3.321 \(\int \frac{1}{(\csc (x)-\sin (x))^{7/2}} \, dx\)

Optimal. Leaf size=118 \[ -\frac{5 \sec ^3(x)}{48 \sqrt{\cos (x) \cot (x)}}+\frac{5 \sec (x)}{192 \sqrt{\cos (x) \cot (x)}}-\frac{5 \sqrt{-\sin (x)} \cot (x) \tan ^{-1}\left (\sqrt{-\sin (x)}\right )}{128 \sqrt{\cos (x) \cot (x)}}-\frac{5 \sqrt{-\sin (x)} \cot (x) \tanh ^{-1}\left (\sqrt{-\sin (x)}\right )}{128 \sqrt{\cos (x) \cot (x)}}+\frac{\tan ^2(x) \sec ^3(x)}{6 \sqrt{\cos (x) \cot (x)}} \]

[Out]

(5*Sec[x])/(192*Sqrt[Cos[x]*Cot[x]]) - (5*Sec[x]^3)/(48*Sqrt[Cos[x]*Cot[x]]) - (5*ArcTan[Sqrt[-Sin[x]]]*Cot[x]
*Sqrt[-Sin[x]])/(128*Sqrt[Cos[x]*Cot[x]]) - (5*ArcTanh[Sqrt[-Sin[x]]]*Cot[x]*Sqrt[-Sin[x]])/(128*Sqrt[Cos[x]*C
ot[x]]) + (Sec[x]^3*Tan[x]^2)/(6*Sqrt[Cos[x]*Cot[x]])

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Rubi [A]  time = 0.179569, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 11, 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, 212, 206, 203} \[ -\frac{5 \sec ^3(x)}{48 \sqrt{\cos (x) \cot (x)}}+\frac{5 \sec (x)}{192 \sqrt{\cos (x) \cot (x)}}-\frac{5 \sqrt{-\sin (x)} \cot (x) \tan ^{-1}\left (\sqrt{-\sin (x)}\right )}{128 \sqrt{\cos (x) \cot (x)}}-\frac{5 \sqrt{-\sin (x)} \cot (x) \tanh ^{-1}\left (\sqrt{-\sin (x)}\right )}{128 \sqrt{\cos (x) \cot (x)}}+\frac{\tan ^2(x) \sec ^3(x)}{6 \sqrt{\cos (x) \cot (x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[x] - Sin[x])^(-7/2),x]

[Out]

(5*Sec[x])/(192*Sqrt[Cos[x]*Cot[x]]) - (5*Sec[x]^3)/(48*Sqrt[Cos[x]*Cot[x]]) - (5*ArcTan[Sqrt[-Sin[x]]]*Cot[x]
*Sqrt[-Sin[x]])/(128*Sqrt[Cos[x]*Cot[x]]) - (5*ArcTanh[Sqrt[-Sin[x]]]*Cot[x]*Sqrt[-Sin[x]])/(128*Sqrt[Cos[x]*C
ot[x]]) + (Sec[x]^3*Tan[x]^2)/(6*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 212

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

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])

Rubi steps

\begin{align*} \int \frac{1}{(\csc (x)-\sin (x))^{7/2}} \, dx &=\int \frac{1}{(\cos (x) \cot (x))^{7/2}} \, dx\\ &=\frac{\left (\sqrt{\cos (x)} \sqrt{\cot (x)}\right ) \int \frac{1}{\cos ^{\frac{7}{2}}(x) \cot ^{\frac{7}{2}}(x)} \, dx}{\sqrt{\cos (x) \cot (x)}}\\ &=\frac{\sec ^3(x) \tan ^2(x)}{6 \sqrt{\cos (x) \cot (x)}}-\frac{\left (5 \sqrt{\cos (x)} \sqrt{\cot (x)}\right ) \int \frac{1}{\cos ^{\frac{7}{2}}(x) \cot ^{\frac{3}{2}}(x)} \, dx}{12 \sqrt{\cos (x) \cot (x)}}\\ &=-\frac{5 \sec ^3(x)}{48 \sqrt{\cos (x) \cot (x)}}+\frac{\sec ^3(x) \tan ^2(x)}{6 \sqrt{\cos (x) \cot (x)}}+\frac{\left (5 \sqrt{\cos (x)} \sqrt{\cot (x)}\right ) \int \frac{\sqrt{\cot (x)}}{\cos ^{\frac{7}{2}}(x)} \, dx}{96 \sqrt{\cos (x) \cot (x)}}\\ &=\frac{5 \sec (x)}{192 \sqrt{\cos (x) \cot (x)}}-\frac{5 \sec ^3(x)}{48 \sqrt{\cos (x) \cot (x)}}+\frac{\sec ^3(x) \tan ^2(x)}{6 \sqrt{\cos (x) \cot (x)}}+\frac{\left (5 \sqrt{\cos (x)} \sqrt{\cot (x)}\right ) \int \frac{\sqrt{\cot (x)}}{\cos ^{\frac{3}{2}}(x)} \, dx}{128 \sqrt{\cos (x) \cot (x)}}\\ &=\frac{5 \sec (x)}{192 \sqrt{\cos (x) \cot (x)}}-\frac{5 \sec ^3(x)}{48 \sqrt{\cos (x) \cot (x)}}+\frac{\sec ^3(x) \tan ^2(x)}{6 \sqrt{\cos (x) \cot (x)}}+\frac{\left (5 \cot (x) \sqrt{-\sin (x)}\right ) \int \frac{\sec (x)}{\sqrt{-\sin (x)}} \, dx}{128 \sqrt{\cos (x) \cot (x)}}\\ &=\frac{5 \sec (x)}{192 \sqrt{\cos (x) \cot (x)}}-\frac{5 \sec ^3(x)}{48 \sqrt{\cos (x) \cot (x)}}+\frac{\sec ^3(x) \tan ^2(x)}{6 \sqrt{\cos (x) \cot (x)}}-\frac{\left (5 \cot (x) \sqrt{-\sin (x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1-x^2\right )} \, dx,x,-\sin (x)\right )}{128 \sqrt{\cos (x) \cot (x)}}\\ &=\frac{5 \sec (x)}{192 \sqrt{\cos (x) \cot (x)}}-\frac{5 \sec ^3(x)}{48 \sqrt{\cos (x) \cot (x)}}+\frac{\sec ^3(x) \tan ^2(x)}{6 \sqrt{\cos (x) \cot (x)}}-\frac{\left (5 \cot (x) \sqrt{-\sin (x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^4} \, dx,x,\sqrt{-\sin (x)}\right )}{64 \sqrt{\cos (x) \cot (x)}}\\ &=\frac{5 \sec (x)}{192 \sqrt{\cos (x) \cot (x)}}-\frac{5 \sec ^3(x)}{48 \sqrt{\cos (x) \cot (x)}}+\frac{\sec ^3(x) \tan ^2(x)}{6 \sqrt{\cos (x) \cot (x)}}-\frac{\left (5 \cot (x) \sqrt{-\sin (x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{-\sin (x)}\right )}{128 \sqrt{\cos (x) \cot (x)}}-\frac{\left (5 \cot (x) \sqrt{-\sin (x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{-\sin (x)}\right )}{128 \sqrt{\cos (x) \cot (x)}}\\ &=\frac{5 \sec (x)}{192 \sqrt{\cos (x) \cot (x)}}-\frac{5 \sec ^3(x)}{48 \sqrt{\cos (x) \cot (x)}}-\frac{5 \tan ^{-1}\left (\sqrt{-\sin (x)}\right ) \cot (x) \sqrt{-\sin (x)}}{128 \sqrt{\cos (x) \cot (x)}}-\frac{5 \tanh ^{-1}\left (\sqrt{-\sin (x)}\right ) \cot (x) \sqrt{-\sin (x)}}{128 \sqrt{\cos (x) \cot (x)}}+\frac{\sec ^3(x) \tan ^2(x)}{6 \sqrt{\cos (x) \cot (x)}}\\ \end{align*}

Mathematica [A]  time = 0.267902, size = 74, normalized size = 0.63 \[ \frac{2 \sqrt [4]{\sin ^2(x)} \sec (x) \left (32 \sec ^4(x)-52 \sec ^2(x)+5\right )+15 \cos (x) \tan ^{-1}\left (\sqrt [4]{\sin ^2(x)}\right )+15 \cos (x) \tanh ^{-1}\left (\sqrt [4]{\sin ^2(x)}\right )}{384 \sqrt [4]{\sin ^2(x)} \sqrt{\cos (x) \cot (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[x] - Sin[x])^(-7/2),x]

[Out]

(15*ArcTan[(Sin[x]^2)^(1/4)]*Cos[x] + 15*ArcTanh[(Sin[x]^2)^(1/4)]*Cos[x] + 2*Sec[x]*(5 - 52*Sec[x]^2 + 32*Sec
[x]^4)*(Sin[x]^2)^(1/4))/(384*Sqrt[Cos[x]*Cot[x]]*(Sin[x]^2)^(1/4))

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Maple [C]  time = 0.227, size = 487, normalized size = 4.1 \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))^(7/2),x)

[Out]

-1/768*2^(1/2)*(-1+cos(x))*(15*I*sin(x)*((-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)*cos(x)^6*EllipticPi(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2+1/2*I,1/2*2^(1/2))+1
5*I*sin(x)*((-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
)*cos(x)^6*EllipticPi(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2-1/2*I,1/2*2^(1/2))-30*I*sin(x)*((-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)*cos(x)^6*EllipticF(((I*cos(
x)+sin(x)-I)/sin(x))^(1/2),1/2*2^(1/2))+15*sin(x)*cos(x)^6*((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)*EllipticPi(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2+1/2*I,1/2*
2^(1/2))-15*sin(x)*cos(x)^6*((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)*EllipticPi(((I*cos(x)+sin(x)-I)/sin(x))^(1/2),1/2-1/2*I,1/2*2^(1/2))-10*2^(1/2)*cos(x)^5+10
*2^(1/2)*cos(x)^4+104*cos(x)^3*2^(1/2)-104*cos(x)^2*2^(1/2)-64*cos(x)*2^(1/2)+64*2^(1/2))*cos(x)*(1+cos(x))^2/
sin(x)^7/(cos(x)^2/sin(x))^(7/2)

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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{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(csc(x)-sin(x))^(7/2),x, algorithm="maxima")

[Out]

integrate((csc(x) - sin(x))^(-7/2), x)

________________________________________________________________________________________

Fricas [A]  time = 2.60421, size = 528, normalized size = 4.47 \begin{align*} -\frac{30 \, \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 )^{7} - 15 \, \cos \left (x\right )^{7} \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 (5 \, \cos \left (x\right )^{4} - 52 \, \cos \left (x\right )^{2} + 32\right )} \sqrt{\frac{\cos \left (x\right )^{2}}{\sin \left (x\right )}} \sin \left (x\right )}{1536 \, \cos \left (x\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(csc(x)-sin(x))^(7/2),x, algorithm="fricas")

[Out]

-1/1536*(30*arctan(2*sqrt(cos(x)^2/sin(x))*sin(x)/(cos(x)*sin(x) - cos(x)))*cos(x)^7 - 15*cos(x)^7*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*(5*cos(x)^4
- 52*cos(x)^2 + 32)*sqrt(cos(x)^2/sin(x))*sin(x))/cos(x)^7

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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))**(7/2),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{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(csc(x)-sin(x))^(7/2),x, algorithm="giac")

[Out]

integrate((csc(x) - sin(x))^(-7/2), x)