3.449 \(\int \frac {\csc ^6(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=132 \[ \frac {3 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{20 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {\csc ^5(e+f x)}{5 b f (b \sec (e+f x))^{3/2}}+\frac {\csc ^3(e+f x)}{10 b f (b \sec (e+f x))^{3/2}}+\frac {3 \csc (e+f x)}{20 b f (b \sec (e+f x))^{3/2}} \]

[Out]

3/20*csc(f*x+e)/b/f/(b*sec(f*x+e))^(3/2)+1/10*csc(f*x+e)^3/b/f/(b*sec(f*x+e))^(3/2)-1/5*csc(f*x+e)^5/b/f/(b*se
c(f*x+e))^(3/2)+3/20*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/b^2
/f/cos(f*x+e)^(1/2)/(b*sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2623, 2625, 3771, 2639} \[ \frac {3 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{20 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {\csc ^5(e+f x)}{5 b f (b \sec (e+f x))^{3/2}}+\frac {\csc ^3(e+f x)}{10 b f (b \sec (e+f x))^{3/2}}+\frac {3 \csc (e+f x)}{20 b f (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6/(b*Sec[e + f*x])^(5/2),x]

[Out]

(3*Csc[e + f*x])/(20*b*f*(b*Sec[e + f*x])^(3/2)) + Csc[e + f*x]^3/(10*b*f*(b*Sec[e + f*x])^(3/2)) - Csc[e + f*
x]^5/(5*b*f*(b*Sec[e + f*x])^(3/2)) + (3*EllipticE[(e + f*x)/2, 2])/(20*b^2*f*Sqrt[Cos[e + f*x]]*Sqrt[b*Sec[e
+ f*x]])

Rule 2623

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(a*(a*Csc[e
+ f*x])^(m - 1)*(b*Sec[e + f*x])^(n + 1))/(f*b*(m - 1)), x] + Dist[(a^2*(n + 1))/(b^2*(m - 1)), Int[(a*Csc[e +
f*x])^(m - 2)*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && Intege
rsQ[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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(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 \frac {\csc ^6(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx &=-\frac {\csc ^5(e+f x)}{5 b f (b \sec (e+f x))^{3/2}}-\frac {3 \int \frac {\csc ^4(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx}{10 b^2}\\ &=\frac {\csc ^3(e+f x)}{10 b f (b \sec (e+f x))^{3/2}}-\frac {\csc ^5(e+f x)}{5 b f (b \sec (e+f x))^{3/2}}-\frac {3 \int \frac {\csc ^2(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx}{20 b^2}\\ &=\frac {3 \csc (e+f x)}{20 b f (b \sec (e+f x))^{3/2}}+\frac {\csc ^3(e+f x)}{10 b f (b \sec (e+f x))^{3/2}}-\frac {\csc ^5(e+f x)}{5 b f (b \sec (e+f x))^{3/2}}+\frac {3 \int \frac {1}{\sqrt {b \sec (e+f x)}} \, dx}{40 b^2}\\ &=\frac {3 \csc (e+f x)}{20 b f (b \sec (e+f x))^{3/2}}+\frac {\csc ^3(e+f x)}{10 b f (b \sec (e+f x))^{3/2}}-\frac {\csc ^5(e+f x)}{5 b f (b \sec (e+f x))^{3/2}}+\frac {3 \int \sqrt {\cos (e+f x)} \, dx}{40 b^2 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}\\ &=\frac {3 \csc (e+f x)}{20 b f (b \sec (e+f x))^{3/2}}+\frac {\csc ^3(e+f x)}{10 b f (b \sec (e+f x))^{3/2}}-\frac {\csc ^5(e+f x)}{5 b f (b \sec (e+f x))^{3/2}}+\frac {3 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{20 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 87, normalized size = 0.66 \[ \frac {\sin (e+f x) \sqrt {b \sec (e+f x)} \left (-4 \csc ^6(e+f x)+6 \csc ^4(e+f x)+\csc ^2(e+f x)+3 \sqrt {\cos (e+f x)} \csc (e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )-3\right )}{20 b^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^6/(b*Sec[e + f*x])^(5/2),x]

[Out]

((-3 + Csc[e + f*x]^2 + 6*Csc[e + f*x]^4 - 4*Csc[e + f*x]^6 + 3*Sqrt[Cos[e + f*x]]*Csc[e + f*x]*EllipticE[(e +
f*x)/2, 2])*Sqrt[b*Sec[e + f*x]]*Sin[e + f*x])/(20*b^3*f)

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (f x + e\right )} \csc \left (f x + e\right )^{6}}{b^{3} \sec \left (f x + e\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(b*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(f*x + e))*csc(f*x + e)^6/(b^3*sec(f*x + e)^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (f x + e\right )^{6}}{\left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(b*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^6/(b*sec(f*x + e))^(5/2), x)

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maple [C]  time = 0.25, size = 923, normalized size = 6.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6/(b*sec(f*x+e))^(5/2),x)

[Out]

1/20/f*(cos(f*x+e)+1)^2*(-1+cos(f*x+e))^2*(6*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*Elli
pticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*cos(f*x+e)^3*sin(f*x+e)+3*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)-6*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)+3*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)*cos(f*x+e)^4*sin(
f*x+e)-3*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I
)*cos(f*x+e)^4*sin(f*x+e)-3*I*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*sin(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(c
os(f*x+e)/(cos(f*x+e)+1))^(1/2)+6*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(-1
+cos(f*x+e))/sin(f*x+e),I)*cos(f*x+e)^2*sin(f*x+e)-6*I*cos(f*x+e)^2*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)+3*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)*cos(f*x+e)^5*sin(f*x+e)+3*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)+3*cos(f*x+e)^5-
3*I*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*cos(f*x+e)*sin(f*x+e)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(
f*x+e)+1))^(1/2)-3*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/si
n(f*x+e),I)*cos(f*x+e)^5*sin(f*x+e)-2*cos(f*x+e)^4-6*cos(f*x+e)^3-2*cos(f*x+e)^2+3*cos(f*x+e))/cos(f*x+e)^3/si
n(f*x+e)^9/(b/cos(f*x+e))^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (f x + e\right )^{6}}{\left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(b*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate(csc(f*x + e)^6/(b*sec(f*x + e))^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\sin \left (e+f\,x\right )}^6\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^6*(b/cos(e + f*x))^(5/2)),x)

[Out]

int(1/(sin(e + f*x)^6*(b/cos(e + f*x))^(5/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**6/(b*sec(f*x+e))**(5/2),x)

[Out]

Timed out

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