3.684 \(\int \frac {(-1+x^2)^{5/2} \csc ^{-1}(x)}{x^3} \, dx\)

Optimal. Leaf size=106 \[ -\frac {7 x \log (x)}{3 \sqrt {x^2}}+\frac {\left (x^2-1\right )^{5/2} \csc ^{-1}(x)}{3 x^2}-\frac {5 \left (x^2-1\right )^{3/2} \csc ^{-1}(x)}{3 x^2}-\frac {5 \sqrt {x^2-1} \csc ^{-1}(x)}{2 x^2}-\frac {5 x \csc ^{-1}(x)^2}{4 \sqrt {x^2}}+\frac {2 x^4+3}{12 x \sqrt {x^2}} \]

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Rubi [A]  time = 0.20, antiderivative size = 133, normalized size of antiderivative = 1.25, number of steps used = 11, number of rules used = 8, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {5243, 4695, 4647, 4641, 30, 14, 266, 43} \[ \frac {x \sqrt {x^2}}{6}+\frac {\sqrt {x^2}}{4 x^3}-\frac {7 \sqrt {x^2} \log (x)}{3 x}+\frac {1}{3} \left (x^2\right )^{3/2} \left (1-\frac {1}{x^2}\right )^{5/2} \csc ^{-1}(x)-\frac {5}{3} \sqrt {x^2} \left (1-\frac {1}{x^2}\right )^{3/2} \csc ^{-1}(x)-\frac {5 \sqrt {1-\frac {1}{x^2}} \csc ^{-1}(x)}{2 \sqrt {x^2}}-\frac {5 \sqrt {x^2} \csc ^{-1}(x)^2}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[((-1 + x^2)^(5/2)*ArcCsc[x])/x^3,x]

[Out]

Sqrt[x^2]/(4*x^3) + (x*Sqrt[x^2])/6 - (5*Sqrt[1 - x^(-2)]*ArcCsc[x])/(2*Sqrt[x^2]) - (5*(1 - x^(-2))^(3/2)*Sqr
t[x^2]*ArcCsc[x])/3 + ((1 - x^(-2))^(5/2)*(x^2)^(3/2)*ArcCsc[x])/3 - (5*Sqrt[x^2]*ArcCsc[x]^2)/(4*x) - (7*Sqrt
[x^2]*Log[x])/(3*x)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4641

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSin[c*x])^
(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && NeQ[n,
-1]

Rule 4647

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(x*Sqrt[d + e*x^2]*(
a + b*ArcSin[c*x])^n)/2, x] + (Dist[Sqrt[d + e*x^2]/(2*Sqrt[1 - c^2*x^2]), Int[(a + b*ArcSin[c*x])^n/Sqrt[1 -
c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(2*Sqrt[1 - c^2*x^2]), Int[x*(a + b*ArcSin[c*x])^(n - 1), x],
x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]

Rule 4695

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n)/(f*(m + 1)), x] + (-Dist[(2*e*p)/(f^2*(m + 1)), Int[(f*x)^
(m + 2)*(d + e*x^2)^(p - 1)*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/
(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcSin[c*x])^(n - 1),
x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]

Rule 5243

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Dist[Sqrt[x
^2]/x, Subst[Int[((e + d*x^2)^p*(a + b*ArcSin[x/c])^n)/x^(m + 2*(p + 1)), x], x, 1/x], x] /; FreeQ[{a, b, c, d
, e, n}, x] && IGtQ[n, 0] && EqQ[c^2*d + e, 0] && IntegerQ[m] && IntegerQ[p + 1/2] && GtQ[e, 0] && LtQ[d, 0]

Rubi steps

\begin {align*} \int \frac {\left (-1+x^2\right )^{5/2} \csc ^{-1}(x)}{x^3} \, dx &=-\frac {\sqrt {x^2} \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^{5/2} \sin ^{-1}(x)}{x^4} \, dx,x,\frac {1}{x}\right )}{x}\\ &=\frac {1}{3} \left (1-\frac {1}{x^2}\right )^{5/2} \left (x^2\right )^{3/2} \csc ^{-1}(x)-\frac {\sqrt {x^2} \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{x^3} \, dx,x,\frac {1}{x}\right )}{3 x}+\frac {\left (5 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^{3/2} \sin ^{-1}(x)}{x^2} \, dx,x,\frac {1}{x}\right )}{3 x}\\ &=-\frac {5}{3} \left (1-\frac {1}{x^2}\right )^{3/2} \sqrt {x^2} \csc ^{-1}(x)+\frac {1}{3} \left (1-\frac {1}{x^2}\right )^{5/2} \left (x^2\right )^{3/2} \csc ^{-1}(x)-\frac {\sqrt {x^2} \operatorname {Subst}\left (\int \frac {(1-x)^2}{x^2} \, dx,x,\frac {1}{x^2}\right )}{6 x}+\frac {\left (5 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{x} \, dx,x,\frac {1}{x}\right )}{3 x}-\frac {\left (5 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \sqrt {1-x^2} \sin ^{-1}(x) \, dx,x,\frac {1}{x}\right )}{x}\\ &=-\frac {5 \sqrt {1-\frac {1}{x^2}} \csc ^{-1}(x)}{2 \sqrt {x^2}}-\frac {5}{3} \left (1-\frac {1}{x^2}\right )^{3/2} \sqrt {x^2} \csc ^{-1}(x)+\frac {1}{3} \left (1-\frac {1}{x^2}\right )^{5/2} \left (x^2\right )^{3/2} \csc ^{-1}(x)-\frac {\sqrt {x^2} \operatorname {Subst}\left (\int \left (1+\frac {1}{x^2}-\frac {2}{x}\right ) \, dx,x,\frac {1}{x^2}\right )}{6 x}+\frac {\left (5 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x}-x\right ) \, dx,x,\frac {1}{x}\right )}{3 x}+\frac {\left (5 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int x \, dx,x,\frac {1}{x}\right )}{2 x}-\frac {\left (5 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {\sin ^{-1}(x)}{\sqrt {1-x^2}} \, dx,x,\frac {1}{x}\right )}{2 x}\\ &=\frac {\sqrt {x^2}}{4 x^3}+\frac {x \sqrt {x^2}}{6}-\frac {5 \sqrt {1-\frac {1}{x^2}} \csc ^{-1}(x)}{2 \sqrt {x^2}}-\frac {5}{3} \left (1-\frac {1}{x^2}\right )^{3/2} \sqrt {x^2} \csc ^{-1}(x)+\frac {1}{3} \left (1-\frac {1}{x^2}\right )^{5/2} \left (x^2\right )^{3/2} \csc ^{-1}(x)-\frac {5 \sqrt {x^2} \csc ^{-1}(x)^2}{4 x}-\frac {7 \sqrt {x^2} \log (x)}{3 x}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 86, normalized size = 0.81 \[ \frac {\sqrt {x^2-1} \left (4 x^2+\csc ^{-1}(x) \left (8 \sqrt {1-\frac {1}{x^2}} x \left (x^2-7\right )-6 \sin \left (2 \csc ^{-1}(x)\right )\right )+48 \log \left (\frac {1}{x}\right )-8 \log (x)-30 \csc ^{-1}(x)^2-3 \cos \left (2 \csc ^{-1}(x)\right )\right )}{24 \sqrt {1-\frac {1}{x^2}} x} \]

Antiderivative was successfully verified.

[In]

Integrate[((-1 + x^2)^(5/2)*ArcCsc[x])/x^3,x]

[Out]

(Sqrt[-1 + x^2]*(4*x^2 - 30*ArcCsc[x]^2 - 3*Cos[2*ArcCsc[x]] + 48*Log[x^(-1)] - 8*Log[x] + ArcCsc[x]*(8*Sqrt[1
- x^(-2)]*x*(-7 + x^2) - 6*Sin[2*ArcCsc[x]])))/(24*Sqrt[1 - x^(-2)]*x)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (-1+x^2\right )^{5/2} \csc ^{-1}(x)}{x^3} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[((-1 + x^2)^(5/2)*ArcCsc[x])/x^3,x]

[Out]

Could not integrate

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fricas [A]  time = 0.96, size = 51, normalized size = 0.48 \[ \frac {2 \, x^{4} - 15 \, x^{2} \operatorname {arccsc}\relax (x)^{2} - 28 \, x^{2} \log \relax (x) + 2 \, {\left (2 \, x^{4} - 14 \, x^{2} - 3\right )} \sqrt {x^{2} - 1} \operatorname {arccsc}\relax (x) + 3}{12 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)^(5/2)*arccsc(x)/x^3,x, algorithm="fricas")

[Out]

1/12*(2*x^4 - 15*x^2*arccsc(x)^2 - 28*x^2*log(x) + 2*(2*x^4 - 14*x^2 - 3)*sqrt(x^2 - 1)*arccsc(x) + 3)/x^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x^{2} - 1\right )}^{\frac {5}{2}} \operatorname {arccsc}\relax (x)}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)^(5/2)*arccsc(x)/x^3,x, algorithm="giac")

[Out]

integrate((x^2 - 1)^(5/2)*arccsc(x)/x^3, x)

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maple [C]  time = 0.59, size = 305, normalized size = 2.88

method	result	size
default	\(-\frac {5 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x \mathrm {arccsc}\relax (x )^{2}}{4 \sqrt {x^{2}-1}}+\frac {\left (i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}-2 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x -2 x^{2}+2\right ) \left (2 \,\mathrm {arccsc}\relax (x )+i\right )}{16 \sqrt {x^{2}-1}\, x^{2}}-\frac {\left (i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}-2 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x +2 x^{2}-2\right ) \left (-i+2 \,\mathrm {arccsc}\relax (x )\right )}{16 \sqrt {x^{2}-1}\, x^{2}}-\frac {14 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x \,\mathrm {arccsc}\relax (x )}{3 \sqrt {x^{2}-1}}+\frac {\left (x^{4}+7 i \sqrt {\frac {x^{2}-1}{x^{2}}}\, x -8 x^{2}+7\right ) \left (2 \,\mathrm {arccsc}\relax (x ) x^{4}+\sqrt {\frac {x^{2}-1}{x^{2}}}\, x^{3}-30 \,\mathrm {arccsc}\relax (x ) x^{2}-7 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x +126 \,\mathrm {arccsc}\relax (x )-7 i\right )}{6 \sqrt {x^{2}-1}\, \left (x^{4}-15 x^{2}+63\right )}+\frac {7 \sqrt {\frac {x^{2}-1}{x^{2}}}\, x \ln \left (\left (\frac {i}{x}+\sqrt {1-\frac {1}{x^{2}}}\right )^{2}-1\right )}{3 \sqrt {x^{2}-1}}\)	\(305\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-1)^(5/2)*arccsc(x)/x^3,x,method=_RETURNVERBOSE)

[Out]

-5/4/(x^2-1)^(1/2)*((x^2-1)/x^2)^(1/2)*x*arccsc(x)^2+1/16/(x^2-1)^(1/2)/x^2*(I*((x^2-1)/x^2)^(1/2)*x^3-2*I*((x
^2-1)/x^2)^(1/2)*x-2*x^2+2)*(2*arccsc(x)+I)-1/16/(x^2-1)^(1/2)/x^2*(I*((x^2-1)/x^2)^(1/2)*x^3-2*I*((x^2-1)/x^2
)^(1/2)*x+2*x^2-2)*(-I+2*arccsc(x))-14/3*I/(x^2-1)^(1/2)*((x^2-1)/x^2)^(1/2)*x*arccsc(x)+1/6/(x^2-1)^(1/2)*(x^
4+7*I*((x^2-1)/x^2)^(1/2)*x-8*x^2+7)*(2*arccsc(x)*x^4+((x^2-1)/x^2)^(1/2)*x^3-30*arccsc(x)*x^2-7*((x^2-1)/x^2)
^(1/2)*x+126*arccsc(x)-7*I)/(x^4-15*x^2+63)+7/3/(x^2-1)^(1/2)*((x^2-1)/x^2)^(1/2)*x*ln((I/x+(1-1/x^2)^(1/2))^2
-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x^{2} - 1\right )}^{\frac {5}{2}} \operatorname {arccsc}\relax (x)}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)^(5/2)*arccsc(x)/x^3,x, algorithm="maxima")

[Out]

integrate((x^2 - 1)^(5/2)*arccsc(x)/x^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {asin}\left (\frac {1}{x}\right )\,{\left (x^2-1\right )}^{5/2}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((asin(1/x)*(x^2 - 1)^(5/2))/x^3,x)

[Out]

int((asin(1/x)*(x^2 - 1)^(5/2))/x^3, 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((x**2-1)**(5/2)*acsc(x)/x**3,x)

[Out]

Timed out

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