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

Optimal. Leaf size=50 \[ \frac{2}{5} \cos ^2(x) \cot (x) \sqrt{\cos (x) \cot (x)}-\frac{16}{15} \cot (x) \sqrt{\cos (x) \cot (x)}-\frac{64}{15} \tan (x) \sqrt{\cos (x) \cot (x)} \]

[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

________________________________________________________________________________________

Rubi [A]  time = 0.111701, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {4397, 4400, 2598, 2594, 2589} \[ \frac{2}{5} \cos ^2(x) \cot (x) \sqrt{\cos (x) \cot (x)}-\frac{16}{15} \cot (x) \sqrt{\cos (x) \cot (x)}-\frac{64}{15} \tan (x) \sqrt{\cos (x) \cot (x)} \]

Antiderivative was successfully verified.

[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 4397

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

Rule 4400

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])

Rule 2598

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*Ta
n[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 2594

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 2589

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]

Rubi steps

\begin{align*} \int (\csc (x)-\sin (x))^{5/2} \, dx &=\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]  time = 0.0781942, size = 29, normalized size = 0.58 \[ -\frac{2}{15} \tan (x) \sqrt{\cos (x) \cot (x)} \left (3 \cos ^2(x)+5 \cot ^2(x)+32\right ) \]

Antiderivative was successfully verified.

[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]  time = 0.118, size = 34, normalized size = 0.7 \begin{align*}{\frac{ \left ( 6\, \left ( \cos \left ( x \right ) \right ) ^{4}+48\, \left ( \cos \left ( x \right ) \right ) ^{2}-64 \right ) \sin \left ( x \right ) }{15\, \left ( \cos \left ( x \right ) \right ) ^{5}} \left ({\frac{ \left ( \cos \left ( x \right ) \right ) ^{2}}{\sin \left ( x \right ) }} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*cos(x)^4+24*cos(x)^2-32)*(cos(x)^2/sin(x))^(5/2)*sin(x)/cos(x)^5

________________________________________________________________________________________

Maxima [B]  time = 1.87634, size = 576, normalized size = 11.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

Fricas [A]  time = 2.26392, size = 103, normalized size = 2.06 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (\csc \left (x\right ) - \sin \left (x\right )\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((csc(x) - sin(x))^(5/2), x)