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

Optimal. Leaf size=110 \[ \frac{4 c^2 \tan (2 a+2 b x)}{5 b \sqrt{c \sec (2 a+2 b x)-c}}-\frac{c \tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{5 b}+\frac{\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b} \]

[Out]

(4*c^2*Tan[2*a + 2*b*x])/(5*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])/(5*b) + ((-c + c*Sec[2*a + 2*b*x])^(3/2)*Tan[2*a + 2*b*x])/(5*b)

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Rubi [A]  time = 0.268276, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {4397, 3798, 3793, 3792} \[ \frac{4 c^2 \tan (2 a+2 b x)}{5 b \sqrt{c \sec (2 a+2 b x)-c}}-\frac{c \tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{5 b}+\frac{\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*c^2*Tan[2*a + 2*b*x])/(5*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])/(5*b) + ((-c + c*Sec[2*a + 2*b*x])^(3/2)*Tan[2*a + 2*b*x])/(5*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 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(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx &=\int \sec ^2(2 a+2 b x) (-c+c \sec (2 a+2 b x))^{3/2} \, dx\\ &=\frac{(-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{5 b}-\frac{3}{5} \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)}{5 b}+\frac{(-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{5 b}+\frac{1}{5} (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)}{5 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)}{5 b}+\frac{(-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.237816, size = 59, normalized size = 0.54 \[ \frac{\cot (a+b x) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} (4 \cot (a+b x) \cot (2 (a+b x))+\sec (2 (a+b x))-2)}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cot[a + b*x]*(-2 + 4*Cot[a + b*x]*Cot[2*(a + b*x)] + Sec[2*(a + b*x)])*(c*Tan[a + b*x]*Tan[2*(a + b*x)])^(3/2
))/(5*b)

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Maple [A]  time = 0.34, size = 85, normalized size = 0.8 \begin{align*}{\frac{2\,\sqrt{2} \left ( 12\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}-15\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}+5 \right ) \cos \left ( bx+a \right ) }{5\,b \left ( 2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1 \right ) \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)^2*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x)

[Out]

2/5*2^(1/2)/b*(12*cos(b*x+a)^4-15*cos(b*x+a)^2+5)*cos(b*x+a)*(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(3/2)/(2*cos(
b*x+a)^2-1)/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)^2*(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 = 2.0322, size = 220, normalized size = 2. \begin{align*} \frac{2 \, \sqrt{2}{\left (5 \, c \tan \left (b x + a\right )^{4} - 5 \, 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}}}{5 \,{\left (b \tan \left (b x + a\right )^{5} - 2 \, 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))^(3/2),x, algorithm="fricas")

[Out]

2/5*sqrt(2)*(5*c*tan(b*x + a)^4 - 5*c*tan(b*x + a)^2 + 2*c)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))/(b*ta
n(b*x + a)^5 - 2*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))**(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)^2*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="giac")

[Out]

Timed out