3.614 \(\int \sec (2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx\)

Optimal. Leaf size=75 \[ \frac{c \tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{3 b}-\frac{4 c^2 \tan (2 a+2 b x)}{3 b \sqrt{c \sec (2 a+2 b x)-c}} \]

[Out]

(-4*c^2*Tan[2*a + 2*b*x])/(3*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (c*Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b
*x])/(3*b)

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Rubi [A]  time = 0.109576, antiderivative size = 75, 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, 3793, 3792} \[ \frac{c \tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{3 b}-\frac{4 c^2 \tan (2 a+2 b x)}{3 b \sqrt{c \sec (2 a+2 b x)-c}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[2*(a + b*x)]*(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2),x]

[Out]

(-4*c^2*Tan[2*a + 2*b*x])/(3*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (c*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 3793

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

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)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx &=\int \sec (2 a+2 b x) (-c+c \sec (2 a+2 b x))^{3/2} \, dx\\ &=\frac{c \sqrt{-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{3 b}-\frac{1}{3} (4 c) \int \sec (2 a+2 b x) \sqrt{-c+c \sec (2 a+2 b x)} \, dx\\ &=-\frac{4 c^2 \tan (2 a+2 b x)}{3 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{c \sqrt{-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.16986, size = 51, normalized size = 0.68 \[ -\frac{\cot (a+b x) (4 \cot (a+b x) \cot (2 (a+b x))-1) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[2*(a + b*x)]*(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2),x]

[Out]

-(Cot[a + b*x]*(-1 + 4*Cot[a + b*x]*Cot[2*(a + b*x)])*(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2))/(3*b)

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Maple [A]  time = 0.429, size = 61, normalized size = 0.8 \begin{align*} -{\frac{2\,\sqrt{2} \left ( 5\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-3 \right ) \cos \left ( bx+a \right ) }{3\,b \left ( \sin \left ( bx+a \right ) \right ) ^{3}} \left ({\frac{c \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1}} \right ) ^{{\frac{3}{2}}}} \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))^(3/2),x)

[Out]

-2/3*2^(1/2)/b*(5*cos(b*x+a)^2-3)*cos(b*x+a)*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(3/2)/sin(b*x+a)^3

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Maxima [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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.99335, size = 165, normalized size = 2.2 \begin{align*} -\frac{2 \, \sqrt{2}{\left (3 \, c \tan \left (b x + a\right )^{2} - 2 \, c\right )} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{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)*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="fricas")

[Out]

-2/3*sqrt(2)*(3*c*tan(b*x + a)^2 - 2*c)*sqrt(-c*tan(b*x + a)^2/(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)*(c*tan(b*x+a)*tan(2*b*x+2*a))**(3/2),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)*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="giac")

[Out]

Timed out