3.435 \(\int \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=102 \[ -\frac {\sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \sec (e+f x)}}{6 b^2 f}-\frac {\csc ^3(e+f x)}{3 b f \sqrt {b \sec (e+f x)}}+\frac {\csc (e+f x)}{6 b f \sqrt {b \sec (e+f x)}} \]

[Out]

1/6*csc(f*x+e)/b/f/(b*sec(f*x+e))^(1/2)-1/3*csc(f*x+e)^3/b/f/(b*sec(f*x+e))^(1/2)-1/6*(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)/b^2/f

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Rubi [A]  time = 0.10, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2623, 2625, 3771, 2641} \[ -\frac {\sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \sec (e+f x)}}{6 b^2 f}-\frac {\csc ^3(e+f x)}{3 b f \sqrt {b \sec (e+f x)}}+\frac {\csc (e+f x)}{6 b f \sqrt {b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4/(b*Sec[e + f*x])^(3/2),x]

[Out]

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

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 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 \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx &=-\frac {\csc ^3(e+f x)}{3 b f \sqrt {b \sec (e+f x)}}-\frac {\int \csc ^2(e+f x) \sqrt {b \sec (e+f x)} \, dx}{6 b^2}\\ &=\frac {\csc (e+f x)}{6 b f \sqrt {b \sec (e+f x)}}-\frac {\csc ^3(e+f x)}{3 b f \sqrt {b \sec (e+f x)}}-\frac {\int \sqrt {b \sec (e+f x)} \, dx}{12 b^2}\\ &=\frac {\csc (e+f x)}{6 b f \sqrt {b \sec (e+f x)}}-\frac {\csc ^3(e+f x)}{3 b f \sqrt {b \sec (e+f x)}}-\frac {\left (\sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{12 b^2}\\ &=\frac {\csc (e+f x)}{6 b f \sqrt {b \sec (e+f x)}}-\frac {\csc ^3(e+f x)}{3 b 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)}}{6 b^2 f}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 62, normalized size = 0.61 \[ \frac {-2 \csc ^3(e+f x)+\csc (e+f x)-\frac {F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{\sqrt {\cos (e+f x)}}}{6 b f \sqrt {b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4/(b*Sec[e + f*x])^(3/2),x]

[Out]

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

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (f x + e\right )} \csc \left (f x + e\right )^{4}}{b^{2} \sec \left (f x + e\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4/(b*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(f*x + e))*csc(f*x + e)^4/(b^2*sec(f*x + e)^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (f x + e\right )^{4}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4/(b*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^4/(b*sec(f*x + e))^(3/2), x)

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maple [C]  time = 0.20, size = 343, normalized size = 3.36 \[ \frac {\left (\cos \left (f x +e \right )+1\right )^{2} \left (-1+\cos \left (f x +e \right )\right )^{2} \left (i \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )+i \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-\left (\cos ^{3}\left (f x +e \right )\right )-\cos \left (f x +e \right )\right )}{6 f \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{7} \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^4/(b*sec(f*x+e))^(3/2),x)

[Out]

1/6/f*(cos(f*x+e)+1)^2*(-1+cos(f*x+e))^2*(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)+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)-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)^3-co
s(f*x+e))/cos(f*x+e)^2/sin(f*x+e)^7/(b/cos(f*x+e))^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (f x + e\right )^{4}}{\left (b \sec \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4/(b*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate(csc(f*x + e)^4/(b*sec(f*x + e))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\sin \left (e+f\,x\right )}^4\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^4*(b/cos(e + f*x))^(3/2)),x)

[Out]

int(1/(sin(e + f*x)^4*(b/cos(e + f*x))^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{4}{\left (e + f x \right )}}{\left (b \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**4/(b*sec(f*x+e))**(3/2),x)

[Out]

Integral(csc(e + f*x)**4/(b*sec(e + f*x))**(3/2), x)

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