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

Optimal. Leaf size=290 \[ -\frac{\sqrt{\sin (2 c+2 d x)} \sec (c+d x) \text{EllipticF}\left (c+d x-\frac{\pi }{4},2\right )}{3 a d \sqrt{e \tan (c+d x)}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a d \sqrt{e}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} a d \sqrt{e}}-\frac{\log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a d \sqrt{e}}+\frac{\log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a d \sqrt{e}}+\frac{2 e (1-\sec (c+d x))}{3 a d (e \tan (c+d x))^{3/2}} \]

[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*e*(1 - Sec[c + d*x]))/(3*a*d*(e*Tan[c + d*x])^(3/2)) - (EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*
Sqrt[Sin[2*c + 2*d*x]])/(3*a*d*Sqrt[e*Tan[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.345878, antiderivative size = 290, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 14, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.56, Rules used = {3888, 3882, 3884, 3476, 329, 211, 1165, 628, 1162, 617, 204, 2614, 2573, 2641} \[ -\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}\right )}{\sqrt{2} a d \sqrt{e}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{e \tan (c+d x)}}{\sqrt{e}}+1\right )}{\sqrt{2} a d \sqrt{e}}-\frac{\log \left (\sqrt{e} \tan (c+d x)-\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a d \sqrt{e}}+\frac{\log \left (\sqrt{e} \tan (c+d x)+\sqrt{2} \sqrt{e \tan (c+d x)}+\sqrt{e}\right )}{2 \sqrt{2} a d \sqrt{e}}+\frac{2 e (1-\sec (c+d x))}{3 a d (e \tan (c+d x))^{3/2}}-\frac{\sqrt{\sin (2 c+2 d x)} \sec (c+d x) F\left (\left .c+d x-\frac{\pi }{4}\right |2\right )}{3 a d \sqrt{e \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*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*e*(1 - Sec[c + d*x]))/(3*a*d*(e*Tan[c + d*x])^(3/2)) - (EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*
Sqrt[Sin[2*c + 2*d*x]])/(3*a*d*Sqrt[e*Tan[c + d*x]])

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rule 3882

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

Rule 3884

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(e*
Cot[c + d*x])^m, x], x] + Dist[b, Int[(e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]

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 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 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 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 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 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 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 2614

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sec (c+d x)) \sqrt{e \tan (c+d x)}} \, dx &=\frac{e^2 \int \frac{-a+a \sec (c+d x)}{(e \tan (c+d x))^{5/2}} \, dx}{a^2}\\ &=\frac{2 e (1-\sec (c+d x))}{3 a d (e \tan (c+d x))^{3/2}}+\frac{2 \int \frac{\frac{3 a}{2}-\frac{1}{2} a \sec (c+d x)}{\sqrt{e \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac{2 e (1-\sec (c+d x))}{3 a d (e \tan (c+d x))^{3/2}}-\frac{\int \frac{\sec (c+d x)}{\sqrt{e \tan (c+d x)}} \, dx}{3 a}+\frac{\int \frac{1}{\sqrt{e \tan (c+d x)}} \, dx}{a}\\ &=\frac{2 e (1-\sec (c+d x))}{3 a d (e \tan (c+d x))^{3/2}}+\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{\sqrt{\sin (c+d x)} \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{\sin (c+d x)}} \, dx}{3 a \sqrt{\cos (c+d x)} \sqrt{e \tan (c+d x)}}\\ &=\frac{2 e (1-\sec (c+d x))}{3 a d (e \tan (c+d x))^{3/2}}+\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 (\sec (c+d x) \sqrt{\sin (2 c+2 d x)}\right ) \int \frac{1}{\sqrt{\sin (2 c+2 d x)}} \, dx}{3 a \sqrt{e \tan (c+d x)}}\\ &=\frac{2 e (1-\sec (c+d x))}{3 a d (e \tan (c+d x))^{3/2}}-\frac{F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{3 a d \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{2 e (1-\sec (c+d x))}{3 a d (e \tan (c+d x))^{3/2}}-\frac{F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{3 a d \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{\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 e (1-\sec (c+d x))}{3 a d (e \tan (c+d x))^{3/2}}-\frac{F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{3 a d \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 e (1-\sec (c+d x))}{3 a d (e \tan (c+d x))^{3/2}}-\frac{F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{3 a d \sqrt{e \tan (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 8.89735, size = 1253, normalized size = 4.32 \[ \frac{4 \sqrt [4]{-1} \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt{\tan (c+d x)}\right ),-1\right ) \sqrt{\tan (c+d x)} \sec ^4(c+d x)}{3 d (\sec (c+d x) a+a) \sqrt{e \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )^{3/2}}+\frac{\cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (\frac{2 \sec ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{3 d}+\frac{2 (-2 \cos (c)+3 \cos (2 c)+3) \cos (d x) \sec (2 c)}{3 d}-\frac{2 \sec (2 c) (3 \sin (2 c)-2 \sin (c)) \sin (d x)}{3 d}-\frac{4}{3 d}\right ) \tan (c+d x) \sec (c+d x)}{(\sec (c+d x) a+a) \sqrt{e \tan (c+d x)}}+\frac{2 e^{-i (c+d x)} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec (2 c) \sqrt{\tan (c+d x)} \sec (c+d x)}{3 d (\sec (c+d x) a+a) \sqrt{e \tan (c+d x)}}+\frac{e^{-2 i c} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (e^{4 i c} \sqrt{-1+e^{4 i (c+d x)}} \tan ^{-1}\left (\sqrt{-1+e^{4 i (c+d x)}}\right )+2 \sqrt{-1+e^{2 i (c+d x)}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec (2 c) \sqrt{\tan (c+d x)} \sec (c+d x)}{2 d \left (-1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a) \sqrt{e \tan (c+d x)}}+\frac{e^{-2 i c} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (\sqrt{-1+e^{4 i (c+d x)}} \tan ^{-1}\left (\sqrt{-1+e^{4 i (c+d x)}}\right )+2 e^{4 i c} \sqrt{-1+e^{2 i (c+d x)}} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec (2 c) \sqrt{\tan (c+d x)} \sec (c+d x)}{2 d \left (-1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a) \sqrt{e \tan (c+d x)}}-\frac{e^{-i (2 c+d x)} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (3 \left (-1+e^{4 i (c+d x)}\right )+e^{4 i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt{1-e^{4 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},e^{4 i (c+d x)}\right )\right ) \sec (2 c) \sqrt{\tan (c+d x)} \sec (c+d x)}{3 d \left (-1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a) \sqrt{e \tan (c+d x)}}+\frac{e^{-i d x} \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \cos ^2\left (\frac{c}{2}+\frac{d x}{2}\right ) \left (e^{2 i (c+2 d x)} \left (-1+e^{2 i c}\right ) \sqrt{1-e^{4 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},e^{4 i (c+d x)}\right )-3 e^{4 i (c+d x)}+3\right ) \sec (2 c) \sqrt{\tan (c+d x)} \sec (c+d x)}{3 d \left (-1+e^{2 i (c+d x)}\right ) (\sec (c+d x) a+a) \sqrt{e \tan (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*Cos[c/2 + (d*x)
/2]^2*Sec[2*c]*Sec[c + d*x]*Sqrt[Tan[c + d*x]])/(3*d*E^(I*(c + d*x))*(a + a*Sec[c + d*x])*Sqrt[e*Tan[c + d*x]]
) + (Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(E^((4*I)*c)*Sqrt[-1 + E^((4*I)*(c + d*
x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d*x))]] + 2*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*A
rcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])*Cos[c/2 + (d*x)/2]^2*Sec[2*c]*Sec[c + d*x]
*Sqrt[Tan[c + d*x]])/(2*d*E^((2*I)*c)*(-1 + E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])*Sqrt[e*Tan[c + d*x]]) +
(Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt
[-1 + E^((4*I)*(c + d*x))]] + 2*E^((4*I)*c)*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTa
nh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])*Cos[c/2 + (d*x)/2]^2*Sec[2*c]*Sec[c + d*x]*Sqr
t[Tan[c + d*x]])/(2*d*E^((2*I)*c)*(-1 + E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])*Sqrt[e*Tan[c + d*x]]) - (Sqr
t[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*Cos[c/2 + (d*x)/2]^2*(3*(-1 + E^((4*I)*(c + d*x
))) + E^((4*I)*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 - E^((4*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, E^
((4*I)*(c + d*x))])*Sec[2*c]*Sec[c + d*x]*Sqrt[Tan[c + d*x]])/(3*d*E^(I*(2*c + d*x))*(-1 + E^((2*I)*(c + d*x))
)*(a + a*Sec[c + d*x])*Sqrt[e*Tan[c + d*x]]) + (Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)
))]*Cos[c/2 + (d*x)/2]^2*(3 - 3*E^((4*I)*(c + d*x)) + E^((2*I)*(c + 2*d*x))*(-1 + E^((2*I)*c))*Sqrt[1 - E^((4*
I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, E^((4*I)*(c + d*x))])*Sec[2*c]*Sec[c + d*x]*Sqrt[Tan[c + d*x]]
)/(3*d*E^(I*d*x)*(-1 + E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])*Sqrt[e*Tan[c + d*x]]) + (Cos[c/2 + (d*x)/2]^2
*Sec[c + d*x]*(-4/(3*d) + (2*(3 - 2*Cos[c] + 3*Cos[2*c])*Cos[d*x]*Sec[2*c])/(3*d) + (2*Sec[c/2 + (d*x)/2]^2)/(
3*d) - (2*Sec[2*c]*(-2*Sin[c] + 3*Sin[2*c])*Sin[d*x])/(3*d))*Tan[c + d*x])/((a + a*Sec[c + d*x])*Sqrt[e*Tan[c
+ d*x]]) + (4*(-1)^(1/4)*Cos[c/2 + (d*x)/2]^2*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]], -1]*Sec[c +
d*x]^4*Sqrt[Tan[c + d*x]])/(3*d*(a + a*Sec[c + d*x])*Sqrt[e*Tan[c + d*x]]*(1 + Tan[c + d*x]^2)^(3/2))

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Maple [C]  time = 0.254, size = 1269, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/a/d*2^(1/2)*(cos(d*x+c)+1)^2*(-1+cos(d*x+c))^2*(3*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))*cos(d*x+c)*((-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)*sin(d*x+c)-3*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))*cos(d*x+c)*((-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)*sin(d*x+c)+3*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))*((-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)*sin(d*x+c)+3*sin(d*x+c)*cos(d*x+c)*((-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)*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))-3*I*sin(d*x+c)*((-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)*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))+3*sin(d*x+c)*cos(d*x+c)*((-
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)*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))-8*sin(d*x+c)*cos(d
*x+c)*((-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)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+3*sin(d*x+c)*((-1+co
s(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)*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))+3*sin(d*x+c)*((-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)*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))-8*sin(d*x+c)*((-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)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+2*cos(d*x+c)^2*2^(1/2)-2*cos(d*x+c)*2
^(1/2))/sin(d*x+c)^5/cos(d*x+c)/(e*sin(d*x+c)/cos(d*x+c))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sec \left (d x + c\right ) + a\right )} \sqrt{e \tan \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{\sqrt{e \tan{\left (c + d x \right )}} \sec{\left (c + d x \right )} + \sqrt{e \tan{\left (c + d x \right )}}}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sec \left (d x + c\right ) + a\right )} \sqrt{e \tan \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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