3.408 \(\int \csc ^2(e+f x) (b \sec (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=98 \[ -\frac{5 b^3 \csc (e+f x)}{3 f \sqrt{b \sec (e+f x)}}+\frac{5 b^2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{3 f}+\frac{2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f} \]

[Out]

(-5*b^3*Csc[e + f*x])/(3*f*Sqrt[b*Sec[e + f*x]]) + (5*b^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*
Sec[e + f*x]])/(3*f) + (2*b*Csc[e + f*x]*(b*Sec[e + f*x])^(3/2))/(3*f)

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Rubi [A]  time = 0.102253, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2626, 2625, 3771, 2641} \[ -\frac{5 b^3 \csc (e+f x)}{3 f \sqrt{b \sec (e+f x)}}+\frac{5 b^2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{3 f}+\frac{2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^2*(b*Sec[e + f*x])^(5/2),x]

[Out]

(-5*b^3*Csc[e + f*x])/(3*f*Sqrt[b*Sec[e + f*x]]) + (5*b^2*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*
Sec[e + f*x]])/(3*f) + (2*b*Csc[e + f*x]*(b*Sec[e + f*x])^(3/2))/(3*f)

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 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 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]

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]

Rubi steps

\begin{align*} \int \csc ^2(e+f x) (b \sec (e+f x))^{5/2} \, dx &=\frac{2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}+\frac{1}{3} \left (5 b^2\right ) \int \csc ^2(e+f x) \sqrt{b \sec (e+f x)} \, dx\\ &=-\frac{5 b^3 \csc (e+f x)}{3 f \sqrt{b \sec (e+f x)}}+\frac{2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}+\frac{1}{6} \left (5 b^2\right ) \int \sqrt{b \sec (e+f x)} \, dx\\ &=-\frac{5 b^3 \csc (e+f x)}{3 f \sqrt{b \sec (e+f x)}}+\frac{2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}+\frac{1}{6} \left (5 b^2 \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \int \frac{1}{\sqrt{\cos (e+f x)}} \, dx\\ &=-\frac{5 b^3 \csc (e+f x)}{3 f \sqrt{b \sec (e+f x)}}+\frac{5 b^2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{3 f}+\frac{2 b \csc (e+f x) (b \sec (e+f x))^{3/2}}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.181185, size = 67, normalized size = 0.68 \[ \frac{b \sin (e+f x) (b \sec (e+f x))^{3/2} \left (-3 \cot ^2(e+f x)+5 \cos ^{\frac{3}{2}}(e+f x) \csc (e+f x) F\left (\left .\frac{1}{2} (e+f x)\right |2\right )+2\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^2*(b*Sec[e + f*x])^(5/2),x]

[Out]

(b*(2 - 3*Cot[e + f*x]^2 + 5*Cos[e + f*x]^(3/2)*Csc[e + f*x]*EllipticF[(e + f*x)/2, 2])*(b*Sec[e + f*x])^(3/2)
*Sin[e + f*x])/(3*f)

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Maple [C]  time = 0.145, size = 202, normalized size = 2.1 \begin{align*}{\frac{ \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}{3\,f \left ( \sin \left ( fx+e \right ) \right ) ^{5}} \left ( 5\,i{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}} \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +5\,i\cos \left ( fx+e \right ){\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( fx+e \right ) \right ) }{\sin \left ( fx+e \right ) }},i \right ) \sqrt{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( fx+e \right ) }{\cos \left ( fx+e \right ) +1}}}\sin \left ( fx+e \right ) -5\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2 \right ) \left ({\frac{b}{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^2*(b*sec(f*x+e))^(5/2),x)

[Out]

1/3/f*(-1+cos(f*x+e))^2*(5*I*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(c
os(f*x+e)+1))^(1/2)*cos(f*x+e)^2*sin(f*x+e)+5*I*cos(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(1/(cos(f
*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-5*cos(f*x+e)^2+2)*cos(f*x+e)*(cos(f*x+e)+1)^2*(b/
cos(f*x+e))^(5/2)/sin(f*x+e)^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (f x + e\right )\right )^{\frac{5}{2}} \csc \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2*(b*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^(5/2)*csc(f*x + e)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b \sec \left (f x + e\right )} b^{2} \csc \left (f x + e\right )^{2} \sec \left (f x + e\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2*(b*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [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(csc(f*x+e)**2*(b*sec(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (f x + e\right )\right )^{\frac{5}{2}} \csc \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2*(b*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^(5/2)*csc(f*x + e)^2, x)