3.409 \(\int \csc ^4(e+f x) (b \sec (e+f x))^{5/2} \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 123, 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 = {2625, 2626, 3771, 2641} \[ -\frac {5 b^3 \csc (e+f x)}{2 f \sqrt {b \sec (e+f x)}}+\frac {5 b^2 \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {b \sec (e+f x)}}{2 f}-\frac {b \csc ^3(e+f x) (b \sec (e+f x))^{3/2}}{3 f}+\frac {b \csc (e+f x) (b \sec (e+f x))^{3/2}}{f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4*(b*Sec[e + f*x])^(5/2),x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.39, size = 79, normalized size = 0.64 \[ \frac {b \sin (e+f x) (b \sec (e+f x))^{3/2} \left (-\left (\cot ^2(e+f x) \left (2 \csc ^2(e+f x)+11\right )\right )+15 \cos ^{\frac {3}{2}}(e+f x) \csc (e+f x) F\left (\left .\frac {1}{2} (e+f x)\right |2\right )+4\right )}{6 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4*(b*Sec[e + f*x])^(5/2),x]

[Out]

(b*(4 - Cot[e + f*x]^2*(11 + 2*Csc[e + f*x]^2) + 15*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])/(6*f)

________________________________________________________________________________________

fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sec \left (f x + e\right )} b^{2} \csc \left (f x + e\right )^{4} \sec \left (f x + e\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}} \csc \left (f x + e\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^(5/2)*csc(f*x + e)^4, x)

________________________________________________________________________________________

maple [C]  time = 0.23, size = 352, normalized size = 2.86 \[ -\frac {\left (-1+\cos \left (f x +e \right )\right )^{2} \left (15 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )+15 i \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )-15 i \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-15 i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )-15 \left (\cos ^{4}\left (f x +e \right )\right )+21 \left (\cos ^{2}\left (f x +e \right )\right )-4\right ) \cos \left (f x +e \right ) \left (\cos \left (f x +e \right )+1\right )^{2} \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}{6 f \sin \left (f x +e \right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^4*(b*sec(f*x+e))^(5/2),x)

[Out]

-1/6/f*(-1+cos(f*x+e))^2*(15*I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)^4*sin(f*x
+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)+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)*cos(f*x+e)^3*sin(f*x+e)-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)*cos(f*x+e)^2*sin(f*x+e)-15*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)-15*cos
(f*x+e)^4+21*cos(f*x+e)^2-4)*cos(f*x+e)*(cos(f*x+e)+1)^2*(b/cos(f*x+e))^(5/2)/sin(f*x+e)^7

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x + e\right )\right )^{\frac {5}{2}} \csc \left (f x + e\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(b*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^(5/2)*csc(f*x + e)^4, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\sin \left (e+f\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cos(e + f*x))^(5/2)/sin(e + f*x)^4,x)

[Out]

int((b/cos(e + f*x))^(5/2)/sin(e + f*x)^4, x)

________________________________________________________________________________________

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)**4*(b*sec(f*x+e))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________