\(\int (\csc (x)-\sin (x))^{5/2} \, dx\) [315]
Optimal result
Integrand size = 11, antiderivative size = 50 \[
\int (\csc (x)-\sin (x))^{5/2} \, dx=-\frac {16}{15} \cot (x) \sqrt {\cos (x) \cot (x)}+\frac {2}{5} \cos ^2(x) \cot (x) \sqrt {\cos (x) \cot (x)}-\frac {64}{15} \sqrt {\cos (x) \cot (x)} \tan (x)
\]
[Out]
-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)
Rubi [A] (verified)
Time = 0.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of
steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {4482, 4485, 2678, 2674, 2669}
\[
\int (\csc (x)-\sin (x))^{5/2} \, dx=\frac {2}{5} \cos ^2(x) \cot (x) \sqrt {\cos (x) \cot (x)}-\frac {16}{15} \cot (x) \sqrt {\cos (x) \cot (x)}-\frac {64}{15} \tan (x) \sqrt {\cos (x) \cot (x)}
\]
[In]
Int[(Csc[x] - Sin[x])^(5/2),x]
[Out]
(-16*Cot[x]*Sqrt[Cos[x]*Cot[x]])/15 + (2*Cos[x]^2*Cot[x]*Sqrt[Cos[x]*Cot[x]])/5 - (64*Sqrt[Cos[x]*Cot[x]]*Tan[
x])/15
Rule 2669
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 2674
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] - Dist[b^2*((m + n - 1)/(n - 1)), Int[(a*Sin[e + f*x])^m*(
b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && IntegersQ[2*m, 2*n] && !(GtQ[m,
1] && !IntegerQ[(m - 1)/2])
Rule 2678
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] + Dist[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 4482
Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]
Rule 4485
Int[(u_.)*((v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> With[{uu = ActivateTrig[u], vv = ActivateTrig[v], ww = Ac
tivateTrig[w]}, Dist[(vv^m*ww^n)^FracPart[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])
Rubi steps \begin{align*}
\text {integral}& = \int (\cos (x) \cot (x))^{5/2} \, dx \\ & = \frac {\sqrt {\cos (x) \cot (x)} \int \cos ^{\frac {5}{2}}(x) \cot ^{\frac {5}{2}}(x) \, dx}{\sqrt {\cos (x)} \sqrt {\cot (x)}} \\ & = \frac {2}{5} \cos ^2(x) \cot (x) \sqrt {\cos (x) \cot (x)}+\frac {\left (8 \sqrt {\cos (x) \cot (x)}\right ) \int \sqrt {\cos (x)} \cot ^{\frac {5}{2}}(x) \, dx}{5 \sqrt {\cos (x)} \sqrt {\cot (x)}} \\ & = -\frac {16}{15} \cot (x) \sqrt {\cos (x) \cot (x)}+\frac {2}{5} \cos ^2(x) \cot (x) \sqrt {\cos (x) \cot (x)}-\frac {\left (32 \sqrt {\cos (x) \cot (x)}\right ) \int \sqrt {\cos (x)} \sqrt {\cot (x)} \, dx}{15 \sqrt {\cos (x)} \sqrt {\cot (x)}} \\ & = -\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) \\
\end{align*}
Mathematica [A] (verified)
Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.58
\[
\int (\csc (x)-\sin (x))^{5/2} \, dx=-\frac {2}{15} \sqrt {\cos (x) \cot (x)} \left (32+3 \cos ^2(x)+5 \cot ^2(x)\right ) \tan (x)
\]
[In]
Integrate[(Csc[x] - Sin[x])^(5/2),x]
[Out]
(-2*Sqrt[Cos[x]*Cot[x]]*(32 + 3*Cos[x]^2 + 5*Cot[x]^2)*Tan[x])/15
Maple [A] (verified)
Time = 2.59 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.58
| | |
| method | result | size |
| | |
| default |
\(\frac {2 \sqrt {\cot \left (x \right ) \cos \left (x \right )}\, \left (3 \cos \left (x \right )^{2} \cot \left (x \right )+24 \cot \left (x \right )-32 \sec \left (x \right ) \csc \left (x \right )\right )}{15}\) |
\(29\) |
| | |
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[In]
int((csc(x)-sin(x))^(5/2),x,method=_RETURNVERBOSE)
[Out]
2/15*(cot(x)*cos(x))^(1/2)*(3*cos(x)^2*cot(x)+24*cot(x)-32*sec(x)*csc(x))
Fricas [A] (verification not implemented)
none
Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.70
\[
\int (\csc (x)-\sin (x))^{5/2} \, dx=\frac {2 \, {\left (3 \, \cos \left (x\right )^{4} + 24 \, \cos \left (x\right )^{2} - 32\right )} \sqrt {\frac {\cos \left (x\right )^{2}}{\sin \left (x\right )}}}{15 \, \cos \left (x\right ) \sin \left (x\right )}
\]
[In]
integrate((csc(x)-sin(x))^(5/2),x, algorithm="fricas")
[Out]
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. \[
\int (\csc (x)-\sin (x))^{5/2} \, dx=\text {Timed out}
\]
[In]
integrate((csc(x)-sin(x))**(5/2),x)
[Out]
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 = 0.42 (sec) , antiderivative size = 427, normalized size of antiderivative = 8.54
\[
\int (\csc (x)-\sin (x))^{5/2} \, dx=\text {Too large to display}
\]
[In]
integrate((csc(x)-sin(x))^(5/2),x, algorithm="maxima")
[Out]
-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*sin(7/2*x) + 3*sin(5/2*x) + 410*sin(3/2*x) + 105*sin(1/2*x))*cos(5/2*arct
an2(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)))*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))*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), co
s(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)*(cos(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]
\[
\int (\csc (x)-\sin (x))^{5/2} \, dx=\int { {\left (\csc \left (x\right ) - \sin \left (x\right )\right )}^{\frac {5}{2}} \,d x }
\]
[In]
integrate((csc(x)-sin(x))^(5/2),x, algorithm="giac")
[Out]
integrate((csc(x) - sin(x))^(5/2), x)
Mupad [F(-1)]
Timed out. \[
\int (\csc (x)-\sin (x))^{5/2} \, dx=\int {\left (\frac {1}{\sin \left (x\right )}-\sin \left (x\right )\right )}^{5/2} \,d x
\]
[In]
int((1/sin(x) - sin(x))^(5/2),x)
[Out]
int((1/sin(x) - sin(x))^(5/2), x)