3.383 \(\int \csc ^4(e+f x) \sqrt {b \sec (e+f x)} \, dx\)

Optimal. Leaf size=95 \[ -\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}-\frac {5 b \csc (e+f x)}{6 f \sqrt {b \sec (e+f x)}}+\frac {5 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \sec (e+f x)}}{6 f} \]

[Out]

-5/6*b*csc(f*x+e)/f/(b*sec(f*x+e))^(1/2)-1/3*b*csc(f*x+e)^3/f/(b*sec(f*x+e))^(1/2)+5/6*(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)/f

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Rubi [A]  time = 0.10, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2625, 3771, 2641} \[ -\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}-\frac {5 b \csc (e+f x)}{6 f \sqrt {b \sec (e+f x)}}+\frac {5 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \sec (e+f x)}}{6 f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4*Sqrt[b*Sec[e + f*x]],x]

[Out]

(-5*b*Csc[e + f*x])/(6*f*Sqrt[b*Sec[e + f*x]]) - (b*Csc[e + f*x]^3)/(3*f*Sqrt[b*Sec[e + f*x]]) + (5*Sqrt[Cos[e
+ f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*Sec[e + f*x]])/(6*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 ^4(e+f x) \sqrt {b \sec (e+f x)} \, dx &=-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {5}{6} \int \csc ^2(e+f x) \sqrt {b \sec (e+f x)} \, dx\\ &=-\frac {5 b \csc (e+f x)}{6 f \sqrt {b \sec (e+f x)}}-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {5}{12} \int \sqrt {b \sec (e+f x)} \, dx\\ &=-\frac {5 b \csc (e+f x)}{6 f \sqrt {b \sec (e+f x)}}-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {1}{12} \left (5 \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx\\ &=-\frac {5 b \csc (e+f x)}{6 f \sqrt {b \sec (e+f x)}}-\frac {b \csc ^3(e+f x)}{3 f \sqrt {b \sec (e+f x)}}+\frac {5 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \sec (e+f x)}}{6 f}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 63, normalized size = 0.66 \[ \frac {\sqrt {b \sec (e+f x)} \left (5 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )-\cot (e+f x) \left (2 \csc ^2(e+f x)+5\right )\right )}{6 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4*Sqrt[b*Sec[e + f*x]],x]

[Out]

((-(Cot[e + f*x]*(5 + 2*Csc[e + f*x]^2)) + 5*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])*Sqrt[b*Sec[e + f*x]
])/(6*f)

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sec \left (f x + e\right )} \csc \left (f x + e\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(f*x + e))*csc(f*x + e)^4, x)

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maple [C]  time = 0.23, size = 335, normalized size = 3.53 \[ -\frac {\left (-1+\cos \left (f x +e \right )\right )^{2} \left (5 i \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )+5 i \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-5 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-5 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right )-5 \left (\cos ^{3}\left (f x +e \right )\right )+7 \cos \left (f x +e \right )\right ) \left (\cos \left (f x +e \right )+1\right )^{2} \sqrt {\frac {b}{\cos \left (f x +e \right )}}}{6 f \sin \left (f x +e \right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^4*(b*sec(f*x+e))^(1/2),x)

[Out]

-1/6/f*(-1+cos(f*x+e))^2*(5*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)+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)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)^2*sin(f*x+e)-5*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)-5*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)-5*cos(f*x+e)^3+7*cos(f*
x+e))*(cos(f*x+e)+1)^2*(b/cos(f*x+e))^(1/2)/sin(f*x+e)^7

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sec(f*x + e))*csc(f*x + e)^4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**4*(b*sec(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(b*sec(e + f*x))*csc(e + f*x)**4, x)

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