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

Optimal. Leaf size=33 \[ \frac{c \tan (2 a+2 b x)}{b \sqrt{c \sec (2 a+2 b x)-c}} \]

[Out]

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

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Rubi [A]  time = 0.0649942, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {4397, 3792} \[ \frac{c \tan (2 a+2 b x)}{b \sqrt{c \sec (2 a+2 b x)-c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4397

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

Rule 3792

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*Cot[e + f*x])/
(f*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 \sec (2 (a+b x)) \sqrt{c \tan (a+b x) \tan (2 (a+b x))} \, dx &=\int \sec (2 a+2 b x) \sqrt{-c+c \sec (2 a+2 b x)} \, dx\\ &=\frac{c \tan (2 a+2 b x)}{b \sqrt{-c+c \sec (2 a+2 b x)}}\\ \end{align*}

Mathematica [A]  time = 0.0853419, size = 30, normalized size = 0.91 \[ \frac{\cot (a+b x) \sqrt{c \tan (a+b x) \tan (2 (a+b x))}}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cot[a + b*x]*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/b

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Maple [A]  time = 0.357, size = 52, normalized size = 1.6 \begin{align*}{\frac{\sqrt{2}\sqrt{4}\cos \left ( bx+a \right ) }{2\,b\sin \left ( bx+a \right ) }\sqrt{{\frac{c \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}}}} \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]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \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]

(2*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*b*sqrt(c)*integrate(-(((cos(8*b*x
+ 8*a)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + sin(8*b*x + 8*a)*sin(4*b*x + 4*a) + sin(4*b*x + 4*a)^2)*cos(1/2
*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) + (cos(4*b*x + 4*a)*sin(8*b*x + 8*a) - cos(8*b*x + 8*a)*sin
(4*b*x + 4*a))*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))*cos(1/2*arctan2(sin(4*b*x + 4*a), co
s(4*b*x + 4*a))) + ((cos(4*b*x + 4*a)*sin(8*b*x + 8*a) - cos(8*b*x + 8*a)*sin(4*b*x + 4*a))*cos(1/2*arctan2(si
n(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) - (cos(8*b*x + 8*a)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + sin(8*b*x
+ 8*a)*sin(4*b*x + 4*a) + sin(4*b*x + 4*a)^2)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))*sin(1
/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a))))/(((cos(8*b*x + 8*a)^2 + 2*cos(8*b*x + 8*a)*cos(4*b*x + 4*a) +
cos(4*b*x + 4*a)^2 + sin(8*b*x + 8*a)^2 + 2*sin(8*b*x + 8*a)*sin(4*b*x + 4*a) + sin(4*b*x + 4*a)^2)*cos(1/2*a
rctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2 + (cos(8*b*x + 8*a)^2 + 2*cos(8*b*x + 8*a)*cos(4*b*x + 4*a)
+ cos(4*b*x + 4*a)^2 + sin(8*b*x + 8*a)^2 + 2*sin(8*b*x + 8*a)*sin(4*b*x + 4*a) + sin(4*b*x + 4*a)^2)*sin(1/2
*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1))^2)*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x +
4*a) + 1)^(1/4)), x) - sqrt(c)*sin(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)))/((cos(4*b*x + 4*a)^
2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(1/4)*b)

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Fricas [A]  time = 2.14543, size = 96, normalized size = 2.91 \begin{align*} \frac{\sqrt{2} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{b \tan \left (b x + a\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]

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

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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: NotImplementedError} \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: NotImplementedError