3.5.48 \(\int \frac {\csc ^4(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\) [448]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.08, 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 = {2703, 2705, 3856, 2719} \begin {gather*} \frac {E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{2 b^2 f \sqrt {\cos (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {\csc ^3(e+f x)}{3 b f (b \sec (e+f x))^{3/2}}+\frac {\csc (e+f x)}{2 b f (b \sec (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2703

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] && Inte
gersQ[2*m, 2*n]

Rule 2705

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(-a)*b*(a*Cs
c[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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 3856

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

________________________________________________________________________________________

Mathematica [A]
time = 0.19, size = 79, normalized size = 0.77 \begin {gather*} \frac {\left (-3+5 \csc ^2(e+f x)-2 \csc ^4(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 )\right ) \sqrt {b \sec (e+f x)} \sin (e+f x)}{6 b^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-3 + 5*Csc[e + f*x]^2 - 2*Csc[e + f*x]^4 + 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])/(6*b^3*f)

________________________________________________________________________________________

Maple [C] Result contains complex when optimal does not.
time = 0.25, size = 623, normalized size = 6.11

method	result	size
default	\(-\frac {\left (\cos \left (f x +e \right )+1\right )^{2} \left (-1+\cos \left (f x +e \right )\right )^{2} \left (3 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 ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-3 i \sin \left (f x +e \right ) \left (\cos ^{3}\left (f x +e \right )\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}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )+3 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 )-3 i \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\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}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}, i\right )-3 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 )+3 i \EllipticE \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 ) \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}}-3 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 )+3 i \sin \left (f x +e \right ) \EllipticE \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}}+3 \left (\cos ^{3}\left (f x +e \right )\right )+2 \left (\cos ^{2}\left (f x +e \right )\right )-3 \cos \left (f x +e \right )\right )}{6 f \cos \left (f x +e \right )^{3} \sin \left (f x +e \right )^{7} \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}\)	\(623\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 175, normalized size = 1.72 \begin {gather*} -\frac {3 \, \sqrt {2} {\left (-i \, \cos \left (f x + e\right )^{2} + i\right )} \sqrt {b} \sin \left (f x + e\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {2} {\left (i \, \cos \left (f x + e\right )^{2} - i\right )} \sqrt {b} \sin \left (f x + e\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - 2 \, {\left (3 \, \cos \left (f x + e\right )^{4} - \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{12 \, {\left (b^{3} f \cos \left (f x + e\right )^{2} - b^{3} f\right )} \sin \left (f x + e\right )} \end {gather*}

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]

-1/12*(3*sqrt(2)*(-I*cos(f*x + e)^2 + I)*sqrt(b)*sin(f*x + e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0
, cos(f*x + e) + I*sin(f*x + e))) + 3*sqrt(2)*(I*cos(f*x + e)^2 - I)*sqrt(b)*sin(f*x + e)*weierstrassZeta(-4,
0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) - 2*(3*cos(f*x + e)^4 - cos(f*x + e)^2)*sqrt(b/c
os(f*x + e)))/((b^3*f*cos(f*x + e)^2 - b^3*f)*sin(f*x + e))

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{4}{\left (e + f x \right )}}{\left (b \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

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]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^4*(b/cos(e + f*x))^(5/2)),x)

[Out]

int(1/(sin(e + f*x)^4*(b/cos(e + f*x))^(5/2)), x)

________________________________________________________________________________________