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

Optimal. Leaf size=422 \[ -\frac {2 \sqrt {2} b \sqrt {\sin (c+d x)} \Pi \left (\frac {b}{a-\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right )\right |-1\right )}{a d \sqrt {a^2-b^2} \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} b \sqrt {\sin (c+d x)} \Pi \left (\frac {b}{a+\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right )\right |-1\right )}{a d \sqrt {a^2-b^2} \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d \sqrt {e}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d \sqrt {e}}-\frac {\log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d \sqrt {e}}+\frac {\log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d \sqrt {e}} \]

[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)

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Rubi [A]  time = 0.56, antiderivative size = 422, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 14, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3892, 3476, 329, 211, 1165, 628, 1162, 617, 204, 2733, 2729, 2907, 1213, 537} \[ -\frac {2 \sqrt {2} b \sqrt {\sin (c+d x)} \Pi \left (\frac {b}{a-\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right )\right |-1\right )}{a d \sqrt {a^2-b^2} \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}+\frac {2 \sqrt {2} b \sqrt {\sin (c+d x)} \Pi \left (\frac {b}{a+\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {\sin (c+d x)+1}}\right )\right |-1\right )}{a d \sqrt {a^2-b^2} \sqrt {-\cos (c+d x)} \sqrt {e \tan (c+d x)}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d \sqrt {e}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d \sqrt {e}}-\frac {\log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d \sqrt {e}}+\frac {\log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d \sqrt {e}} \]

Antiderivative was successfully verified.

[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 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 211

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 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 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 2729

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 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 2907

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 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 3892

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*} \int \frac {1}{(a+b \sec (c+d x)) \sqrt {e \tan (c+d x)}} \, dx &=\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 \operatorname {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) \operatorname {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 {\operatorname {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}+\frac {\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 (2 \sqrt {2} b \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \sqrt {\sin (c+d x)}\right ) \operatorname {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 ) \operatorname {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 {\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 {\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 {\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 \sqrt {e}}-\frac {\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 \sqrt {e}}-\frac {\left (2 \sqrt {2} b \left (1-\frac {a}{\sqrt {a^2-b^2}}\right ) \sqrt {\sin (c+d x)}\right ) \operatorname {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 ) \operatorname {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 \Pi \left (\frac {b}{a-\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right )\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 \Pi \left (\frac {b}{a+\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right )\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 {\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 \sqrt {e}}-\frac {\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 \sqrt {e}}\\ &=-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d \sqrt {e}}+\frac {\tan ^{-1}\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 \Pi \left (\frac {b}{a-\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right )\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 \Pi \left (\frac {b}{a+\sqrt {a^2-b^2}};\left .\sin ^{-1}\left (\frac {\sqrt {-\cos (c+d x)}}{\sqrt {1+\sin (c+d x)}}\right )\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*}

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Mathematica [C]  time = 12.80, size = 246, normalized size = 0.58 \[ \frac {4 \left (\left (b^2-a^2\right ) \Pi \left (-i;\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {1}{2} (c+d x)\right )}}\right )\right |-1\right )+a^2 \left (-\Pi \left (i;\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {1}{2} (c+d x)\right )}}\right )\right |-1\right )\right )-b^2 \Pi \left (-\frac {\sqrt {a+b}}{\sqrt {a-b}};\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {1}{2} (c+d x)\right )}}\right )\right |-1\right )-b^2 \Pi \left (\frac {\sqrt {a+b}}{\sqrt {a-b}};\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {1}{2} (c+d x)\right )}}\right )\right |-1\right )+a (a-b) F\left (\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {1}{2} (c+d x)\right )}}\right )\right |-1\right )+b^2 \Pi \left (i;\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {1}{2} (c+d x)\right )}}\right )\right |-1\right )\right )}{a d \left (a^2-b^2\right ) \sqrt {\tan \left (\frac {1}{2} (c+d x)\right )-1} \sqrt {\cot \left (\frac {1}{2} (c+d x)\right )+1} \sqrt {e \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(a*(a - b)*EllipticF[ArcSin[1/Sqrt[Tan[(c + d*x)/2]]], -1] + (-a^2 + b^2)*EllipticPi[-I, ArcSin[1/Sqrt[Tan[
(c + d*x)/2]]], -1] - a^2*EllipticPi[I, ArcSin[1/Sqrt[Tan[(c + d*x)/2]]], -1] + b^2*EllipticPi[I, ArcSin[1/Sqr
t[Tan[(c + d*x)/2]]], -1] - b^2*EllipticPi[-(Sqrt[a + b]/Sqrt[a - b]), ArcSin[1/Sqrt[Tan[(c + d*x)/2]]], -1] -
b^2*EllipticPi[Sqrt[a + b]/Sqrt[a - b], ArcSin[1/Sqrt[Tan[(c + d*x)/2]]], -1]))/(a*(a^2 - b^2)*d*Sqrt[1 + Cot
[(c + d*x)/2]]*Sqrt[-1 + Tan[(c + d*x)/2]]*Sqrt[e*Tan[c + d*x]])

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \sqrt {e \tan \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 1.73, size = 2313, normalized size = 5.48 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*(2*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(a^2-b^2)^(3/2)*a-(a^2-b^2)^(3/2)
*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+(a^2-b^2)^(3/2)*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))*b-(a^2-b^2)^(3/2)*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+(a^2-b^2)^(3/2)*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))*b+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))*b^3+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))*b^3+(a^2-b^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^3-(a^2-b^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))*b^3+(a^2-b^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^3+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^2*b^2-4*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^3-2*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^2*b^2+4*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^3-2*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2)
)*(a^2-b^2)^(1/2)*a^3-(a^2-b^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))*b^3+4*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(a^2-b^2)^(1/2)*a^2*b-2*Ellip
ticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(a^2-b^2)^(1/2)*a*b^2-2*(a^2-b^2)^(1/2)*Ellipti
cPi(((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^2-I*(a^2-b^2
)^(3/2)*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))*b-I*(a^2-b^2)^(1/2)*Ell
ipticPi(((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^3-I*(a^2-b^2)^(3/2)*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+I*(a^2-b^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^3-I*(a^2-b^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))*b^3-2*(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))*a*b^2-3*(a^2-b^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^2*b+3*(a^2-b^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*b^2-3*(a^2-b^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^2*b+3*(a^2-b^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*b^2+I*(a^2-b^2)^(3/2)*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+I*(a^2-b^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))*b^3+I*(a^2-b^2)^(3/2)*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))*b+3*I*(a^2-b^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^
2*b-3*I*(a^2-b^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*b^2-3
*I*(a^2-b^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^2*b+3*I*(a
^2-b^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*b^2+2*EllipticP
i(((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))*b^4-2*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))*b^4)*((-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
)*(-1+cos(d*x+c))/sin(d*x+c)^2/cos(d*x+c)*(1+cos(d*x+c))^2/(e*sin(d*x+c)/cos(d*x+c))^(1/2)*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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sec \left (d x + c\right ) + a\right )} \sqrt {e \tan \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {e \tan {\left (c + d x \right )}} \left (a + b \sec {\left (c + d x \right )}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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