Integrand size = 8, antiderivative size = 63 \[ \int x^3 \csc ^{-1}(x)^2 \, dx=\frac {x^2}{12}+\frac {1}{3} \sqrt {1-\frac {1}{x^2}} x \csc ^{-1}(x)+\frac {1}{6} \sqrt {1-\frac {1}{x^2}} x^3 \csc ^{-1}(x)+\frac {1}{4} x^4 \csc ^{-1}(x)^2+\frac {\log (x)}{3} \] Output:
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)
Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.67 \[ \int x^3 \csc ^{-1}(x)^2 \, dx=\frac {1}{12} \left (x^2+2 \sqrt {1-\frac {1}{x^2}} x \left (2+x^2\right ) \csc ^{-1}(x)+3 x^4 \csc ^{-1}(x)^2+4 \log (x)\right ) \] Input:
Integrate[x^3*ArcCsc[x]^2,x]
Output:
(x^2 + 2*Sqrt[1 - x^(-2)]*x*(2 + x^2)*ArcCsc[x] + 3*x^4*ArcCsc[x]^2 + 4*Lo g[x])/12
Time = 0.41 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {5746, 4245, 3042, 4673, 3042, 4672, 3042, 25, 3956}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 \csc ^{-1}(x)^2 \, dx\) |
\(\Big \downarrow \) 5746 |
\(\displaystyle -\int \sqrt {1-\frac {1}{x^2}} x^5 \csc ^{-1}(x)^2d\csc ^{-1}(x)\) |
\(\Big \downarrow \) 4245 |
\(\displaystyle \frac {1}{4} x^4 \csc ^{-1}(x)^2-\frac {1}{2} \int x^4 \csc ^{-1}(x)d\csc ^{-1}(x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} x^4 \csc ^{-1}(x)^2-\frac {1}{2} \int \csc ^{-1}(x) \csc \left (\csc ^{-1}(x)\right )^4d\csc ^{-1}(x)\) |
\(\Big \downarrow \) 4673 |
\(\displaystyle \frac {1}{2} \left (-\frac {2}{3} \int x^2 \csc ^{-1}(x)d\csc ^{-1}(x)+\frac {x^2}{6}+\frac {1}{3} \sqrt {1-\frac {1}{x^2}} x^3 \csc ^{-1}(x)\right )+\frac {1}{4} x^4 \csc ^{-1}(x)^2\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (-\frac {2}{3} \int \csc ^{-1}(x) \csc \left (\csc ^{-1}(x)\right )^2d\csc ^{-1}(x)+\frac {x^2}{6}+\frac {1}{3} \sqrt {1-\frac {1}{x^2}} x^3 \csc ^{-1}(x)\right )+\frac {1}{4} x^4 \csc ^{-1}(x)^2\) |
\(\Big \downarrow \) 4672 |
\(\displaystyle \frac {1}{2} \left (-\frac {2}{3} \left (\int \sqrt {1-\frac {1}{x^2}} xd\csc ^{-1}(x)-\sqrt {1-\frac {1}{x^2}} x \csc ^{-1}(x)\right )+\frac {x^2}{6}+\frac {1}{3} \sqrt {1-\frac {1}{x^2}} x^3 \csc ^{-1}(x)\right )+\frac {1}{4} x^4 \csc ^{-1}(x)^2\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \left (-\frac {2}{3} \left (\int -\tan \left (\csc ^{-1}(x)+\frac {\pi }{2}\right )d\csc ^{-1}(x)-\sqrt {1-\frac {1}{x^2}} x \csc ^{-1}(x)\right )+\frac {x^2}{6}+\frac {1}{3} \sqrt {1-\frac {1}{x^2}} x^3 \csc ^{-1}(x)\right )+\frac {1}{4} x^4 \csc ^{-1}(x)^2\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (-\frac {2}{3} \left (-\int \tan \left (\csc ^{-1}(x)+\frac {\pi }{2}\right )d\csc ^{-1}(x)-\sqrt {1-\frac {1}{x^2}} x \csc ^{-1}(x)\right )+\frac {x^2}{6}+\frac {1}{3} \sqrt {1-\frac {1}{x^2}} x^3 \csc ^{-1}(x)\right )+\frac {1}{4} x^4 \csc ^{-1}(x)^2\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {1}{4} x^4 \csc ^{-1}(x)^2+\frac {1}{2} \left (\frac {x^2}{6}-\frac {2}{3} \left (\log \left (\frac {1}{x}\right )-\sqrt {1-\frac {1}{x^2}} x \csc ^{-1}(x)\right )+\frac {1}{3} \sqrt {1-\frac {1}{x^2}} x^3 \csc ^{-1}(x)\right )\) |
Input:
Int[x^3*ArcCsc[x]^2,x]
Output:
(x^4*ArcCsc[x]^2)/4 + (x^2/6 + (Sqrt[1 - x^(-2)]*x^3*ArcCsc[x])/3 - (2*(-( Sqrt[1 - x^(-2)]*x*ArcCsc[x]) + Log[x^(-1)]))/3)/2
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
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] + Simp[(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]
Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp [(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1) *Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
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] + (-Simp[b^2*d*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x] + S imp[b^2*((n - 2)/(n - 1)) Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2]
Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[- (c^(m + 1))^(-1) Subst[Int[(a + b*x)^n*Csc[x]^(m + 1)*Cot[x], x], x, ArcC sc[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (GtQ[n , 0] || LtQ[m, -1])
Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89
method | result | size |
default | \(\frac {x^{4} \operatorname {arccsc}\left (x \right )^{2}}{4}+\frac {x^{3} \operatorname {arccsc}\left (x \right ) \sqrt {\frac {x^{2}-1}{x^{2}}}}{6}+\frac {x^{2}}{12}+\frac {\sqrt {\frac {x^{2}-1}{x^{2}}}\, \operatorname {arccsc}\left (x \right ) x}{3}-\frac {\ln \left (\frac {1}{x}\right )}{3}\) | \(56\) |
Input:
int(x^3*arccsc(x)^2,x,method=_RETURNVERBOSE)
Output:
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*((x ^2-1)/x^2)^(1/2)*arccsc(x)*x-1/3*ln(1/x)
Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.56 \[ \int x^3 \csc ^{-1}(x)^2 \, dx=\frac {1}{4} \, x^{4} \operatorname {arccsc}\left (x\right )^{2} + \frac {1}{6} \, {\left (x^{2} + 2\right )} \sqrt {x^{2} - 1} \operatorname {arccsc}\left (x\right ) + \frac {1}{12} \, x^{2} + \frac {1}{3} \, \log \left (x\right ) \] Input:
integrate(x^3*arccsc(x)^2,x, algorithm="fricas")
Output:
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)
\[ \int x^3 \csc ^{-1}(x)^2 \, dx=\int x^{3} \operatorname {acsc}^{2}{\left (x \right )}\, dx \] Input:
integrate(x**3*acsc(x)**2,x)
Output:
Integral(x**3*acsc(x)**2, x)
Time = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.51 \[ \int x^3 \csc ^{-1}(x)^2 \, dx=\frac {1}{4} \, x^{4} \operatorname {arccsc}\left (x\right )^{2} + \frac {2 \, x^{4} \arctan \left (1, \sqrt {x + 1} \sqrt {x - 1}\right ) + 2 \, x^{2} \arctan \left (1, \sqrt {x + 1} \sqrt {x - 1}\right ) + {\left (x^{2} + 2 \, \log \left (x^{2}\right )\right )} \sqrt {x + 1} \sqrt {x - 1} - 4 \, \arctan \left (1, \sqrt {x + 1} \sqrt {x - 1}\right )}{12 \, \sqrt {x + 1} \sqrt {x - 1}} \] Input:
integrate(x^3*arccsc(x)^2,x, algorithm="maxima")
Output:
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)*s qrt(x - 1) - 4*arctan2(1, sqrt(x + 1)*sqrt(x - 1)))/(sqrt(x + 1)*sqrt(x - 1))
Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (49) = 98\).
Time = 0.15 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.68 \[ \int x^3 \csc ^{-1}(x)^2 \, dx=\frac {1}{4} \, x^{4} \arcsin \left (\frac {1}{x}\right )^{2} + \frac {1}{12} \, x^{2} {\left (\frac {2}{x^{2}} + 1\right )} + \frac {1}{48} \, {\left (x^{3} {\left (\sqrt {-\frac {1}{x^{2}} + 1} - 1\right )}^{3} + 9 \, x {\left (\sqrt {-\frac {1}{x^{2}} + 1} - 1\right )} - \frac {9 \, x^{2} {\left (\sqrt {-\frac {1}{x^{2}} + 1} - 1\right )}^{2} + 1}{x^{3} {\left (\sqrt {-\frac {1}{x^{2}} + 1} - 1\right )}^{3}}\right )} \arcsin \left (\frac {1}{x}\right ) - \frac {1}{6} \, \log \left (\frac {1}{x^{2}}\right ) \] Input:
integrate(x^3*arccsc(x)^2,x, algorithm="giac")
Output:
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))
Timed out. \[ \int x^3 \csc ^{-1}(x)^2 \, dx=\int x^3\,{\mathrm {asin}\left (\frac {1}{x}\right )}^2 \,d x \] Input:
int(x^3*asin(1/x)^2,x)
Output:
int(x^3*asin(1/x)^2, x)
\[ \int x^3 \csc ^{-1}(x)^2 \, dx=\int \mathit {acsc} \left (x \right )^{2} x^{3}d x \] Input:
int(x^3*acsc(x)^2,x)
Output:
int(acsc(x)**2*x**3,x)