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]

-16/15*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)

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Rubi [A]  time = 0.11, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 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]

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 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 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])

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*}

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Mathematica [A]  time = 0.08, 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

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fricas [A]  time = 1.72, size = 35, normalized size = 0.70 \[ \frac {2 \, {\left (3 \, \cos \relax (x)^{4} + 24 \, \cos \relax (x)^{2} - 32\right )} \sqrt {\frac {\cos \relax (x)^{2}}{\sin \relax (x)}}}{15 \, \cos \relax (x) \sin \relax (x)} \]

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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (\csc \relax (x) - \sin \relax (x)\right )}^{\frac {5}{2}}\,{d x} \]

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)

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maple [A]  time = 0.28, size = 34, normalized size = 0.68 \[ \frac {2 \left (3 \left (\cos ^{4}\relax (x )\right )+24 \left (\cos ^{2}\relax (x )\right )-32\right ) \left (\frac {\cos ^{2}\relax (x )}{\sin \relax (x )}\right )^{\frac {5}{2}} \sin \relax (x )}{15 \cos \relax (x )^{5}} \]

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

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maxima [B]  time = 0.56, size = 427, normalized size = 8.54 \[ \text {result too large to display} \]

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))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int {\left (\frac {1}{\sin \relax (x)}-\sin \relax (x)\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/sin(x) - sin(x))^(5/2),x)

[Out]

int((1/sin(x) - sin(x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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