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

Optimal. Leaf size=415 \[ \frac {2 \sqrt {2} b \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \Pi \left (-\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}\right )\right |-1\right )}{a d \sqrt {a-b} \sqrt {a+b} \sqrt {\sin (c+d x)}}-\frac {2 \sqrt {2} b \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \Pi \left (\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}\right )\right |-1\right )}{a d \sqrt {a-b} \sqrt {a+b} \sqrt {\sin (c+d x)}}-\frac {\sqrt {e} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}-\frac {\sqrt {e} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d} \]

[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]  time = 0.70, antiderivative size = 415, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 16, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3890, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2733, 2730, 2906, 2905, 490, 1213, 537} \[ \frac {2 \sqrt {2} b \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \Pi \left (-\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}\right )\right |-1\right )}{a d \sqrt {a-b} \sqrt {a+b} \sqrt {\sin (c+d x)}}-\frac {2 \sqrt {2} b \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)} \Pi \left (\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {\cos (c+d x)+1}}\right )\right |-1\right )}{a d \sqrt {a-b} \sqrt {a+b} \sqrt {\sin (c+d x)}}-\frac {\sqrt {e} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}-\frac {\sqrt {e} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d} \]

Antiderivative was successfully verified.

[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 204

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

Rule 297

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 329

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 490

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), In
t[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 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), 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 617

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 628

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 1162

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 1165

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 1213

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 2730

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 2733

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 2905

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 2906

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 3476

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 3890

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*} \int \frac {\sqrt {e \tan (c+d x)}}{a+b \sec (c+d x)} \, dx &=\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 \operatorname {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) \operatorname {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 \operatorname {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}+\frac {e \operatorname {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} \operatorname {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} \operatorname {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 \operatorname {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 \operatorname {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 ) \operatorname {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} \operatorname {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} \operatorname {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 ) \operatorname {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 ) \operatorname {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} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \tan ^{-1}\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 ) \operatorname {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 ) \operatorname {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} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {\sqrt {e} \tan ^{-1}\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)} \Pi \left (-\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )\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)} \Pi \left (\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\frac {\sqrt {\sin (c+d x)}}{\sqrt {1+\cos (c+d x)}}\right )\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]  time = 5.48, size = 224, normalized size = 0.54 \[ -\frac {4 \sqrt {\tan \left (\frac {1}{2} (c+d x)\right )} \csc (c+d x) \sqrt {e \tan (c+d x)} (a \cos (c+d x)+b) \left (\frac {b \left (\Pi \left (\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\sqrt {\tan \left (\frac {1}{2} (c+d x)\right )}\right )\right |-1\right )-\Pi \left (-\frac {\sqrt {a-b}}{\sqrt {a+b}};\left .\sin ^{-1}\left (\sqrt {\tan \left (\frac {1}{2} (c+d x)\right )}\right )\right |-1\right )\right )}{\sqrt {a-b} \sqrt {a+b}}-i \Pi \left (-i;\left .\sin ^{-1}\left (\sqrt {\tan \left (\frac {1}{2} (c+d x)\right )}\right )\right |-1\right )+i \Pi \left (i;\left .\sin ^{-1}\left (\sqrt {\tan \left (\frac {1}{2} (c+d x)\right )}\right )\right |-1\right )\right )}{a d \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(b + a*Cos[c + d*x])*Csc[c + d*x]*((-I)*EllipticPi[-I, ArcSin[Sqrt[Tan[(c + d*x)/2]]], -1] + I*EllipticPi[
I, ArcSin[Sqrt[Tan[(c + d*x)/2]]], -1] + (b*(-EllipticPi[-(Sqrt[a - b]/Sqrt[a + b]), ArcSin[Sqrt[Tan[(c + d*x)
/2]]], -1] + EllipticPi[Sqrt[a - b]/Sqrt[a + b], ArcSin[Sqrt[Tan[(c + d*x)/2]]], -1]))/(Sqrt[a - b]*Sqrt[a + b
]))*Sqrt[Tan[(c + d*x)/2]]*Sqrt[e*Tan[c + d*x]])/(a*d*Sqrt[Cos[c + d*x]*Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x
]))

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e \tan \left (d x + c\right )}}{b \sec \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [B]  time = 1.75, size = 859, normalized size = 2.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c
))/sin(d*x+c))^(1/2)*(e*sin(d*x+c)/cos(d*x+c))^(1/2)*(1+cos(d*x+c))^2*(-1+cos(d*x+c))*(I*EllipticPi(((1-cos(d*
x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a-I*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c)
)^(1/2),1/2-1/2*I,1/2*2^(1/2))*b-I*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/
2))*a+I*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*b-(a^2-b^2)^(1/2)*Ellip
ticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))+(a^2-b^2)^(1/2
)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),-(a-b)/(-a+b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))+Ellip
ticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*a-b*EllipticPi
(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),(a-b)/(a-b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))+EllipticPi(((1-cos(
d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),-(a-b)/(-a+b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))*a-b*EllipticPi(((1-cos(d*x
+c)+sin(d*x+c))/sin(d*x+c))^(1/2),-(a-b)/(-a+b+((a-b)*(a+b))^(1/2)),1/2*2^(1/2))-EllipticPi(((1-cos(d*x+c)+sin
(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a+EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/
2-1/2*I,1/2*2^(1/2))*b-EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a+Ellipt
icPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*b)/sin(d*x+c)^3*2^(1/2)*b/((a^2-b^2)^
(1/2)-a+b)/((a^2-b^2)^(1/2)+a-b)/a

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e \tan \left (d x + c\right )}}{b \sec \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\cos \left (c+d\,x\right )\,\sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}}{b+a\,\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {e \tan {\left (c + d x \right )}}}{a + b \sec {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________