\(\int (\csc (x)-\sin (x))^{5/2} \, dx\) [208]

Optimal result

Integrand size = 11, antiderivative size = 50 \[ \int (\csc (x)-\sin (x))^{5/2} \, dx=-\frac {16}{15} \cot (x) \sqrt {\cos (x) \cot (x)}+\frac {2}{5} \cos ^2(x) \cot (x) \sqrt {\cos (x) \cot (x)}-\frac {64}{15} \sqrt {\cos (x) \cot (x)} \tan (x) \] Output:

-16/15*cot(x)*(cos(x)*cot(x))^(1/2)+2/5*cos(x)^2*cot(x)*(cos(x)*cot(x))^(1 
/2)-64/15*(cos(x)*cot(x))^(1/2)*tan(x)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.58 \[ \int (\csc (x)-\sin (x))^{5/2} \, dx=-\frac {2}{15} \sqrt {\cos (x) \cot (x)} \left (32+3 \cos ^2(x)+5 \cot ^2(x)\right ) \tan (x) \] Input:

Integrate[(Csc[x] - Sin[x])^(5/2),x]
 

Output:

(-2*Sqrt[Cos[x]*Cot[x]]*(32 + 3*Cos[x]^2 + 5*Cot[x]^2)*Tan[x])/15
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.52, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.909, Rules used = {3042, 4897, 3042, 4900, 3042, 3078, 3042, 3074, 3042, 3069}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (\csc (x)-\sin (x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (\csc (x)-\sin (x))^{5/2}dx\)

\(\Big \downarrow \) 4897

\(\displaystyle \int (\cos (x) \cot (x))^{5/2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (\cos (x) \cot (x))^{5/2}dx\)

\(\Big \downarrow \) 4900

\(\displaystyle \frac {\sqrt {\cos (x) \cot (x)} \int \cos ^{\frac {5}{2}}(x) \cot ^{\frac {5}{2}}(x)dx}{\sqrt {\cos (x)} \sqrt {\cot (x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cos (x) \cot (x)} \int \sin \left (x+\frac {\pi }{2}\right )^{5/2} \left (-\tan \left (x+\frac {\pi }{2}\right )\right )^{5/2}dx}{\sqrt {\cos (x)} \sqrt {\cot (x)}}\)

\(\Big \downarrow \) 3078

\(\displaystyle \frac {\sqrt {\cos (x) \cot (x)} \left (\frac {8}{5} \int \sqrt {\cos (x)} \cot ^{\frac {5}{2}}(x)dx+\frac {2}{5} \cos ^{\frac {5}{2}}(x) \cot ^{\frac {3}{2}}(x)\right )}{\sqrt {\cos (x)} \sqrt {\cot (x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cos (x) \cot (x)} \left (\frac {8}{5} \int \sqrt {\sin \left (x+\frac {\pi }{2}\right )} \left (-\tan \left (x+\frac {\pi }{2}\right )\right )^{5/2}dx+\frac {2}{5} \cos ^{\frac {5}{2}}(x) \cot ^{\frac {3}{2}}(x)\right )}{\sqrt {\cos (x)} \sqrt {\cot (x)}}\)

\(\Big \downarrow \) 3074

\(\displaystyle \frac {\sqrt {\cos (x) \cot (x)} \left (\frac {8}{5} \left (-\frac {4}{3} \int \sqrt {\cos (x)} \sqrt {\cot (x)}dx-\frac {2}{3} \sqrt {\cos (x)} \cot ^{\frac {3}{2}}(x)\right )+\frac {2}{5} \cos ^{\frac {5}{2}}(x) \cot ^{\frac {3}{2}}(x)\right )}{\sqrt {\cos (x)} \sqrt {\cot (x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\cos (x) \cot (x)} \left (\frac {8}{5} \left (-\frac {4}{3} \int \sqrt {\sin \left (x+\frac {\pi }{2}\right )} \sqrt {-\tan \left (x+\frac {\pi }{2}\right )}dx-\frac {2}{3} \sqrt {\cos (x)} \cot ^{\frac {3}{2}}(x)\right )+\frac {2}{5} \cos ^{\frac {5}{2}}(x) \cot ^{\frac {3}{2}}(x)\right )}{\sqrt {\cos (x)} \sqrt {\cot (x)}}\)

\(\Big \downarrow \) 3069

\(\displaystyle \frac {\sqrt {\cos (x) \cot (x)} \left (\frac {2}{5} \cos ^{\frac {5}{2}}(x) \cot ^{\frac {3}{2}}(x)+\frac {8}{5} \left (-\frac {2}{3} \sqrt {\cos (x)} \cot ^{\frac {3}{2}}(x)-\frac {8 \sqrt {\cos (x)}}{3 \sqrt {\cot (x)}}\right )\right )}{\sqrt {\cos (x)} \sqrt {\cot (x)}}\)

Input:

Int[(Csc[x] - Sin[x])^(5/2),x]
 

Output:

(Sqrt[Cos[x]*Cot[x]]*((2*Cos[x]^(5/2)*Cot[x]^(3/2))/5 + (8*((-8*Sqrt[Cos[x 
]])/(3*Sqrt[Cot[x]]) - (2*Sqrt[Cos[x]]*Cot[x]^(3/2))/3))/5))/(Sqrt[Cos[x]] 
*Sqrt[Cot[x]])
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3069

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Simp[(-b)*(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f* 
m)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n - 1, 0]
 

rule 3074

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[b*(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(n 
 - 1))), x] - Simp[b^2*((m + n - 1)/(n - 1))   Int[(a*Sin[e + f*x])^m*(b*Ta 
n[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && In 
tegersQ[2*m, 2*n] &&  !(GtQ[m, 1] &&  !IntegerQ[(m - 1)/2])
 

rule 3078

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_.), x_Symbol] :> Simp[(-b)*(a*Sin[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/( 
f*m)), x] + Simp[a^2*((m + n - 1)/m)   Int[(a*Sin[e + f*x])^(m - 2)*(b*Tan[ 
e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1 
] && EqQ[n, 1/2])) && IntegersQ[2*m, 2*n]
 

rule 4897

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]
 

rule 4900

Int[(u_.)*((v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> With[{uu = ActivateTri 
g[u], vv = ActivateTrig[v], ww = ActivateTrig[w]}, Simp[(vv^m*ww^n)^FracPar 
t[p]/(vv^(m*FracPart[p])*ww^(n*FracPart[p]))   Int[uu*vv^(m*p)*ww^(n*p), x] 
, x]] /; FreeQ[{m, n, p}, x] &&  !IntegerQ[p] && ( !InertTrigFreeQ[v] ||  ! 
InertTrigFreeQ[w])
 

Maple [A] (verified)

Time = 1.43 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.58

method	result	size
default	\(\frac {2 \sqrt {\cos \left (x \right ) \cot \left (x \right )}\, \left (3 \cos \left (x \right )^{2} \cot \left (x \right )+24 \cot \left (x \right )-32 \sec \left (x \right ) \csc \left (x \right )\right )}{15}\)	\(29\)

Input:

int((csc(x)-sin(x))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/15*(cos(x)*cot(x))^(1/2)*(3*cos(x)^2*cot(x)+24*cot(x)-32*sec(x)*csc(x))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.70 \[ \int (\csc (x)-\sin (x))^{5/2} \, dx=\frac {2 \, {\left (3 \, \cos \left (x\right )^{4} + 24 \, \cos \left (x\right )^{2} - 32\right )} \sqrt {\frac {\cos \left (x\right )^{2}}{\sin \left (x\right )}}}{15 \, \cos \left (x\right ) \sin \left (x\right )} \] Input:

integrate((csc(x)-sin(x))^(5/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

2/15*(3*cos(x)^4 + 24*cos(x)^2 - 32)*sqrt(cos(x)^2/sin(x))/(cos(x)*sin(x))
 

Sympy [F(-1)]

Timed out. \[ \int (\csc (x)-\sin (x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate((csc(x)-sin(x))**(5/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (38) = 76\).

Time = 0.22 (sec) , antiderivative size = 427, normalized size of antiderivative = 8.54 \[ \int (\csc (x)-\sin (x))^{5/2} \, dx=\text {Too large to display} \] Input:

integrate((csc(x)-sin(x))^(5/2),x, algorithm="maxima")
 

Output:

-1/60*(((3*cos(15/2*x) + 105*cos(11/2*x) - 410*cos(7/2*x) - 3*cos(5/2*x) + 
 410*cos(3/2*x) - 105*cos(1/2*x) + 3*sin(15/2*x) + 105*sin(11/2*x) - 410*s 
in(7/2*x) + 3*sin(5/2*x) + 410*sin(3/2*x) + 105*sin(1/2*x))*cos(5/2*arctan 
2(sin(x), cos(x) - 1)) - (3*cos(15/2*x) + 105*cos(11/2*x) - 410*cos(7/2*x) 
 - 3*cos(5/2*x) + 410*cos(3/2*x) - 105*cos(1/2*x) - 3*sin(15/2*x) - 105*si 
n(11/2*x) + 410*sin(7/2*x) - 3*sin(5/2*x) - 410*sin(3/2*x) - 105*sin(1/2*x 
))*sin(5/2*arctan2(sin(x), cos(x) - 1)))*cos(5/2*arctan2(sin(x), cos(x) + 
1)) - ((3*cos(15/2*x) + 105*cos(11/2*x) - 410*cos(7/2*x) - 3*cos(5/2*x) + 
410*cos(3/2*x) - 105*cos(1/2*x) - 3*sin(15/2*x) - 105*sin(11/2*x) + 410*si 
n(7/2*x) - 3*sin(5/2*x) - 410*sin(3/2*x) - 105*sin(1/2*x))*cos(5/2*arctan2 
(sin(x), cos(x) - 1)) + (3*cos(15/2*x) + 105*cos(11/2*x) - 410*cos(7/2*x) 
- 3*cos(5/2*x) + 410*cos(3/2*x) - 105*cos(1/2*x) + 3*sin(15/2*x) + 105*sin 
(11/2*x) - 410*sin(7/2*x) + 3*sin(5/2*x) + 410*sin(3/2*x) + 105*sin(1/2*x) 
)*sin(5/2*arctan2(sin(x), cos(x) - 1)))*sin(5/2*arctan2(sin(x), cos(x) + 1 
)))/((cos(x)^4 + sin(x)^4 + 2*(cos(x)^2 + 1)*sin(x)^2 - 2*cos(x)^2 + 1)*(c 
os(x)^2 + sin(x)^2 + 2*cos(x) + 1)^(1/4)*(cos(x)^2 + sin(x)^2 - 2*cos(x) + 
 1)^(1/4))
 

Giac [F]

\[ \int (\csc (x)-\sin (x))^{5/2} \, dx=\int { {\left (\csc \left (x\right ) - \sin \left (x\right )\right )}^{\frac {5}{2}} \,d x } \] Input:

integrate((csc(x)-sin(x))^(5/2),x, algorithm="giac")
 

Output:

integrate((csc(x) - sin(x))^(5/2), x)
 

Mupad [F(-1)]

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

int((1/sin(x) - sin(x))^(5/2),x)
 

Output:

int((1/sin(x) - sin(x))^(5/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int (\csc (x)-\sin (x))^{5/2} \, dx=-2 \left (\int \sqrt {\csc \left (x \right )-\sin \left (x \right )}\, \csc \left (x \right ) \sin \left (x \right )d x \right )+\int \sqrt {\csc \left (x \right )-\sin \left (x \right )}\, \csc \left (x \right )^{2}d x +\int \sqrt {\csc \left (x \right )-\sin \left (x \right )}\, \sin \left (x \right )^{2}d x \] Input:

int((csc(x)-sin(x))^(5/2),x)
 

Output:

 - 2*int(sqrt(csc(x) - sin(x))*csc(x)*sin(x),x) + int(sqrt(csc(x) - sin(x) 
)*csc(x)**2,x) + int(sqrt(csc(x) - sin(x))*sin(x)**2,x)