3.382 \(\int \csc ^2(e+f x) \sqrt {b \sec (e+f x)} \, dx\)

Optimal. Leaf size=62 \[ \frac {\sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \sec (e+f x)}}{f}-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}} \]

[Out]

-b*csc(f*x+e)/f/(b*sec(f*x+e))^(1/2)+(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2
*e),2^(1/2))*cos(f*x+e)^(1/2)*(b*sec(f*x+e))^(1/2)/f

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Rubi [A]  time = 0.06, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2625, 3771, 2641} \[ \frac {\sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \sec (e+f x)}}{f}-\frac {b \csc (e+f x)}{f \sqrt {b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*Csc[e + f*x])/(f*Sqrt[b*Sec[e + f*x]])) + (Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*
x]])/f

Rule 2625

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*(m - 1)), x] + Dist[(a^2*(m + n - 2))/(m - 1), Int[(a*Csc[e +
f*x])^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && IntegersQ[2*m, 2*n] &&
!GtQ[n, m]

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

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Mathematica [A]  time = 0.12, size = 47, normalized size = 0.76 \[ \frac {\sqrt {b \sec (e+f x)} \left (\sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )-\cot (e+f x)\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-Cot[e + f*x] + Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])*Sqrt[b*Sec[e + f*x]])/f

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sec \left (f x + e\right )} \csc \left (f x + e\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.20, size = 184, normalized size = 2.97 \[ \frac {\left (-1+\cos \left (f x +e \right )\right )^{2} \left (i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-\cos \left (f x +e \right )\right ) \left (\cos \left (f x +e \right )+1\right )^{2} \sqrt {\frac {b}{\cos \left (f x +e \right )}}}{f \sin \left (f x +e \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-1+cos(f*x+e))^2*(I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(-1+cos(f*x+e)
)/sin(f*x+e),I)*sin(f*x+e)*cos(f*x+e)+I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I
*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)-cos(f*x+e))*(cos(f*x+e)+1)^2*(b/cos(f*x+e))^(1/2)/sin(f*x+e)^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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