3.272 \(\int \frac{\sec (a+b x)}{\sqrt{\csc (a+b x)}} \, dx\)

Optimal. Leaf size=31 \[ \frac{\tan ^{-1}\left (\sqrt{\csc (a+b x)}\right )}{b}+\frac{\tanh ^{-1}\left (\sqrt{\csc (a+b x)}\right )}{b} \]

[Out]

ArcTan[Sqrt[Csc[a + b*x]]]/b + ArcTanh[Sqrt[Csc[a + b*x]]]/b

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Rubi [A]  time = 0.0287734, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {2621, 329, 212, 206, 203} \[ \frac{\tan ^{-1}\left (\sqrt{\csc (a+b x)}\right )}{b}+\frac{\tanh ^{-1}\left (\sqrt{\csc (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]/Sqrt[Csc[a + b*x]],x]

[Out]

ArcTan[Sqrt[Csc[a + b*x]]]/b + ArcTanh[Sqrt[Csc[a + b*x]]]/b

Rule 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 206

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

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.0338167, size = 50, normalized size = 1.61 \[ -\frac{\sqrt{\sin (a+b x)} \sqrt{\csc (a+b x)} \left (\tan ^{-1}\left (\sqrt{\sin (a+b x)}\right )-\tanh ^{-1}\left (\sqrt{\sin (a+b x)}\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]/Sqrt[Csc[a + b*x]],x]

[Out]

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

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Maple [A]  time = 1.172, size = 48, normalized size = 1.6 \begin{align*} -{\frac{1}{2\,b}\ln \left ( \sqrt{\sin \left ( bx+a \right ) }-1 \right ) }+{\frac{1}{2\,b}\ln \left ( \sqrt{\sin \left ( bx+a \right ) }+1 \right ) }-{\frac{1}{b}\arctan \left ( \sqrt{\sin \left ( bx+a \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)/csc(b*x+a)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.48677, size = 55, normalized size = 1.77 \begin{align*} \frac{2 \, \arctan \left (\frac{1}{\sqrt{\sin \left (b x + a\right )}}\right ) + \log \left (\frac{1}{\sqrt{\sin \left (b x + a\right )}} + 1\right ) - \log \left (\frac{1}{\sqrt{\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)/csc(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

1/2*(2*arctan(1/sqrt(sin(b*x + a))) + log(1/sqrt(sin(b*x + a)) + 1) - log(1/sqrt(sin(b*x + a)) - 1))/b

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Fricas [B]  time = 1.31492, size = 275, normalized size = 8.87 \begin{align*} -\frac{2 \, \arctan \left (\frac{\sin \left (b x + a\right ) - 1}{2 \, \sqrt{\sin \left (b x + a\right )}}\right ) - \log \left (\frac{\cos \left (b x + a\right )^{2} + \frac{4 \,{\left (\cos \left (b x + a\right )^{2} - \sin \left (b x + a\right ) - 1\right )}}{\sqrt{\sin \left (b x + a\right )}} - 6 \, \sin \left (b x + a\right ) - 2}{\cos \left (b x + a\right )^{2} + 2 \, \sin \left (b x + a\right ) - 2}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)/csc(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

-1/4*(2*arctan(1/2*(sin(b*x + a) - 1)/sqrt(sin(b*x + a))) - log((cos(b*x + a)^2 + 4*(cos(b*x + a)^2 - sin(b*x
+ a) - 1)/sqrt(sin(b*x + a)) - 6*sin(b*x + a) - 2)/(cos(b*x + a)^2 + 2*sin(b*x + a) - 2)))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)/csc(b*x+a)**(1/2),x)

[Out]

Integral(sec(a + b*x)/sqrt(csc(a + b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (b x + a\right )}{\sqrt{\csc \left (b x + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)/csc(b*x+a)^(1/2),x, algorithm="giac")

[Out]

integrate(sec(b*x + a)/sqrt(csc(b*x + a)), x)