3.142 \(\int \frac{1}{\sqrt{a-a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=87 \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}-\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d} \]

[Out]

(2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(Sqrt[a]*d) - (Sqrt[2]*ArcTan[(Sqrt[a]*Tan[c + d*x
])/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(Sqrt[a]*d)

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Rubi [A]  time = 0.0705244, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {3776, 3774, 203, 3795} \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}-\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a - a*Sec[c + d*x]],x]

[Out]

(2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(Sqrt[a]*d) - (Sqrt[2]*ArcTan[(Sqrt[a]*Tan[c + d*x
])/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(Sqrt[a]*d)

Rule 3776

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

Rule 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 203

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

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a-a \sec (c+d x)}} \, dx &=\frac{\int \sqrt{a-a \sec (c+d x)} \, dx}{a}+\int \frac{\sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,\frac{a \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{d}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,\frac{a \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{d}\\ &=\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}-\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{a} d}\\ \end{align*}

Mathematica [C]  time = 0.45565, size = 127, normalized size = 1.46 \[ -\frac{i \left (-1+e^{i (c+d x)}\right ) \left (\sinh ^{-1}\left (e^{i (c+d x)}\right )-\sqrt{2} \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+\tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )}{d \sqrt{1+e^{2 i (c+d x)}} \sqrt{a-a \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a - a*Sec[c + d*x]],x]

[Out]

((-I)*(-1 + E^(I*(c + d*x)))*(ArcSinh[E^(I*(c + d*x))] - Sqrt[2]*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1
+ E^((2*I)*(c + d*x))])] + ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/(d*Sqrt[1 + E^((2*I)*(c + d*x))]*Sqrt[a -
a*Sec[c + d*x]])

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Maple [A]  time = 0.143, size = 119, normalized size = 1.4 \begin{align*} -{\frac{\sqrt{2} \left ( -1+\cos \left ( dx+c \right ) \right ) }{d\sin \left ( dx+c \right ) } \left ( \sqrt{2}\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) +2\,\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) \right ){\frac{1}{\sqrt{{\frac{a \left ( -1+\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}}}{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*2^(1/2)*(-1+cos(d*x+c))*(2^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+2*arctan(1/2*2^(1/2)*(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)))/(a*(-1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.83487, size = 778, normalized size = 8.94 \begin{align*} \left [\frac{\sqrt{2} a \sqrt{-\frac{1}{a}} \log \left (-\frac{2 \, \sqrt{2}{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} -{\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) - 2 \, \sqrt{-a} \log \left (\frac{2 \,{\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} -{\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right )}{2 \, a d}, \frac{\sqrt{2} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) - 2 \, \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right )}{a d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(2)*a*sqrt(-1/a)*log(-(2*sqrt(2)*(cos(d*x + c)^2 + cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x +
c))*sqrt(-1/a) - (3*cos(d*x + c) + 1)*sin(d*x + c))/((cos(d*x + c) - 1)*sin(d*x + c))) - 2*sqrt(-a)*log((2*(c
os(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)) - (2*a*cos(d*x + c) + a)*sin(d*
x + c))/sin(d*x + c)))/(a*d), (sqrt(2)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c))*cos(d*x
+ c)/(sqrt(a)*sin(d*x + c))) - 2*sqrt(a)*arctan(sqrt((a*cos(d*x + c) - a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*
sin(d*x + c))))/(a*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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