3.384 \(\int \csc ^6(e+f x) \sqrt{b \sec (e+f x)} \, dx\)

Optimal. Leaf size=123 \[ -\frac{b \csc ^5(e+f x)}{5 f \sqrt{b \sec (e+f x)}}-\frac{3 b \csc ^3(e+f x)}{10 f \sqrt{b \sec (e+f x)}}-\frac{3 b \csc (e+f x)}{4 f \sqrt{b \sec (e+f x)}}+\frac{3 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{4 f} \]

[Out]

(-3*b*Csc[e + f*x])/(4*f*Sqrt[b*Sec[e + f*x]]) - (3*b*Csc[e + f*x]^3)/(10*f*Sqrt[b*Sec[e + f*x]]) - (b*Csc[e +
f*x]^5)/(5*f*Sqrt[b*Sec[e + f*x]]) + (3*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(4
*f)

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Rubi [A]  time = 0.137841, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2625, 3771, 2641} \[ -\frac{b \csc ^5(e+f x)}{5 f \sqrt{b \sec (e+f x)}}-\frac{3 b \csc ^3(e+f x)}{10 f \sqrt{b \sec (e+f x)}}-\frac{3 b \csc (e+f x)}{4 f \sqrt{b \sec (e+f x)}}+\frac{3 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6*Sqrt[b*Sec[e + f*x]],x]

[Out]

(-3*b*Csc[e + f*x])/(4*f*Sqrt[b*Sec[e + f*x]]) - (3*b*Csc[e + f*x]^3)/(10*f*Sqrt[b*Sec[e + f*x]]) - (b*Csc[e +
f*x]^5)/(5*f*Sqrt[b*Sec[e + f*x]]) + (3*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(4
*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 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 ^6(e+f x) \sqrt{b \sec (e+f x)} \, dx &=-\frac{b \csc ^5(e+f x)}{5 f \sqrt{b \sec (e+f x)}}+\frac{9}{10} \int \csc ^4(e+f x) \sqrt{b \sec (e+f x)} \, dx\\ &=-\frac{3 b \csc ^3(e+f x)}{10 f \sqrt{b \sec (e+f x)}}-\frac{b \csc ^5(e+f x)}{5 f \sqrt{b \sec (e+f x)}}+\frac{3}{4} \int \csc ^2(e+f x) \sqrt{b \sec (e+f x)} \, dx\\ &=-\frac{3 b \csc (e+f x)}{4 f \sqrt{b \sec (e+f x)}}-\frac{3 b \csc ^3(e+f x)}{10 f \sqrt{b \sec (e+f x)}}-\frac{b \csc ^5(e+f x)}{5 f \sqrt{b \sec (e+f x)}}+\frac{3}{8} \int \sqrt{b \sec (e+f x)} \, dx\\ &=-\frac{3 b \csc (e+f x)}{4 f \sqrt{b \sec (e+f x)}}-\frac{3 b \csc ^3(e+f x)}{10 f \sqrt{b \sec (e+f x)}}-\frac{b \csc ^5(e+f x)}{5 f \sqrt{b \sec (e+f x)}}+\frac{1}{8} \left (3 \sqrt{\cos (e+f x)} \sqrt{b \sec (e+f x)}\right ) \int \frac{1}{\sqrt{\cos (e+f x)}} \, dx\\ &=-\frac{3 b \csc (e+f x)}{4 f \sqrt{b \sec (e+f x)}}-\frac{3 b \csc ^3(e+f x)}{10 f \sqrt{b \sec (e+f x)}}-\frac{b \csc ^5(e+f x)}{5 f \sqrt{b \sec (e+f x)}}+\frac{3 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{b \sec (e+f x)}}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.441914, size = 73, normalized size = 0.59 \[ \frac{\sqrt{b \sec (e+f x)} \left (15 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )-\cot (e+f x) \left (4 \csc ^4(e+f x)+6 \csc ^2(e+f x)+15\right )\right )}{20 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^6*Sqrt[b*Sec[e + f*x]],x]

[Out]

((-(Cot[e + f*x]*(15 + 6*Csc[e + f*x]^2 + 4*Csc[e + f*x]^4)) + 15*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]
)*Sqrt[b*Sec[e + f*x]])/(20*f)

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Maple [C]  time = 0.181, size = 485, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6*(b*sec(f*x+e))^(1/2),x)

[Out]

1/20/f*(-1+cos(f*x+e))^2*(15*I*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos
(f*x+e)^5*sin(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)+15*I*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(cos(f*x+e)/(cos(
f*x+e)+1))^(1/2)*cos(f*x+e)^4*sin(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)-30*I*cos(f*x+e)^3*sin(f*x+e)*(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)-30*I*EllipticF(I*(-1+cos
(f*x+e))/sin(f*x+e),I)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)^2*sin(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)+15*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)-15*cos(f*x+e)^5+15*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)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)+36*cos(f*x+e)^3-25*cos(f*x+e))*(cos(f*x+e)+1)^2*(b/cos(f*x+e))^(1/2)/sin(
f*x+e)^9

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right )} \csc \left (f x + e\right )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6*(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sec(f*x + e))*csc(f*x + e)^6, 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 )} \csc \left (f x + e\right )^{6}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6*(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(f*x + e))*csc(f*x + e)^6, 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)**6*(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (f x + e\right )} \csc \left (f x + e\right )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6*(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(f*x + e))*csc(f*x + e)^6, x)