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

Optimal. Leaf size=46 \[ \frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d} \]

[Out]

(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 = 0.0359379, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3795, 203} \[ \frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

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

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
]

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])

Rubi steps

\begin{align*} \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx &=-\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{\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 [A]  time = 0.0467547, size = 64, normalized size = 1.39 \[ \frac{\sqrt{2} \tan (c+d x) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )}{d \sqrt{1-\sec (c+d x)} \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d
*x])])

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Maple [B]  time = 0.127, size = 95, normalized size = 2.1 \begin{align*}{\frac{1}{ad}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\ln \left ({\frac{1}{\sin \left ( dx+c \right ) } \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) +1 \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )}{\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(sec(d*x+c)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.22211, size = 428, normalized size = 9.3 \begin{align*} \left [\frac{\sqrt{2} \sqrt{-\frac{1}{a}} \log \left (-\frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{2 \, d}, -\frac{\sqrt{2} \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 )}{\sqrt{a} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 9.77395, size = 88, normalized size = 1.91 \begin{align*} \frac{\sqrt{2} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} d \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*d*sgn(tan(1/2
*d*x + 1/2*c)^2 - 1))