3.7.49 \(\int x^3 \csc ^{-1}(x)^2 \, dx\) [649]

Optimal. Leaf size=63 \[ \frac {x^2}{12}+\frac {1}{3} \sqrt {1-\frac {1}{x^2}} x \csc ^{-1}(x)+\frac {1}{6} \sqrt {1-\frac {1}{x^2}} x^3 \csc ^{-1}(x)+\frac {1}{4} x^4 \csc ^{-1}(x)^2+\frac {\log (x)}{3} \]

[Out]

1/12*x^2+1/4*x^4*arccsc(x)^2+1/3*ln(x)+1/3*x*arccsc(x)*(1-1/x^2)^(1/2)+1/6*x^3*arccsc(x)*(1-1/x^2)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5331, 3843, 4270, 4269, 3556} \begin {gather*} \frac {1}{4} x^4 \csc ^{-1}(x)^2+\frac {x^2}{12}+\frac {1}{3} \sqrt {1-\frac {1}{x^2}} x \csc ^{-1}(x)+\frac {1}{6} \sqrt {1-\frac {1}{x^2}} x^3 \csc ^{-1}(x)+\frac {\log (x)}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCsc[x]^2,x]

[Out]

x^2/12 + (Sqrt[1 - x^(-2)]*x*ArcCsc[x])/3 + (Sqrt[1 - x^(-2)]*x^3*ArcCsc[x])/6 + (x^4*ArcCsc[x]^2)/4 + Log[x]/
3

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3843

Int[Cot[(a_.) + (b_.)*(x_)^(n_.)]^(q_.)*Csc[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[(-x^
(m - n + 1))*(Csc[a + b*x^n]^p/(b*n*p)), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Csc[a + b*x^n]^p, x], x]
/; FreeQ[{a, b, p}, x] && IntegerQ[n] && GeQ[m, n] && EqQ[q, 1]

Rule 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4270

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(-b^2)*(c + d*x)*Cot[e + f*x]
*((b*Csc[e + f*x])^(n - 2)/(f*(n - 1))), x] + (Dist[b^2*((n - 2)/(n - 1)), Int[(c + d*x)*(b*Csc[e + f*x])^(n -
2), x], x] - Simp[b^2*d*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x]) /; FreeQ[{b, c, d, e, f}, x] &&
GtQ[n, 1] && NeQ[n, 2]

Rule 5331

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[-(c^(m + 1))^(-1), Subst[Int[(a + b*
x)^n*Csc[x]^(m + 1)*Cot[x], x], x, ArcCsc[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (G
tQ[n, 0] || LtQ[m, -1])

Rubi steps

\begin {align*} \int x^3 \csc ^{-1}(x)^2 \, dx &=-\text {Subst}\left (\int x^2 \cot (x) \csc ^4(x) \, dx,x,\csc ^{-1}(x)\right )\\ &=\frac {1}{4} x^4 \csc ^{-1}(x)^2-\frac {1}{2} \text {Subst}\left (\int x \csc ^4(x) \, dx,x,\csc ^{-1}(x)\right )\\ &=\frac {x^2}{12}+\frac {1}{6} \sqrt {1-\frac {1}{x^2}} x^3 \csc ^{-1}(x)+\frac {1}{4} x^4 \csc ^{-1}(x)^2-\frac {1}{3} \text {Subst}\left (\int x \csc ^2(x) \, dx,x,\csc ^{-1}(x)\right )\\ &=\frac {x^2}{12}+\frac {1}{3} \sqrt {1-\frac {1}{x^2}} x \csc ^{-1}(x)+\frac {1}{6} \sqrt {1-\frac {1}{x^2}} x^3 \csc ^{-1}(x)+\frac {1}{4} x^4 \csc ^{-1}(x)^2-\frac {1}{3} \text {Subst}\left (\int \cot (x) \, dx,x,\csc ^{-1}(x)\right )\\ &=\frac {x^2}{12}+\frac {1}{3} \sqrt {1-\frac {1}{x^2}} x \csc ^{-1}(x)+\frac {1}{6} \sqrt {1-\frac {1}{x^2}} x^3 \csc ^{-1}(x)+\frac {1}{4} x^4 \csc ^{-1}(x)^2+\frac {\log (x)}{3}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 42, normalized size = 0.67 \begin {gather*} \frac {1}{12} \left (x^2+2 \sqrt {1-\frac {1}{x^2}} x \left (2+x^2\right ) \csc ^{-1}(x)+3 x^4 \csc ^{-1}(x)^2+4 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCsc[x]^2,x]

[Out]

(x^2 + 2*Sqrt[1 - x^(-2)]*x*(2 + x^2)*ArcCsc[x] + 3*x^4*ArcCsc[x]^2 + 4*Log[x])/12

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Maple [A]
time = 0.05, size = 56, normalized size = 0.89

method	result	size
default	\(\frac {x^{4} \mathrm {arccsc}\left (x \right )^{2}}{4}+\frac {x^{3} \mathrm {arccsc}\left (x \right ) \sqrt {\frac {x^{2}-1}{x^{2}}}}{6}+\frac {x^{2}}{12}+\frac {\mathrm {arccsc}\left (x \right ) \sqrt {\frac {x^{2}-1}{x^{2}}}\, x}{3}-\frac {\ln \left (\frac {1}{x}\right )}{3}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccsc(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/4*x^4*arccsc(x)^2+1/6*x^3*arccsc(x)*((x^2-1)/x^2)^(1/2)+1/12*x^2+1/3*arccsc(x)*((x^2-1)/x^2)^(1/2)*x-1/3*ln(
1/x)

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Maxima [A]
time = 2.07, size = 95, normalized size = 1.51 \begin {gather*} \frac {1}{4} \, x^{4} \operatorname {arccsc}\left (x\right )^{2} + \frac {2 \, x^{4} \arctan \left (1, \sqrt {x + 1} \sqrt {x - 1}\right ) + 2 \, x^{2} \arctan \left (1, \sqrt {x + 1} \sqrt {x - 1}\right ) + {\left (x^{2} + 2 \, \log \left (x^{2}\right )\right )} \sqrt {x + 1} \sqrt {x - 1} - 4 \, \arctan \left (1, \sqrt {x + 1} \sqrt {x - 1}\right )}{12 \, \sqrt {x + 1} \sqrt {x - 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccsc(x)^2,x, algorithm="maxima")

[Out]

1/4*x^4*arccsc(x)^2 + 1/12*(2*x^4*arctan2(1, sqrt(x + 1)*sqrt(x - 1)) + 2*x^2*arctan2(1, sqrt(x + 1)*sqrt(x -
1)) + (x^2 + 2*log(x^2))*sqrt(x + 1)*sqrt(x - 1) - 4*arctan2(1, sqrt(x + 1)*sqrt(x - 1)))/(sqrt(x + 1)*sqrt(x
- 1))

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Fricas [A]
time = 0.55, size = 35, normalized size = 0.56 \begin {gather*} \frac {1}{4} \, x^{4} \operatorname {arccsc}\left (x\right )^{2} + \frac {1}{6} \, {\left (x^{2} + 2\right )} \sqrt {x^{2} - 1} \operatorname {arccsc}\left (x\right ) + \frac {1}{12} \, x^{2} + \frac {1}{3} \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccsc(x)^2,x, algorithm="fricas")

[Out]

1/4*x^4*arccsc(x)^2 + 1/6*(x^2 + 2)*sqrt(x^2 - 1)*arccsc(x) + 1/12*x^2 + 1/3*log(x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \operatorname {acsc}^{2}{\left (x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*acsc(x)**2,x)

[Out]

Integral(x**3*acsc(x)**2, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (49) = 98\).
time = 1.17, size = 106, normalized size = 1.68 \begin {gather*} \frac {1}{4} \, x^{4} \arcsin \left (\frac {1}{x}\right )^{2} + \frac {1}{12} \, x^{2} {\left (\frac {2}{x^{2}} + 1\right )} + \frac {1}{48} \, {\left (x^{3} {\left (\sqrt {-\frac {1}{x^{2}} + 1} - 1\right )}^{3} + 9 \, x {\left (\sqrt {-\frac {1}{x^{2}} + 1} - 1\right )} - \frac {9 \, x^{2} {\left (\sqrt {-\frac {1}{x^{2}} + 1} - 1\right )}^{2} + 1}{x^{3} {\left (\sqrt {-\frac {1}{x^{2}} + 1} - 1\right )}^{3}}\right )} \arcsin \left (\frac {1}{x}\right ) - \frac {1}{6} \, \log \left (\frac {1}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccsc(x)^2,x, algorithm="giac")

[Out]

1/4*x^4*arcsin(1/x)^2 + 1/12*x^2*(2/x^2 + 1) + 1/48*(x^3*(sqrt(-1/x^2 + 1) - 1)^3 + 9*x*(sqrt(-1/x^2 + 1) - 1)
- (9*x^2*(sqrt(-1/x^2 + 1) - 1)^2 + 1)/(x^3*(sqrt(-1/x^2 + 1) - 1)^3))*arcsin(1/x) - 1/6*log(x^(-2))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int x^3\,{\mathrm {asin}\left (\frac {1}{x}\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^3*asin(1/x)^2, x)

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