Integrand size = 10, antiderivative size = 51 \[ \int x^2 \coth ^{-1}\left (\sqrt {x}\right ) \, dx=\frac {\sqrt {x}}{3}+\frac {x^{3/2}}{9}+\frac {x^{5/2}}{15}+\frac {1}{3} x^3 \coth ^{-1}\left (\sqrt {x}\right )-\frac {\text {arctanh}\left (\sqrt {x}\right )}{3} \] Output:
1/3*x^(1/2)+1/9*x^(3/2)+1/15*x^(5/2)+1/3*x^3*arccoth(x^(1/2))-1/3*arctanh( x^(1/2))
Time = 0.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.16 \[ \int x^2 \coth ^{-1}\left (\sqrt {x}\right ) \, dx=\frac {1}{90} \left (30 \sqrt {x}+10 x^{3/2}+6 x^{5/2}+30 x^3 \coth ^{-1}\left (\sqrt {x}\right )+15 \log \left (1-\sqrt {x}\right )-15 \log \left (1+\sqrt {x}\right )\right ) \] Input:
Integrate[x^2*ArcCoth[Sqrt[x]],x]
Output:
(30*Sqrt[x] + 10*x^(3/2) + 6*x^(5/2) + 30*x^3*ArcCoth[Sqrt[x]] + 15*Log[1 - Sqrt[x]] - 15*Log[1 + Sqrt[x]])/90
Time = 0.20 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6453, 60, 60, 60, 73, 219}
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.
\(\displaystyle \int x^2 \coth ^{-1}\left (\sqrt {x}\right ) \, dx\) |
\(\Big \downarrow \) 6453 |
\(\displaystyle \frac {1}{3} x^3 \coth ^{-1}\left (\sqrt {x}\right )-\frac {1}{6} \int \frac {x^{5/2}}{1-x}dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{6} \left (\frac {2 x^{5/2}}{5}-\int \frac {x^{3/2}}{1-x}dx\right )+\frac {1}{3} x^3 \coth ^{-1}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{6} \left (-\int \frac {\sqrt {x}}{1-x}dx+\frac {2 x^{5/2}}{5}+\frac {2 x^{3/2}}{3}\right )+\frac {1}{3} x^3 \coth ^{-1}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{6} \left (-\int \frac {1}{(1-x) \sqrt {x}}dx+\frac {2 x^{5/2}}{5}+\frac {2 x^{3/2}}{3}+2 \sqrt {x}\right )+\frac {1}{3} x^3 \coth ^{-1}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{6} \left (-2 \int \frac {1}{1-x}d\sqrt {x}+\frac {2 x^{5/2}}{5}+\frac {2 x^{3/2}}{3}+2 \sqrt {x}\right )+\frac {1}{3} x^3 \coth ^{-1}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{6} \left (-2 \text {arctanh}\left (\sqrt {x}\right )+\frac {2 x^{5/2}}{5}+\frac {2 x^{3/2}}{3}+2 \sqrt {x}\right )+\frac {1}{3} x^3 \coth ^{-1}\left (\sqrt {x}\right )\) |
Input:
Int[x^2*ArcCoth[Sqrt[x]],x]
Output:
(x^3*ArcCoth[Sqrt[x]])/3 + (2*Sqrt[x] + (2*x^(3/2))/3 + (2*x^(5/2))/5 - 2* ArcTanh[Sqrt[x]])/6
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : > Simp[x^(m + 1)*((a + b*ArcCoth[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcCoth[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x ], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 ] && IntegerQ[m])) && NeQ[m, -1]
Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(\frac {x^{3} \operatorname {arccoth}\left (\sqrt {x}\right )}{3}+\frac {x^{\frac {5}{2}}}{15}+\frac {x^{\frac {3}{2}}}{9}+\frac {\sqrt {x}}{3}+\frac {\ln \left (\sqrt {x}-1\right )}{6}-\frac {\ln \left (\sqrt {x}+1\right )}{6}\) | \(42\) |
default | \(\frac {x^{3} \operatorname {arccoth}\left (\sqrt {x}\right )}{3}+\frac {x^{\frac {5}{2}}}{15}+\frac {x^{\frac {3}{2}}}{9}+\frac {\sqrt {x}}{3}+\frac {\ln \left (\sqrt {x}-1\right )}{6}-\frac {\ln \left (\sqrt {x}+1\right )}{6}\) | \(42\) |
parts | \(\frac {x^{3} \operatorname {arccoth}\left (\sqrt {x}\right )}{3}+\frac {x^{\frac {5}{2}}}{15}+\frac {x^{\frac {3}{2}}}{9}+\frac {\sqrt {x}}{3}+\frac {\ln \left (\sqrt {x}-1\right )}{6}-\frac {\ln \left (\sqrt {x}+1\right )}{6}\) | \(42\) |
Input:
int(x^2*arccoth(x^(1/2)),x,method=_RETURNVERBOSE)
Output:
1/3*x^3*arccoth(x^(1/2))+1/15*x^(5/2)+1/9*x^(3/2)+1/3*x^(1/2)+1/6*ln(x^(1/ 2)-1)-1/6*ln(x^(1/2)+1)
Time = 0.10 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.75 \[ \int x^2 \coth ^{-1}\left (\sqrt {x}\right ) \, dx=\frac {1}{6} \, {\left (x^{3} - 1\right )} \log \left (\frac {x + 2 \, \sqrt {x} + 1}{x - 1}\right ) + \frac {1}{45} \, {\left (3 \, x^{2} + 5 \, x + 15\right )} \sqrt {x} \] Input:
integrate(x^2*arccoth(x^(1/2)),x, algorithm="fricas")
Output:
1/6*(x^3 - 1)*log((x + 2*sqrt(x) + 1)/(x - 1)) + 1/45*(3*x^2 + 5*x + 15)*s qrt(x)
\[ \int x^2 \coth ^{-1}\left (\sqrt {x}\right ) \, dx=\int x^{2} \operatorname {acoth}{\left (\sqrt {x} \right )}\, dx \] Input:
integrate(x**2*acoth(x**(1/2)),x)
Output:
Integral(x**2*acoth(sqrt(x)), x)
Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.80 \[ \int x^2 \coth ^{-1}\left (\sqrt {x}\right ) \, dx=\frac {1}{3} \, x^{3} \operatorname {arcoth}\left (\sqrt {x}\right ) + \frac {1}{15} \, x^{\frac {5}{2}} + \frac {1}{9} \, x^{\frac {3}{2}} + \frac {1}{3} \, \sqrt {x} - \frac {1}{6} \, \log \left (\sqrt {x} + 1\right ) + \frac {1}{6} \, \log \left (\sqrt {x} - 1\right ) \] Input:
integrate(x^2*arccoth(x^(1/2)),x, algorithm="maxima")
Output:
1/3*x^3*arccoth(sqrt(x)) + 1/15*x^(5/2) + 1/9*x^(3/2) + 1/3*sqrt(x) - 1/6* log(sqrt(x) + 1) + 1/6*log(sqrt(x) - 1)
Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (31) = 62\).
Time = 0.12 (sec) , antiderivative size = 164, normalized size of antiderivative = 3.22 \[ \int x^2 \coth ^{-1}\left (\sqrt {x}\right ) \, dx=\frac {2 \, {\left (\frac {45 \, {\left (\sqrt {x} + 1\right )}^{4}}{{\left (\sqrt {x} - 1\right )}^{4}} - \frac {90 \, {\left (\sqrt {x} + 1\right )}^{3}}{{\left (\sqrt {x} - 1\right )}^{3}} + \frac {140 \, {\left (\sqrt {x} + 1\right )}^{2}}{{\left (\sqrt {x} - 1\right )}^{2}} - \frac {70 \, {\left (\sqrt {x} + 1\right )}}{\sqrt {x} - 1} + 23\right )}}{45 \, {\left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1} - 1\right )}^{5}} + \frac {2 \, {\left (\frac {3 \, {\left (\sqrt {x} + 1\right )}^{5}}{{\left (\sqrt {x} - 1\right )}^{5}} + \frac {10 \, {\left (\sqrt {x} + 1\right )}^{3}}{{\left (\sqrt {x} - 1\right )}^{3}} + \frac {3 \, {\left (\sqrt {x} + 1\right )}}{\sqrt {x} - 1}\right )} \log \left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1}\right )}{3 \, {\left (\frac {\sqrt {x} + 1}{\sqrt {x} - 1} - 1\right )}^{6}} \] Input:
integrate(x^2*arccoth(x^(1/2)),x, algorithm="giac")
Output:
2/45*(45*(sqrt(x) + 1)^4/(sqrt(x) - 1)^4 - 90*(sqrt(x) + 1)^3/(sqrt(x) - 1 )^3 + 140*(sqrt(x) + 1)^2/(sqrt(x) - 1)^2 - 70*(sqrt(x) + 1)/(sqrt(x) - 1) + 23)/((sqrt(x) + 1)/(sqrt(x) - 1) - 1)^5 + 2/3*(3*(sqrt(x) + 1)^5/(sqrt( x) - 1)^5 + 10*(sqrt(x) + 1)^3/(sqrt(x) - 1)^3 + 3*(sqrt(x) + 1)/(sqrt(x) - 1))*log((sqrt(x) + 1)/(sqrt(x) - 1))/((sqrt(x) + 1)/(sqrt(x) - 1) - 1)^6
Time = 3.67 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.61 \[ \int x^2 \coth ^{-1}\left (\sqrt {x}\right ) \, dx=\frac {x^3\,\mathrm {acoth}\left (\sqrt {x}\right )}{3}-\frac {\mathrm {acoth}\left (\sqrt {x}\right )}{3}+\frac {\sqrt {x}}{3}+\frac {x^{3/2}}{9}+\frac {x^{5/2}}{15} \] Input:
int(x^2*acoth(x^(1/2)),x)
Output:
(x^3*acoth(x^(1/2)))/3 - acoth(x^(1/2))/3 + x^(1/2)/3 + x^(3/2)/9 + x^(5/2 )/15
Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.59 \[ \int x^2 \coth ^{-1}\left (\sqrt {x}\right ) \, dx=\frac {\mathit {acoth} \left (\sqrt {x}\right ) x^{3}}{3}-\frac {\mathit {acoth} \left (\sqrt {x}\right )}{3}-\frac {\sqrt {x}\, x^{2}}{15}-\frac {\sqrt {x}\, x}{9}-\frac {\sqrt {x}}{3} \] Input:
int(x^2*acoth(x^(1/2)),x)
Output:
(15*acoth(sqrt(x))*x**3 - 15*acoth(sqrt(x)) - 3*sqrt(x)*x**2 - 5*sqrt(x)*x - 15*sqrt(x))/45