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

Optimal. Leaf size=129 \[ \frac{\tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{3 b c}+\frac{2 \tan (2 a+2 b x)}{3 b \sqrt{c \sec (2 a+2 b x)-c}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{\sqrt{2} b \sqrt{c}} \]

[Out]

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

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Rubi [A]  time = 0.35947, antiderivative size = 129, 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, 3795, 207} \[ \frac{\tan (2 a+2 b x) \sqrt{c \sec (2 a+2 b x)-c}}{3 b c}+\frac{2 \tan (2 a+2 b x)}{3 b \sqrt{c \sec (2 a+2 b x)-c}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{\sqrt{2} b \sqrt{c}} \]

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]

-(ArcTanh[(Sqrt[c]*Tan[2*a + 2*b*x])/(Sqrt[2]*Sqrt[-c + c*Sec[2*a + 2*b*x]])]/(Sqrt[2]*b*Sqrt[c])) + (2*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*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 3795

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

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.392146, size = 89, normalized size = 0.69 \[ \frac{\cos ^2(a+b x) \csc (2 (a+b x)) \sqrt{c \tan (a+b x) \tan (2 (a+b x))} \left (3 \sqrt{\tan ^2(a+b x)-1} \tan ^{-1}\left (\sqrt{\tan ^2(a+b x)-1}\right )+2 \sec (2 (a+b x))+2\right )}{3 b c} \]

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]

(Cos[a + b*x]^2*Csc[2*(a + b*x)]*(2 + 2*Sec[2*(a + b*x)] + 3*ArcTan[Sqrt[-1 + Tan[a + b*x]^2]]*Sqrt[-1 + Tan[a
+ b*x]^2])*Sqrt[c*Tan[a + b*x]*Tan[2*(a + b*x)]])/(3*b*c)

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Maple [B]  time = 0.506, size = 673, normalized size = 5.2 \begin{align*} \text{result too large to display} \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]

-1/24*2^(1/2)/b*4^(1/2)/(3+2*2^(1/2))^2/(-3+2*2^(1/2))^2*(-1+cos(b*x+a))*(12*cos(b*x+a)^4*ln(-2*(cos(b*x+a)^2*
((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)^2+cos(b*x+a)-((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1
/2)+1)/sin(b*x+a)^2)+12*arctanh(1/2*4^(1/2)*(2*cos(b*x+a)^2-3*cos(b*x+a)+1)/sin(b*x+a)^2/((2*cos(b*x+a)^2-1)/(
cos(b*x+a)+1)^2)^(1/2))*cos(b*x+a)^4+8*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*cos(b*x+a)^4+8*((2*cos(b*x+
a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*cos(b*x+a)^3-12*cos(b*x+a)^2*ln(-2*(cos(b*x+a)^2*((2*cos(b*x+a)^2-1)/(cos(b*x+
a)+1)^2)^(1/2)-2*cos(b*x+a)^2+cos(b*x+a)-((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+1)/sin(b*x+a)^2)-12*arcta
nh(1/2*4^(1/2)*(2*cos(b*x+a)^2-3*cos(b*x+a)+1)/sin(b*x+a)^2/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))*cos(b
*x+a)^2+3*ln(-2*(cos(b*x+a)^2*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)-2*cos(b*x+a)^2+cos(b*x+a)-((2*cos(b*
x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+1)/sin(b*x+a)^2)+3*arctanh(1/2*4^(1/2)*(2*cos(b*x+a)^2-3*cos(b*x+a)+1)/sin(b
*x+a)^2/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)))/(2*cos(b*x+a)^2-1)^2/sin(b*x+a)/((2*cos(b*x+a)^2-1)/(cos
(b*x+a)+1)^2)^(1/2)/(c*sin(b*x+a)^2/(2*cos(b*x+a)^2-1))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (2 \, b x + 2 \, a\right )^{3}}{\sqrt{c \tan \left (2 \, b x + 2 \, a\right ) \tan \left (b x + a\right )}}\,{d x} \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]

integrate(sec(2*b*x + 2*a)^3/sqrt(c*tan(2*b*x + 2*a)*tan(b*x + a)), x)

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Fricas [A]  time = 2.37962, size = 749, normalized size = 5.81 \begin{align*} \left [\frac{\frac{3 \, \sqrt{2}{\left (c \tan \left (b x + a\right )^{3} - c \tan \left (b x + a\right )\right )} \log \left (\frac{\tan \left (b x + a\right )^{3} - \frac{2 \, \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}{\left (\tan \left (b x + a\right )^{2} - 1\right )}}{\sqrt{c}} - 2 \, \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{3}}\right )}{\sqrt{c}} - 8 \, \sqrt{2} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{12 \,{\left (b c \tan \left (b x + a\right )^{3} - b c \tan \left (b x + a\right )\right )}}, -\frac{3 \, \sqrt{2}{\left (c \tan \left (b x + a\right )^{3} - c \tan \left (b x + a\right )\right )} \sqrt{-\frac{1}{c}} \arctan \left (\frac{\sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}{\left (\tan \left (b x + a\right )^{2} - 1\right )} \sqrt{-\frac{1}{c}}}{\tan \left (b x + a\right )}\right ) + 4 \, \sqrt{2} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{6 \,{\left (b c \tan \left (b x + a\right )^{3} - b c \tan \left (b x + a\right )\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]

[1/12*(3*sqrt(2)*(c*tan(b*x + a)^3 - c*tan(b*x + a))*log((tan(b*x + a)^3 - 2*sqrt(-c*tan(b*x + a)^2/(tan(b*x +
a)^2 - 1))*(tan(b*x + a)^2 - 1)/sqrt(c) - 2*tan(b*x + a))/tan(b*x + a)^3)/sqrt(c) - 8*sqrt(2)*sqrt(-c*tan(b*x
+ a)^2/(tan(b*x + a)^2 - 1)))/(b*c*tan(b*x + a)^3 - b*c*tan(b*x + a)), -1/6*(3*sqrt(2)*(c*tan(b*x + a)^3 - c*
tan(b*x + a))*sqrt(-1/c)*arctan(sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1))*(tan(b*x + a)^2 - 1)*sqrt(-1/c)/t
an(b*x + a)) + 4*sqrt(2)*sqrt(-c*tan(b*x + a)^2/(tan(b*x + a)^2 - 1)))/(b*c*tan(b*x + a)^3 - b*c*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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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]

Exception raised: TypeError