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

Optimal. Leaf size=72 \[ \frac{\tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{3 b}-\frac{c \tan (2 a+2 b x)}{3 b \sqrt{c \sec (2 a+2 b x)-c}} \]

[Out]

-(c*Tan[2*a + 2*b*x])/(3*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x])/(
3*b)

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Rubi [A]  time = 0.198944, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {4397, 3798, 3792} \[ \frac{\tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{3 b}-\frac{c \tan (2 a+2 b x)}{3 b \sqrt{c \sec (2 a+2 b x)-c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*Tan[2*a + 2*b*x])/(3*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x])/(
3*b)

Rule 4397

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

Rule 3798

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*m)/(b*(m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x
] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

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(2 (a+b x)) \sqrt{c \tan (a+b x) \tan (2 (a+b x))} \, dx &=\int \sec ^2(2 a+2 b x) \sqrt{-c+c \sec (2 a+2 b x)} \, dx\\ &=\frac{\sqrt{-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{3 b}-\frac{1}{3} \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)}{3 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{\sqrt{-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.181313, size = 44, normalized size = 0.61 \[ \frac{\sqrt{c \tan (a+b x) \tan (2 (a+b x))} (\tan (2 (a+b x))-\cot (a+b x))}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]]*(-Cot[a + b*x] + Tan[2*(a + b*x)]))/(3*b)

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Maple [A]  time = 0.445, size = 78, normalized size = 1.1 \begin{align*} -{\frac{\sqrt{2}\sqrt{4}\cos \left ( bx+a \right ) \left ( 4\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-3 \right ) }{6\,b\sin \left ( bx+a \right ) \left ( 2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1 \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)^2*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x)

[Out]

-1/6*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)*(4*cos(b*x+a)^2-3)/sin(b*x+a)/(2*c
os(b*x+a)^2-1)

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

[Out]

-2/3*(6*(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1)^(3/4)*b*sqrt(c)*integrate(-(((cos(1
2*b*x + 12*a)*cos(4*b*x + 4*a) + 2*cos(8*b*x + 8*a)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + sin(12*b*x + 12*a)
*sin(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(12*b*x + 12*a) + 2*cos(4*b*x + 4*a)*sin(8*b*x + 8*a) - cos(1
2*b*x + 12*a)*sin(4*b*x + 4*a) - 2*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(3/2*arctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a))) + ((cos(4*b*x + 4*a)*sin(12*b*x + 12*a
) + 2*cos(4*b*x + 4*a)*sin(8*b*x + 8*a) - cos(12*b*x + 12*a)*sin(4*b*x + 4*a) - 2*cos(8*b*x + 8*a)*sin(4*b*x +
4*a))*cos(1/2*arctan2(sin(4*b*x + 4*a), -cos(4*b*x + 4*a) - 1)) - (cos(12*b*x + 12*a)*cos(4*b*x + 4*a) + 2*co
s(8*b*x + 8*a)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + sin(12*b*x + 12*a)*sin(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(3/2*ar
ctan2(sin(4*b*x + 4*a), cos(4*b*x + 4*a))))/(((2*(2*cos(8*b*x + 8*a) + cos(4*b*x + 4*a))*cos(12*b*x + 12*a) +
cos(12*b*x + 12*a)^2 + 4*cos(8*b*x + 8*a)^2 + 4*cos(8*b*x + 8*a)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + 2*(2*
sin(8*b*x + 8*a) + sin(4*b*x + 4*a))*sin(12*b*x + 12*a) + sin(12*b*x + 12*a)^2 + 4*sin(8*b*x + 8*a)^2 + 4*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))^
2 + (2*(2*cos(8*b*x + 8*a) + cos(4*b*x + 4*a))*cos(12*b*x + 12*a) + cos(12*b*x + 12*a)^2 + 4*cos(8*b*x + 8*a)^
2 + 4*cos(8*b*x + 8*a)*cos(4*b*x + 4*a) + cos(4*b*x + 4*a)^2 + 2*(2*sin(8*b*x + 8*a) + sin(4*b*x + 4*a))*sin(1
2*b*x + 12*a) + sin(12*b*x + 12*a)^2 + 4*sin(8*b*x + 8*a)^2 + 4*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(3/2*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)^(3/4)*b)

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Fricas [A]  time = 2.03655, size = 159, normalized size = 2.21 \begin{align*} -\frac{\sqrt{2} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}{\left (3 \, \tan \left (b x + a\right )^{2} - 1\right )}}{3 \,{\left (b \tan \left (b x + a\right )^{3} - b \tan \left (b x + a\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(3*tan(b*x + a)^2 - 1)/(b*tan(b*x + a)^3 - 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)**2*(c*tan(b*x+a)*tan(2*b*x+2*a))**(1/2),x)

[Out]

Timed out

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

[Out]

Timed out