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

Optimal. Leaf size=148 \[ -\frac{76 c^2 \tan (2 a+2 b x)}{105 b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{5/2}}{7 b c}+\frac{2 \tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{35 b}+\frac{19 c \tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{105 b} \]

[Out]

(-76*c^2*Tan[2*a + 2*b*x])/(105*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (19*c*Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a
+ 2*b*x])/(105*b) + (2*(-c + c*Sec[2*a + 2*b*x])^(3/2)*Tan[2*a + 2*b*x])/(35*b) + ((-c + c*Sec[2*a + 2*b*x])^
(5/2)*Tan[2*a + 2*b*x])/(7*b*c)

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Rubi [A]  time = 0.347543, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {4397, 3800, 4001, 3793, 3792} \[ -\frac{76 c^2 \tan (2 a+2 b x)}{105 b \sqrt{c \sec (2 a+2 b x)-c}}+\frac{\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{5/2}}{7 b c}+\frac{2 \tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{35 b}+\frac{19 c \tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{105 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-76*c^2*Tan[2*a + 2*b*x])/(105*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (19*c*Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a
+ 2*b*x])/(105*b) + (2*(-c + c*Sec[2*a + 2*b*x])^(3/2)*Tan[2*a + 2*b*x])/(35*b) + ((-c + c*Sec[2*a + 2*b*x])^
(5/2)*Tan[2*a + 2*b*x])/(7*b*c)

Rule 4397

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

Rule 3800

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 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 ^3(2 (a+b x)) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \, dx &=\int \sec ^3(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))^{5/2} \tan (2 a+2 b x)}{7 b c}+\frac{2 \int \sec (2 a+2 b x) (-c+c \sec (2 a+2 b x))^{3/2} \left (\frac{5 c}{2}+c \sec (2 a+2 b x)\right ) \, dx}{7 c}\\ &=\frac{2 (-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{35 b}+\frac{(-c+c \sec (2 a+2 b x))^{5/2} \tan (2 a+2 b x)}{7 b c}+\frac{19}{35} \int \sec (2 a+2 b x) (-c+c \sec (2 a+2 b x))^{3/2} \, dx\\ &=\frac{19 c \sqrt{-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{105 b}+\frac{2 (-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{35 b}+\frac{(-c+c \sec (2 a+2 b x))^{5/2} \tan (2 a+2 b x)}{7 b c}-\frac{1}{105} (76 c) \int \sec (2 a+2 b x) \sqrt{-c+c \sec (2 a+2 b x)} \, dx\\ &=-\frac{76 c^2 \tan (2 a+2 b x)}{105 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{19 c \sqrt{-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{105 b}+\frac{2 (-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{35 b}+\frac{(-c+c \sec (2 a+2 b x))^{5/2} \tan (2 a+2 b x)}{7 b c}\\ \end{align*}

Mathematica [A]  time = 0.220233, size = 73, normalized size = 0.49 \[ -\frac{\cot (a+b x) (c \tan (a+b x) \tan (2 (a+b x)))^{3/2} \left (76 \cot (a+b x) \cot (2 (a+b x))-15 \sec ^2(2 (a+b x))+24 \sec (2 (a+b x))-28\right )}{105 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cot[a + b*x]*(-28 + 76*Cot[a + b*x]*Cot[2*(a + b*x)] + 24*Sec[2*(a + b*x)] - 15*Sec[2*(a + b*x)]^2)*(c*Tan[a
+ b*x]*Tan[2*(a + b*x)])^(3/2))/(105*b)

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Maple [A]  time = 0.398, size = 95, normalized size = 0.6 \begin{align*} -{\frac{2\,\sqrt{2} \left ( 416\, \left ( \cos \left ( bx+a \right ) \right ) ^{6}-728\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}+455\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-105 \right ) \cos \left ( bx+a \right ) }{105\,b \left ( 2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1 \right ) ^{2} \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)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x)

[Out]

-2/105*2^(1/2)/b*(416*cos(b*x+a)^6-728*cos(b*x+a)^4+455*cos(b*x+a)^2-105)*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)^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)^3*(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 = 2.05706, size = 290, normalized size = 1.96 \begin{align*} -\frac{2 \, \sqrt{2}{\left (105 \, c \tan \left (b x + a\right )^{6} - 140 \, c \tan \left (b x + a\right )^{4} + 133 \, c \tan \left (b x + a\right )^{2} - 38 \, c\right )} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{105 \,{\left (b \tan \left (b x + a\right )^{7} - 3 \, b \tan \left (b x + a\right )^{5} + 3 \, 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)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="fricas")

[Out]

-2/105*sqrt(2)*(105*c*tan(b*x + a)^6 - 140*c*tan(b*x + a)^4 + 133*c*tan(b*x + a)^2 - 38*c)*sqrt(-c*tan(b*x + a
)^2/(tan(b*x + a)^2 - 1))/(b*tan(b*x + a)^7 - 3*b*tan(b*x + a)^5 + 3*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)**3*(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)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(3/2),x, algorithm="giac")

[Out]

Timed out