\(\int \frac {\sec (2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx\) [622]

Optimal result

Integrand size = 29, antiderivative size = 55 \[ \int \frac {\sec (2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{\sqrt {2} b \sqrt {c}} \]

[Out]

-1/2*arctanh(1/2*c^(1/2)*tan(2*b*x+2*a)*2^(1/2)/(-c+c*sec(2*b*x+2*a))^(1/2))/b*2^(1/2)/c^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {4482, 3880, 213} \[ \int \frac {\sec (2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {c \sec (2 a+2 b x)-c}}\right )}{\sqrt {2} b \sqrt {c}} \]

[In]

Int[Sec[2*(a + b*x)]/Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]

[Out]

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

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[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
]

Rule 4482

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\sec (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}} \, dx \\ & = -\frac {\text {Subst}\left (\int \frac {1}{-2 c+x^2} \, dx,x,-\frac {c \tan (2 a+2 b x)}{\sqrt {-c+c \sec (2 a+2 b x)}}\right )}{b} \\ & = -\frac {\text {arctanh}\left (\frac {\sqrt {c} \tan (2 a+2 b x)}{\sqrt {2} \sqrt {-c+c \sec (2 a+2 b x)}}\right )}{\sqrt {2} b \sqrt {c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.16 \[ \int \frac {\sec (2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\frac {\arctan \left (\sqrt {-1+\tan ^2(a+b x)}\right ) \sqrt {-1+\tan ^2(a+b x)} \tan (2 (a+b x))}{2 b \sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \]

[In]

Integrate[Sec[2*(a + b*x)]/Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]],x]

[Out]

(ArcTan[Sqrt[-1 + Tan[a + b*x]^2]]*Sqrt[-1 + Tan[a + b*x]^2]*Tan[2*(a + b*x)])/(2*b*Sqrt[c*Tan[a + b*x]*Tan[2*
(a + b*x)]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(208\) vs. \(2(46)=92\).

Time = 3.51 (sec) , antiderivative size = 209, normalized size of antiderivative = 3.80

method	result	size
default	\(\frac {\sqrt {2}\, \sin \left (x b +a \right ) \left (\ln \left (\frac {2 \cos \left (x b +a \right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}+2 \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}-4 \cos \left (x b +a \right )-2}{1+\cos \left (x b +a \right )}\right )-\operatorname {arctanh}\left (\frac {2 \cos \left (x b +a \right )-1}{\left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\right )\right ) \sqrt {4}}{8 b \left (1+\cos \left (x b +a \right )\right ) \sqrt {\frac {c \sin \left (x b +a \right )^{2}}{2 \cos \left (x b +a \right )^{2}-1}}\, \sqrt {\frac {2 \cos \left (x b +a \right )^{2}-1}{\left (1+\cos \left (x b +a \right )\right )^{2}}}}\)	\(209\)

[In]

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

[Out]

1/8*2^(1/2)/b*sin(b*x+a)*(ln(2*(cos(b*x+a)*((2*cos(b*x+a)^2-1)/(1+cos(b*x+a))^2)^(1/2)+((2*cos(b*x+a)^2-1)/(1+
cos(b*x+a))^2)^(1/2)-2*cos(b*x+a)-1)/(1+cos(b*x+a)))-arctanh((2*cos(b*x+a)-1)/(1+cos(b*x+a))/((2*cos(b*x+a)^2-
1)/(1+cos(b*x+a))^2)^(1/2)))/(1+cos(b*x+a))/(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)/((2*cos(b*x+a)^2-1)/(1+c
os(b*x+a))^2)^(1/2)*4^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.65 \[ \int \frac {\sec (2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\left [\frac {\sqrt {2} \log \left (\frac {\tan \left (b x + a\right )^{3} - \frac {2 \, \sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )}}{\sqrt {c}} - 2 \, \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right )}{4 \, b \sqrt {c}}, -\frac {\sqrt {2} \sqrt {-\frac {1}{c}} \arctan \left (\frac {\sqrt {-\frac {c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}} {\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt {-\frac {1}{c}}}{\tan \left (b x + a\right )}\right )}{2 \, b}\right ] \]

[In]

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

[Out]

[1/4*sqrt(2)*log((tan(b*x + a)^3 - 2*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)/sqrt(c)
- 2*tan(b*x + a))/tan(b*x + a)^3)/(b*sqrt(c)), -1/2*sqrt(2)*sqrt(-1/c)*arctan(sqrt(-c*tan(b*x + a)^2/(tan(b*x
+ a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(-1/c)/tan(b*x + a))/b]

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec (2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sec (2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\int { \frac {\sec \left (2 \, b x + 2 \, a\right )}{\sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )}} \,d x } \]

[In]

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

[Out]

integrate(sec(2*b*x + 2*a)/sqrt(c*tan(2*b*x + 2*a)*tan(b*x + a)), x)

Giac [F]

\[ \int \frac {\sec (2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\int { \frac {\sec \left (2 \, b x + 2 \, a\right )}{\sqrt {c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )}} \,d x } \]

[In]

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

[Out]

integrate(sec(2*b*x + 2*a)/sqrt(c*tan(2*b*x + 2*a)*tan(b*x + a)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (2 (a+b x))}{\sqrt {c \tan (a+b x) \tan (2 (a+b x))}} \, dx=\int \frac {1}{\cos \left (2\,a+2\,b\,x\right )\,\sqrt {c\,\mathrm {tan}\left (a+b\,x\right )\,\mathrm {tan}\left (2\,a+2\,b\,x\right )}} \,d x \]

[In]

int(1/(cos(2*a + 2*b*x)*(c*tan(a + b*x)*tan(2*a + 2*b*x))^(1/2)),x)

[Out]

int(1/(cos(2*a + 2*b*x)*(c*tan(a + b*x)*tan(2*a + 2*b*x))^(1/2)), x)