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

Optimal. Leaf size=110 \[ \frac{\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}+\frac{2 \tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{15 b}+\frac{7 c \tan (2 a+2 b x)}{15 b \sqrt{c \sec (2 a+2 b x)-c}} \]

[Out]

(7*c*Tan[2*a + 2*b*x])/(15*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (2*Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x
])/(15*b) + ((-c + c*Sec[2*a + 2*b*x])^(3/2)*Tan[2*a + 2*b*x])/(5*b*c)

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Rubi [A]  time = 0.2762, 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, 3800, 4001, 3792} \[ \frac{\tan (2 a+2 b x) (c \sec (2 a+2 b x)-c)^{3/2}}{5 b c}+\frac{2 \tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{15 b}+\frac{7 c \tan (2 a+2 b x)}{15 b \sqrt{c \sec (2 a+2 b x)-c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*c*Tan[2*a + 2*b*x])/(15*b*Sqrt[-c + c*Sec[2*a + 2*b*x]]) + (2*Sqrt[-c + c*Sec[2*a + 2*b*x]]*Tan[2*a + 2*b*x
])/(15*b) + ((-c + c*Sec[2*a + 2*b*x])^(3/2)*Tan[2*a + 2*b*x])/(5*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 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)) \sqrt{c \tan (a+b x) \tan (2 (a+b x))} \, dx &=\int \sec ^3(2 a+2 b x) \sqrt{-c+c \sec (2 a+2 b x)} \, dx\\ &=\frac{(-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{5 b c}+\frac{2 \int \sec (2 a+2 b x) \sqrt{-c+c \sec (2 a+2 b x)} \left (\frac{3 c}{2}+c \sec (2 a+2 b x)\right ) \, dx}{5 c}\\ &=\frac{2 \sqrt{-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{15 b}+\frac{(-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{5 b c}+\frac{7}{15} \int \sec (2 a+2 b x) \sqrt{-c+c \sec (2 a+2 b x)} \, dx\\ &=\frac{7 c \tan (2 a+2 b x)}{15 b \sqrt{-c+c \sec (2 a+2 b x)}}+\frac{2 \sqrt{-c+c \sec (2 a+2 b x)} \tan (2 a+2 b x)}{15 b}+\frac{(-c+c \sec (2 a+2 b x))^{3/2} \tan (2 a+2 b x)}{5 b c}\\ \end{align*}

Mathematica [A]  time = 0.18082, size = 62, normalized size = 0.56 \[ \frac{(5 \cos (a+b x)+2 \cos (5 (a+b x))) \csc (a+b x) \sec ^2(2 (a+b x)) \sqrt{c \tan (a+b x) \tan (2 (a+b x))}}{15 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*Cos[a + b*x] + 2*Cos[5*(a + b*x)])*Csc[a + b*x]*Sec[2*(a + b*x)]^2*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/
(15*b)

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Maple [A]  time = 0.466, size = 88, normalized size = 0.8 \begin{align*}{\frac{\sqrt{2}\sqrt{4}\cos \left ( bx+a \right ) \left ( 32\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}-40\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}+15 \right ) }{30\,b\sin \left ( bx+a \right ) \left ( 2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}-1 \right ) ^{2}}\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)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x)

[Out]

1/30*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)*(32*cos(b*x+a)^4-40*cos(b*x+a)^2+1
5)/sin(b*x+a)/(2*cos(b*x+a)^2-1)^2

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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))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.09083, size = 216, normalized size = 1.96 \begin{align*} \frac{\sqrt{2}{\left (15 \, \tan \left (b x + a\right )^{4} - 10 \, \tan \left (b x + a\right )^{2} + 7\right )} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{15 \,{\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)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="fricas")

[Out]

1/15*sqrt(2)*(15*tan(b*x + a)^4 - 10*tan(b*x + a)^2 + 7)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))/(b*tan(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)**3*(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)^3*(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="giac")

[Out]

Timed out