\(\int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx\) [141]

Optimal result

Integrand size = 22, antiderivative size = 48 \[ \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a} d} \]

[Out]

-arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a-a*sec(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3880, 209} \[ \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a} d} \]

[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 209

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

Rule 3880

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*} \text {integral}& = -\frac {2 \text {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} \arctan \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] (verified)

Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.29 \[ \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right ) \tan (c+d x)}{d \sqrt {1+\sec (c+d x)} \sqrt {a-a \sec (c+d x)}} \]

[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 - a*Sec[c +
d*x]]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(82\) vs. \(2(39)=78\).

Time = 0.94 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.73

method	result	size
default	\(\frac {\sqrt {2}\, \sin \left (d x +c \right ) \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-a \left (\sec \left (d x +c \right )-1\right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(83\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 161, normalized size of antiderivative = 3.35 \[ \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx=\left [\frac {\sqrt {2} \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right )}{2 \, d}, \frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a} d}\right ] \]

[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)*(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)))/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)]

Sympy [F]

\[ \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\sqrt {- a \left (\sec {\left (c + d x \right )} - 1\right )}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )}{\sqrt {-a \sec \left (d x + c\right ) + a}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.40 \[ \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{\sqrt {a} d \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \]

[In]

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

[Out]

sqrt(2)*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(sqrt(a)*d*sgn(tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x
+ 1/2*c))*sgn(cos(d*x + c)))

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,\sqrt {a-\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]

[In]

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

[Out]

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