3.320 \(\int \frac{1}{(\csc (x)-\sin (x))^{5/2}} \, dx\)

Optimal. Leaf size=99 \[ -\frac{3 \tan (x)}{16 \sqrt{\cos (x) \cot (x)}}-\frac{3 \cos (x) \tan ^{-1}\left (\sqrt{-\sin (x)}\right )}{32 \sqrt{-\sin (x)} \sqrt{\cos (x) \cot (x)}}+\frac{3 \cos (x) \tanh ^{-1}\left (\sqrt{-\sin (x)}\right )}{32 \sqrt{-\sin (x)} \sqrt{\cos (x) \cot (x)}}+\frac{\tan (x) \sec ^2(x)}{4 \sqrt{\cos (x) \cot (x)}} \]

[Out]

(-3*ArcTan[Sqrt[-Sin[x]]]*Cos[x])/(32*Sqrt[Cos[x]*Cot[x]]*Sqrt[-Sin[x]]) + (3*ArcTanh[Sqrt[-Sin[x]]]*Cos[x])/(
32*Sqrt[Cos[x]*Cot[x]]*Sqrt[-Sin[x]]) - (3*Tan[x])/(16*Sqrt[Cos[x]*Cot[x]]) + (Sec[x]^2*Tan[x])/(4*Sqrt[Cos[x]
*Cot[x]])

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Rubi [A]  time = 0.151284, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.909, Rules used = {4397, 4400, 2597, 2599, 2601, 2564, 329, 298, 203, 206} \[ -\frac{3 \tan (x)}{16 \sqrt{\cos (x) \cot (x)}}-\frac{3 \cos (x) \tan ^{-1}\left (\sqrt{-\sin (x)}\right )}{32 \sqrt{-\sin (x)} \sqrt{\cos (x) \cot (x)}}+\frac{3 \cos (x) \tanh ^{-1}\left (\sqrt{-\sin (x)}\right )}{32 \sqrt{-\sin (x)} \sqrt{\cos (x) \cot (x)}}+\frac{\tan (x) \sec ^2(x)}{4 \sqrt{\cos (x) \cot (x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[x] - Sin[x])^(-5/2),x]

[Out]

(-3*ArcTan[Sqrt[-Sin[x]]]*Cos[x])/(32*Sqrt[Cos[x]*Cot[x]]*Sqrt[-Sin[x]]) + (3*ArcTanh[Sqrt[-Sin[x]]]*Cos[x])/(
32*Sqrt[Cos[x]*Cot[x]]*Sqrt[-Sin[x]]) - (3*Tan[x])/(16*Sqrt[Cos[x]*Cot[x]]) + (Sec[x]^2*Tan[x])/(4*Sqrt[Cos[x]
*Cot[x]])

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 2597

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sin[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n + 1)), x] - Dist[(n + 1)/(b^2*(m + 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] && LtQ[n, -1] && NeQ[m + n + 1, 0] && Integer
sQ[2*m, 2*n] &&  !(EqQ[n, -3/2] && EqQ[m, 1])

Rule 2599

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(b*(a*Sin[e
+ f*x])^(m + 2)*(b*Tan[e + f*x])^(n - 1))/(a^2*f*(m + n + 1)), x] + Dist[(m + 2)/(a^2*(m + n + 1)), Int[(a*Sin
[e + f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && NeQ[m + n + 1, 0]
&& IntegersQ[2*m, 2*n]

Rule 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x
]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{(\csc (x)-\sin (x))^{5/2}} \, dx &=\int \frac{1}{(\cos (x) \cot (x))^{5/2}} \, dx\\ &=\frac{\left (\sqrt{\cos (x)} \sqrt{\cot (x)}\right ) \int \frac{1}{\cos ^{\frac{5}{2}}(x) \cot ^{\frac{5}{2}}(x)} \, dx}{\sqrt{\cos (x) \cot (x)}}\\ &=\frac{\sec ^2(x) \tan (x)}{4 \sqrt{\cos (x) \cot (x)}}-\frac{\left (3 \sqrt{\cos (x)} \sqrt{\cot (x)}\right ) \int \frac{1}{\cos ^{\frac{5}{2}}(x) \sqrt{\cot (x)}} \, dx}{8 \sqrt{\cos (x) \cot (x)}}\\ &=-\frac{3 \tan (x)}{16 \sqrt{\cos (x) \cot (x)}}+\frac{\sec ^2(x) \tan (x)}{4 \sqrt{\cos (x) \cot (x)}}-\frac{\left (3 \sqrt{\cos (x)} \sqrt{\cot (x)}\right ) \int \frac{1}{\sqrt{\cos (x)} \sqrt{\cot (x)}} \, dx}{32 \sqrt{\cos (x) \cot (x)}}\\ &=-\frac{3 \tan (x)}{16 \sqrt{\cos (x) \cot (x)}}+\frac{\sec ^2(x) \tan (x)}{4 \sqrt{\cos (x) \cot (x)}}-\frac{(3 \cos (x)) \int \sec (x) \sqrt{-\sin (x)} \, dx}{32 \sqrt{\cos (x) \cot (x)} \sqrt{-\sin (x)}}\\ &=-\frac{3 \tan (x)}{16 \sqrt{\cos (x) \cot (x)}}+\frac{\sec ^2(x) \tan (x)}{4 \sqrt{\cos (x) \cot (x)}}+\frac{(3 \cos (x)) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1-x^2} \, dx,x,-\sin (x)\right )}{32 \sqrt{\cos (x) \cot (x)} \sqrt{-\sin (x)}}\\ &=-\frac{3 \tan (x)}{16 \sqrt{\cos (x) \cot (x)}}+\frac{\sec ^2(x) \tan (x)}{4 \sqrt{\cos (x) \cot (x)}}+\frac{(3 \cos (x)) \operatorname{Subst}\left (\int \frac{x^2}{1-x^4} \, dx,x,\sqrt{-\sin (x)}\right )}{16 \sqrt{\cos (x) \cot (x)} \sqrt{-\sin (x)}}\\ &=-\frac{3 \tan (x)}{16 \sqrt{\cos (x) \cot (x)}}+\frac{\sec ^2(x) \tan (x)}{4 \sqrt{\cos (x) \cot (x)}}+\frac{(3 \cos (x)) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{-\sin (x)}\right )}{32 \sqrt{\cos (x) \cot (x)} \sqrt{-\sin (x)}}-\frac{(3 \cos (x)) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{-\sin (x)}\right )}{32 \sqrt{\cos (x) \cot (x)} \sqrt{-\sin (x)}}\\ &=-\frac{3 \tan ^{-1}\left (\sqrt{-\sin (x)}\right ) \cos (x)}{32 \sqrt{\cos (x) \cot (x)} \sqrt{-\sin (x)}}+\frac{3 \tanh ^{-1}\left (\sqrt{-\sin (x)}\right ) \cos (x)}{32 \sqrt{\cos (x) \cot (x)} \sqrt{-\sin (x)}}-\frac{3 \tan (x)}{16 \sqrt{\cos (x) \cot (x)}}+\frac{\sec ^2(x) \tan (x)}{4 \sqrt{\cos (x) \cot (x)}}\\ \end{align*}

Mathematica [A]  time = 0.532499, size = 69, normalized size = 0.7 \[ -\frac{\sin (x) \tan (x) \sqrt{\cos (x) \cot (x)} \left (-3 \tan ^{-1}\left (\sqrt [4]{\sin ^2(x)}\right )+3 \tanh ^{-1}\left (\sqrt [4]{\sin ^2(x)}\right )+\sin ^2(x)^{3/4} (3 \cos (2 x)-5) \sec ^4(x)\right )}{32 \sin ^2(x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[x] - Sin[x])^(-5/2),x]

[Out]

-(Sqrt[Cos[x]*Cot[x]]*Sin[x]*(-3*ArcTan[(Sin[x]^2)^(1/4)] + 3*ArcTanh[(Sin[x]^2)^(1/4)] + (-5 + 3*Cos[2*x])*Se
c[x]^4*(Sin[x]^2)^(3/4))*Tan[x])/(32*(Sin[x]^2)^(3/4))

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Maple [C]  time = 0.191, size = 382, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*2^(1/2)*(-1+cos(x))*(3*I*cos(x)^4*EllipticPi(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((
-I*cos(x)+sin(x)+I)/sin(x))^(1/2)*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-I*(-1+cos(x))/sin(x))^(1/2)-3*I*cos(x)^
4*EllipticPi(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-I*cos(x)+sin(x)+I)/sin(x))^(1/2)*((I
*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-I*(-1+cos(x))/sin(x))^(1/2)+3*cos(x)^4*EllipticPi(((I*cos(x)+sin(x)-I)/sin(x
))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((-I*cos(x)+sin(x)+I)/sin(x))^(1/2)*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-I*(-1
+cos(x))/sin(x))^(1/2)+3*cos(x)^4*EllipticPi(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-I*co
s(x)+sin(x)+I)/sin(x))^(1/2)*((I*cos(x)+sin(x)-I)/sin(x))^(1/2)*(-I*(-1+cos(x))/sin(x))^(1/2)-6*cos(x)^3*2^(1/
2)+6*cos(x)^2*2^(1/2)+8*cos(x)*2^(1/2)-8*2^(1/2))*cos(x)*(1+cos(x))^2/sin(x)^5/(cos(x)^2/sin(x))^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\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(1/(csc(x)-sin(x))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.67216, size = 512, normalized size = 5.17 \begin{align*} -\frac{6 \, \arctan \left (\frac{2 \, \sqrt{\frac{\cos \left (x\right )^{2}}{\sin \left (x\right )}} \sin \left (x\right )}{\cos \left (x\right ) \sin \left (x\right ) - \cos \left (x\right )}\right ) \cos \left (x\right )^{5} - 3 \, \cos \left (x\right )^{5} \log \left (\frac{\cos \left (x\right )^{3} - 5 \, \cos \left (x\right )^{2} -{\left (\cos \left (x\right )^{2} + 6 \, \cos \left (x\right ) + 4\right )} \sin \left (x\right ) - 4 \,{\left (\cos \left (x\right )^{2} -{\left (\cos \left (x\right ) + 1\right )} \sin \left (x\right ) - 1\right )} \sqrt{\frac{\cos \left (x\right )^{2}}{\sin \left (x\right )}} - 2 \, \cos \left (x\right ) + 4}{\cos \left (x\right )^{3} + 3 \, \cos \left (x\right )^{2} -{\left (\cos \left (x\right )^{2} - 2 \, \cos \left (x\right ) - 4\right )} \sin \left (x\right ) - 2 \, \cos \left (x\right ) - 4}\right ) - 8 \,{\left (3 \, \cos \left (x\right )^{4} - 7 \, \cos \left (x\right )^{2} + 4\right )} \sqrt{\frac{\cos \left (x\right )^{2}}{\sin \left (x\right )}}}{128 \, \cos \left (x\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(csc(x)-sin(x))^(5/2),x, algorithm="fricas")

[Out]

-1/128*(6*arctan(2*sqrt(cos(x)^2/sin(x))*sin(x)/(cos(x)*sin(x) - cos(x)))*cos(x)^5 - 3*cos(x)^5*log((cos(x)^3
- 5*cos(x)^2 - (cos(x)^2 + 6*cos(x) + 4)*sin(x) - 4*(cos(x)^2 - (cos(x) + 1)*sin(x) - 1)*sqrt(cos(x)^2/sin(x))
- 2*cos(x) + 4)/(cos(x)^3 + 3*cos(x)^2 - (cos(x)^2 - 2*cos(x) - 4)*sin(x) - 2*cos(x) - 4)) - 8*(3*cos(x)^4 -
7*cos(x)^2 + 4)*sqrt(cos(x)^2/sin(x)))/cos(x)^5

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\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(1/(csc(x)-sin(x))^(5/2),x, algorithm="giac")

[Out]

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