3.273 \(\int \sqrt {\csc (a+b x)} \sec ^2(a+b x) \, dx\)

Optimal. Leaf size=61 \[ \frac {\sec (a+b x)}{b \sqrt {\csc (a+b x)}}+\frac {\sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{b} \]

[Out]

sec(b*x+a)/b/csc(b*x+a)^(1/2)-(sin(1/2*a+1/4*Pi+1/2*b*x)^2)^(1/2)/sin(1/2*a+1/4*Pi+1/2*b*x)*EllipticF(cos(1/2*
a+1/4*Pi+1/2*b*x),2^(1/2))*csc(b*x+a)^(1/2)*sin(b*x+a)^(1/2)/b

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Rubi [A]  time = 0.05, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2626, 3771, 2641} \[ \frac {\sec (a+b x)}{b \sqrt {\csc (a+b x)}}+\frac {\sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Csc[a + b*x]]*Sec[a + b*x]^2,x]

[Out]

Sec[a + b*x]/(b*Sqrt[Csc[a + b*x]]) + (Sqrt[Csc[a + b*x]]*EllipticF[(a - Pi/2 + b*x)/2, 2]*Sqrt[Sin[a + b*x]])
/b

Rule 2626

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*b*(a*Csc[
e + f*x])^(m - 1)*(b*Sec[e + f*x])^(n - 1))/(f*(n - 1)), x] + Dist[(b^2*(m + n - 2))/(n - 1), Int[(a*Csc[e + f
*x])^m*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && IntegersQ[2*m, 2*n]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \sqrt {\csc (a+b x)} \sec ^2(a+b x) \, dx &=\frac {\sec (a+b x)}{b \sqrt {\csc (a+b x)}}+\frac {1}{2} \int \sqrt {\csc (a+b x)} \, dx\\ &=\frac {\sec (a+b x)}{b \sqrt {\csc (a+b x)}}+\frac {1}{2} \left (\sqrt {\csc (a+b x)} \sqrt {\sin (a+b x)}\right ) \int \frac {1}{\sqrt {\sin (a+b x)}} \, dx\\ &=\frac {\sec (a+b x)}{b \sqrt {\csc (a+b x)}}+\frac {\sqrt {\csc (a+b x)} F\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {\sin (a+b x)}}{b}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 49, normalized size = 0.80 \[ \frac {\sec (a+b x)+\frac {F\left (\left .\frac {1}{4} (2 a+2 b x-\pi )\right |2\right )}{\sqrt {\sin (a+b x)}}}{b \sqrt {\csc (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Csc[a + b*x]]*Sec[a + b*x]^2,x]

[Out]

(Sec[a + b*x] + EllipticF[(2*a - Pi + 2*b*x)/4, 2]/Sqrt[Sin[a + b*x]])/(b*Sqrt[Csc[a + b*x]])

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {\csc \left (b x + a\right )} \sec \left (b x + a\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(csc(b*x + a))*sec(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(csc(b*x + a))*sec(b*x + a)^2, x)

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maple [A]  time = 0.19, size = 123, normalized size = 2.02 \[ \frac {\sqrt {\left (\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}\, \left (\sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, \EllipticF \left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )+2 \sin \left (b x +a \right )\right )}{2 \sqrt {-\sin \left (b x +a \right ) \left (\sin \left (b x +a \right )-1\right ) \left (\sin \left (b x +a \right )+1\right )}\, \cos \left (b x +a \right ) \sqrt {\sin \left (b x +a \right )}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(cos(b*x+a)^2*sin(b*x+a))^(1/2)*((sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*Elliptic
F((sin(b*x+a)+1)^(1/2),1/2*2^(1/2))+2*sin(b*x+a))/(-sin(b*x+a)*(sin(b*x+a)-1)*(sin(b*x+a)+1))^(1/2)/cos(b*x+a)
/sin(b*x+a)^(1/2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(csc(b*x + a))*sec(b*x + a)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\sqrt {\frac {1}{\sin \left (a+b\,x\right )}}}{{\cos \left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {\csc {\left (a + b x \right )}} \sec ^{2}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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