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

Optimal. Leaf size=55 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{\sqrt{2} b \sqrt{c}} \]

[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]))

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Rubi [A]  time = 0.0761244, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {4397, 3795, 207} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{\sqrt{2} b \sqrt{c}} \]

Antiderivative was successfully verified.

[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 4397

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

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 207

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

Rubi steps

\begin{align*} \int \frac{\sec (2 (a+b x))}{\sqrt{c \tan (a+b x) \tan (2 (a+b x))}} \, dx &=\int \frac{\sec (2 a+2 b x)}{\sqrt{-c+c \sec (2 a+2 b x)}} \, dx\\ &=-\frac{\operatorname{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{\tanh ^{-1}\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]  time = 0.147745, size = 64, normalized size = 1.16 \[ \frac{\tan ^{-1}\left (\sqrt{\tan ^2(a+b x)-1}\right ) \sqrt{\tan ^2(a+b x)-1} \tan (2 (a+b x))}{2 b \sqrt{c \tan (a+b x) \tan (2 (a+b x))}} \]

Antiderivative was successfully verified.

[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)]])

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Maple [B]  time = 0.388, size = 236, normalized size = 4.3 \begin{align*}{\frac{\sqrt{2}\sqrt{4} \left ( \cos \left ( bx+a \right ) +1 \right ) }{8\,b\sin \left ( bx+a \right ) c}\sqrt{{\frac{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}}}\sqrt{{\frac{c \left ( 1- \left ( \cos \left ( bx+a \right ) \right ) ^{2} \right ) }{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}}} \left ( \ln \left ( -2\,{\frac{1}{ \left ( \sin \left ( bx+a \right ) \right ) ^{2}} \left ( \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sqrt{{\frac{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}}}-2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}+\cos \left ( bx+a \right ) -\sqrt{{\frac{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}}}+1 \right ) } \right ) +{\it Artanh} \left ({\frac{\sqrt{4} \left ( 2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-3\,\cos \left ( bx+a \right ) +1 \right ) }{2\, \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{\frac{1}{\sqrt{{\frac{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}}}}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*2^(1/2)/b*4^(1/2)*(cos(b*x+a)+1)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*(c*(1-cos(b*x+a)^2)/(2*cos(b*
x+a)^2-1))^(1/2)*(ln(-2*(cos(b*x+a)^2*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)^2+cos(b*x+a)-((
2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+1)/sin(b*x+a)^2)+arctanh(1/2*4^(1/2)*(2*cos(b*x+a)^2-3*cos(b*x+a)+1)
/sin(b*x+a)^2/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)))/sin(b*x+a)/c

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.27136, size = 386, normalized size = 7.02 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[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]

Exception raised: TypeError