\(\int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx\) [315]

Optimal result

Integrand size = 25, antiderivative size = 422 \[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx=-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d \sqrt {e}}+\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d \sqrt {e}}-\frac {\log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d \sqrt {e}}+\frac {\log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d \sqrt {e}}-\frac {2 \sqrt {2} b \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \sqrt {a^2-b^2} d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} b \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \sqrt {a^2-b^2} d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}} \]

[Out]

-1/2*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a/d*2^(1/2)/e^(1/2)+1/2*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1
/2)/e^(1/2))/a/d*2^(1/2)/e^(1/2)-1/4*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a/d*2^(1/2)/e
^(1/2)+1/4*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a/d*2^(1/2)/e^(1/2)-2*b*EllipticPi((-co
s(d*x+c))^(1/2)/(1+sin(d*x+c))^(1/2),b/(a-(a^2-b^2)^(1/2)),I)*2^(1/2)*sin(d*x+c)^(1/2)/a/d/(a^2-b^2)^(1/2)/(-c
os(d*x+c))^(1/2)/(e*tan(d*x+c))^(1/2)+2*b*EllipticPi((-cos(d*x+c))^(1/2)/(1+sin(d*x+c))^(1/2),b/(a+(a^2-b^2)^(
1/2)),I)*2^(1/2)*sin(d*x+c)^(1/2)/a/d/(a^2-b^2)^(1/2)/(-cos(d*x+c))^(1/2)/(e*tan(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 422, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3977, 3557, 335, 217, 1179, 642, 1176, 631, 210, 2812, 2808, 2986, 1227, 551} \[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx=-\frac {2 \sqrt {2} b \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{a d \sqrt {a^2-b^2} \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} b \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right ),-1\right )}{a d \sqrt {a^2-b^2} \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d \sqrt {e}}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d \sqrt {e}}-\frac {\log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d \sqrt {e}}+\frac {\log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d \sqrt {e}} \]

[In]

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

[Out]

-(ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]]/(Sqrt[2]*a*d*Sqrt[e])) + ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c
+ d*x]])/Sqrt[e]]/(Sqrt[2]*a*d*Sqrt[e]) - Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c + d*x]]]/
(2*Sqrt[2]*a*d*Sqrt[e]) + Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]]/(2*Sqrt[2]*a*d*Sq
rt[e]) - (2*Sqrt[2]*b*EllipticPi[b/(a - Sqrt[a^2 - b^2]), ArcSin[Sqrt[-Cos[c + d*x]]/Sqrt[1 + Sin[c + d*x]]],
-1]*Sqrt[Sin[c + d*x]])/(a*Sqrt[a^2 - b^2]*d*Sqrt[-Cos[c + d*x]]*Sqrt[e*Tan[c + d*x]]) + (2*Sqrt[2]*b*Elliptic
Pi[b/(a + Sqrt[a^2 - b^2]), ArcSin[Sqrt[-Cos[c + d*x]]/Sqrt[1 + Sin[c + d*x]]], -1]*Sqrt[Sin[c + d*x]])/(a*Sqr
t[a^2 - b^2]*d*Sqrt[-Cos[c + d*x]]*Sqrt[e*Tan[c + d*x]])

Rule 210

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

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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 2808

Int[Sqrt[(g_.)*tan[(e_.) + (f_.)*(x_)]]/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[Sqrt[Cos[e +
f*x]]*(Sqrt[g*Tan[e + f*x]]/Sqrt[Sin[e + f*x]]), Int[Sqrt[Sin[e + f*x]]/(Sqrt[Cos[e + f*x]]*(a + b*Sin[e + f*
x])), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2812

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[g^(2*
IntPart[p])*(g*Cot[e + f*x])^FracPart[p]*(g*Tan[e + f*x])^FracPart[p], Int[(a + b*Sin[e + f*x])^m/(g*Tan[e + f
*x])^p, x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&  !IntegerQ[p]

Rule 2986

Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]))
, x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Dist[2*Sqrt[2]*d*((b + q)/(f*q)), Subst[Int[1/((d*(b + q) + a*x^2
)*Sqrt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]], x] - Dist[2*Sqrt[2]*d*((b - q)/(f*q
)), Subst[Int[1/((d*(b - q) + a*x^2)*Sqrt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]],
x]] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3977

Int[1/(Sqrt[cot[(c_.) + (d_.)*(x_)]*(e_.)]*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))), x_Symbol] :> Dist[1/a, Int
[1/Sqrt[e*Cot[c + d*x]], x], x] - Dist[b/a, Int[1/(Sqrt[e*Cot[c + d*x]]*(b + a*Sin[c + d*x])), x], x] /; FreeQ
[{a, b, c, d, e}, x] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {1}{\sqrt {e \tan (c+d x)}} \, dx}{a}-\frac {b \int \frac {1}{(b+a \cos (c+d x)) \sqrt {e \tan (c+d x)}} \, dx}{a} \\ & = \frac {e \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (e^2+x^2\right )} \, dx,x,e \tan (c+d x)\right )}{a d}-\frac {b \int \frac {\sqrt {e \cot (c+d x)}}{b+a \cos (c+d x)} \, dx}{a \sqrt {e \cot (c+d x)} \sqrt {e \tan (c+d x)}} \\ & = \frac {(2 e) \text {Subst}\left (\int \frac {1}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}-\frac {\left (b \sqrt {\sin (c+d x)}\right ) \int \frac {\sqrt {-\cos (c+d x)}}{(b+a \cos (c+d x)) \sqrt {\sin (c+d x)}} \, dx}{a \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}} \\ & = \frac {\text {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}+\frac {\text {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}-\frac {\left (2 \sqrt {2} b \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\left (-a+\sqrt {a^2-b^2}+b x^2\right ) \sqrt {1-x^4}} \, dx,x,\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right )}{a d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}-\frac {\left (2 \sqrt {2} b \left (1+\frac {a}{\sqrt {a^2-b^2}}\right ) \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\left (-a-\sqrt {a^2-b^2}+b x^2\right ) \sqrt {1-x^4}} \, dx,x,\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right )}{a d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}} \\ & = \frac {\text {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a d}+\frac {\text {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a d}-\frac {\text {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}+2 x}{-e-\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d \sqrt {e}}-\frac {\text {Subst}\left (\int \frac {\sqrt {2} \sqrt {e}-2 x}{-e+\sqrt {2} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d \sqrt {e}}-\frac {\left (2 \sqrt {2} b \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (-a+\sqrt {a^2-b^2}+b x^2\right )} \, dx,x,\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right )}{a d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}-\frac {\left (2 \sqrt {2} b \left (1+\frac {a}{\sqrt {a^2-b^2}}\right ) \sqrt {\sin (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (-a-\sqrt {a^2-b^2}+b x^2\right )} \, dx,x,\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right )}{a d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}} \\ & = -\frac {\log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d \sqrt {e}}+\frac {\log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d \sqrt {e}}-\frac {2 \sqrt {2} b \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \sqrt {a^2-b^2} d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} b \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \sqrt {a^2-b^2} d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d \sqrt {e}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d \sqrt {e}} \\ & = -\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d \sqrt {e}}+\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d \sqrt {e}}-\frac {\log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d \sqrt {e}}+\frac {\log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d \sqrt {e}}-\frac {2 \sqrt {2} b \operatorname {EllipticPi}\left (\frac {b}{a-\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \sqrt {a^2-b^2} d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} b \operatorname {EllipticPi}\left (\frac {b}{a+\sqrt {a^2-b^2}},\arcsin \left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right ),-1\right ) \sqrt {\sin (c+d x)}}{a \sqrt {a^2-b^2} d \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 38.30 (sec) , antiderivative size = 1475, normalized size of antiderivative = 3.50 \[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx=\frac {(b+a \cos (c+d x)) \sec (c+d x) \sqrt {\tan (c+d x)} \left (\frac {2 \sec ^3(c+d x) \left (a+b \sqrt {1+\tan ^2(c+d x)}\right ) \left (-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) a \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+\log \left (\sqrt {a^2-b^2}-(1+i) \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\tan (c+d x)}+i b \tan (c+d x)\right )-\log \left (\sqrt {a^2-b^2}+(1+i) \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\tan (c+d x)}+i b \tan (c+d x)\right )\right )}{\sqrt {b} \left (a^2-b^2\right )^{3/4}}+\frac {5 b \left (-a^2+b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},-\tan ^2(c+d x),\frac {b^2 \tan ^2(c+d x)}{a^2-b^2}\right ) \sqrt {\tan (c+d x)} \sqrt {1+\tan ^2(c+d x)}}{\left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},-\tan ^2(c+d x),\frac {b^2 \tan ^2(c+d x)}{a^2-b^2}\right )+2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},-\frac {1}{2},2,\frac {9}{4},-\tan ^2(c+d x),\frac {b^2 \tan ^2(c+d x)}{a^2-b^2}\right )+\left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2(c+d x),\frac {b^2 \tan ^2(c+d x)}{a^2-b^2}\right )\right ) \tan ^2(c+d x)\right ) \left (a^2-b^2 \left (1+\tan ^2(c+d x)\right )\right )}\right )}{(b+a \cos (c+d x)) \left (1+\tan ^2(c+d x)\right )^2}+\frac {\cos (2 (c+d x)) \sec ^3(c+d x) \left (a+b \sqrt {1+\tan ^2(c+d x)}\right ) \left (-\frac {20 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{a}+\frac {20 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{a}+\frac {(10-10 i) \left (a^2-2 b^2\right ) \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )}{a \sqrt {b} \left (a^2-b^2\right )^{3/4}}-\frac {(10-10 i) \left (a^2-2 b^2\right ) \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )}{a \sqrt {b} \left (a^2-b^2\right )^{3/4}}-\frac {10 \sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{a}+\frac {10 \sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{a}+\frac {(5-5 i) \left (a^2-2 b^2\right ) \log \left (\sqrt {a^2-b^2}-(1+i) \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\tan (c+d x)}+i b \tan (c+d x)\right )}{a \sqrt {b} \left (a^2-b^2\right )^{3/4}}-\frac {(5-5 i) \left (a^2-2 b^2\right ) \log \left (\sqrt {a^2-b^2}+(1+i) \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\tan (c+d x)}+i b \tan (c+d x)\right )}{a \sqrt {b} \left (a^2-b^2\right )^{3/4}}-\frac {8 b \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2(c+d x),\frac {b^2 \tan ^2(c+d x)}{a^2-b^2}\right ) \tan ^{\frac {5}{2}}(c+d x)}{-a^2+b^2}-\frac {200 b \left (-a^2+b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},-\tan ^2(c+d x),\frac {b^2 \tan ^2(c+d x)}{a^2-b^2}\right ) \sqrt {\tan (c+d x)}}{\sqrt {1+\tan ^2(c+d x)} \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},-\tan ^2(c+d x),\frac {b^2 \tan ^2(c+d x)}{a^2-b^2}\right )+2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},-\tan ^2(c+d x),\frac {b^2 \tan ^2(c+d x)}{a^2-b^2}\right )+\left (-a^2+b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},-\tan ^2(c+d x),\frac {b^2 \tan ^2(c+d x)}{a^2-b^2}\right )\right ) \tan ^2(c+d x)\right ) \left (-a^2+b^2 \left (1+\tan ^2(c+d x)\right )\right )}\right )}{20 (b+a \cos (c+d x)) \left (1-\tan ^2(c+d x)\right ) \left (1+\tan ^2(c+d x)\right )}\right )}{2 d (a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \]

[In]

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

[Out]

((b + a*Cos[c + d*x])*Sec[c + d*x]*Sqrt[Tan[c + d*x]]*((2*Sec[c + d*x]^3*(a + b*Sqrt[1 + Tan[c + d*x]^2])*(((-
1/8 + I/8)*a*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqr
t[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)] + Log[Sqrt[a^2 - b^2] - (1 + I)*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Tan
[c + d*x]] + I*b*Tan[c + d*x]] - Log[Sqrt[a^2 - b^2] + (1 + I)*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Tan[c + d*x]] +
I*b*Tan[c + d*x]]))/(Sqrt[b]*(a^2 - b^2)^(3/4)) + (5*b*(-a^2 + b^2)*AppellF1[1/4, -1/2, 1, 5/4, -Tan[c + d*x]^
2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Sqrt[Tan[c + d*x]]*Sqrt[1 + Tan[c + d*x]^2])/((5*(a^2 - b^2)*AppellF1[1/4
, -1/2, 1, 5/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + 2*(2*b^2*AppellF1[5/4, -1/2, 2, 9/4, -Tan
[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, 9/4, -Tan[c + d*x]^2, (b^2*
Tan[c + d*x]^2)/(a^2 - b^2)])*Tan[c + d*x]^2)*(a^2 - b^2*(1 + Tan[c + d*x]^2)))))/((b + a*Cos[c + d*x])*(1 + T
an[c + d*x]^2)^2) + (Cos[2*(c + d*x)]*Sec[c + d*x]^3*(a + b*Sqrt[1 + Tan[c + d*x]^2])*((-20*Sqrt[2]*ArcTan[1 -
Sqrt[2]*Sqrt[Tan[c + d*x]]])/a + (20*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/a + ((10 - 10*I)*(a^2 -
2*b^2)*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) - ((1
0 - 10*I)*(a^2 - 2*b^2)*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 - b^2)^(1/4)])/(a*Sqrt[b]*(a^2 -
b^2)^(3/4)) - (10*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/a + (10*Sqrt[2]*Log[1 + Sqrt[2]*
Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/a + ((5 - 5*I)*(a^2 - 2*b^2)*Log[Sqrt[a^2 - b^2] - (1 + I)*Sqrt[b]*(a^2 -
b^2)^(1/4)*Sqrt[Tan[c + d*x]] + I*b*Tan[c + d*x]])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) - ((5 - 5*I)*(a^2 - 2*b^2)*Lo
g[Sqrt[a^2 - b^2] + (1 + I)*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Tan[c + d*x]] + I*b*Tan[c + d*x]])/(a*Sqrt[b]*(a^2
- b^2)^(3/4)) - (8*b*AppellF1[5/4, 1/2, 1, 9/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Tan[c + d*x
]^(5/2))/(-a^2 + b^2) - (200*b*(-a^2 + b^2)*AppellF1[1/4, 1/2, 1, 5/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(
a^2 - b^2)]*Sqrt[Tan[c + d*x]])/(Sqrt[1 + Tan[c + d*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, -Tan[c + d
*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)] + 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, -Tan[c + d*x]^2, (b^2*Tan[c + d
*x]^2)/(a^2 - b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2
)])*Tan[c + d*x]^2)*(-a^2 + b^2*(1 + Tan[c + d*x]^2)))))/(20*(b + a*Cos[c + d*x])*(1 - Tan[c + d*x]^2)*(1 + Ta
n[c + d*x]^2))))/(2*d*(a + b*Sec[c + d*x])*Sqrt[e*Tan[c + d*x]])

Maple [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1918 vs. \(2 (343 ) = 686\).

Time = 3.58 (sec) , antiderivative size = 1919, normalized size of antiderivative = 4.55

method	result	size
default	\(\text {Expression too large to display}\)	\(1919\)

[In]

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

[Out]

1/2/d*2^(1/2)/((a^2-b^2)^(1/2)-a+b)/((a^2-b^2)^(1/2)+a-b)/(a^2-b^2)^(1/2)/(a-b)/a*(3*I*EllipticPi((csc(d*x+c)-
cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*a^2*b-3*I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2
),1/2-1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*a*b^2-3*I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^
(1/2))*(a^2-b^2)^(1/2)*a^2*b+3*I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2-b^2)^(
1/2)*a*b^2-3*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*a*b^2+3*Ellipti
cPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*a^2*b-3*EllipticPi((csc(d*x+c)-cot(
d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*a*b^2+2*(a^2-b^2)^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+
c)+1)^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*a*b^2+2*(a^2-b^2)^(1/2)*EllipticPi((csc(d*x+c)-cot(d*
x+c)+1)^(1/2),-(a-b)/(-a+b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*a*b^2-I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2
),1/2-1/2*I,1/2*2^(1/2))*(a^2-b^2)^(3/2)*b-I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))
*(a^2-b^2)^(1/2)*a^3-I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2-b^2)^(3/2)*a-I*E
llipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*b^3+I*EllipticPi((csc(d*x+c)-
cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2-b^2)^(3/2)*a+I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-
1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*b^3+I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^
2-b^2)^(3/2)*b-2*EllipticF((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*(a^2-b^2)^(3/2)*a+2*EllipticF((csc(d*x
+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*(a^2-b^2)^(1/2)*a^3+EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,
1/2*2^(1/2))*(a^2-b^2)^(3/2)*a-EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2-b^2)^(3/
2)*b-EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*a^3+EllipticPi((csc(d*x
+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*b^3+EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),
1/2+1/2*I,1/2*2^(1/2))*(a^2-b^2)^(3/2)*a-EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^
2-b^2)^(3/2)*b-EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*a^3+EllipticP
i((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*b^3-2*(a^2-b^2)^(1/2)*EllipticPi((csc
(d*x+c)-cot(d*x+c)+1)^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b^3-2*(a^2-b^2)^(1/2)*EllipticPi((csc
(d*x+c)-cot(d*x+c)+1)^(1/2),-(a-b)/(-a+b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*b^3-2*EllipticPi((csc(d*x+c)-cot(d*
x+c)+1)^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*a^2*b^2+4*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2
),(a-b)/(a-b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*a*b^3+2*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),-(a-b)/(-a+b
+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*a^2*b^2-4*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),-(a-b)/(-a+b+((a-b)*(a
+b))^(1/2)),1/2*2^(1/2))*a*b^3+I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(a^2-b^2)^(
1/2)*a^3-4*EllipticF((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*(a^2-b^2)^(1/2)*a^2*b+2*EllipticF((csc(d*x+c
)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*(a^2-b^2)^(1/2)*a*b^2+3*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*
I,1/2*2^(1/2))*(a^2-b^2)^(1/2)*a^2*b+2*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),-(a-b)/(-a+b+((a-b)*(a+b))^(
1/2)),1/2*2^(1/2))*b^4-2*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2
))*b^4)*(cot(d*x+c)-csc(d*x+c))^(1/2)*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)/(e*tan(d
*x+c))^(1/2)*(1+sec(d*x+c))

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate(1/(a+b*sec(d*x+c))/(e*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx=\int \frac {1}{\sqrt {e \tan {\left (c + d x \right )}} \left (a + b \sec {\left (c + d x \right )}\right )}\, dx \]

[In]

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

[Out]

Integral(1/(sqrt(e*tan(c + d*x))*(a + b*sec(c + d*x))), x)

Maxima [F]

\[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \sqrt {e \tan \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

integrate(1/((b*sec(d*x + c) + a)*sqrt(e*tan(d*x + c))), x)

Giac [F]

\[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx=\int { \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \sqrt {e \tan \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

integrate(1/((b*sec(d*x + c) + a)*sqrt(e*tan(d*x + c))), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx=\int \frac {\cos \left (c+d\,x\right )}{\sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}\,\left (b+a\,\cos \left (c+d\,x\right )\right )} \,d x \]

[In]

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

[Out]

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