\(\int \frac {1}{\csc ^{\frac {5}{2}}(a+b x)} \, dx\) [15]

Optimal result

Integrand size = 10, antiderivative size = 67 \[ \int \frac {1}{\csc ^{\frac {5}{2}}(a+b x)} \, dx=-\frac {2 \cos (a+b x)}{5 b \csc ^{\frac {3}{2}}(a+b x)}+\frac {6 \sqrt {\csc (a+b x)} E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {\sin (a+b x)}}{5 b} \]

[Out]

-2/5*cos(b*x+a)/b/csc(b*x+a)^(3/2)-6/5*(sin(1/2*a+1/4*Pi+1/2*b*x)^2)^(1/2)/sin(1/2*a+1/4*Pi+1/2*b*x)*EllipticE
(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))*csc(b*x+a)^(1/2)*sin(b*x+a)^(1/2)/b

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3854, 3856, 2719} \[ \int \frac {1}{\csc ^{\frac {5}{2}}(a+b x)} \, dx=\frac {6 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{5 b}-\frac {2 \cos (a+b x)}{5 b \csc ^{\frac {3}{2}}(a+b x)} \]

[In]

Int[Csc[a + b*x]^(-5/2),x]

[Out]

(-2*Cos[a + b*x])/(5*b*Csc[a + b*x]^(3/2)) + (6*Sqrt[Csc[a + b*x]]*EllipticE[(a - Pi/2 + b*x)/2, 2]*Sqrt[Sin[a
+ b*x]])/(5*b)

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 3854

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Csc[c + d*x])^(n + 1)/(b*d*n)), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

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*} \text {integral}& = -\frac {2 \cos (a+b x)}{5 b \csc ^{\frac {3}{2}}(a+b x)}+\frac {3}{5} \int \frac {1}{\sqrt {\csc (a+b x)}} \, dx \\ & = -\frac {2 \cos (a+b x)}{5 b \csc ^{\frac {3}{2}}(a+b x)}+\frac {1}{5} \left (3 \sqrt {\csc (a+b x)} \sqrt {\sin (a+b x)}\right ) \int \sqrt {\sin (a+b x)} \, dx \\ & = -\frac {2 \cos (a+b x)}{5 b \csc ^{\frac {3}{2}}(a+b x)}+\frac {6 \sqrt {\csc (a+b x)} E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {\sin (a+b x)}}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\csc ^{\frac {5}{2}}(a+b x)} \, dx=-\frac {2 \sqrt {\csc (a+b x)} \left (3 E\left (\left .\frac {1}{4} (-2 a+\pi -2 b x)\right |2\right ) \sqrt {\sin (a+b x)}+\cos (a+b x) \sin ^2(a+b x)\right )}{5 b} \]

[In]

Integrate[Csc[a + b*x]^(-5/2),x]

[Out]

(-2*Sqrt[Csc[a + b*x]]*(3*EllipticE[(-2*a + Pi - 2*b*x)/4, 2]*Sqrt[Sin[a + b*x]] + Cos[a + b*x]*Sin[a + b*x]^2
))/(5*b)

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 142, normalized size of antiderivative = 2.12

method	result	size
default	\(\frac {\frac {2 \sin \left (x b +a \right )^{4}}{5}-\frac {2 \sin \left (x b +a \right )^{2}}{5}-\frac {6 \sqrt {\sin \left (x b +a \right )+1}\, \sqrt {-2 \sin \left (x b +a \right )+2}\, \sqrt {-\sin \left (x b +a \right )}\, \operatorname {EllipticE}\left (\sqrt {\sin \left (x b +a \right )+1}, \frac {\sqrt {2}}{2}\right )}{5}+\frac {3 \sqrt {\sin \left (x b +a \right )+1}\, \sqrt {-2 \sin \left (x b +a \right )+2}\, \sqrt {-\sin \left (x b +a \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (x b +a \right )+1}, \frac {\sqrt {2}}{2}\right )}{5}}{\cos \left (x b +a \right ) \sqrt {\sin \left (x b +a \right )}\, b}\)	\(142\)

[In]

int(1/csc(b*x+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

(2/5*sin(b*x+a)^4-2/5*sin(b*x+a)^2-6/5*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*Ellipt
icE((sin(b*x+a)+1)^(1/2),1/2*2^(1/2))+3/5*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*Ell
ipticF((sin(b*x+a)+1)^(1/2),1/2*2^(1/2)))/cos(b*x+a)/sin(b*x+a)^(1/2)/b

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.27 \[ \int \frac {1}{\csc ^{\frac {5}{2}}(a+b x)} \, dx=\frac {3 \, \sqrt {2 i} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 3 \, \sqrt {-2 i} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right ) + \frac {2 \, {\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )}}{\sqrt {\sin \left (b x + a\right )}}}{5 \, b} \]

[In]

integrate(1/csc(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

1/5*(3*sqrt(2*I)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(b*x + a) + I*sin(b*x + a))) + 3*sqrt(-2*I
)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(b*x + a) - I*sin(b*x + a))) + 2*(cos(b*x + a)^3 - cos(b*
x + a))/sqrt(sin(b*x + a)))/b

Sympy [F]

\[ \int \frac {1}{\csc ^{\frac {5}{2}}(a+b x)} \, dx=\int \frac {1}{\csc ^{\frac {5}{2}}{\left (a + b x \right )}}\, dx \]

[In]

integrate(1/csc(b*x+a)**(5/2),x)

[Out]

Integral(csc(a + b*x)**(-5/2), x)

Maxima [F]

\[ \int \frac {1}{\csc ^{\frac {5}{2}}(a+b x)} \, dx=\int { \frac {1}{\csc \left (b x + a\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/csc(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)^(-5/2), x)

Giac [F]

\[ \int \frac {1}{\csc ^{\frac {5}{2}}(a+b x)} \, dx=\int { \frac {1}{\csc \left (b x + a\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/csc(b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)^(-5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\csc ^{\frac {5}{2}}(a+b x)} \, dx=\int \frac {1}{{\left (\frac {1}{\sin \left (a+b\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int(1/(1/sin(a + b*x))^(5/2),x)

[Out]

int(1/(1/sin(a + b*x))^(5/2), x)