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

Optimal. Leaf size=32 \[ \frac {\tanh ^{-1}\left (\sqrt {\csc (a+b x)}\right )}{b}-\frac {\tan ^{-1}\left (\sqrt {\csc (a+b x)}\right )}{b} \]

[Out]

-arctan(csc(b*x+a)^(1/2))/b+arctanh(csc(b*x+a)^(1/2))/b

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Rubi [A]  time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.00, 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, 298, 203, 206} \[ \frac {\tanh ^{-1}\left (\sqrt {\csc (a+b x)}\right )}{b}-\frac {\tan ^{-1}\left (\sqrt {\csc (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 298

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

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 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])

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 47, normalized size = 1.47 \[ \frac {\sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} \left (\tan ^{-1}\left (\sqrt {\sin (a+b x)}\right )+\tanh ^{-1}\left (\sqrt {\sin (a+b x)}\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Csc[a + b*x]]*Sec[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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fricas [B]  time = 0.82, size = 95, normalized size = 2.97 \[ \frac {2 \, \arctan \left (\frac {\sin \left (b x + a\right ) - 1}{2 \, \sqrt {\sin \left (b x + a\right )}}\right ) + \log \left (\frac {\cos \left (b x + a\right )^{2} + \frac {4 \, {\left (\cos \left (b x + a\right )^{2} - \sin \left (b x + a\right ) - 1\right )}}{\sqrt {\sin \left (b x + a\right )}} - 6 \, \sin \left (b x + a\right ) - 2}{\cos \left (b x + a\right )^{2} + 2 \, \sin \left (b x + a\right ) - 2}\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^(1/2)*sec(b*x+a),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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\csc \left (b x + a\right )} \sec \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.18, size = 28, normalized size = 0.88 \[ \frac {\arctanh \left (\sqrt {\sin }\left (b x +a \right )\right )}{b}+\frac {\arctan \left (\sqrt {\sin }\left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.35, size = 41, normalized size = 1.28 \[ -\frac {2 \, \arctan \left (\frac {1}{\sqrt {\sin \left (b x + a\right )}}\right ) - \log \left (\frac {1}{\sqrt {\sin \left (b x + a\right )}} + 1\right ) + \log \left (\frac {1}{\sqrt {\sin \left (b x + a\right )}} - 1\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^(1/2)*sec(b*x+a),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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\sqrt {\frac {1}{\sin \left (a+b\,x\right )}}}{\cos \left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/sin(a + b*x))^(1/2)/cos(a + b*x),x)

[Out]

int((1/sin(a + b*x))^(1/2)/cos(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\csc {\left (a + b x \right )}} \sec {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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