e+f x)}}{a+b \tan (e+f x)} \, dx\) [606]

Optimal. Leaf size=324 \[ -\frac {\sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right )^{3/4} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt {d \sec (e+f x)}}{\left (a^2+b^2\right )^{3/4} f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a \cot (e+f x) \Pi \left (-\frac {b}{\sqrt {a^2+b^2}};\left .\text {ArcSin}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a \cot (e+f x) \Pi \left (\frac {b}{\sqrt {a^2+b^2}};\left .\text {ArcSin}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right ) \sqrt {d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}} \]

[Out]

-arctan((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*b^(1/2)*(d*sec(f*x+e))^(1/2)/(a^2+b^2)^(3/4)/f/(sec(f*x+
e)^2)^(1/4)-arctanh((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*b^(1/2)*(d*sec(f*x+e))^(1/2)/(a^2+b^2)^(3/4)
/f/(sec(f*x+e)^2)^(1/4)+a*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)/(a^2+b^2)/f/(sec(f*x+e)^2)^(1/4)+a*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)/(a^2+b^2)/f/(sec(f*x+e)^2)^(1/4)

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Rubi [A]
time = 0.24, antiderivative size = 324, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3593, 761, 410, 109, 418, 1227, 551, 455, 65, 218, 214, 211} \begin {gather*} \frac {a \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 )}{f \left (a^2+b^2\right ) \sqrt [4]{\sec ^2(e+f x)}}+\frac {a \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 )}{f \left (a^2+b^2\right ) \sqrt [4]{\sec ^2(e+f x)}}-\frac {\sqrt {b} \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 )}{f \left (a^2+b^2\right )^{3/4} \sqrt [4]{\sec ^2(e+f x)}}-\frac {\sqrt {b} \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 )}{f \left (a^2+b^2\right )^{3/4} \sqrt [4]{\sec ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[b]*ArcTan[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*Sqrt[d*Sec[e + f*x]])/((a^2 + b^2)^(3/4)
*f*(Sec[e + f*x]^2)^(1/4))) - (Sqrt[b]*ArcTanh[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*Sqrt[d*Sec[
e + f*x]])/((a^2 + b^2)^(3/4)*f*(Sec[e + f*x]^2)^(1/4)) + (a*Cot[e + f*x]*EllipticPi[-(b/Sqrt[a^2 + b^2]), Arc
Sin[(Sec[e + f*x]^2)^(1/4)], -1]*Sqrt[d*Sec[e + f*x]]*Sqrt[-Tan[e + f*x]^2])/((a^2 + b^2)*f*(Sec[e + f*x]^2)^(
1/4)) + (a*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])/((a^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 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 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 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 {\sqrt {d \sec (e+f x)}}{a+b \tan (e+f x)} \, dx &=\frac {\sqrt {d \sec (e+f x)} \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 {\sqrt {d \sec (e+f x)} \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 \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 {\sqrt {d \sec (e+f x)} \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 \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 {\left (2 b \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 \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 {\left (b \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 (b \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 \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 \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 {\sqrt {b} \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)}}{\left (a^2+b^2\right )^{3/4} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\sqrt {b} \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)}}{\left (a^2+b^2\right )^{3/4} f \sqrt [4]{\sec ^2(e+f x)}}+\frac {\left (a \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 \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 {\sqrt {b} \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)}}{\left (a^2+b^2\right )^{3/4} f \sqrt [4]{\sec ^2(e+f x)}}-\frac {\sqrt {b} \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)}}{\left (a^2+b^2\right )^{3/4} f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a \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)}}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}+\frac {a \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)}}{\left (a^2+b^2\right ) f \sqrt [4]{\sec ^2(e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 44.54, size = 4648, normalized size = 14.35 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2]^4]*Sqrt[d*Sec[e + f*x]]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(3/2)*(-1 + T
an[(e + f*x)/2]^2)*(EllipticF[ArcSin[Tan[(e + f*x)/2]], -1] + (((-2*I)*b*Sqrt[a^2 + b^2]*EllipticF[ArcSin[Sqrt
[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(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*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2]
+ a*(-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*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2])*Sqrt[I*Cos[e + f*x] - Sin[e + f*x]]*Sqrt[Cos[e +
f*x]*(Cos[e + f*x] + I*Sin[e + f*x])]*(I + Tan[(e + f*x)/2])^2)/(Sqrt[2]*(a - I*b)*Sqrt[a^2 + b^2]*Sqrt[Cos[e
+ f*x]*Sec[(e + f*x)/2]^4])))/(a*f*Sqrt[Sec[(e + f*x)/2]^2]*(a + b*Tan[e + f*x])*((-2*Sqrt[Sec[(e + f*x)/2]^2
]*Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2]^4]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(3/2)*Tan[(e + f*x)/2]*(EllipticF[Ar
cSin[Tan[(e + f*x)/2]], -1] + (((-2*I)*b*Sqrt[a^2 + b^2]*EllipticF[ArcSin[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*
x]]/Sqrt[2]], 2] + a*(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*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(-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*Cos[e + f*x]
+ Sin[e + f*x]]/Sqrt[2]], 2])*Sqrt[I*Cos[e + f*x] - Sin[e + f*x]]*Sqrt[Cos[e + f*x]*(Cos[e + f*x] + I*Sin[e +
f*x])]*(I + Tan[(e + f*x)/2])^2)/(Sqrt[2]*(a - I*b)*Sqrt[a^2 + b^2]*Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2]^4])))/a
+ (Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2]^4]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(3/2)*Tan[(e + f*x)/2]*(-1 + Tan[(
e + f*x)/2]^2)*(EllipticF[ArcSin[Tan[(e + f*x)/2]], -1] + (((-2*I)*b*Sqrt[a^2 + b^2]*EllipticF[ArcSin[Sqrt[1 -
I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(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*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*
(-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]), Arc
Sin[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2])*Sqrt[I*Cos[e + f*x] - Sin[e + f*x]]*Sqrt[Cos[e + f*x
]*(Cos[e + f*x] + I*Sin[e + f*x])]*(I + Tan[(e + f*x)/2])^2)/(Sqrt[2]*(a - I*b)*Sqrt[a^2 + b^2]*Sqrt[Cos[e + f
*x]*Sec[(e + f*x)/2]^4])))/(a*Sqrt[Sec[(e + f*x)/2]^2]) - ((Cos[(e + f*x)/2]^2*Sec[e + f*x])^(3/2)*(-(Sec[(e +
f*x)/2]^4*Sin[e + f*x]) + 2*Cos[e + f*x]*Sec[(e + f*x)/2]^4*Tan[(e + f*x)/2])*(-1 + Tan[(e + f*x)/2]^2)*(Elli
pticF[ArcSin[Tan[(e + f*x)/2]], -1] + (((-2*I)*b*Sqrt[a^2 + b^2]*EllipticF[ArcSin[Sqrt[1 - I*Cos[e + f*x] + Si
n[e + f*x]]/Sqrt[2]], 2] + a*(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*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(-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*Cos[e
+ f*x] + Sin[e + f*x]]/Sqrt[2]], 2])*Sqrt[I*Cos[e + f*x] - Sin[e + f*x]]*Sqrt[Cos[e + f*x]*(Cos[e + f*x] + I*
Sin[e + f*x])]*(I + Tan[(e + f*x)/2])^2)/(Sqrt[2]*(a - I*b)*Sqrt[a^2 + b^2]*Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2]
^4])))/(a*Sqrt[Sec[(e + f*x)/2]^2]*Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2]^4]) - (2*Sqrt[Cos[e + f*x]*Sec[(e + f*x)
/2]^4]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(3/2)*(-1 + Tan[(e + f*x)/2]^2)*((((-2*I)*b*Sqrt[a^2 + b^2]*EllipticF
[ArcSin[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(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*Cos[e + f*x] + Sin[e + f*x]]/S
qrt[2]], 2] + a*(-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*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2])*Sec[(e + f*x)/2]^2*Sqrt[I*Cos[e + f*
x] - Sin[e + f*x]]*Sqrt[Cos[e + f*x]*(Cos[e + f*x] + I*Sin[e + f*x])]*(I + Tan[(e + f*x)/2]))/(Sqrt[2]*(a - I*
b)*Sqrt[a^2 + b^2]*Sqrt[Cos[e + f*x]*Sec[(e + f*x)/2]^4]) + (((-2*I)*b*Sqrt[a^2 + b^2]*EllipticF[ArcSin[Sqrt[1
- I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(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*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] +
a*(-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]), A
rcSin[Sqrt[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2])*(-Cos[e + f*x] - I*Sin[e + f*x])*Sqrt[Cos[e + f*x]
*(Cos[e + f*x] + I*Sin[e + f*x])]*(I + Tan[(e + f*x)/2])^2)/(2*Sqrt[2]*(a - I*b)*Sqrt[a^2 + b^2]*Sqrt[Cos[e +
f*x]*Sec[(e + f*x)/2]^4]*Sqrt[I*Cos[e + f*x] - Sin[e + f*x]]) + (((-2*I)*b*Sqrt[a^2 + b^2]*EllipticF[ArcSin[Sq
rt[1 - I*Cos[e + f*x] + Sin[e + f*x]]/Sqrt[2]], 2] + a*(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^...

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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. 3130 vs. \(2 (276 ) = 552\).
time = 0.81, size = 3131, normalized size = 9.66

method	result	size
default	\(\text {Expression too large to display}\)	\(3131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sec(f*x+e))^(1/2)/(a+b*tan(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/2/f*(d/cos(f*x+e))^(1/2)*(cos(f*x+e)+1)^2*(cos(f*x+e)-1)*(-4*I*b*a^3*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(c
os(f*x+e)+1))^(1/2)*EllipticPi(I*(cos(f*x+e)-1)/sin(f*x+e),-1/(-b+(a^2+b^2)^(1/2))^2*a^2,I)*(b*((a^2+b^2)^(1/2
)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2
*b^3)/a^4)^(1/2)*(a^2+b^2)^(1/2)+4*I*b*a^3*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Elliptic
Pi(I*(cos(f*x+e)-1)/sin(f*x+e),-1/(b+(a^2+b^2)^(1/2))^2*a^2,I)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2
*a^2*b+2*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*(a^2+b^2)^(1
/2)-4*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(b
*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/
2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*a^5-4*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(
I*(cos(f*x+e)-1)/sin(f*x+e),I)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(-b*((a
^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*a^3*b^2+(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)
*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+
e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(-b*((
a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*a^4*b-(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*
ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+
1)^2)^(1/2)-1)/sin(f*x+e)^2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(-b*((a^2
+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*a^4*b-(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(
2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2
)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*a^4*b+(a^2+b^2)^(3/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(b*((a^2+b^2)
^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*arctanh(1/2*(cos(f*x+e)-1)*((a^2+b^2)^(1/2)*cos(f*x
+e)*b-cos(f*x+e)*a^2-cos(f*x+e)*b^2-b*(a^2+b^2)^(1/2)+b^2)/sin(f*x+e)^2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(
-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)/a^2)*b^2+(a^2+b^2)^(3/2)*(-cos(f*x+e)/
(cos(f*x+e)+1)^2)^(1/2)*arctanh(1/2*(cos(f*x+e)-1)*((a^2+b^2)^(1/2)*cos(f*x+e)*b+cos(f*x+e)*a^2+cos(f*x+e)*b^2
-b*(a^2+b^2)^(1/2)-b^2)/sin(f*x+e)^2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^
(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)/a^2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1
/2)*b^2-(a^2+b^2)^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a
^2*b+2*b^3)/a^4)^(1/2)*arctanh(1/2*(cos(f*x+e)-1)*((a^2+b^2)^(1/2)*cos(f*x+e)*b-cos(f*x+e)*a^2-cos(f*x+e)*b^2-
b*(a^2+b^2)^(1/2)+b^2)/sin(f*x+e)^2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^
(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)/a^2)*b^4-(a^2+b^2)^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*arctanh(1/2*
(cos(f*x+e)-1)*((a^2+b^2)^(1/2)*cos(f*x+e)*b+cos(f*x+e)*a^2+cos(f*x+e)*b^2-b*(a^2+b^2)^(1/2)-b^2)/sin(f*x+e)^2
/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)/
a^2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*b^4-(-cos(f*x+e)/(cos(f*x+e)+1)^
2)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*arctanh(1/2*(cos(f*x+e)-1)*((
a^2+b^2)^(1/2)*cos(f*x+e)*b-cos(f*x+e)*a^2-cos(f*x+e)*b^2-b*(a^2+b^2)^(1/2)+b^2)/sin(f*x+e)^2/(-cos(f*x+e)/(co
s(f*x+e)+1)^2)^(1/2)/(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)/a^2)*a^4*b-(-cos
(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*arctan
h(1/2*(cos(f*x+e)-1)*((a^2+b^2)^(1/2)*cos(f*x+e)*b-cos(f*x+e)*a^2-cos(f*x+e)*b^2-b*(a^2+b^2)^(1/2)+b^2)/sin(f*
x+e)^2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)
^(1/2)/a^2)*a^2*b^3+(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*arctanh(1/2*(cos(f*x+e)-1)*((a^2+b^2)^(1/2)*cos(f*x+e
)*b+cos(f*x+e)*a^2+cos(f*x+e)*b^2-b*(a^2+b^2)^(1/2)-b^2)/sin(f*x+e)^2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(b*
((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)/a^2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)
^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*a^4*b+(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*arctanh(1/2*(cos(f*x+e)-1)*((a
^2+b^2)^(1/2)*cos(f*x+e)*b+cos(f*x+e)*a^2+cos(f*x+e)*b^2-b*(a^2+b^2)^(1/2)-b^2)/sin(f*x+e)^2/(-cos(f*x+e)/(cos
(f*x+e)+1)^2)^(1/2)/(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)/a^2)*(-b*((a^2+b^2
)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*a^2*b^3)/sin(f*x+e)^2/(a^2+b^2)/(-b+(a^2+b^2)^(1/2
))/a^2/(b+(a^2+b^2)^(1/2))/(b*((a^2+b^2)^(1/2)*...

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

[Out]

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

[Out]

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

[Out]

integrate(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 {\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}}{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))^(1/2)/(a + b*tan(e + f*x)),x)

[Out]

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

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