\(\int \frac {\sqrt {e \tan (c+d x)}}{a+b \sec (c+d x)} \, dx\) [314]

Optimal result

Integrand size = 25, antiderivative size = 415 \[ \int \frac {\sqrt {e \tan (c+d x)}}{a+b \sec (c+d x)} \, dx=-\frac {\sqrt {e} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {\sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {2 \sqrt {2} b \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (-\frac {\sqrt {a-b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right ),-1\right ) \sqrt {e \tan (c+d x)}}{a \sqrt {a-b} \sqrt {a+b} d \sqrt {\sin (c+d x)}}-\frac {2 \sqrt {2} b \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {\sqrt {a-b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right ),-1\right ) \sqrt {e \tan (c+d x)}}{a \sqrt {a-b} \sqrt {a+b} d \sqrt {\sin (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 415, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3975, 3557, 335, 303, 1176, 631, 210, 1179, 642, 2812, 2809, 2985, 2984, 504, 1227, 551} \[ \int \frac {\sqrt {e \tan (c+d x)}}{a+b \sec (c+d x)} \, dx=\frac {2 \sqrt {2} b \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \operatorname {EllipticPi}\left (-\frac {\sqrt {a-b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}\right ),-1\right )}{a d \sqrt {a-b} \sqrt {a+b} \sqrt {\sin (c+d x)}}-\frac {2 \sqrt {2} b \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \operatorname {EllipticPi}\left (\frac {\sqrt {a-b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}\right ),-1\right )}{a d \sqrt {a-b} \sqrt {a+b} \sqrt {\sin (c+d x)}}-\frac {\sqrt {e} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}-\frac {\sqrt {e} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d} \]

[In]

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

[Out]

-((Sqrt[e]*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a*d)) + (Sqrt[e]*ArcTan[1 + (Sqrt[2]*S
qrt[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*Ta
n[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) + (2*Sqrt[2]*b*Sqrt[Cos[c + d*x]]*EllipticPi[-(Sqrt[a - b]/Sqrt[a + b]), ArcSin[Sqrt[Sin[c + d*
x]]/Sqrt[1 + Cos[c + d*x]]], -1]*Sqrt[e*Tan[c + d*x]])/(a*Sqrt[a - b]*Sqrt[a + b]*d*Sqrt[Sin[c + d*x]]) - (2*S
qrt[2]*b*Sqrt[Cos[c + d*x]]*EllipticPi[Sqrt[a - b]/Sqrt[a + b], ArcSin[Sqrt[Sin[c + d*x]]/Sqrt[1 + Cos[c + d*x
]]], -1]*Sqrt[e*Tan[c + d*x]])/(a*Sqrt[a - b]*Sqrt[a + b]*d*Sqrt[Sin[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 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{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 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 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 2809

Int[1/(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(g_)*tan[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[
e + f*x]]/(Sqrt[Cos[e + f*x]]*Sqrt[g*Tan[e + f*x]]), Int[Sqrt[Cos[e + f*x]]/(Sqrt[Sin[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 2984

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

Rule 2985

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[(d_)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x
_)])), x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]], Int[Sqrt[g*Cos[e + f*x]]/(Sqrt[Sin[e + f*x]]
*(a + b*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f, g}, 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 3975

Int[Sqrt[cot[(c_.) + (d_.)*(x_)]*(e_.)]/(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/a, Int[Sqr
t[e*Cot[c + d*x]], x], x] - Dist[b/a, Int[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 \sqrt {e \tan (c+d x)} \, dx}{a}-\frac {b \int \frac {\sqrt {e \tan (c+d x)}}{b+a \cos (c+d x)} \, dx}{a} \\ & = \frac {e \text {Subst}\left (\int \frac {\sqrt {x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a d}-\frac {\left (b \sqrt {e \cot (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \frac {1}{(b+a \cos (c+d x)) \sqrt {e \cot (c+d x)}} \, dx}{a} \\ & = \frac {(2 e) \text {Subst}\left (\int \frac {x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}-\frac {\left (b \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \frac {\sqrt {\sin (c+d x)}}{\sqrt {-\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a \sqrt {\sin (c+d x)}} \\ & = -\frac {e \text {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}+\frac {e \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 (b \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \, dx}{a \sqrt {\sin (c+d x)}} \\ & = \frac {\sqrt {e} \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}+\frac {\sqrt {e} \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}+\frac {e \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 {e \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 {\left (4 \sqrt {2} b \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-x^4} \left (a+b+(-a+b) x^4\right )} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{a d \sqrt {\sin (c+d x)}} \\ & = \frac {\sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {\sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {\sqrt {e} \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}-\frac {\sqrt {e} \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}-\frac {\left (2 \sqrt {2} b \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {a+b}-\sqrt {a-b} x^2\right ) \sqrt {1-x^4}} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{a \sqrt {a-b} d \sqrt {\sin (c+d x)}}+\frac {\left (2 \sqrt {2} b \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {a+b}+\sqrt {a-b} x^2\right ) \sqrt {1-x^4}} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{a \sqrt {a-b} d \sqrt {\sin (c+d x)}} \\ & = -\frac {\sqrt {e} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {\sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {\left (2 \sqrt {2} b \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {a+b}-\sqrt {a-b} x^2\right )} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{a \sqrt {a-b} d \sqrt {\sin (c+d x)}}+\frac {\left (2 \sqrt {2} b \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {a+b}+\sqrt {a-b} x^2\right )} \, dx,x,\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )}{a \sqrt {a-b} d \sqrt {\sin (c+d x)}} \\ & = -\frac {\sqrt {e} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {\sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {2 \sqrt {2} b \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (-\frac {\sqrt {a-b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right ),-1\right ) \sqrt {e \tan (c+d x)}}{a \sqrt {a-b} \sqrt {a+b} d \sqrt {\sin (c+d x)}}-\frac {2 \sqrt {2} b \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {\sqrt {a-b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right ),-1\right ) \sqrt {e \tan (c+d x)}}{a \sqrt {a-b} \sqrt {a+b} d \sqrt {\sin (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 = 5.98 (sec) , antiderivative size = 646, normalized size of antiderivative = 1.56 \[ \int \frac {\sqrt {e \tan (c+d x)}}{a+b \sec (c+d x)} \, dx=-\frac {\cos (c+d x) \left (a+b \sqrt {\sec ^2(c+d x)}\right ) \sqrt {e \tan (c+d x)} \left (6 \sqrt {2} \left (a^2-b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-6 \sqrt {2} a^2 \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+6 \sqrt {2} b^2 \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )-(6+6 i) \sqrt {b} \left (a^2-b^2\right )^{3/4} \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )+(6+6 i) \sqrt {b} \left (a^2-b^2\right )^{3/4} \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-3 \sqrt {2} a^2 \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+3 \sqrt {2} b^2 \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+3 \sqrt {2} a^2 \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-3 \sqrt {2} b^2 \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+(3+3 i) \sqrt {b} \left (a^2-b^2\right )^{3/4} \log \left (\sqrt {a^2-b^2}-(1+i) \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\tan (c+d x)}+i b \tan (c+d x)\right )-(3+3 i) \sqrt {b} \left (a^2-b^2\right )^{3/4} \log \left (\sqrt {a^2-b^2}+(1+i) \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\tan (c+d x)}+i b \tan (c+d x)\right )+8 a b \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},-\tan ^2(c+d x),\frac {b^2 \tan ^2(c+d x)}{a^2-b^2}\right ) \tan ^{\frac {3}{2}}(c+d x)\right )}{12 a \left (a^2-b^2\right ) d (b+a \cos (c+d x)) \sqrt {\tan (c+d x)}} \]

[In]

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

[Out]

-1/12*(Cos[c + d*x]*(a + b*Sqrt[Sec[c + d*x]^2])*Sqrt[e*Tan[c + d*x]]*(6*Sqrt[2]*(a^2 - b^2)*ArcTan[1 - Sqrt[2
]*Sqrt[Tan[c + d*x]]] - 6*Sqrt[2]*a^2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + 6*Sqrt[2]*b^2*ArcTan[1 + Sqrt[2
]*Sqrt[Tan[c + d*x]]] - (6 + 6*I)*Sqrt[b]*(a^2 - b^2)^(3/4)*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a
^2 - b^2)^(1/4)] + (6 + 6*I)*Sqrt[b]*(a^2 - b^2)^(3/4)*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Tan[c + d*x]])/(a^2 -
b^2)^(1/4)] - 3*Sqrt[2]*a^2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + 3*Sqrt[2]*b^2*Log[1 - Sqrt[2]
*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + 3*Sqrt[2]*a^2*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - 3*Sqr
t[2]*b^2*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + (3 + 3*I)*Sqrt[b]*(a^2 - b^2)^(3/4)*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]] - (3 + 3*I)*Sqrt[b]*(a^2 -
b^2)^(3/4)*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]] + 8*
a*b*AppellF1[3/4, 1/2, 1, 7/4, -Tan[c + d*x]^2, (b^2*Tan[c + d*x]^2)/(a^2 - b^2)]*Tan[c + d*x]^(3/2)))/(a*(a^2
- b^2)*d*(b + a*Cos[c + d*x])*Sqrt[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. 819 vs. \(2 (328 ) = 656\).

Time = 2.54 (sec) , antiderivative size = 820, normalized size of antiderivative = 1.98

method	result	size
default	\(\text {Expression too large to display}\)	\(820\)

[In]

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

[Out]

1/d*2^(1/2)*b/((a^2-b^2)^(1/2)-a+b)/((a^2-b^2)^(1/2)+a-b)/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-I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*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+I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*
2^(1/2))*b-EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*a+EllipticPi((csc(d*x+c)-cot(d*x+
c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*b-EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*a+Ellip
ticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*b-(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^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))+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*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))+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*Ell
ipticPi((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)))*(cot(d*x+c)-csc(d*x+c)
)^(1/2)*(2-2*csc(d*x+c)+2*cot(d*x+c))^(1/2)*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(-e/((1-cos(d*x+c))^2*csc(d*x+c)^2
-1)*(-cot(d*x+c)+csc(d*x+c)))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)/((1-cos(d*x+c))^3*csc(d*x+c)^3+cot(d*x+c
)-csc(d*x+c))^(1/2)/((1-cos(d*x+c))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*csc(d*x+c))^(1/2)

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((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)