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

Optimal. Leaf size=88 \[ \frac{\tan (2 a+2 b x)}{b \sqrt{c \sec (2 a+2 b x)-c}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{\sqrt{2} b \sqrt{c}} \]

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

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Rubi [A]  time = 0.237719, antiderivative size = 88, 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, 3795, 207} \[ \frac{\tan (2 a+2 b x)}{b \sqrt{c \sec (2 a+2 b x)-c}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \tan (2 a+2 b x)}{\sqrt{2} \sqrt{c \sec (2 a+2 b x)-c}}\right )}{\sqrt{2} b \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

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

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

Mathematica [A]  time = 0.257214, size = 67, normalized size = 0.76 \[ \frac{\left (\sqrt{\tan ^2(a+b x)-1} \tan ^{-1}\left (\sqrt{\tan ^2(a+b x)-1}\right )+2\right ) \tan (2 (a+b x))}{2 b \sqrt{c \tan (a+b x) \tan (2 (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.467, size = 478, normalized size = 5.4 \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)^2/(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x)

[Out]

1/4*2^(1/2)/b*(cos(b*x+a)*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)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)
+cos(b*x+a)*((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2)*arctanh(1/2*4^(1/2)*(2*cos(b*x+a)^2-3*cos(b*x+a)+1)/si
n(b*x+a)^2/((2*cos(b*x+a)^2-1)/(cos(b*x+a)+1)^2)^(1/2))+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)*((2*cos(b*x+a
)^2-1)/(cos(b*x+a)+1)^2)^(1/2)+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)/(cos(b*x+a)+1)^2)^(1/2)+4*cos(b*x+a))*sin(b*x+a)/(2*cos(b*x
+a)^2-1)/(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 )^{2}}{\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)^2/(c*tan(b*x+a)*tan(2*b*x+2*a))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.40652, size = 626, normalized size = 7.11 \begin{align*} \left [\frac{\sqrt{2} \sqrt{c} \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 ) \tan \left (b x + a\right ) + 4 \, \sqrt{2} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{4 \, b c \tan \left (b x + a\right )}, -\frac{\sqrt{2} c \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 ) \tan \left (b x + a\right ) - 2 \, \sqrt{2} \sqrt{-\frac{c \tan \left (b x + a\right )^{2}}{\tan \left (b x + a\right )^{2} - 1}}}{2 \, b c \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))^(1/2),x, algorithm="fricas")

[Out]

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

[Out]

Exception raised: TypeError