∫ sec(a + bx) sec(2a + 2bx)dx

3.64 \(\int \sec (a+b x) \sec (2 a+2 b x) \, dx\)

Optimal. Leaf size=35 \[ \frac{\sqrt{2} \tanh ^{-1}\left (\sqrt{2} \sin (a+b x)\right )}{b}-\frac{\tanh ^{-1}(\sin (a+b x))}{b} \]

[Out]

-(ArcTanh[Sin[a + b*x]]/b) + (Sqrt[2]*ArcTanh[Sqrt[2]*Sin[a + b*x]])/b

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Rubi [A] time = 0.0333557, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {4364, 1093, 207} \[ \frac{\sqrt{2} \tanh ^{-1}\left (\sqrt{2} \sin (a+b x)\right )}{b}-\frac{\tanh ^{-1}(\sin (a+b x))}{b} \]

Antiderivative was successfully veriﬁed.

[In]

Int[Sec[a + b*x]*Sec[2*a + 2*b*x],x]

[Out]

-(ArcTanh[Sin[a + b*x]]/b) + (Sqrt[2]*ArcTanh[Sqrt[2]*Sin[a + b*x]])/b

Rule 4364

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_), x_Symbol] :> With[{d = FreeFactors[Sin[c*(a + b*x)], x]}, Dist

[d/(b*c), Subst[Int[SubstFor[(1 - d^2*x^2)^((n - 1)/2), Sin[c*(a + b*x)]/d, u, x], x], x, Sin[c*(a + b*x)]/d],

x] /; FunctionOfQ[Sin[c*(a + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] && (

EqQ[F, Cos] || EqQ[F, cos])

Rule 1093

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/(b/

2 - q/2 + c*x^2), x], x] - Dist[c/q, Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*

a*c, 0] && PosQ[b^2 - 4*a*c]

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 \sec (a+b x) \sec (2 a+2 b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{1-3 x^2+2 x^4} \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{1}{-2+2 x^2} \, dx,x,\sin (a+b x)\right )}{b}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{-1+2 x^2} \, dx,x,\sin (a+b x)\right )}{b}\\ &=-\frac{\tanh ^{-1}(\sin (a+b x))}{b}+\frac{\sqrt{2} \tanh ^{-1}\left (\sqrt{2} \sin (a+b x)\right )}{b}\\ \end{align*}

Mathematica [A] time = 0.0369994, size = 35, normalized size = 1. \[ \frac{\sqrt{2} \tanh ^{-1}\left (\sqrt{2} \sin (a+b x)\right )}{b}-\frac{\tanh ^{-1}(\sin (a+b x))}{b} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sec[a + b*x]*Sec[2*a + 2*b*x],x]

[Out]

-(ArcTanh[Sin[a + b*x]]/b) + (Sqrt[2]*ArcTanh[Sqrt[2]*Sin[a + b*x]])/b

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Maple [A] time = 0.045, size = 48, normalized size = 1.4 \begin{align*}{\frac{{\it Artanh} \left ( \sin \left ( bx+a \right ) \sqrt{2} \right ) \sqrt{2}}{b}}-{\frac{\ln \left ( 1+\sin \left ( bx+a \right ) \right ) }{2\,b}}+{\frac{\ln \left ( \sin \left ( bx+a \right ) -1 \right ) }{2\,b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)*sec(2*b*x+2*a),x)

[Out]

arctanh(sin(b*x+a)*2^(1/2))*2^(1/2)/b-1/2/b*ln(1+sin(b*x+a))+1/2/b*ln(sin(b*x+a)-1)

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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(b*x+a)*sec(2*b*x+2*a),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B] time = 2.4085, size = 193, normalized size = 5.51 \begin{align*} \frac{\sqrt{2} \log \left (-\frac{2 \, \cos \left (b x + a\right )^{2} - 2 \, \sqrt{2} \sin \left (b x + a\right ) - 3}{2 \, \cos \left (b x + a\right )^{2} - 1}\right ) - \log \left (\sin \left (b x + a\right ) + 1\right ) + \log \left (-\sin \left (b x + a\right ) + 1\right )}{2 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)*sec(2*b*x+2*a),x, algorithm="fricas")

[Out]

1/2*(sqrt(2)*log(-(2*cos(b*x + a)^2 - 2*sqrt(2)*sin(b*x + a) - 3)/(2*cos(b*x + a)^2 - 1)) - log(sin(b*x + a) +

1) + log(-sin(b*x + a) + 1))/b

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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec{\left (a + b x \right )} \sec{\left (2 a + 2 b x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)*sec(2*b*x+2*a),x)

[Out]

Integral(sec(a + b*x)*sec(2*a + 2*b*x), x)

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Giac [B] time = 2.54691, size = 1280, normalized size = 36.57 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)*sec(2*b*x+2*a),x, algorithm="giac")

[Out]

1/2*(sqrt(2)*log(abs(2*tan(1/2*b*x + 2*a)*tan(1/2*a)^6 + 12*tan(1/2*b*x + 2*a)*tan(1/2*a)^5 - 2*tan(1/2*a)^6 -

30*tan(1/2*b*x + 2*a)*tan(1/2*a)^4 + 12*tan(1/2*a)^5 - 40*tan(1/2*b*x + 2*a)*tan(1/2*a)^3 + 30*tan(1/2*a)^4 +

30*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 - 40*tan(1/2*a)^3 + 12*tan(1/2*b*x + 2*a)*tan(1/2*a) - 30*tan(1/2*a)^2 - 2

*sqrt(2)*(tan(1/2*a)^6 + 3*tan(1/2*a)^4 + 3*tan(1/2*a)^2 + 1) - 2*tan(1/2*b*x + 2*a) + 12*tan(1/2*a) + 2)/abs(

2*tan(1/2*b*x + 2*a)*tan(1/2*a)^6 + 12*tan(1/2*b*x + 2*a)*tan(1/2*a)^5 - 2*tan(1/2*a)^6 - 30*tan(1/2*b*x + 2*a

)*tan(1/2*a)^4 + 12*tan(1/2*a)^5 - 40*tan(1/2*b*x + 2*a)*tan(1/2*a)^3 + 30*tan(1/2*a)^4 + 30*tan(1/2*b*x + 2*a

)*tan(1/2*a)^2 - 40*tan(1/2*a)^3 + 12*tan(1/2*b*x + 2*a)*tan(1/2*a) - 30*tan(1/2*a)^2 + 2*sqrt(2)*(tan(1/2*a)^

6 + 3*tan(1/2*a)^4 + 3*tan(1/2*a)^2 + 1) - 2*tan(1/2*b*x + 2*a) + 12*tan(1/2*a) + 2)) + sqrt(2)*log(abs(2*tan(

1/2*b*x + 2*a)*tan(1/2*a)^6 - 12*tan(1/2*b*x + 2*a)*tan(1/2*a)^5 + 2*tan(1/2*a)^6 - 30*tan(1/2*b*x + 2*a)*tan(

1/2*a)^4 + 12*tan(1/2*a)^5 + 40*tan(1/2*b*x + 2*a)*tan(1/2*a)^3 - 30*tan(1/2*a)^4 + 30*tan(1/2*b*x + 2*a)*tan(

1/2*a)^2 - 40*tan(1/2*a)^3 - 12*tan(1/2*b*x + 2*a)*tan(1/2*a) + 30*tan(1/2*a)^2 - 2*sqrt(2)*(tan(1/2*a)^6 + 3*

tan(1/2*a)^4 + 3*tan(1/2*a)^2 + 1) - 2*tan(1/2*b*x + 2*a) + 12*tan(1/2*a) - 2)/abs(2*tan(1/2*b*x + 2*a)*tan(1/

2*a)^6 - 12*tan(1/2*b*x + 2*a)*tan(1/2*a)^5 + 2*tan(1/2*a)^6 - 30*tan(1/2*b*x + 2*a)*tan(1/2*a)^4 + 12*tan(1/2

*a)^5 + 40*tan(1/2*b*x + 2*a)*tan(1/2*a)^3 - 30*tan(1/2*a)^4 + 30*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 - 40*tan(1/2

*a)^3 - 12*tan(1/2*b*x + 2*a)*tan(1/2*a) + 30*tan(1/2*a)^2 + 2*sqrt(2)*(tan(1/2*a)^6 + 3*tan(1/2*a)^4 + 3*tan(

1/2*a)^2 + 1) - 2*tan(1/2*b*x + 2*a) + 12*tan(1/2*a) - 2)) - 2*log(abs(tan(1/2*b*x + 2*a)*tan(1/2*a)^3 + 3*tan

(1/2*b*x + 2*a)*tan(1/2*a)^2 - tan(1/2*a)^3 - 3*tan(1/2*b*x + 2*a)*tan(1/2*a) + 3*tan(1/2*a)^2 - tan(1/2*b*x +

2*a) + 3*tan(1/2*a) - 1)) + 2*log(abs(tan(1/2*b*x + 2*a)*tan(1/2*a)^3 - 3*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 + t

an(1/2*a)^3 - 3*tan(1/2*b*x + 2*a)*tan(1/2*a) + 3*tan(1/2*a)^2 + tan(1/2*b*x + 2*a) - 3*tan(1/2*a) - 1)))/b

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